Article Reference: 7F60239AB9EC990E
Date: Sat, 19 Oct 1996 03:09:45 -0500
Original Subject: Re: 115/230 VAC switching - how do they do it? (I thought it would be easy!)
Phil Wasson wrote:
>
> I thought this would be simple: Design a low cost 15VDC 150ma
> unregulated power supply that will run on 115 or 230 VAC. It must use a
> 2 VA tranformer due to size requirements.
>
> Many companies make a suitable transformer that has two primaries and
> two secondaries. The secondaries are connected in parallel for 15V, 170
> ma output. To switch between 115 and 230 VAC, the primaries are
> connected in either parallel or series. This requires a 3PDT switch.
Nope, see below.
> All the slide switches I can find for this purpose are SPDT or DPDT.
> My question is, how are products designed using a DPDT switch for this
> purpose?
> --
> Phil Wasson
>
> pwasson@pacbell.net
How about this:
L1--+-----------------)||
\ 120 )||
\ +-------)||
\ 120 | 240 ||
\ O__O O ||
\/ | ||
/\ __ | ||
/ O O O ||
/ | ||
/ +-------)||
/ 120 )||
N--+-----------------)||
Gene
Article Reference: BF797C2DBD9D5FAD
Date: 23 Aug 1996 19:03:22 -0400
Original Subject: Re: Help my rusty math please
From: no@email.ads (Christopher E.Giese) >Here is one charge pump circuit. It works best if the square-wave >oscillator has a low output impedance, and if Schottky diodes are >used: > VCC > O > | > _|_ > \ / > __V__) > | | + ( | / > VCC | | | |\ | > O .----+ +------+---| >|------+---------> 2 * VCC > | | | | |/ | | > | | | | + / + | + > ______|_____ | ___|___ >| | | _______ >| square- | | | >| wave +---+ ___|___ >| oscillator | | _____ >| | | ___ >|____________| | + | | / . > | | | | | /| > | '---+ +-----+------|< |------+---------> -2 * VCC > | | | | | \| | > ___|___ + | | | / | > _____ _|_ ___|___ > ___ \ / _______ > . __V__) + | + > ( | ___|___ > | _____ > ___|___ ___ > _____ . > ___ > . >(I love ASCII art!) More capacitor-diode stages can be added to get >higher output voltages. I don't know how efficient it is. >Hope this helps. Actually, I think the -2 * VCC is -1 * VCC, unless the oscillator output swings below ground. Bill
Article Reference: 815B9013C58FA331
Date: Sat, 16 Nov 1996 15:42:14 -0800
Original Subject: Re: Mosfet Information
I assume you want to run the mosfet in the region where current
is essentially independent of drain source voltage. What about making
a current source out of it with a source resistor. I've done an ASCII
sketch below of an AC coupled amplifier. For DC and lower frequency
applications we use an opamp looking at the current viewing resistor
and driving the gate. This turns the mosfet into a controlled current
source (sink?). In your case you would need a low power (single
supply maybe ?) op amp.
By the way, if you're trying to drive a power mosfet from only a 5V
supply make sure it is a "Logic Level" device (i.e the gate threshold
is low enough to ensure that 5V can drive it fully on.)
If the picture below comes out funny try to line it up so that every
line starts with an x.
x _____
x |
x |
x 5V______ RRR
x | RRR (load)
x | RRR
x | RRR ||
x RRR | ||
x RRR (R1) @------------||-------
x RRR | || Output
x || RRR |--| || (AC)
x || | | |
x Signal - -|| ----@----------| | Mosfet
x || | |--|
x || | |
x | ------@--------------
x (C1) \ | | | |
x \ | | RRR |
x \ |---- RRR (R2) |
x npn / | RRR _______
x / | RRR _______
x |/_ | | (C2)
x | | |
x _|_______________|_____________|_____
x 0V
x
Notes:
Quiescent drain current approx: 0.6V/R2
Load must ensure that there is an appropriate DC voltage
across the mosfet with this quiescent current flowing.
Input impedance at signal frequencies R1 (can be large).
R1*C1 timeconstant must be large w.r.t. to signal frequencies
Article Reference: 7ECF16C17FF79F1B
Date: 24 Jun 1996 14:22:36 GMT
Original Subject: Re: Help Discharging Capacitor Properly!
dmcfadye@kea.bc.ca (Doug McFadyen) wrote:
>
>I need to make a small mod to my audio AMP (changing the Phono/turntable
>input so that it doesn't pre-amplify the input signal).
>
>There are four 50V 4,700uF caps in the final amplifer stage that like to
>bite, long after the power is disconnected.
>
>Is the correct procedure for discharging a cap to just short it at the base
>of its leads? Or, should a resistor be used to "bleed" off the power slowly?
>If I should use a resistor, what value is good? What wattage?
I do not recommend shorting the caps. You probably have enough stored
energy to melt and launch small bits of metal from the leads. You might
hurt an eye or some other component.
I have used the following active discharge circuit for perminently
connected power supply caps. I am only showing the barest of details.
Plus DC power ----+----------------------+
| |
D1 e |
| |/-------+--------- load
+----+--------| pnp |
| | b|\ |
C1 R1 c| C2 (big cap)
| | | |
ground --+----+----------+------+
The pnp is actually a darlington.
The time constant R1,C1 is short, so that on power up C1 is only 1 diode
drop below the dc power and the pnp never turns on. On loss of power,
R1C1 time constant determines how fast the big cap discharges.
You might play with resisters and parallel diodes in the base circuit.
Don't forget the base current during discharge.
Opinions expressed herein are my own and may not represent those of my employer.
Article Reference: 403D3A2C56DCEE2F
Date: 7 Jul 1996 17:38:40 GMT
Original Subject: Re: Ultrasonic transmitter
eric.c-side@worldnet.att.net (Eric Thosteson) wrote:
>
> I am trying to find/design a driver circuit for use in a high
> frequency underwater ultrasonic transducer....The piezos I want to
> drive operate at 5 MHz with an impedance around 5 ohms
> at that frequency.
5 Ohms seems pretty low. Most transducers I have seen had some __ty
Ohms, but a few had internal matching networks. BTW, the actual
resonance frequency may deviate +-10% from the nominal frequency
for most heads and 10% away from resonance you get completely
different impedance values.
The transducers must have a quite good efficiency. If you measure
their S11 with a network analyzer you get a much smoother curve
if you press a finger against the active surface or if you wet the
surface with a drop of water.
> I would like to hit the crystals with
> around 100 volts with a short pulse width, around 10
> microseconds.
That's ok, but 10 usec seems unneccessarily long. It will
probably be best to approximate a 5 MHz halfwave i.e. 100 ns.
> This poses an interesting problem,
> because the power requirements should be minimal.
> (100 volts across 5 ohms sounds large, but because of
> the small pulse width, the majority of the time, the
> circuit is idle.) The precise pulse width provided by the
> circuit is of the utmost importance.
>
> I have seen the problem solved before by generating
> a 5 MHz waveform at low voltage, then using a hand
> wound toroid to bump up the voltage. (Commercially
> available transformers with these demands seem
> to be nonexistant - but if anyone knows of any...)
Just take some Amidon cores and have someone wrap them.
The transformer itself is probably less of a problem than
the source that has to drive it. If you choose a voltage
step-up ratio os, say, 1:10, the impedance that has to be
driven will sink to 5/3.3 Ohms.
> I would like to find a more elegant method of doing this,
> as hand wound toroids at this frequency have very
> few wraps,
That's an advantage if you have to wrap them. :-)
> and bumping them totally changes their characteristics.
> If anyone can suggest an alternative, knows anyone
> who has attempted this, knows of a source for any
You may want to try an avalanche transistor. That's a
normal low-power transistor, operated under unusual
conditions. (see the text books)
250 V (Yes!) --------------------*--------
|
\ 100K
/
\
| ----------------------
*----------------------------
| ----------------------
50pF 100R C | some meters of
TTL-gate----||----\/\/----*----B GND coax cable
| E
/ |
100R \ |
/ |
\------*-------- out
|
/
\ 50R
/
|
GND
(first time i used netscape as a CAD tool.)
The low cost 2N3904 is reported to work well as an avalanche
transistor. In the quiescent state, the collector will be at
250 V and the coax line will be charged through the 100k resistor.
A rising edge at the ttl input will bring the transistor to
immediate breakdown. The charge contained in the cable will
flow to the output. You will see there a positive pulse with
150 V amplitude and a risetime of 1 ns. The pulse duration
will be 2 times the propagation time through the coax cable.
When the cable has lost it's charge, the transistor will switch
off again and the cable will recharge slowly through the 100K-R.
If you don't need a rectangular waveform you can substitute a
capacitor for the coax cable. If you make the capacitor too
large you may burn the transistor.( > 5nF is risky.) The
capacitor must have good rf properties.
There are way to cascade several avalanche transistors if you
want even more pulse power:
W. Pfeiffer: Erzeugung hoher Impulsspannungen mit
Avalanche-Transistoren in Kaskadenschaltung
Internationale Elektronische Rundschau, November 1973
It's written in german, sorry.
Mister Pfeiffer produced pulses of upto 1.6 KV into a 125 Ohm load
and upto 4.2 KV into 20 pF. Risetime was abt. 5ns.
The pulse repetition rate is limited, but this won't hurt in
your application.
Gerhard
Article Reference: 60C71045F069AAD2
Date: 14 Nov 1998 19:18:46 GMT
Original Subject: Re: Help - constructing an audio signal phase delay.`
This is the circuit that many old guitar phase padals were based on. Ian Ken Smith <kensmith@rahul.net> wrote in article <72arkj$8ma$1@samba.rahul.net>... > In article <7272ee$k47$1@news1-alterdial.uu.net>, > Geo <Gmarutz@teleweb.net> wrote: > >Is it possible to design a phase delay circuit for audio that would allow me > >to vary the phase of a 4Vp-p signal 0-180deg without affecting the signals > >amplitude? Possibly user input through a multi-turn potentiometer? > > > >If you have some ideas please share them. > > > ASCII ART: > > R1 R2 > --------/\/\/\/--------------\/\/\/\---------- > ! ! ! > ! ! ! > ! !---!- \ ! > ! ! \ ! > ! ! ---------- > ! R3 ! / > !-----/\/\/\-----------!+ / > ! > ----- > ----- C1 > ! > GND > > If R1=R2 this circuit has equal gain at all frequencies. At the frequency > given by 1/ (2 * PI * R3 * C1) the phase shift is 90 degrees lagging. > > -- > -- > kensmith@rahul.net forging knowledge > >
Article Reference: 862D6CABE82373F2
Date: 10 Feb 1996 19:15:13 GMT
Original Subject: Re: Audio relay trigger info needed
Justin Olexy (jut@netaxs.com) penned:
: Lo all... I'm fixing up my project for the Duracell electronics
: competition, and I need a bit of help with an improvement I am making. I
: need to be able to trigger a relay (or other near-zero resistance switch)
: from the audio input of a microphone or small speaker. I have made a
: 2-transistor (NPN) amplifier which helps, but the audio into the
: microphone (using small speaker) has to be very loud (actually, I have to
: tap on the speaker to get the relay to trigger). I need some way to
: either boost the audio input, or another way to connect a circuit using
: audio. Any ideas? Thanks
Speakers work OK as microphones, but they're low impedance. If you put a
transformer between the speaker and amp, then the impdeance and voltage
will be higher. The transformer from Rat Snack, with a 1K primary and 8
ohm secondary will work. But the 8 ohm secondary is connected to the
speaker and the primary to the amp. This might not increase the voltage
enough, since it seems from your description that you have an amp with
very poor sensitivity. You may need another stage of amplification ahead
of it.
Don't worry. A single transistor with the emitter grounded, a 10k
resistor from the collecctor to the positive supply.. Oh, yeah, I forgot
that most people don't understand tech talk. Here.
----
Basic Single Stage Amplifier
A Schematic Ver. 951229a
Legend: ) = No Connection
+ = Connection
All resistors are 1/4 W 5%.
All capacitors are 16V or more.
J1 Audio Input
+--------------------( O )
| |
| |
| |
| ----- C1 .05 to .1 uF
| -----
| R2 47K | + R1 470K
+-------/\/\/\---------+-------------/\/\/\------------+
| | |
| | |
| | |
| | Q1 NPN |
| | 2N3904 |
| ------- |
| R4 470 E / \ C R3 10K |
+------/\/\/\-----/ \-----+-----/\/\/\---------+
| | |
| | |
| | + |
| ----- C2 .1 uF |
| ----- |
| | |
| | |
| R5 50K | |
+---------------/\/\/\/\/\--------+ +
| /|\ C3 47 uF |
| | || + |
+--------------------)------------------||-------------+
| | || |
| | |
+------------------( O ) |
| J2 Audio Output |
| - 9VDC + |
C1 .05 uf to .1 uF 50 or more volts
C2 0.1 uF 50V or more
Q1 2N3904 NPN low noise audio transistor
J1, J2 Jack to accomodate the equipment and cables
R1 470K 5% 1/4W
R2 47K 5% 1/4W
R3 10K 5% 1/4W
R4 470 ohm 5% 1/4W
R5 50K Audio Taper Potentiometer
The unbypassed R4 emitter resistor causes the imput impedeance to be
high, and it causes a loss of gain. The amplifier stage cannot have a
gain greater than the ratio of R3 to R4, so 10000/470 is about 20.
The coupling capacitors C1 and C2 are fairly small, about 1/10 uF.
This reduces the gain at low frequencies. The capacitors can be
larger for Hi Fi use. Anywhere from 1 uF to 10 uF should be OK.
The input impedance is several tens of Kohms. The output impedance
depends somewhat on the setting or the R5 level pot. It can't be
higher than 15 K, but it can be as low as the pot setting. The DC
supply should be well filtered to keep any hum and noise out of the
audio. If a lower impedance is needed to drive long lengths of cable
or a pair of earphones, then an emitter follower stage should be added
to increase the output current.
With a maximum peak-to-peak voltage at the collector of approx 8V, the
RMS output voltage should be about 2.5 to 2.8 VRMS maximum.
Total circuit current drain on the battery is less than a half
milliamp, probably around 400 uA.
----
: --
: jut <> Justin Olexy <> "http://www.voicenet.com/~jut1"
--
#======P=G=P==k=e=y==a=v=a=i=l=a=b=l=e==u=p=o=n==r=e=q=u=e=s=t======#
| John Lundgren - Elec Tech - Info Tech Svcs. | jlundgre@delta1 |
| Rancho Santiago Community College District | .deltanet.com |
| 17th St at Bristol \ Santa Ana, CA 92706 | "http://rsc.rancho"|
| My opinions are my own, and not my employer's. | .cc.ca.us |
| I have gone out to look for myself.. If I should |
| return before I get back, hold me until I get here. |
| "You can flame your brains out -- it won't take long." |
Article Reference: 06562142CAA45AC8
Date: 24 May 1996 09:24:35 GMT
Original Subject: Re: Making a battery backup.
Colin MacKenzie (cmackenzie@mail.ips.ca) wrote:
-> I feel foolish, I have never fooled around with battery backup systems
-> but it is required for a project I have designed. I would like to
-> recharge the battery while the power is on. When the power goes off
-> the battery has to switch over quickly since it is powering a
-> microprocessor. That the circuits used in emergency lights are
-> probably simple but do they switch over fast enough?
-> How do I go about doing this?
The 'standard' way of doing this involves a battery, cap, and a few diodes.
resistor diodes
Vin O-------*------------|>|----------*-------------O Vout
| |
\--/\/\/\----|>|---*--|>|-/
|
-----
--- Battery
-----
---
|
-----
--- Ground
-
Note: -|>|- is a diode with the anode on the left.
Basically, the way it works is when power is applied, it passes thru the
top diode during normal operation, and charges the battery thru the
resistor and diode. When power fails, the Vin side voltage falls below
the battery voltage, and the battery then takes over the load on the
Vout side thru the diode. The switching is instantaneous, and probably
faster than a few uS. Add a 100uf or so cap between Vout and ground to
take care of any diode switching transients.
You can remove some of the diodes if you don't need 'em; i.e. if you're
taking the output of a rectifier and running it to the Vin terminal,
the diode(s) in the rectifier will stop any appriciable current flowing.
I've used this circuit with much success in my designs.
PS: forgive the non-lucid theory of operation- I've just finished routing
the Board From Hell. ;-)
--
-----------------------------------------------------------------------
| \_ \_ \_\_\_ \_ \_ \_ \_ \_ B5(6) t w g+ s-- r- |
| \_\_ \_\_ \_\_ \_ \_\_\_ &h48,&h65,&h6C,&h6C,&h6F,&h21 |
| \_ \_ \_\_\_ \_ \_ \_ \_ __ khorton@tech.iupui.edu __ |
| "It's 5:50 AM. Do you know \/__Can *YOU* write 8085__\/ |
| where your stack pointer is?" \/ assembly? \/ |
-----------------------------------------------------------------------
Article Reference: ECB5564C01722894
Date: Wed, 2 Dec 1998 17:15:14 GMT
Original Subject: Re: DC motor control
Oh, and put protection diodes across Q1 & Q2! (for Q1, cathode towards +Vmotor; for Q2, anode towards -Vmotor). And experiment with the resistor values as needed. Rich Sulin wrote: > > We Gates Herr Holger, > > Holger Metschulat wrote: > > > Hi, > > > > I want to build a bidirectional motor control for a DC motor with one pin > > connected to GND, so I need a dual power supply (+/-Uo). How can I control > > this motor using only 2 digital (+5V/0V) lines (enable/disable and > > forward/backward)? > > > > -- > > Gruss * Holger Metschulat > > Holger * e-mail: homer@sgs.wh.tu-darmstadt.de > > * "http://www.sgs.wh.tu-darmstadt.de/homer" > > ** "Verstaerker schwingen immer, Oszillatoren nie!" (Murphy) ** > > How about this: > +Vmotor > | > / > |/ > FORWARD-------[R1]----+---[R3]---+--| Q1 > | |K |\> > -5V---[R5]----+ [D1] | > 1+---+6 | |A | > DISABLE--[R2]--+ +--+--+ GND--+ +----[D.C. MOTOR]--+ > 2|OC1| | |K | | > +--+ +--+ | [D2] | | > | +---+8 | | |A | GND > | | | | |/> > GND GND +-[R4]--+--| Q2 > |\ > \ > | > -Vmotor > > Where Q1=NPN darlington power, e.g. D40K2; Q2=PNP darlington power, e.g. D41K2; OC1=Optocoupler H11F1; R1=1k; R2=180; R3=2k; R4=1k; R5=2k; D1=D2=1N4148. > > But really, it would be nicer not to have one end of the motor grounded, so you could use existing H-bridge driver ICs. > -- > +----------------------------------------- > | Richard Sulin > | > | Please remove the anti-spam stars in > | my e-mail address in order to reply. > | > | thus: richs <at> dcdu <dot> com > +----------------------------------------- -- +----------------------------------------- | Richard Sulin | | Please remove the anti-spam stars in | my e-mail address in order to reply. | | thus: richs <at> dcdu <dot> com +-----------------------------------------
Article Reference: 834FE0BEA01E9D16
Date: Wed, 22 May 1996 18:33:32 -0700
Original Subject: Re: a good video black level clamp
Peter Gregson wrote:
>
> I'm looking for a simple GOOD black level clamp circuit for clamping a
> video signal. Currently, I use the black level timing indicator out of
> an LM1881 to indicate when the video signal is displaying the black
> level. I then use a diode-switched sample and hold to capture the DC
> black level, save this on a holding capacitor, and input this signal
> into an op-amp subtractor (the other input is a desired clamp level).
> The output of this is fed back to the bias circuit at the video
> input so as to achieve the desired clamped video signal.
>
> Peter H. Gregson, Ph.D, P.Eng,
> Computer Vision and Image Processing Lab
> Department of Electrical Engineering
> Technical University of Nova Scotia
> Halifax, Nova Scotia
> B3J-2X4
> telephone: (902) 420-7783
> FAX: (902) 422-7535
> e-mail: gregson@tuns.ca
The way I have seen this done is to clamp the actual video to the level
you want. (Or maybe I made this up?) If I remember right...
Coupling
Cap
Video In -----|(--+-------- Clamped video
|
FET switch [ ]--- Signal from LM1881
|
|
GND (Or what ever voltage you want)
If I understand what you are trying to do, I think this should work.
BTW, the LM1881 data sheet says if you have noisy video, use the RC
filter they show for the video. I think it was a 510K ressistor and .1 uf
or something like that. I ALWAYS use the filter.
Jeff Sampson jeffs39@skypoint.com
Article Reference: 100B4962673C8180
Date: 19 Apr 1996 13:55:41 GMT
Original Subject: Re: HELP designing a simple switch circut!
Well, how about red when the fuse blows, and off when the fuse is good?
To do that, wire a resister, (1 to 10 Kohms depending on the led current
rating) to the side of the fuse that will be positive when the fuse blows.
Wire the resister to the led. Wire the other side of the led to the other
side of the fuse. Now, when the fuse is ok, there is almost no voltage
drop across the fuse, so the led is off. It the fuse blows, and its normal
load is connected, then the greatly reduced load current (1 to 12 milliamp
depending on the resister selected) will flow thru the led and light it up.
If you wire this circuit backward, you may destroy the led. To prevent that,
you can wire another led across that led but in the opposite direction.
Now, no matter how you wire it up, one led will be on, and the other will be
protected.
led
+-------->|---------+
| |
+---------|<--------+-------Resister-----+
| led |
| |
=========+===== FUSE =============================+========
OR
Battery ====+=====FUSE=============+===================Load===Ground
| |
+----->|-----Resister--+----->|-----Resister -----Ground
red green
led led
Green only, fuse ok
Red and Green, fuse bad
None, power off
Opinions expressed herein are my own and may not represent those of my employer.
Article Reference: B4FEA3D1EC28DB13
Date: Sat, 20 Apr 1996 21:15:57 -0700
Original Subject: Re: HELP designing a simple switch circut!
Replying to a reply because a missed the first post.
In article <4l45t3$qs3@news.fonorola.net> marka@spots.ab.ca (Mark Achtemichuk)
writes:
> The basic idea is that a dual colored LED (red and green) lights green when it
> receives a 12V lead. When that 12volt lead disappears, the LED must turn red.
> If I remember correctly the LED's work by switching their polarity for the
> different colors. This is where I run into trouble.
> The circut will be used to test fuses in my car. When a fuse blows, the LED
> will turn from green to red. The circut will have a constant 12 volt supply as
> well as the 12 volt probe. The LED's are rated at 3 volts so I assume it would
> be easiest to put a current limiting resistor on them vs. building a power
> supply.
> Can anyone help me with this? I would appreciate anything anyone has to offer.
If the led really changes color by switching its polarity, what about this
circuit:
fuse
O============================O
| |
| |
| 12V |
| R1 + | | - |
+-----/\/\-------||||--------+
| | | |
| |
| 6V |
| R2 LED + | - |
+----/\/\-----|<|-----||-----+
|
When the fuse is OK, R1 and the 12V battery can be ignored, and
the LED will be conducting (and, hopefully, green). Adjust R2
to limit the current into the LED to a proper value.
When the fuse is blown, the 12V battery forms a serie circuit
with the 6V battery and the total voltage makes the LED reverse
biased (and, hopefully, red). You can make R1 very large
because only a little current is needed when the LED and
reverse biased; and a large R1 will reduce the power consumption
of R1 when the fuse is OK.
Note that it is the first time I heard about LED changing color
with polarity (I am a newbie). I assumed that this type of LED
operates like normal LEDs. Maybe this assumption is not true:
it seems strange to me that the very little current in the
reversed biased LED can light it.
Any comments/question/critics would be appreciated.
Louis-Antoine Blais-Morin
Electrical Engineering
labm@graf.polymtl.ca Ecole Polytechnique de Montreal
lblais@ee.ubc.ca University of British Columbia
Article Reference: A8AF2156777F4F06
Date: Mon, 17 Jun 1996 15:31:50 GMT
Original Subject: Re: [Q] HELP on Single-Supple Op-Amp Biasing......
On Sun, 16 Jun 1996 23:12:08 GMT, aleung@engr.ukans.edu (Artur Leung)
wrote:
>Hi,
>
> I would like to receive some help about dc-biasing the single
>supply op-amp.
> The one I am using is the LM324N from National Semiconductor and
>my primary supply voltage will be +12V. Assume that there is an input
>signal of 1V peak sine wave. In the data sheet National Semi. states
>that the input voltage cannot go lower than -0.3VDC. If I want to use
>the LM324 in active band-pass filters, non-inverting amp., and
>inverting amp, what should I do to dc-bias the input voltage source?
>The information given by National Semi.'s data sheet is limited. It
>only suggests using a reference voltage which is half of the supply
>voltage, but I don't know where to put the reference voltage.
> Thank you very much for the help in advance.
Have you tried using a capacitor-coupled input?
-o- V+
|
<
> R1
<
| |`.
| | `.
Signal Cin o-------|+ `.
In o-||----o | >---o----- To other stages
| ,-|- ,' | (DC biased)
< | | ,' |
> R2 | |,' |
< | |
| o---/\/\-----'
| |
| <
| >
| <
| |
-o-----o- Gnd
In this case (a non-inverting input stage), R1 and R2 determine the
DC bias point for the rest of the circuit, and the AC portion of the
input signal is coupled (added on) to the DC component.
Cin needs to be large enough to allow the lowest signal frequency to
pass -- Cin forms an RC high-pass filter with R1 and R2, with the 3dB
cutoff frequency at 1/(2 pi Cin (R1 // R2)).
The output of this stage will already be DC-biased, and other stages
downstream can be dc-coupled from this one.
-dr
Article Reference: 04B9A86D0E920500
Date: Fri, 23 Aug 1996 05:22:49 GMT
Original Subject: Re: Help my rusty math please
Thus spake jwisnia@aol.com (JWISNIA) in sci.electronics.misc:
>Today I was trying to show someone that the total energy of the I^2*R
>losses when resistively charging a capacitor from a dc source is exactly
>equal to the energy stored in the capacitor after it has charged for an
[snip]
>I'd appreciate someone helping me out here, my calculus is rusty.....and
>never was very good anyway!
/ + Vr -
___/ ___________/\/\/\/\____
_|_ -----> I R |
/ + \ ___|___
| | V C _______
\_-_/ |
|_____________________________|
<---- I
When the switch is closed (time t=0), the capacitor acts as a short
circuit. Current is V/R. Once the cap charges up, current flow is 0.
The charging process is exponential, so instantaneous current is:
V -t/RC
I=--- * e
R
(This means I'm waving my hand, instead of actually doing the
integration!)
Instantaneous voltage across resistor is:
-t/RC
Vr= V * e
Instantaneous power dissipated in resistor is voltage * current:
V * V -2t/RC
Pr= ----- * e
R
Integrate this to get total energy dissipated in resistor:
oo /
| V * V -2t/RC
Er= | ----- * e dt
| R
0 /
_ _
| |
V * V * R * C | -oo 0 |
= ------------- * | e - e |
R * (-2) | |
|_ _|
_ _
V * V * C | |
= ---------- * | 0 - 1 |
(-2) |_ _|
V * V * C
= ---------
2
The capacitor charges to voltage V, so the energy it stores is also:
V * V * C
Ec = ---------
2
>BTW, we got into this while discussing "charge pump" voltage multiplying
>power supplies. I'd read an article in a recent mag which claimed they
Here is one charge pump circuit. It works best if the square-wave
oscillator has a low output impedance, and if Schottky diodes are
used:
VCC
O
|
_|_
\ /
__V__)
| | + ( | /
VCC | | | |\ |
O .----+ +------+---| >|------+---------> 2 * VCC
| | | | |/ | |
| | | | + / + | +
______|_____ | ___|___
| | | _______
| square- | | |
| wave +---+ ___|___
| oscillator | | _____
| | | ___
|____________| | + | | / .
| | | | | /|
| '---+ +-----+------|< |------+---------> -2 * VCC
| | | | | \| |
___|___ + | | | / |
_____ _|_ ___|___
___ \ / _______
. __V__) + | +
( | ___|___
| _____
___|___ ___
_____ .
___
.
(I love ASCII art!) More capacitor-diode stages can be added to get
higher output voltages. I don't know how efficient it is.
Hope this helps.
>Thanks guys...
>Jeff Wisnia W1BSV
>Winchester, MA.
> "Common sense isn't very common..."
--
_ _ _ _ _ _ | Sales rushes in
|_| |_| |_| / |_| | | where Engineering
_| |_ |_ /_ |_ | @ nconnect.net | fears to tread.
Article Reference: DFAF0857DAD60FF1
Date: Mon, 22 Jan 1996 07:22:27 GMT
Original Subject: Re: Controlling LED intensity
On Thu, 18 Jan 1996 10:05:52 +0000, Simon Watson <simon.watson@dur.ac.uk>
wrote:
>Hi! If anyone could help me with ideas/suggestions for solving the
>following problem I'd be really grateful.
>Basically, I need to use the output of a DAC to control the intensity of
>two infra-red LEDs (only one lit at a time). Each LED will only be
>pulsed for around 10uS, allowing forward currents of up to 1A (from the
>datasheet), with the whole thing under control of a microprocessor. The
>DAC is a 10-bit voltage output device, allowing for a linear resolution
>of 1mA or so. An important requirement is that each intensity level is
>repeatable, so is some sort of temperature compensation required?
>What's the best way of achieving this? Any help will be much
>appreciated.
Here's a wild shot in the dark. I'm a complete newbie, but this might
*just* work...
Basic circuit:
DAC output -> matched (monolithic) transistor current mirror ->
current multiplier -> current-controlled LED.
---o----o-------------o-----------------o--- Vcc
| | | |
> | | |
R1 < | | |
> / \ |
| |v v| _|_
o--| Q1 Q2 |--. \ /
| |\ (pnp) (pnp) /| | _V_ \\
| \ / | |
| | | | | LED
o----|-------------|----' |
| | | .-------------'
o------|----o | |
| | `--( )
From DAC < < ( ) 1:20
> > R3 | Current multiplier
R2 < < |
| | |
| | |
-o----o-----------------o------------------ Gnd
The DAC voltage will control the Ic of Q1, which will then cause Q2 to
reflect the same current. That current is then multiplied by the current
multiplier (TLxxx series or something or other ... don't really know) through
the LED.
Q1 and Q2 should be a monolithic pair (to compensate for temperature
variation), and R3 should be 1% tolerance. As long as you can control the
variation of the DAC's voltage over temperature, the current through the LED
should be pretty much proportional to the input voltage.
Typical values (for 0-5V DAC output, 10V Vcc):
R1 : 1.5k Ohm
R2 : 20k Ohm
R3 : 100 Ohm (1%)
This will give a 0-1A current swing through the LED for 0-5V swing from the
DAC. Max. dissipation of Q1 is 500mW, Q1+Q2 dissipates 1W.
Comments anyone? I'm still trying to figure this out myself.
/+_,--___ _|
++( P \_\ \ ---------------------------------
|++ U ) ) \_\ Dave Rahardja at ORU, Tulsa, OK
-++ ( P /__/\:: ---------------------------------
/++++__--~ |/
Article Reference: 56D5D261E5069AED
Date: 11 Jun 1996 14:14:29 GMT
Original Subject: Re: Constant Current Source (1-5 mA)
Geoff Raynak <graynak@u.washington.edu> wrote:
>Sorry for the confusion. I meant to say that I tried the circuits
>provided by H&H but they don't seem to be _precision_ in the sense that
>they cannot maintain the load against the resistance. I am trying to
>sense geometry changes in a tube filled with saline. Using a resistance
>method, i need a constant current source (very constant) to measure
>resistance-induced voltage changes without any current-induced voltage
>changes.
Basic circuit
<-- constant current
|\ d
Vref --------|+ \ g |--------- Load ----- Vcc
| > -----------| | Q1
+--|- / |--+
| |/ s | transister = n-channel mosfet
+-------------------------+
|
R1
|
gnd
This circuit provides an approximatly constant current of Vref/R1
1.A resister (1K) may be required between op-amp output and fet gate to
surpress oscilations.
2. A resister (100K) from the fet gate to ground to insure the fet is
off when power is applied could be useful.
3. Use diff amp to sense load voltage.
Opinions expressed herein are my own and may not represent those of my employer.
Article Reference: ECB5564CF4B57AD9
Date: Thu, 3 Dec 1998 01:05:26 GMT
Original Subject: Re: Simple crossfader circuit?
Hello Mr. Rogers,
This is the classic way to do it:
Left in-----[R1]--+--[R2]--+--[R3]------+
| | +-----\ |
| +--|- \ |
[P1]<--+ | |Op-Amp |-+-- Out
| | | +|+ /
| GND | |+-----/
Right In ---[R4]--+--[R5]--+ |
GND
where R1=R2=R4=R5=10k, R3=30k, and P1=10k "W"-taper pot (a linear
taper is OK for testing).
Michael Rogers wrote:
>
> Can anybody show me a simple circuit to fade between two audio sources?
> I know how to make a simple mixer with an op amp and how to calculate
> the gain on each channel, but since I always want to fade out channel 1
> when I fade in channel 2 (and vice versa) I would like to be able to
> have both controls on one pot. Is there a more elegant way than using a
> dual-ganged pot?
>
> Thanks in advance.
>
> Michael Rogers
--
+-----------------------------------------
| Richard Sulin
|
| Please remove the anti-spam stars in
| my e-mail address in order to reply.
|
| thus: richs <at> dcdu <dot> com
+-----------------------------------------
Article Reference: 9DAB051521336DB0
Date: Fri, 24 Jul 1998 03:02:44 GMT
Original Subject: Re: Current Sensing? Any ideas?
In article <35b9a4f0.90061240@news.giganews.com>, none@this.address (Cyber Bandit ¥ßûÿ) wrote: > Probelm: > > I am trying to sense a current in an AC line (which is run through a > circuit board) The current is from 20ma > 8A, 120VAC @ 60Hz. > > To make matters worse, it is on a different phase of power than the > circuit board is running on. Any help would be appreciated, oh, BTW, > I must keep the cost at about $1.50 for all parts, so that rules out > the current Xfmr. > Hi, May be you can try this circuit : Rs I ----------- Power line -----+----/\/\/\/\-+------------<--------------- | | R1 |/ | +------------+-----/\/\/\------|/ --------------------------------- | |/ \ + - / | \ / \ \ / / R2 \ / \ \ / / \/ | | | +--------------------- Vo | <----------------------| You can calculate Rs, R1, R2 according to your current and your Amp-Op. Regards, -------- Michael A. Purwoadi Group for Instrumentation and Control Technology
Article Reference: 5B680B4E0BE5E522
curve tracer
Date: 28 Sep 1996 00:47:35 GMT
Original Subject: Re: Curve Tracer?
In article <588.6844T1120T285@concentric.net> who3@concentric.net (who3) writes: > Hoping this is the right place. > I am looking for information on what i think is called a "curve-tracer" > BUT I'm really not sure at all of the name of the device. It's function > is to test transistors & diodes, but also works with capasitors. > it is a simple circuit when used with an O-scope produces a > lazious(sp?) pattern, a circle, or a calapsed circle that shows > the knee of a junction graphically on the scope. I know the circuit > was published in 2 magazines, but have no way of finding which > or when or where. > Please, any direction/correction would be much appreciated. > Thanks > Wes Curve tracer design: ------------------- The design of a curve tracer is simple in principle (description here for bipolar transistors): For the horizontal (collector supply) you need a variable ramp generator. If your scope has a sweep output, then you can derive it from this - if you are not interested in frequency response, an audio amplifier may be adequate with a volume control to adjust the amplitude. For the base drive you need a programmable current source capable of putting out a series of constant currents for the base drive. Here, a counter driving a D/A set up for a current output mode. Use the trigger output or sweep output of the scope to increment the counter so that it sequences through a set of say, 10 current settings. Then, you need some way of sensing collector current to drive the vertical channel - a small series resistor in the emitter circuit, for example. For simple diode tests, you can just use a variable AC voltage source like a variable isolation transformer with a current limiting resistor resistor across the diode. X of the scope goes to the output of the transformer. Y of the scope goes across the diode under test. See the section: "Quick and dirty curve tracer". Then, you can jazz it up with microprocessor controlled on-screen display. Quick and dirty curve tracer: ---------------------------- I threw the following circuit together in about 10 minutes. With minor modifications, it is capable of displaying V-I curves for diodes, zeners, transistors, resistors, capacitors, inductors, etc. I used a 12 VAC transformer just because it was handy. You can use anything you like as long as you understand the safety implications of higher voltages and make sure the components you use can withstand the power that might be dissipated in them if the Device Under Test (DUT) is a dead short. In addition, it is bad form to blow out the DUT while testing it! A signal generator driving a small audio transformer could also be used if it is desired to test components at frequencies other than 60 (or 50) Hz. o-------+ R1 )|| +------/\/\/\------+-------o Horizontal input of scope )||( | (Voltage display) )||( o Adjustable AC )||( + from Variac )||( DUT )||( - )||( o )||( R2 | )|| +--+---/\/\/\------+-------o Scope ground o-------+ | T1 +-----------------------o Vertical scope input (inverted) (Current display) CAUTION: turn down the intensity of the scope so the spot is just barely visible so that when there is no input, you don't end up drilling a hole in the face of the CRT! R1: Current limiting and phase shift control. I used 500 ohms which works well for small signal semiconductors and capacitors around 1 uF. R2: Current sense. I used 10 ohms and put the scope on the one of the .1, .2, or .5 V/cm ranges. T1: Small power transformer. I used the 12 VAC wall wart from an obsolete modem. This will supply a voltage of up to about 17 V peak to your DUT. A Variac provides a convenient method of adjusting the voltage applied to the DUT. Modify these values (selector switches might be nice) for your needs. Try this with a 5 V zener diode first since you will know what that should look like - just as in the textbook! For transistors, a source of base current is needed. You can be fancy or simple. For a simple source, I used a variable 0 to 15 V power supply and a current limiting resistor. Since we know that the voltage drop across the B-E junction is fairly constant at around .7 V (for silicon), the output of the supply can be calibrated in terms of base current. --- sam
Article Reference: 8FFB1F8996BE1578
DC restorer
Date: Fri, 17 May 1996 19:13:37 GMT
Original Subject: Re: LED Voltage..Look at this circuit
"James P. Meyer" <jimbob@acpub.duke.edu> wrote: |On Thu, 16 May 1996, James P. Meyer wrote: |> On 16 May 1996, Clyde Manning wrote: |> |> > I imagine you will get a lot of answers to this. |> |> Some right, and some wrong.... 8-) | Oops.... One of mine was wrong too. In particular, this part: |> > |> > If you are using an AC input to the opamp you risk damaging the opamp |> > every time the waveform goes negative. Some opamps become very unhappy ( |> > regards to Dr. Budak at Colorado State University) when their input swing |> > below their negative rail. |> |> The post clearly states that he's using plus and minus 9 volt |> supplies. Eighteen volts, center tapped. AC would have to be larger |> than 18 volts P-P to swing below the negative rail. | Looking back, I see that he's probably only got +9 volts instead |of a center tapped supply. In that case, he would need to be concerned |about biasing the input so that the AC at the output could swing between |+9 and ground. The circuit looks like a gain-of-one buffer, so he should |make sure that the input has a no-signal bias voltage of about 4.5 volts |as well as the AC input. |Jim If he has an AC input signal, he could get around this input problem by putting a DC-Restorer on the input to the Op-Amp. This consists of a capacitor and a low drop diode such as a germanium (small signal diode). This will shift his AC signal so that it will "ride" on top of the DC axis. Not perfect but this may suit his needs. There is a slight dip below the 0V DC axis but not much. I had a -0.1V total dip in an actual application with a 6Vp-p squarewave input signal but most opamps that are designed to run rail-to-rail can tolerate up to -0.3V before latch-up which is fine. My circuit was slightly different than below but the idea is close enough. The circuit looks like this: || |\ >-------||-----+----------+ \ || | | \ 0.001uF --- | /---+--- Ge diode / \ ----- / | --- | |/ | | |__________| to GND This only works as well as long as the input resistance to the OP-Amp very is high. If you start connecting other circuitry around it the performance degrades dramatically. It is best to connect as he has, a hi-Z input Op-Amp in the unity gain configuration as he has it for the best results. If you do do this Greg, check it out on a scope. If you have any problems I'll let you know what I did differently. -Vince vhadley@ee.utah.edu
Article Reference: 6DDBEC69F0150A50
detect either logic transition
Date: Thu, 25 Jan 1996 13:33:04 UNDEFINED
Original Subject: Re: Help with design
In article <4e34s3$ok5@nyx.cs.du.edu> slindsay@nyx.cs.du.edu (Steve Lindsay) writes: >From: slindsay@nyx.cs.du.edu (Steve Lindsay) >Subject: Help with design >Date: 23 Jan 1996 10:11:31 -0700 >I have eight external lines any one of them can be high or >low (+5v high or 0v low). When any one of them change >state I would like another line to break connection for just a >very short time and then reconnect. The reconnect part >has to be automatic. (It will be causing a hardware >interrupt). This one line needs to reconnect right away >before any of the other lines change state. The eight lines >will change state at about 5000 blips a sec. >Anyone have any ideas what I could do this with? Is there >a chip that will take in eight inputs and if any change it >would output something happened on just one line? >Any ideas for the reconnect part for the interrupt line? It >needs to do switching. It would normally be closed but >when it was told to open it would only open and then shut >right away automatically. >Thanks for any help! What you can use to detect a line-change is the following: __ line o---------\\ \ | __ || >-----> to 8-input nor-gate -|__|---//__/ | exclusive-or gate --- --- | ___ /// I think the resistor can be something like 10K, the capacitor 4.7nF So each line needs one ex-or gate, one capacitor and one resistor. The idea is that a line-change will make the inputs of the ex-or to differ from each other for a short time (about 50 microseconds), and the resulting High-pulse on the output can be nor'd (or you could even use 8 diodes for a wired-or. I don't really understand what you mean by breaking and reconnecting the interrupt line; most interrupt lines need to be pulled down or something like that to generate an interrupt. Here that can easily be accomplished by using a transistor after the 8-input or- or nor-gate. Success, Richard Richard Rasker Calslaan 54-11 7522 MG Enschede Holland tel. +31-(0)53-4350834
Article Reference: BA79C16DF46BDA9F
dimming flourescents
Date: 01 May 1996 21:12:55 GMT
Original Subject: Re: Another dimmer Q: flourescents?
In article <mdevour.433.000C633F@sinbad.ph.gmr.com> mdevour@sinbad.ph.gmr.com (M. G. Devour) writes: <snip> > 'Meantime... Is there technology for dimming flourescent tubes? What? Where? > How much? How is it done? Is it possible to slow start them? > Thanks in advance. Here is some info I pulled off of one of these groups on diming fluorescents: --- sam Dimming fluorescents 1: ---------------------- (From: Peter Miller (p.miller@elec.gla.ac.uk)) I've tinkered a bit... the trick is to keep the filaments at the ends of the tubes warm. You will NOT be able to dim down to zero - probably about 25% at best. Here is a possible circuit: Dimmer +--+ Choke AC Live/hot o--------| |--^^^^^^^^--+--------------+ +--+ | | | | +---------------------+ | | | | +-+ +---------------+ +-------+ | )|( | | | Filament )|( | -|---|- Transformer )| +------------------+ | ^^^ | Tube 2 * 6.3V 2A )| | | )| | | )| +------------------+ | ^^^ | )|( | -|---|- )|( | | | +-+ +---------------+ +-------- | | | | AC Neutral/cold o--+---------------------+--------------+ The lamp must be in a earthed / grounded reflector fitting. The metal end caps of the tube must be connected to the reflector. The dimmer MUST be a 'hard fired' dimmer capable of operating an inductive load. The choke is a standard type for the tube in use. Play with an inexpensive everyday tube before using expensive aquaria ones. with a 240V supply a 4 ft 40W tube operates ok. The main difficulty with this circuit is in getting the tube to start - starting is greatly helped by a grounded reflector fitting and connecting the metal end caps of the tube to the reflector (Dont ask - it works!). The transformer can be a standard valve filament transformer - use a separate transformer for each end of the tube if you are unsure of the insulation between the secondaries of any transformer that you buy. As the tube draws less current the voltage across it rises, turning up the heat in the filaments. At start up, maximum voltage is across the lamp and so the filaments are fully on. All dimming ballasts/chokes use some scheme to add extra heat to the filaments at dim running. An undimmed tube draws enough current to keep the filaments warm by itself. There is no glow starter or other starting device in the circuit, so the lamps tend to come on smoothly with no flickering. Shorter tubes are easier to start. New slimline tubes are a real pain to start. Dimming fluorescents 2: ---------------------- (From: Bruce G Bostwick (llbgb@utxdp.dp.utexas.edu)) This applies to rapid start fixtures. If the fixture says "RAPID START" somewhere on it or on the package it came in, the internal schematic will be roughly as follows: || +-------+---------o AC line H o---------+ ||( +----+ to both pins on )||( ( <-6.3VAC one end )||( +--------------o )||( )|| +=---- 2kVAC* -------------------------+ )||( | )||( +--------------o | )||( ( <-6.3VAC to both pins on | AC line N o---------+ ||( +----+ the other end | || +-------+---------o +-+ V Most of the 48" 40-watt "shop light" type of bulbs use this. *The high voltage winding of the secondary is on a branch magnetic circuit that limits the output current to the mercury discharge. Open circuit voltage will be in the kilovolt range, while the voltage across a lighted tube will be somewhat less than that and _exceedingly_ non-sinusoidal. If you're using the big bulbs (F96T12's for example) the ballast will only have the high voltage winding and the cathodes are heated by ionic bombardment from the mercury arc. These take a bit longer to light up when the power is turned on. If you want instantaneous on/off control, I'd suggest using 4-footers and linking up two ballasts in such a way that the cathode heaters are driven from one which is always on, and the arc is driven from the other which is turned on and off as you desire. They won't last too long that way, but it will work better for show effects. The cathodes could be driven from a pair of low-voltage filament transformers, but be sure to isolate them well -- or you could use a ballast with a blown HV secondary ... Another suggestion: Use solid-state relays to drive the ballast primaries. These are fairly cheap and provide clean current-zero-crossing switching even with very reactive loads (I've used them for such! ;-) and provide a neat and rugged way to connect the lights to logic controls such as your computer -- great for light sequencing.
Article Reference: 0E19D978276CC927
diode clamp
Date: 10 Sep 1996 21:37:55 GMT
Original Subject: Re: Thevenin equiv. circuit question
In article <50ka4q$9ac_002@bzn08.imt.net>, ronman@imt.net (ron newman) wrote: > Could somebody explain in an intuitive way how to generate the thevenin > equivalent circuit for something with an input, an output, and a DC offset > voltage? > > What it is is a diode clamp. Signal goes through a 1K resistor, but the > output is tied to a diode, which in turn is tied to 5 VDC. But this 5V > comes from a voltage divider: the middle of a 2.2K and a 1K resistor, with 15V > applied at the 2.2K end. > > I'm trying to construct the Thevenin equivalent for the ENTIRE circuit. > I know that you figure Vopencircuit and Ishortcircuit, but at what points? > At the 5VDC offset? At the output, which is AC, so constantly changing? > > Huh?? > Thanks. > > ***************************************************************** > Ron Newman > Troubadour Technology > "http://www.imt.net/~ronman/animal.htm" > ronman@imt.net It sounds like your circuit looks something like this: 15V 1K | Vin ---/\/\/\-------|-------- Vout / | \ 2.2K | / ----|>|---------------| (4.7V) diode / \ 1K / | gnd The diode is non-linear, so you can't come up one Thevenin equivalent for all cases. But there are basically only two cases to consider: when the diode conducts, and when it doesn't. When the diode is not conducting, i.e. for Vin < 4.7 + 0.6, then your Thevenin circuit is trivial: RT = 1K VT = Vin ---/\/\/\--------- Vout For the more interesting case of the diode conducting (Vin > 4.7 + 0.6), then the diode can be approximated by a DC battery of about 0.6V. The 2.2K and 1K network can be combined into a Thevenin equivalent of: 1 1 * 2.2 VT = ----- 15 = 4.7V and RT = -------- = 690 ohms 1+2.2 1 + 2.2 The circuit now can be redrawn: 1K Vin ---/\/\/\-------|-------- Vout | | | | 690 ---||||-------/\/\/--- 4.7V +| |- 0.6V Combining the battery and the 4.7V: 1K Vin ---/\/\/\-------|-------- Vout | | 690 -----/\/\/--- 5.3V This reduces to the final Thevenin equivalent circuit: RT VT ---/\/\/\------- Vout where: 1 * .69 RT = -------- = 408 ohms 1 + .69 and .69 VT = ------- (Vin - 5.3) + 5.3 (remember, only for Vin > 5.3V) 1 + .69 or VT = .41 Vin + 3.1 For example, if Vin = 6.0V, then VT = .41 * 6.0 + 3.1 = 5.56V : 408 5.56V ---/\/\/\------- Vout
Article Reference: 6BD1F1C1F6A35CDC
draw power from RS-232 control lines
Date: 30 Apr 1996 11:18:41 GMT
Original Subject: Re: Power from RS-232
Daniel P. Siu (dps@west.net) wrote: : I need to derive some power (fairly low, a few mA at 5V) from a RS-232 : serial connection. I remember there are ways to do this (in fact, : some techniques are used in serial mice). Any help is appreciated. Here's a circuit I have used (it doesn't guzzle as much current as a 78L05 which is obviously a much simpler solution): +----> DSR 2 X 1N4148 | +-------------+----< DTR +-----------------+-------------+ | +------+----< RTS | |C | _|_ _|_ | | _____ |/ | \ / \ / +----> CTS +---[_____]--+--| BC550 | --- --- | | 10k | |\ | | | +----> DCD | | |E +-----+------+-----------> V++ +__|__ | +----+---------------------+-----------> VDD _____ +-----+ | _____ | +5V 220u | | | +---[_____]---+ | | | C| 22k | +__|__ | __|__ \| BC547 | _____ /// _____ |---------------------+ 47u | 100n | /| | | | E| _____ | | | _|_ +---[_____]---+ /// /// \ /^ | 18k --- | LO-I RED LED | | | | /// /// Strapping DTR/DSR and CTS/RTS/DCD is just to implement a "null-modem" connection, the power comes from DTR and RTS. On a typical PC about 5mA can be drawn from the regulator. David -- david.tait@man.ac.uk
Article Reference: 66EB6EE5DB583808
electronic load switching
Date: 18 Mar 1996 17:09:30 GMT
Original Subject: Re: Parallel port interface
Tom Klepl <tk@enterprise.ca> wrote: >I want to build an interface between a model car (2 motors) and the >computer paralell port, which will allow me to control the model car >with the computer. I have been able to toggle the state of each pin of >the paralell port, but I don't know how to build the electronics to >control the motors. I believe I need transistors to be used as >switches, but I don't know which ones to get and how to build the >circuit. A couple questios, do you need speed control? Do you need to reverse the motor? I give a couple very basic circuits below to control a power circuit from a digital signal for simple on/off... 5 to 20VDC | _|_ This circuit is a basic way | | to turn on relays. LED's or | |Load lamps with a digital line. It | | is good for about 0.5A depending |___| on the transistor. Use a | 2N2222 for starters. | / 10K |/ Digital Out ------/\/\/-------| NPN transistor | |\| | \ / | 10K \ | / Gnd \ / | | Gnd If you are powering an inductive load you need to protect the circuit against the inductive kick that can be generated when turning the load off. Inductive loads are generally anything that has a coil of wire, including relays or DC motors. A diode across the load is the usual solution... +DC | |________ | | | | __|__ | | | | | | __|__ Load | | / \ diode | | /___\ 1N4148 for small relays | | | 1N4001 for larger stuff |_____| | | | | | |________| | | to transistor switch For more current use a power MOSFET in place of the transistor. MOSFET's have a lower 'on resistance'. This means it will run cooler at higher current levels than a bipolar transistor. 5 to 20VDC | _|_ | | | |Load | | |___| | | |D 10K |--- Digital Out ------/\/\/-------| MOSFET |->- |S | | Gnd If you want to completely isolate and protect the computer port against accidental damage an opto-isolator is a good idea... +5V | | / \ R2 / 470 R1 \ 470 |_________ Isolated digital out Digital ---\/\/\------- | to transistor switch. | / _|_ |/ \ / -> | The LED and transistor are sold in --- |\| a single package as a opto-isolator. | \ Higher voltages than 5V can be used | | on the output, but R2 must be raised Gnd --------------- | to keep the current down. Gnd Just a couple ideas to play with... Have fun. --------------------------------------<>-------------------------------------- Andrew Cooper Embedded Systems Designer mailto:acooper@pcc1.com Power Convertibles Corporation Tucson, AZ "http://www.pcc1.com" A man whose only tool is a hammer, sees every problem as a nail. -Unknown It is only myself who is reponsible for the above drivel, any opinion is mine, any dissagreement is with me, don't bother to bring the corporation into it. --------------------------------------<>--------------------------------------
Article Reference: A37E511A69AFC2EE
float battery charger
Date: 19 Sep 1996 05:19:25 GMT
Original Subject: Re: Gel Cell charging IC?
In article <51mum0$86h@viking.mpr.ca>, Bob Wilson <bwilson@mpr.ca> wrote: [.. much good stuff..] I've used the LM317 and a few parts to make chargers and have good luck with them. For example: I ------- O R1 VI----------------| 317 |--------------/\/\/\/\/\---------- ------- | | | A | / / | | R2 \ R3 \ | | / / | | \ \ | | | | Q1 PNP | | | \| | | | |---------------- ---------------/| | / \ R4 / \ | GND R2 and R4 sets the "float" voltage. R3 sets the "Fast charge" voltage. R1 saturates Q1 and sets the current limit. Add bypasses: from U1-I to Gnd from U1-A to U1-O from U1-O to Gnd -- -- kensmith@rahul.net forging knowledge
Article Reference: F32E556EF97591EE
following pre-regulator
Date: Fri, 19 Jul 1996 17:17:17 BST
Original Subject: Re: How do I make a 80V power supply with 317's and short ci...
In article <31EEC08C.7F1A@eduserv.rug.ac.be>, Vandenbruaene Jon <jjvdenbr@eduserv.rug.ac.be> wrote: > > I am making a +80/-80 power supply with 317's and 337's. It must give > a current of about 1A max. How do I make it short-circuit proof, > because when the output is a short, the input terminal is at 80 V > and the output at 0 V, and the 317 can only have a 40 V input-output > difference so it will be broken. (I only use voltages between > 80V and 40V, so it is not really a "normal" power supply) > > I also want to put in a adjusting current-limiter-function, so I can > change the max. current from 100mA to 1A. What you need is a 'following pre-regulator. The attached UUE/GIF file shows the idea. The two resistors define a point half way between input voltage and output. The power transistor 'follows' this point so that it shares voltage and power with the regulator proper. You can put current limit around this follower. If you need more details please contact me. <iimage A> RiscOS filetype : 695 -- /| Richard Torrens - 4qd@argonet.co.uk / | / | 4 Q D / | | We manufacture / /| | MOSFET controllers for battery operated motors / / | | See us on "http://www.argonet.co.uk/users/4qd" /_/__| |____ our www site contains FAQ sheet on motors & controllers /_____ ____\ and a selection of interesting circuit diagrams / _ \| | _ \ | | | | | | | | Phone/fax +44 1638 741 930 | |_| | | |_| | \__\_\ |____/ We use an Acorn RISC-PC 32 bit RISC computer
Article Reference: F3123CC9960A2ABF
frequency doubler
Date: 11 Jul 1996 22:07:52 GMT
Original Subject: Re: Frequency Doubler
Nicholas Gray (cncgsc@tevm2.nsc.com) wrote: : John Albert wrote: : > : > Does anybody have any ideas on how to build a : > frequency doubler. I would like to take a square : > wave signal (0 to +5v) and double the input : > frequency. The input frequency can vary between : > 1Khz and 500khz giving 2Khz to 1Mhz output. : The most simple way to do this is to use an exclucive-or (XOR) gate and : an inverter or two: : _____ : INPUT -------------| | : | | | : | |\ | XOR |----- OUTPUT : --| >O-----| | : |/ |_____| : This circuit produces rather narrow output pulses. You can make these : pulses wider by using two inverters in series where I show one inverter. This approach will not be practical for the huge frequency range he wants. Even a relatively narrow glitch at 2KHz will be longer than the entire period at 1MHz. The only sensible alternative is to use a simple PLL (like the 74HC4046) and a divide by 2 flip flop in the feedback loop. All you need is the 74HC4046, a 74HC74, 2 resistors and a cap. The output is a 5% duty cycle at all frequencies. Beats the heck out of a glitch pulse output. Bob.
Article Reference: 93E7FBECA08BC856
gate feedback crystal oscillator
Date: Sun, 17 Dec 1995 22:21:49 UNDEFINED
Original Subject: Re: Simple Crystal and Clock Qustions
In article <4aps73$sgd@hecate.umd.edu> LEFRAK-B-25@ writes: >From: LEFRAK-B-25@ >Subject: Simple Crystal and Clock Qustions >Date: 14 Dec 1995 18:59:47 GMT >I am trying to use a National Semiconductor ADC1038 A/D converter. I want to >attach an external clock to the Cclk input. My question is how do I hook up a >simple crystal to this input. If this is trivial please forgive me, I am just a >hobbyist...Below is my plan is it right, kinda right or WRONG... >I got a CY3.57Mhz crystal. I will take two 15pf caps and place them between >ground and each lead of the cystal. I will also place a 1Meg resistor between >each lead of crystal. >Now I take one side of the crystal and put it into a CMOS inverter. I take the >output of this inverter and tie it together with the other side of the crystal >and then put both through another inverter. >I think I can then take the output of the second inverter and put it into the >Cclk input. >Any help would be greatly appreciated... >Joe >jkemp@bss1.umd.edu 14 MHz clock: Aa you pointed out correctly, a crystal must be used with some active elements to function as an oscillator; a complete schematic: |\ |\ _ _ ----| >o----------| >o----> out _| |_| |_ 3.57MHz | |/ | |/ | ___ | |----|___|----| | 1M | | | | | _ | |1K5 | | | | |----|| ||----| | |_| | | crystal | --- --- --- 33pF --- 18pF | | ___ ___ /// /// The 1K5 resistor limits the power dissipated in the crystal. So your plan was (almost completely) correct. Success, Richard Richard Rasker Calslaan 54-11 7522 MG Enschede
Article Reference: 0FD2C185F76C446C
gate snubber
Date: 25 Mar 1996 01:10:10 GMT
Original Subject: Re: About Mosfet voltage. (I screwed up!!)
The correct configuration is as follows To plus supply | __|__ | L | | O | | A | | D | |_____| | |___________ | | __|___/ | / / \ 15v | /---\ | | | _|_ | 45v \ / | ----/ | / | ___| Drain | | | | |_______|Gate (note the FET is N-Channel in this case.) _|_ | 15v \ / | -----/ |___ Source / | | | | | | | | | | | | | | |___________| | | Typically grounded
Article Reference: CCCA178269F29DED
generate HV with TV flyback
Date: Tue, 9 Jan 1996 16:43:29 GMT
Original Subject: Re: TV Flyback QUESTION
In article <DKwstC.90z@animal.inescn.pt> jml@bee.inescn.pt (Justino M. R. Lourenco) writes: > Hello ! > I would like to know if i can connect directly the primary of a TV > Flyback Transformer to 110 V a.c. to get H.V. > The primary of the TV flyback has several primary Windings , how > to deal with that when connecting to 110 V A.C. No. It requires a high frequency waveform to provide any substantial output. Furthermore, it WILL melt down if you try 110 VAC 60 Hz without current limiting. None of the windings are suitable. If you want to play around with a flyback to generate high voltage, try the following or email me for a dual 555 based HV generator. Both use low voltage DC for the primary drive power. This is safer for you and will be less costly in the long run as you blow up fewer expensive parts trying to get it to work! --- sam Simple High Voltage Generator - 12 V in, 12,000 V out inverter. --------------------------------------------------------------- *VCC Q1 +-----------------C || o | C || | B |/ C C || | +------| 2N3055 C || | | |\ E 5T C || C-------|>|----------o +HV | | | C || C HV Diode, usually | | -_- C || C built in | | C || C +--|--------------------------C || C | | Q2 _-_ C || C | | | C || C Secondary (HV) winding, | | B |/ E 5T C || C intact. | | ----| 2N3055 C || C | | | |\ C C || C | | | | C || C | | | +-----------------C || C | | | || C | | ------------------------C || C-------------------o -HV | | 2T C || | | +----------C || | | | 2T C || T1 - Flyback transformer from BW | +--------------------------C || or color TV or computer monitor. | | | R1 | R2 +----------/\/\/\--+--/\/\/\---+ 110 27 _|_ 2W 5W - Read in Entirety! 1. Obtain flyback transformer with known good HV secondary winding. primary may be left intact if it is known to be in good condition - non shorted. A flyback removed due to failure may be used if it was the primary that failed and the primary turns can be removed without damaging the HV secondary or losing the secondary return connection! Flybacks fail in both ways (primary and secondary). 2. Wind 10 turn center tapped drive winding and 4 turn centertapped feedback winding using #16-20 guage insulated wire. Make sure both halves of each coil are wound in same direction. 3. VCC should typically be in the range 12-24 volts at a couple of amps. Circuit should start oscillating at around 5 volts VCC or so. If you do not get any HV out, interchange the connections to the transistor bases. Heat sinks are advised for the transistors. Be aware of the capability of your flyback (BW monitors up to 15KV, color up to 30 KV). You risk destroying the secondary windings and/or HV rectifier if you get carried away. Running this on 24 volts will probably cause an internal arc-over in a small flyback, at which point you start over with more caution and a new flyback. 4. Actual output will depend on turns ratio of the flyback you have. For typical monochrome computer monitors or video display terminals, you should be able to get around 12,000 volts with 12 volts input. I made one from a dead MacPlus flyback from which I removed the (dead) primary windings. 5. Frequency of operation will be in the KHz to 10s of KHz range depending on VCC, load, and specific flyback characteristics. 6. You can experiment with # turns, resistor values, etc. to optimize operation and power output for you needs. 7. CAUTION: contact with output will be painful, though probably not particularly dangerous due to low (a few mA) current availability. HOWEVER, if you add a high voltage capacitor to store the charge, don't even think about going near the HV! Parts list: Q1, Q2 - 2N3055 or similar NPN power transistors (reverse polarity of VCC if using PNP transistors.) Maximum stress on transistors are about 2-3 times VCC. Heat sinks will be needed for continuous operation. R1 - 110 ohms, 2W resistor. This provides base current to get circuit started. R2 - 27 ohms, 5W resistor. This provides return path for base feedback during operation. T1 - Flyback transformer from/for BW TV, color TV, or computer monitor modified according to text above. Most modern flybacks include built-in HV rectifier diode(s) so output without additional components will be high voltage positive pulses. Note: this kind of flyback transformer drives the CRT directly and uses its glass envelope as the high voltage filter capacitor. (A foot square piece of 1/8" Plexiglas with Aluminum foil plates makes an adequate filter capacitor.) Wire - a couple of feet of #16-#20 hookup wire, magnet wire, or any other insulated wire for home made primaries. Use electrical tape to fix windings to core. Wind feedback winding on top of drive winding.
Article Reference: CFDC9D65A310CF47
impedance matcher
Date: 17 Mar 1996 21:00:25 GMT
Original Subject: Re: Impedance matching
Workshop (workshop@pcm.co.za) wrote: > Hi there > This is to all those who know anything about impedance matching! > I have a circuit using a ne602 ic.The input impedance is is quoted > as bieng 1.5k.I have a Phillips application note (AN1993) > that uses this ic.The input to the mixer of the ic is from a 50 ohm source. > The 50 ohm source has been matched to the ic by the following circuit. > To pin 1 of > | mixer > | > |----------- > | | > | | > === 47p | > === () > | ()0.28uH > To 50 ohm ____| () To pin 2 of mixer > source === () | > === 220p | | > | | | > | | | > ------------ | > |_______________| > | > ===100nF > === > | > GND > Can anyone out there tell me how these values are calculated.The frequency of interest > is 45Mhz I believe the mixer has a differential input. The 100nF cap just shorts pin 2 of the mixer to ground so that the input is single-ended. You can take that cap out of the picture. There are formulas for calculating the values of the rest components, based on the assumption that the bandwidth of the matching network B < fo/10, where fo = center freq of the circuit. |------------------- Rt (>R2) | | | | === C1 | === () L | () R2 ____| () === () === C2 | | | | | ------------ | --- - 1. Calculate: Qt = fo/B C = 1/(2*pi*B*Rt) L = 1/((2*pi*fo)^2*C) N^2 = Rt/R2 2. Calculate: Qp = sqrt(Qt^2/N^2 - 1) 3. C1 and C2 are given by: C2 = Qp/(2*pi*fo*R2) Cse = C2*(Qp^2+1)/Qp^2 C1 = Cse*C/(Cse - C) Regards -T L
Article Reference: 8813DC4DB7AA6D1C
interface serial EEPROM to parallel port
Date: Fri, 08 Nov 1996 10:40:39 GMT
Original Subject: Re: Dividing a bidirectional line in two!!
Stefan Piatidis <ib93_spi@it.kth.se> wrote: >I want to divide a bidirectional data line into two groups of >EEPROMS. >I have a controll signal that can state to which of these two >groups, it should be activated too. >I.e. i am looking for some kind of 1 to 2 device. But not a >multiplexer, if there isnt a bidirectional one. >I've got some hints on that there are devices called analog >switches. Maybe they could work. > > > ----------- > I I > I------I EEPROM I >bidir.dataline ---- BDL1 I I I >---------------I I------ I I > I XX I ----------- >control signal I I BDL2 >---------------I I------ ----------- > ---- I I I > I------I EEPROM I > I I > I I > ----------- <snip> Two suggestions: 1. If you ABSOLUTELY must use I2C devices, you might be able to use a analog mux or bidirectional I/O device to handle the data line switching. I don't recall the exact part number, but I beleive there are a number of 4000-series CMOS devices that can perform this function. The CD4016 "Quad bilateral switch" might work for your application. 2. Instead of using I2C serial EEPROMS, consider using the 93Cxx serial EEPROMs instead. These devices use a 4-wire serial bus (clock, data in, data out, select) and would be considerably easier to drive using a PC parallel port. Since the data out pins of these devices employ open-collector outputs you could tie the clock, data in and data out pins together and use two digital outputs to (seperately) drive the select signals of each EEPROM, like this: /----------------------o-------------- - - - | | | /--------------------)--o----------- - - - | | | | | | /------------------)--)--o-------- - - - | | | | | | | | | 93C46 | | | 93C46 Clock (from PC) | | | |---------| | | | |---------| -------------------------(-(-o--|CLK | | | \--|CLK | | | | | | | | | Data out (from PC) | | | | | | | | -------------------------(-o----|DI | | \-----|DI | | | | | | | Data in (to PC) | | | | | | -------------------------o------|DO | \--------|DO | | | | | Select #1 (from PC) | | | | --------------------------------|CS | /--|CS | |---------| | |---------| Select #2 (from PC) | ------------------------------------------------------/ This approach has a number of advantages: - It does not require the use of any support hardware (other than the EEPROMs themselves and perhaps a power source). - No bidirectional signal lines are needed, thus it will work on ANY PC parallel port. - Multiple devices can easily be added. If you are driving a large number of devices, a 3-to-8 demultiplexer could be added to drive the multiple SELECT pins with just three output lines. You might even be able to power the EEPROM device(s) (irregardless of type) by using the remaining data outputs as power sources. If you use this approach, make sure to bypass the power supply pins with capacitors (suggest a 0.1 uF and 4.7 uF tantalum in parallel on each device) to handle peak current demands seen when a write cycle occurs.
Article Reference: D9B887DFF23401CF
inverting amplifier
Date: 28 Jun 96 04:34:27 GMT
Original Subject: Our Op-amp Problem Solved
The original post: > Could someone help me figure out how does this circuit work? This is > directly from Art of Electronics, but i just can't understand, how the > gain was calculated. It was said to be 100, whereas I get 102?? I have taken the liberty of redrawing the circuit and writing a detailed solution. First, the circuit: 100k ohm A 100k ohm *------/\/\/\-----*-----/\/\/\----* | | | | | | | |\ | | | | \ | | | | \ | | 100k ohm | | \ | | Vin *----/\/\/\----*-| - \ | | | \-------------------------* Vout | / | | / | *--| + / | | | / | | | / | | |/ | \ \ / / \ 51k ohm \ 1k ohm / / \ \ / / | | | | | | ----- ----- --- --- - - My solution: By the ideal op-amp assumptions, there is no current through the 51k ohm, so the voltage level at the inverting input terminal is zero (ground). Then, nodal analysis at the node I have designated "A". Summing currents away from A: (VA - 0) VA (VA - Vout) -------- + ---- + ----------- = 0 100k 1k 100k By conservation, the 'source' current through the 100k between Vin and the inverting terminal is the negate of the 'feedback' current through the 100k between the inverting terminal and A. Thus, the term (VA - 0) Vin -------- = - ------ 100k 100k so that VA = -Vin. Substituting this: ( -Vin - 0 ) -Vin ( -Vin - Vout ) ------------ + ---- + ---------------- = 0 100k 1k 100k Multiplying by 100k and rearranging, one winds up with Vout = -102 Vin, for a gain magnitude of 102. -- Paul Speicher speicher@pollux.math.iastate.edu
Article Reference: 51222CE878F8E9FB
IR beam detector
Date: Sun, 26 May 1996 02:16:29 GMT
Original Subject: Re: Help: need simple circuit to use photodiode
fly@rm3.flashnet.it (Stefano Mosca) wrote: >Hi world, >I need a simple circuit to use a photodiode to detect an IR beam, such >as that coming out from a remote control, so that I can try and see >the shape of the signal on an oscilloscope. No problem... How about this one.... 15V o | BW 34 ------ 56pF /\ ----||------ / \ | 68k | ---- o-/\/\/\---o 10k | | | -/\/\/\-- | | |\ | | | --------o---|-\ | 1k | |\ | | >---o-||--/\/\/\-o-|-\ | ----|+/ 1nF | >--o---o Output | |/ ----|+/ | | |/ --- | --- The photodiod is BW34 or any simular. The Op.amps are TL084. Greetings Joerg __________ \'|'/ Joerg Oulabi \\\|/// Home 0202 611600 o====LASER=Lab=====>>O<<=== 42399 Wuppertal ///|\\\ Lab. 02191 672280 /.|.\ jou@mail.wupper.de
Article Reference: 580BE531E51BEE57
IR detector circuit
Date: 25 May 1996 10:40:55 GMT
Original Subject: Re: Help: need simple circuit to use photodiode
In article <4o4qnr$m3a@news.flashnet.it> fly@rm3.flashnet.it (Stefano Mosca) writes: > I need a simple circuit to use a photodiode to detect an IR beam, such > as that coming out from a remote control, so that I can try and see > the shape of the signal on an oscilloscope. > Thanks in advance > Stefano Mosca Try this: --- sam IR detector circuit: ------------------- This IR detector may be used for testing of IR remote controls, CD player laserdiodes, and other low level near IR emitters. Alternatives to the use of this circuit for some applications include an IR detector card (available at Radio Shack - sensitivity from 600 to 1500 nm, catalog number 276-1099, $5.99) or a camcorder having a CCD type image sensor which may have some sensitivity to IR. Component values are not critical. Purchase photodiode sensitive to near IR - 750-900 um or salvage from optocoupler or photosensor. Dead computer mice, not the furry kind, usually contain IR sensitive photodiodes. For convenience, use a 9V battery for power. Even a weak one will work fine. Construct so that LED does not illuminate the photodiode! The detected signal may be monitored across the transistor with an oscilloscope. Vcc (+9 V) >-------+---------+ | | | \ / / R3 \ R1 \ 500 / 3.3K / \ __|__ | _\_/_ LED1 Visible LED __|__ | IR ----> _/_\_ PD1 +--------> Scope monitor point Sensor | | Photodiode | B |/ C +-------| Q1 2N3904 | |\ E \ | / R2 +--------> GND \ 27K | / | | | GND >--------+---------+ _|_ -
Article Reference: 8F2B1C2BF2223C4C
IR remote detection with phototransistor
Date: 3 Jun 1996 16:53:53 GMT
Original Subject: Re: Help: need simple circuit to use photodiode
In article <4onrqd$4ra@news02.deltanet.com> jlundgre@delta1.deltanet.com (John Lundgren) writes: > >The pieces of the puzzle below are missing. What transistor for Q1? An >NPN or PNP? It might be possible to tell from the circuit, if the emitter >and collector were labeled. But since they are not, then it's anyone's >guess. > >John Whitmore (whit@hipress.phys.washington.edu) wrote: >: In article <4o4qnr$m3a@news.flashnet.it> fly@rm3.flashnet.it (Stefano Mosca) writes: >: >Hi world, >: >I need a simple circuit to use a photodiode to detect an IR beam ... >: To get broadband response from a phototransistor, it is >: necessary to use it with a low load impedance: a grounded-base >: transistor amplifier is ideal. > >: +V----+--------->GND >: | >: R1 >: | >: +---------> oscilloscope input >: | >: | >: | >: \ +----C1----+ >: | | | >: Q1 +-------+----R2----+---- (-V) >: | | >: / R2 >: | | >: \ +V >: | >: Q2 + (phototransistor or photodiode) >: | >: / >: | >: -V > >: using a small-signal transistor as Q1, a 9V battery for +V/-V, >: R2 circa 100kohm, C1 circa 100 pF, R1=50 ohms > Since I specified 'grounded-base amplifier', it WAS in fact unambiguous that the Q1 had to be an NPN transistor. I'd recommend a general-purpose type, like PN2222A. Things might work better with more complex biasing (like a parallel RC in series with the Q2 collector, to ensure current limiting), but ASCII art is SO limiting... John Whitmore
Article Reference: 94D74F43D40538F9
lamp dimmer circuit
Date: 20 Nov 1996 20:35:07 GMT
Original Subject: Re: Light dimmer schematics?
Here is one reverse engineered from something in my junk box: --- sam : Repair FAQs: "http://www.paranoia.com/~filipg/REPAIR/Repair.html" This is the type of common light dimmer widely available at hardware stores and home centers. The 3-way type can be used to replace a regular 3-way switch used to control a light from two separate locations. However, note that it is not possible to use 2 of these 3-way dimmers to have independent control from 2 locations - only one can be a dimmer. S1 is part of the control assembly which includes R1. For a normal dimmer, S1 is just an SPST switch. While designed for incandescent or heating loads only, these will generally work to some extent with universal motors as well as fluorescent lamps down to about 30 to 50 percent brightness. Long term reliability is unknown for these non-supported applications. Black 1 o--------o \ S1 o----+------------+-----------+ | | | Black 2 o--------o | R1 \ | | 220 K /<-+ | | \ | | | | | | | +--+ | | | | | R2 / | C1 _|_ 47 K \ | .047 uF --- / __|__ TH1 | | _\/\_ SC141B | +---|>| / | 200 V | | |<|--- | | C2 _|_ D1 | | .062 uF --- Diac | | | | Red o-----------------+---CCCCCC---+-----------+ L1 40 T #18, 2 layers 1/4" x 1" ferrite core
Article Reference: 272660418FB17F4F
lamp driver
Date: Tue, 28 May 1996 22:54:34 GMT
Original Subject: Re: Need basic Light Sequencing circuit
On Sun, 26 May 1996 06:28:03 -0400, Lou <louisv@prolog.net> wrote: >Hope this does not sound too silly for those of you >who are not automobile enthusiasts, but... I saw a number of tips on how to make the sequencer. For drivers I would suggest SIPMOS fet's from Siemens. A device like the BUZ10 switches 20 Amps without problems, and causes less powerloss than a mediocre relay. If you think 20 Amps is way to much, consider the cold resistance of a bulb, which is much lower. If the output of your sequencer is TTL-compatible you can drive the BUZ with nothing but a small resistor (47 Ohm or so). Even the diode you'll see in 'normal' transistor output stages is built into the fet, but a small R-C network won't do any harm. 0.47uF and 47 Ohm placed over the load __ |----(load)--- +12V to sequencer --|__|----||<+ 47 |-+---+ | =ground --- Ruben
Article Reference: AC553E40D1F0EB2F
LED blinker
Date: 6 Nov 1995 03:49:28 -0500
Original Subject: Re: Flashing LED?
> Could someone tell me how to make a simple flashing LED circuit please? > thanks > PS: Is there a FAQ for this group? Something like this will work with just about any two transistors (one NPN and one PNP). If the LED never lights, reduce the value of R1. If the LED stays on, increase the value of R1. To run at a lower voltage, or higher brightness, reduce R3. To increase the LED on time, increase R2 or C. Use a low leakage capacitor. Tantalums work well, electrolytics may leak too much. Bill +12 +12 | | | | \ | *R1 / 5.6 Meg | \ / E / B |/ | +--------------------- | PNP (2N3906) | / C |\ | B |/ \ C +---| NPN (2N3904) | | |\ | | \ E | | | - + | +------ | ----||---/\/\/\/--------+ | C R2 | | 0.33uf 82K \ | / R3= 680 Ohms | \ | | GND __V__ LED | | GND
Article Reference: 38BC06627BA4B346
LED flasher
Date: 18 Nov 1996 07:29:00 GMT
Original Subject: Re: need a simple schematic
On 16 Nov 1996 19:07:05 GMT, jaz321@aol.com <jaz321@aol.com> wrote: > I'm a novice trying to learn a few things on my own. I want to try to >make my own circuit board. I need a simple plan or schematic to show how >to make a LED blink. I have many capacitors, transistors, resistors,etc. >I would probably use a 9-volt battery to power it. > You can use the following circuit to flash two leds alternatly. +--------------+---------------+--------------+---- +3 to +9 volts |a | | |a D1 R3 R4 D2 |c | | |c | | | | R1 | | R2 | | | + | | +---------------|------C2------+ | + | | | +-----C1-------|---------------+ | c| | | |c \ b | | b / Q1 |------------+ +------------| Q2 / \ e| |e +---------------------------------------------+----- Gnd R1,R2 220 ohms R3,R4 100k ohms Q1,Q2 any NPN 2N2222, 2N3906, BC108, etc D1,D2 LEDs C1,C2 22uF increase to slow flash rate If you want to only flash one LED replace the other with about and diode. If you want to use pnp transistors just switch the positive and negative from the battery, flip polerized caps and turn the LEDs around. You may want to go to radio shack and buy one of their electronics books. They have many little circuits like this one and some describe the basic fundaments of electronics. --- Brian Mihulka "A lot of people are afraid of bmihulka@engrs.unl.edu heights. Not me. I'm afraid of "http://engrs.unl.edu/~bmihulka" widths" -- Steven Wright PGP Key fingerprint = DF 4F CB C0 B4 90 00 F4 C8 75 8F 0E 6A F9 2A 7E
Article Reference: 05907106924B2C84
light bulb flasher
Date: 18 Feb 1996 01:26:35 GMT
Original Subject: Re: need simple flashing LED
Richard Rasker (r.e.rasker@student.utwente.nl) penned: : In article <4g0r1k$a70@cybernews.cyberus.ca> mthompson@cyberus.ca (Matt Thompson) writes: : >From: mthompson@cyberus.ca (Matt Thompson) : >Subject: need simple flashing LED : >Date: 16 Feb 1996 02:44:04 GMT : >Can someone tell me how to build a simple cicuit that would envolve the : >flashing of a LED. I do have some electronics background so I will be : >able to build it if I have the schematic. : >Thanks Try this one: --------cut here------- Flasher, uses #1860 bulb, with Photocell, for Barricade A Schematic from a real flasher in a real barricade - by John Lundgren - 9-4-95 Legend: ) = No Connection + = Connection All resistors are 1/4 W 5%. P1 /--\ silicon photocell (turns it off during daytime) +------( @ )----+ | \--/ | | | | R1 75 k | R2 2.2 k +------/\/\/\----+------/\/\/\----------+ | | | | +----------)----------------------+ | | | R3 160 k | | | NPN | +----/\/\/\----+ | | TO-92 | | | | | | | | ======= | R4 R5 | | E / \ | 2.7 k 5.6 k +-----)---- / \--+--/\/\/\------+----/\/\/\----+ | | | | | | C1 10 uF 16V tant. | | | | || + | | | +------||-------+ | | | || | | | | | | PNP | | | | TO-92 | | | ======= | | /---\ | / \ E | +--------( @ )------+------------/ \---------+ | \---/ | | I1 # 1860 lamp | | | | | | | O - 6 VDC (2 lantern cells in parallel) + O C1 10 uF 16V tantalum electrolytic (determines flash rate) Q1 2N3904 NPN gen'l purp. transistor Q2 2N2907 or PN2907 PNP audio gen'l purp transistor P1 Silicon photocell (looks like a LED) I1 #1860 lamp 6V, low current R1 75K 5% 1/4W R2 2.2K 5% 1/4W R3 160K 5% 1/4W R4 2.7k 5% 1/4W R5 5.6K 5% 1/4W --------cut here------------- Also try Filip G's web page. It's at "http://www.paranoia.com/~filipg" where there are circuits for the LM3909 flasher chip. Also, try other links to other sites/pages. : The very simplest is to buy a flashing LED; they come in all normal LED : colors, and have supply voltages of 5 and 12 V. Price (here): about $1.50. : If you want to build a flashing circuit yourself, you can use a variety of : devices and circuits; the simplest is the LM3909 which uses 1 external : capacitor to flash a LED on just a 1.5V battery. Also widely used: NE555 or : ICM7555 circuits where the 7555 is the CMOS version of the NE555. If you want : to make 6 LED's flash independently with just one IC you can use a CMOS 40106 : (hex Schmitt-trigger/inverter) and a handful of capacitors & resistors. : Success, : Richard : Richard Rasker : Calslaan 54-11 : 7522 MG Enschede : Holland : tel. +31-(0)53-4350834 -- #======P=G=P==k=e=y==a=v=a=i=l=a=b=l=e==u=p=o=n==r=e=q=u=e=s=t======# | John Lundgren - Elec Tech - Info Tech Svcs. | jlundgre@delta1 | | Rancho Santiago Community College District | .deltanet.com | | 17th St at Bristol \ Santa Ana, CA 92706 | "http://rsc.rancho"| | My opinions are my own, and not my employer's. | .cc.ca.us | | I have gone out to look for myself.. If I should | | return before I get back, hold me until I get here. | | "You can flame your brains out -- it won't take long." | #===T=u=z=l=a==C=o=m=p=a=n=y=.=.===t=h=r=e=e='=s==L=e==C=r=o=w=d=!==#
Article Reference: 9BE0AE1846F1A1EC
Lockout pushbutton encoder
Date: 1 Mar 1996 07:21:33 GMT
Original Subject: Re: "Pushbutton" effect, digitally
In article <4h5lu7$eea@nntp.Stanford.EDU>, meric@harris.Stanford.EDU (Meric Ozcan) wrote: >In article <4h3a95$92e@hobbes.cc.uga.edu> mcovingt@ai.uga.edu (Michael Covington) writes: >>Is there a simple, clever digital circuit that does the following? >> >>Given N pushbuttons and N digital outputs (probably N=8), when you >>press a button, the corresponding output goes active and all the >>other outputs are cleared. Then it remains in that state until you >>press another button. >> >>I have a solution involving a priority encoder, a monostable, and >>an octal latch to hold the output, and that's only 3 chips plus 4 >>analog components, but is there something simpler that I'm overlooking? >> >>I think we can assume that only one button will be pressed at a time. >>If not, the priority encoder can be included in the circuit. >> >>It doesn't absolutely have to be a logic circuit, either... ultimately >>we're controlling relays. >> >>-- >>Michael A. Covington "http://www.ai.uga.edu/faculty/covington/" Assuming each relay has a spare NO contact available and that the power supply voltage is +12V, then the following circuit will be hard to beat for simplity. It uses , in addition to any number of relays and switches, 2 PNP transistors (2N2907A/2N3906) , 3 resistors (1K/100) and 2 rectifiers (1N4002). 1. All relay coils are connected on one side to 0V 2. One side of each spare NO contact is connected to the other side of the coil. The other side of all spare contacts are connected together to the collector of a 2N2907 transistor Q1 with the Q1 emitter connected to +V. 3. A 2N3906 transistor, Q2, has the collector connected to the base of Q1 and the emitter of Q2 is connected t +V. 4. A 1K resistor is connected from the collector of Q2 to 0V and a 1K resistor from the base of Q2 to +V 5. One side of each switch is connected to it's corresponding relay coil and the other sides are connected together and are connected to two series connected rectifiers. 6. the common side of the switches is also connected through a 100 ohm resistor to the base of Q2. ______________________________________ +12V 1N4002 | | | 1N4002 1K e | |----100--|--b 2N3906 e | c------b 2N2907 | | c | 1K |__________________ | | | | | | | | |______________|________|__________|_______ | | | | | | =PB1 =NO =PB2 =NO | | | | | | |____| |____| | | | | ( )k1 ( )k2 | | | |________|__________|______ 0V
Article Reference: 2CF239E51B71E72E
low power transmitter
Date: Tue, 20 Aug 1996 16:19:10 -0700
Original Subject: Re: Q: 2.5(mH?) RFC in 50MHz xtal osc.
Joshua Andrews wrote: > > Hi all, > I hope that this schematic can make my question easier to understand. > This type drawing is something I've never tried. > Here's the deal, this should be a VERY low power RF Xmitter for a > cheap-walkie talkie, fox hunt. My problem is I can't believe that a > 2.5mH choke is the correct value. If I used a solenoid coil it would > be HUGE, isn't it more like 2.5uH? > Thanks, > Joshua > > ANT. > \I/ 2.5(mH?) RFC > _______________________|___________________CCCC_______ > _|_ | | > XXXtal-49.860MHz |________ | > _ |_ | _|_ ~90p V-Cap. | > | |_____________________|/ ^|^ | > | | |\ |_____________________| > | | 2N2222 \e | | | > | / | | | | > | \ R-10K __________| _|_ +___|___ / > | / | | ^|^.001uF === \ > | \ _|_ | | _______ / R-47K > | | ^|^ .001uF / | - === \ > | | | \470R | 9V | | > | | | / | | | > | | | \ | | | > | | | | | | | > | |____________|__________|______|_____________| | > |________________________________________________________| Hi all, I'm sure my first try made everything clear but just in case...
Article Reference: EC960653266F3E34
Lunchbox transmitter
Date: Wed, 10 Jul 96 18:55:50 GMT
Original Subject: Transmitter Question (Involves Transistor Antenna) - lunch.txt [01/01]
*********************************************************** * MEMBER NAME: LUNCHBOX * * *********************************************************** <%>[%>[%>[%>[%>[%>[%>[%>[%>[%>[%>[%>[%>[%>[%>[%>[%>[%>[%> <%> <%> <%> Making the <%> <%> <%> <%> Lunch Box <%> <%> ===== === <%> <%> <%> <%> Written, Typed and Created by: Dr. D-Code <%> <%> <%> <%>[%>[%>[%>[%>[%>[%>[%>[%>[%>[%>[%>[%>[%>[%>[%>[%>[%>[%> Introduction ============ The Lunch Box is a VERY simple transmitter which can be handy for all sorts of things. It is quite small and can easily be put in a number of places. I have successfully used it for tapping fones, getting inside info, blackmail and other such things. The possibilities are endless. I will also include the plans for an equally small receiver for your newly made toy. Use it for just about anything. You can also make the transmitter and receiver together in one box and use it as a walkie talkie. Materials you will need ======================= (1) 9 volt battery with battery clip (1) 25-mfd, 15 volt electrolytic capacitor (2) .0047 mfd capacitors (1) .022 mfd capacitor (1) 51 pf capacitor (1) 365 pf variable capacitor (1) Transistor antenna coil (1) 2N366 transistor (1) 2N464 transistor (1) 100k resistor (1) 5.6k resistor (1) 10k resistor (1) 2meg potentiometer with SPST switch Some good wire, solder, soldering iron, board to put it on, box (optional) Schematic for The Lunch Box =========================== This may get a tad confusing but just print it out and pay attention. [!] ! 51 pf ! ---+---- ------------base collector ! )( 2N366 +----+------/\/\/----GND 365 pf () emitter ! ! )( ! ! +-------- ---+---- ! ! ! ! ! ! ! GND / .022mfd ! ! 10k\ ! ! ! / GND +------------------------emitter ! ! ! 2N464 / .0047 ! base collector 2meg \----+ ! ! +--------+ ! / ! GND ! ! ! GND ! ! ! +-------------+.0047+--------------------+ ! ! ! +--25mfd-----+ -----------------------------------------+ ! ! microphone +--/\/\/-----+ ---------------------------------------------+ 100k ! ! GND---->/<---------------------!+!+!+---------------+ switch Battery from 2meg pot. Notes about the schematic ========================= 1. GND means ground 2. The GND near the switch and the GND by the 2meg potentiometer should be connected. 3. Where you see: )( () )( it is the transistor antenna coil with 15 turns of regular hook-up wire around it. 4. The middle of the loop on the left side (the left of "()") you should run a wire down to the "+" which has nothing attached to it. There is a .0047 capacitor on the correct piece of wire. 5. For the microphone use a magnetic earphone (1k to 2k). 6. Where you see "[!]" is the antenna. Use about 8 feet of wire to broadcast approx 300ft. Part 15 of the FCC rules and regulation says you can't broadcast over 300 feet without a license. (Hahaha). Use more wire for an antenna for longer distances. (Attach it to the black wire on the fone line for about a 250 foot antenna!) Operation of the Lunch Box ========================== This transmitter will send the signals over the AM radio band. You use the variable capacitor to adjust what freq. you want to use. Find a good unused freq. down at the lower end of the scale and you're set. Use the 2 meg pot. to adjust gain. Just fuck with it until you get what sounds good. The switch on the 2meg is for turning the Lunch Box on and off. When everything is adjusted, turn on an AM radio adjust it to where you think the signal is. Have a friend say some shit thru the Box and tune in to it. That's all there is to it. The plans for a simple receiver are shown below: The Lunch Box receiver ====================== (1) 9 volt battery with battery clip (1) 365 pf variable capacitor (1) 51 pf capacitor (1) 1N38B diode (1) Transistor antenna coil (1) 2N366 transistor (1) SPST toggle switch (1) 1k to 2k magnetic earphone Schematic for receiver ====================== [!] ! 51 pf ! +----+----+ ! ! ) 365 pf (----+ ! ) ! ! +---------+---GND ! +---*>!----base collector----- diode 2N366 earphone emitter +----- ! ! GND ! - + - battery + GND------>/<------------+ switch Closing statement ================= This two devices can be built for under a total of $10.00. Not too bad. Using these devices in illegal ways is your option. If you get caught, I accept NO responsibility for your actions. This can be a lot of fun if used correctly. Hook it up to the red wire (I think) on the fone line and it will send the conversation over the air waves. If you have any problems or are confused, leave me mail on:Hi-Times=702/832/7469 Warez House=702/827/9273 ______________________________________________________________________________ Sysops of other systems may use the file as long as none of it is altered. ______________________________________________________________________________ This has been a High Mountain Hackers Production- (c) 1985 by HMH Industries ______________________________________________________________________________
Article Reference: 1EDC44E780AB3341
mixer analysis
Date: Wed, 13 Mar 1996 21:03:54 -0600
Original Subject: Re: Impedance matching
Workshop wrote: > To pin 1 of > | mixer > | > |----------- > | | > | | > === 47p | > === () > | ()0.28uH > To 50 ohm ____| () To pin 2 of mixer > source === () | > === 220p | | > | | | > | | | > ------------ | > |_______________| > | > ===100nF > === > | > GND > Can anyone out there tell me how these values are calculated.The frequency of interest > is 45Mhz As long as the loaded Q is reasonable (say at least 10) then the circuit behaves as a transformer at its resonance. The 100nF to ground is merely an AC short to ground so that the DC at pin 1 isn't shorted to ground through the inductor. It does not play a role in the transformer action. Again, the following is only true at or near resonace. First, the transformer voltage ratio is approximately n = ( 47 / (47 + 220) ) ==> 0.176 Thus the impedance ratio is n**2 ==> 0.031. So, the 1.5K at pin 1 will be transformed down to 1500*0.031 = 46ohms. That is close enough to 50 ohms for all practical purposes. Recall that I said earlier that the circuit must be resonant. The inductor is chosen to resonate the series capacitor network. 47pF in series with 220pF is 38.7pF. The resonant frequency of 38.7pF and 280nH is about 49MHz -- close enough again for practical purposes. There is also usually a few pF of stray capacitance an pin 1 which will lower the resonance some. Normally it would be included in the calculation. As a check, the loaded Q is calculated to be about 17 so the initial assumption of Q>10 holds and the approximations are valid. That's all there is to it!
Article Reference: FDCF9991D3237018
multi-color blown indicator
Date: 24 Apr 1996 17:07:01 GMT
Original Subject: Re: HELP designing a simple switch circut!
In message <4l45t3$qs3@news.fonorola.net> - marka@spots.ab.ca (Mark Achtemichuk ) writes: :> :>I would like to ask anyones help out there in designing a simple circut as :>follows: :> :>The basic idea is that a dual colored LED (red and green) lights green when it :>receives a 12V lead. When that 12volt lead disappears, the LED must turn red. :>If I remember correctly the LED's work by switching their polarity for the :>different colors. This is where I run into trouble. :> :>The circut will be used to test fuses in my car. When a fuse blows, the LED :>will turn from green to red. The circut will have a constant 12 volt supply as :>well as the 12 volt probe. The LED's are rated at 3 volts so I assume it would :>be easiest to put a current limiting resistor on them vs. building a power :>supply. :> :>Can anyone help me with this? I would appreciate anything anyone has to offer. :> You can reach me at: marka@spots.ab.ca :> :>Thanks again. I work better with car engines than electronics <grin>. :> Mark, OK, after reviewing the other posts, permit me to offer my suggestion. I see what you are looking for is a probe which can be run along a series of fuses to determine if they are good or not. The simplest circuit that will meet your description is: 1Kohm 1/4 watt |-------/\/\/\------- To +12 Volts (vehicle battery) -----<---|>|-----| Probe tip and |-------/\/\/\------- To 0 volts (vehicle ground) bi-color LED 1Kohm 1/4 watt The end of a fuse will be at one of three states - +12, (power side of the fuse or load side of a good fuse); ground (load side of a bad fuse with load on); or open circuit (load side of a bad fuse with load off). When the probe is touched to +12 current will flow through the LED to the resistor junction, causing the LED to glow green. When the probe is touched to ground current will flow from the battery through the LED to ground and the LED will glow red. When the probe is touching anything not electrically connected the LED will not glow. I would build it in a clear plastic probe handle so you can watch the LED and the probe tip at the same time. I haven't experimented with LED brightness, it may be desireable to reduce the resistor values to increase brightness. Just watch the power dissipation of the resistors and the LED. (I think most LEDs can handle up to 100 ma. That translates to 120 ohm 2 watt resistors). Next, thanks to all who made the other suggestions. Several of them gave me the ideas for this design. Lastly, Sam, would you care to check my work? Bill
Article Reference: 954527A9FF6168B1
op-amp rectifier:001
Date: Sun, 21 Jul 1996 19:24:26 +0100
Original Subject: Re: No loss, small signal, full wave rectifier based on an op-amp?
In article <19JUL199617484372@ubvms.cc.buffalo.edu>, NEIL GANDLER <v064mb9k@ubvms.cc.buffalo.edu> writes > > > I would like to make an operational amplifier based circuit >that performs full wave rectification on a 1v p-p audio signal. >I would like to keep voltage loss to a minimal (simple diode rectifier >it out). Can anyone tell me of a simple way to use an op-amp as a >no loss full wave recitifier? Distortion is not a concern. I am >feeding the fullwave rectified audio signal into a peak detector. > > Neil Gandler > > > > This is an old one I used +--------2R-----------------+---10R----+ | /--------+---R-----\ | | /-------+ | | | +----2R----+ | | | _____ | | | | |\ | | /^\ D | +----||----+ -|-\ | | |\ ----- | | | | >--+--R-+-|-\ | | /| | | |\ | ----|+/ | >---+---|< |--+-R-+-|-\ | |/ /--|+/ | \| | >-----+----- output | |/ D /---|+/ | | |/ -+-gnd------------------ this is also my first try at drawing in ascii.... probably my last too. Probably a good idea to have a buffer (unity gain perhaps at the output). You can find this in most basic books but this one works well at about 100kHz with TL074 op amp. I also use a resistor pack as the internal matching is pretty good and it can take up minimal pcb space. As for the diodes any small signal ones are OK, I used 1N4148. The capacitor is a small value depending on frequency response and smoothing required. Hope this is a help, even if it was a slow response... Too many beers last night.. Good luck -- Tony Hardman
Article Reference: 5AD0E66044C51FF7
op-amp rectifier:002
Date: Mon, 18 Nov 1996 19:11:58 -0800
Original Subject: Re: Precision rectifier circuits.
Kevin Gaughan wrote: > > I'm using a textbook op amp precision rectifier circuit( see below) > to half wave rectify an input signal. > With a reasonably fast TL084 OP amp the output starts to look bad at as > low as 10kHz input. The problem being that at zero crossing the opamp has > to slew from + a diode drop to - a diode drop. This gives a kink in my > waveform around zero. I am going to try using schottky diodes and a > faster op amp but does any one know a better circuit? > > (If the picture looks funny try lining it up so that every line starts > with an x) > > x --------[RRRR]--------@---- to buffer > x | | > x | |\ _|_ > x | | \ /_\ > x in--[RRRR]----@-------|- \ | > x | | )-------@ > x | ----|+ / | > x | | | / | > x | | |/ _|_ > x | _|_ /_\ > x | - | > x | _____________________| > x I have built one for a limiter at the radio sation where I work. It worked well up to 20kHz. Add a diode on the output of your circuit to match the drop of the diode in the feedback loop. Also the input resistor needs to be slightly larger than the feedback resistor to match the resistance of the diode. I put a variable resistor in the feedback loop to adjust for unity gain. If you want to get extremely high prescision, the load after the output diode needs to match the load of the feedback diode. Place a variable resistor from the output diode to ground and drive the non-inverting input of a second op-amp in a unity gain config. You should be able to achieve rectification tracking error of less than 50mV. Mike
Article Reference: 733AF1606693407C
oscillator circuit
Date: 17 May 1996 20:32:44 GMT
Original Subject: Re: Q: Crystal circuit capacitors ?
The oscilation frequency of crystal oscilators is effected by the load capacitance of the crystal. The crystal manufacturer specifies the required load capacitance to get the stated frequency. Changing from the specified load capacitance will simply change the frequency of your oscilator a small amount. Probably much less than 0.1%, probably less than 0.01% Use the values that the ic maker recomends to get it running, then if you want tweak the freq, you can vary the caps. You will undoubtably have stray capacitance on the order of 10-20 pf, so its hard to design for a specific capacitance. Hint: the circuit will look something like this: | || C1 output| -----------+---------||-----Vss | | || | | | +--+--+ | | | | | |crystal | | | | +--+--+ C2 | | || input | -----------+---------||-----Vss | || | if you make C1 = twice C2, then you will get some voltage gain from input to output. This can help. Opinions expressed herein are my own and may not represent those of my employer.
Article Reference: 61AC107A58C6EBB8
output FET protection
Date: 22 Mar 1996 03:22:19 GMT
Original Subject: Re: About Mosfet voltage.
ba221@FreeNet.Carleton.CA (Gaetan Mailloux) wrote: > > >I have a sine wave inverter circuit , it's a mosfet output , it's work at >24 volts, they don't give any particular mosfet number, but they just say >that it need a 40 amp. mosfet, I have some MTP60N06HD it's mosfet with a >Vdss of 60 volts max and 60 amp max. > >Since it's an inverter who work at 24 volts, is a 60 volt mosfet are >quite OK for that 24 volts inverter ?? > >Any potential problems ? > >Thank > >Gaetan > By the way the mosfet version of the circuit shown below can be used to protect a switching element such as a power FET. Don't try it with a bipolar. You may see secondary break down failure demonstrated. This circuit makes it darn near imposible to destroy a FET from low energy power line spikes or inductive kickback. The 45 volt zener is selected to be VDS(max) minus 15 volts. Thus, in the case of the 60 volt parts above use a zener of less that 45 volts. | |___________ | | __|___/ __|__ / / \ 15v | L | /---\ | O | | | A | _|_ | D | 45v \ / |_____| ----/ | / | ___| Drain | | | | |_______|Gate (note the FET is N-Channel in this case.) _|_ | 15v \ / | -----/ |___ Source / | | | | | | | | | | | | | | |___________| | |
Article Reference: D8BE34407490AABC
passive 2 pole crossover
Date: 21 Sep 1998 17:08:26 GMT
Original Subject: Re: Passive crossover design
Rene Nieuwburg wrote: > > Someone can give me information how to design/calculate a passive > crossover network for a speaker system. I would like to a 12db/octave > filter with a higher crossover frequency as you normally see on the > market, around 4 khz > > Please CC to my Email address Rough & ready way to design HP or LP half of 12 dB / octave crossover, where configurations are as shown: HP: LP: || .-. .-. .-. PA Out ---||------ PA Out --- () () ------ || | | | | ---*--- ---*--- | | Tweeter | | Woofer ) _|_ | _|_ < | |/ === | |/ ) |___|\ | |___|\ < | | | ) | | | | | | | ---*--- ---*--- | | __|__ __|__ ___ ___ _ _ Assume that the impedance of the speaker at the desired crossover frequency, in each case, will be close to its DC coil resistance. This is not a spectacularly accurate assumption in all cases, but won't be too far off. Then, pick L and C for each case so that the reactance of these components is equal in magnitude to the speaker impedance at the crossover frequency (C = 1 / (2 * pi * f * R) and L = R / (2 * pi * f)). This will give you high and low pass filters that are within "spitting distance" of being critically damped. Picking vaues in this fashion will give you a crossover that will work well enough for most practical purposes. For a more accurate and finicky design, suggest you make careful measurements of speakers' impedances vs frequency, come up with model for each, have a go at filter transfer functions, either paper, pencil, and algebra, or with Spice, MatLab, MathCAD et al. Good Luck! W Letendre Dir Eng NEAT
Article Reference: 62A9F20046EE1405
phase shift circuit
Date: 16 Sep 1996 17:56:41 GMT
Original Subject: Re: Zero dB Phase Advance Cct - Possible?
Iain Hunter (via Chris.Garnett@dcs.warwick.ac.uk) wrote: > I would like a phase ADVANCE cct that gives a 90 degree phase > shift with 0dB gain at 10Hz. It must be DC coupled. Iain, you can indeed get a 90 degree phase advance at 10 Hz with a flat frequency response from DC to ultrasound, using a single opamp and a few parts. This circuit gives unity gain, non-nverting at the higher frequencies, with phase advance gradually increasing as frequency decreases, reaching 180 degrees (inverting) at DC. You can set the circuit to get 90 degrees advance wherever you like. R2 R3 IN o--+---/\/\/\/---+-------/\/\/\/-------+--------o OUT | | | --- | |\ | C --- +----|-\ U1 | | | \_____________| | | / +------------------|+/ For very low distortion, U1 should | |/ preferably an op amp with good \ common mode characteristics. / The TL080 series is good. R1 \ / Note that the circuit as shown \ is using positive and negative | power supply rails. ----- --- - R2 = R3; their values are not critical, maybe around 10 to 100K, but for precise results they should be matched. You get 90 degrees of shift when X(c) = R1. In a simple RC circuit, this would produce 45 degrees of advance, but in this active circuit it produces 90 degrees of advance from input to output. The R1 value is also not critical, although you may want to use a value around 1/2 that of R2 and R3 to balance out offset current. Then select C according to the value of R1. R1(in K ohms) * C (in uF) = 15.9 for 90 a degree circuit shift at at 10 Hz. So if R1 = 10K, then C = 1.59 uF; and if R1 = 47K, then C = .338 uF, etc. You can make R1 variable to set the phase shift to exactly 90 degrees at 10 Hz. For more phase shift network reading enjoyment, Christopher Trask (ctrask@primenet.com) once posted this list of references on one of the newsgroups: > 1. Shirley, Frederick R., "Shift Phase Independent of Fre- > quency," Electronic Design, Sept 1, 1970, pp. 62-66. > > 2. Albersheim, Walter J., "Computation Methods for Broad- > Band 90 deg. Phase-Difference Networks," IEEE TCT, Vol > CT-16, May 1969, pp. 189-196. > > 3. Darlington, Sidney, "Realization of a Constant Phase > Difference," Bell System Technical Journal," Jan 1950, > pp. 94-104. Best of luck, Iain. Bob Bruhns, WA3WDR, bbruhns@li.net
Article Reference: 83503751A24F0385
phase shift techniques
Date: 8 Jul 1996 03:08:09 GMT
Original Subject: circuit sugestion need
IN ----+--\/\/\/\/\--+----\/\/\/\/\---+ | | | | | |\ | | C +-----|-\ | | | >-------+---- OUT +--------||---+-----|+/ | |/ \ / \ R3 / \ | ----- --- - This circuit will shift phase from 180 degrees at DC to 0 degrees at very high frequencies. Frequency response is flat, gain is unity. At the frequency where X(c) = R3, the phase shift is +90 degrees. The phase shift is not uniform, but this can be managed as follows. Wideband 90 degree phase shift networks use several of these cascaded one after another, in two separate chains, with their 90 degree frequencies spaced according to some formula which I do not know. The spacing allows the nonuniformity of the high frequency side of one network to compensate for the nonuniformity of the low frequency side of the next network in the chain. With several networks in each chain, this results in a constantly changing phase shift over a wide frequency range, with constant gain. It is the DIFFERERNCE between the outputs of the two chains which has a relative phase shift of 90 degrees. The 90 degree frequencies of the two chains are staggered. Sweeping from DC up, the chain with the lowest "90 degree" frequency has progressively more shift until it reaches a little more than 90 degrees, whereupon the other chain is beginning to shift phase, and from then on until the highest "90 degree" network frequency, the phase difference between the two chains tracks at 90 degrees relative. Above that frequency, the phase difference between the chains gradually falls off. I seem to recall that there were three or four networks in each chain for a 100 to 10,000 Hz, 90 degree PSN. There are a few other ways to accomplish a 90 degree wideband phase shift. An older circuit by R. Dome covered 150 to 5000 Hz, if I recall, with several (5?) of the following cascaded networks in each of two chains, using vacuum tubes. Again, the 90 degree phase shift exists between the outputs of the two chains. B+ ^ | \ / \ R1 / \ / | +---------+ | | ----- --- C IN ---------- - - --- ----- |--------- OUT ' | \ | / | \ R3 | / | \ | | +-------+ | \ / \ / R2 \ / \ | ----- --- - And, there is a passive, multiple cross-linked network of numerous resistors and capacitors, which can produce a decent differential 90 degree shift over a desired range - but I'm not going to try to create an ASCII rendition of that! It's in recent issues of the ARRL handbook (ARRL, Newington, CT), although it turns out that this design is not as forgiving of component tolerances as the handbooks said. Evidently, it wants matched caps and matched resistors, but their exact values are not ultra-critical. Finally, I've read about a wideband 90 degree PSN which uses a multiple-tapped delay line and several shift networks. Unfortunately, I have never examined the design. Bob Bruhns, WA3WDR, bbruhns@li.net
Article Reference: AAE36C3FF5F9C545
phono preamplifier
Date: Mon, 23 Nov 1998 19:38:26 -0800
Original Subject: Re: Preamp help - preamp.gif (0/1)
> >"Kevin F." <fiduccia@dreamscape.com> wrote: > >Thanks for replying. I need an input from a guitar, or dynamic mic, about > >15k impedance. > >Thanks, > >Kev :-) > >>>"Kevin F." <fiduccia@dreamscape.com> wrote: > >>>Does anyone have a simple schematic for a preamp with bass and treble > >>>controls, that uses 2-3 transistors? > >>>Thanks, > >>>Kev :-) leonard@nospam.com wrote: > Here is a good old preamp - can have 70db s/n. > Old transister stuff... So is this one. It's a phono preamp, but you can change it to a mic preamp. All you do is remove the 2.7 nF 10% and the .01 uF 10% capacitors and use the 470k as the level control. The smaller the value, the lower the gain. RIAA Magnetic Phono Preamplifier (one channel shown) ASCII Schematic by John Lundgren 8-22-98 rev 980822a Magnetic Phono Input +-----------------------( O ) | | | | | 22 uF ===== | 16 V ----- | + | | 180K | | +---/\/\/\-----+ | | | | | | | | ----- | | E / \ 62K +--/\/\/\--)-----+----/ Q1 \-----+----/\/\/\--------+ | 220 ohm | | NPN | | | | | | Approx | | | +-----------+ | 1.75 VDC | | | | | | | | | \ | | | | | 30K / 2.7 nF | | | | | \ 10% ----- | | | | / ----- | | | | | | | | | | | | | | | | +-----------+ | | | | | | | | | | | .01 uF \ 470K | | | | ----- 10% / | | | | ----- \ | | | | | / | | | | | | | | | | +-----------+------)------+ Approx | | | | | 11 VDC | | | | | | | | ----- | | | 1 K | 82 ohm E / \ | 10K | +---/\/\/\-+---------/\/\/\-----/ Q2 \--+--/\/\/\---+ | | NPN | | | | | | |100 uF 6V | | | | || + | | | +---||-----+ | | | || | | | ----- | | 4.7K E / \ | +-------------/\/\/\-------+-----------/ Q3 \-------+ | | NPN | | + | | | 10 uF ----- | | 25 V ===== | | | | | | || + | +----> | <--------||------------+ | | || 220 uF | | | 35VDC | +------------------------( O ) | | Output Jack | | | ----- Ground +24VDC | --- well filtered O - Legend: ) = No Connection + = Connection > = Wire continues in direction of arrowhead All resistors are 1/4 W 5% unless otherwise marked. All capacitors are 16V or more. Observe polarity! Note: This must be viewed with a monospaced font such as Courier. Parts List Q1 2N5088 or 2N3904 NPN low noise audio transistor Q2 2N3904 or 2N2222 or PN2222 NPN audio gen'l purp transistor Q3 2N3904 or 2N2222 or PN2222 NPN audio gen'l purp transistor This is a two transistor series shunt feedback circuit with an emitter follower low impedance output stage to drive the amplifier. Notice that this basic circuit has no volume control, so you should have a level control on the following amplifier or mixer. For High Fidelity, certain construction details should be observed. For low noise, the 180K and 62K resistors should be metal film resistors. For low hum, the common should only be grounded to the external circuitry at one point to prevent ground loops. So the shield of the input and output RCA jacks should be isolated from the chassis if the chassis is metal. The Q1 transistor should be a low noise type. The 2N3904 is spec'd for low noise, but the 2N5088 is a better choice. To reduce the input impedance to the usual 47k that is required for a magnetic phono cartridge, an 82K metal film resistor should be put across the input jack. It is not shown here. -- Please remove NO and SPAM from my email addr to reply. -- @@@@@@@@@@T@h@e@@I@n@t@e@r@n@e@t@@w@a@s@@c@o@o@l@@u@n@t@i@l@@@@@@@@ @ John Lundgren Elec. Tech, Info. Tech. Svcs.| lundgrej <AT> mail @ @ Rancho Santiago Community College District | .rancho.cc.ca.us @ @ 17th St. at Bristol \ Santa Ana, CA 92706 |www.rancho.cc.ca.us @ @ My opinions are mine, NOT my employer's. |PGP key avail on req@ @ "Everything is InterTwingled" - Ted Nelson in ComputerLib @ @@@@@@@@@e@v@e@r@y@o@n@e@@f@o@u@n@d@@o@u@t@@a@b@o@u@t@@i@t@@@@@@@@@ Read what the FTC has to say about Junk Email at URL "http://www.ftc.gov/bcp/conline/pubs/online/inbox.htm"
Article Reference: E39B379495E6F64C
photoflash scematic
Date: Sun, 17 Mar 1996 02:13:18 GMT
Original Subject: Another photo strobe schematic
Since this question has come up more than once, here is another example of a small electronic flash design - This time in ASCII. --- sam Photoflash from pocket camera (model unknown): --------------------------------------------- This schematic was traced from an electronic flash unit from an inexpensive pocket camera (110 film size probably). Errors in transcription are possible. I have not entirely determined how the Ready Light functions as it is not in the usual place monitoring the energy storage capacitor voltage. It seems to operate on the principle that once nearly full charge is reached and the invertor is not being heavily loaded, enough drive voltage is available from an auxiliary winding on the invertor transformer to light the LED. It is also interesting that the trigger circuit dumps charge into the trigger capacitor instead of the other way around but the effect is the same. Invertor Flashtube +------------------------------+---------------------+--+--------+---+ | 1 K Ready LED | | | | | | +--/\/\-----+--|<|-----+ | ______ On | +-+ +-+ | Batt. _ | | | | | \___| )||( | 3 V ___ | +------|--/\/\/---+ | | __ Off )||( +| 2-AA _ | ||(2 .4 | 10 | Energy | | )||( _|_ ___ | || +-------------+ | Storage | +-------+---+ ||( | | | | | ||(5 .2 | | +| 280 uF | | ||( || | +---+ || +------+ | __|__ 330 V | Fire -| ||( || | | ||(1 | | _____ | (Shutter) | +--|| | +---+ ||( | | | | +-----+ Trigger || | 3)||( 142 -|47 uF | -| | | | || _ | <.1 )||( _|_ 6.3 | | | \ _|_ Trigger |_|_| )||( ___ V | | | 1 M / ___ .02 uF | +-+ || +-+ | | | | \ | 400 V -| C| 4 6 | +| | | | / | | B|/ | | | | | | | | +--| 2SD879 +--------------|<|--+----------------+-----+--------------+ | |\ | | HV Rect. | | E| | | | | +-------------+------|------------------+ | | +-------------------------+ Note: the polarities of the 47 uF capacitor and/or LED do not seem to make sense but these were confirmed on a working flash removed from a dead camera. There is almost no DC voltage across the capacitor at any time. Invertor transformer winding resistances measured with a Radio Shack DMM. Primary resistance was below .1 ohms. | | ---+--- are connected; ---|--- and ------- are NOT connected. | |
Article Reference: F5FA1BD80CAB3ECB
power on reset
Date: 6 Aug 1996 20:29:27 -0400
Original Subject: Re: PIC 16C84 Start up problems
When it doubt, try this simple power-on circuit: ^ Vcc | | \ / \ 10K 1/4w / \ / | +-------------------> to reset pin | | + --- --- 100uf 16v electrolytic gives ~1 second reset pulse | | | ---- --- - This is a generic reset circuit that I have used with just about everything.
Article Reference: F3C627827EC18C4B
radoi button circuit
Date: Sun, 14 Jan 1996 19:05:09 UNDEFINED
Original Subject: Re: Radio button circuit
In article <4d3v6f$5fa@henge2.henge.com> words@henge.com (Michael O'Connor) writes: >From: words@henge.com (Michael O'Connor) >Subject: Radio button circuit >Date: 11 Jan 1996 21:24:31 GMT >I am looking for a digital circuit that emulates the selector buttons >on a radio. A pulse at one of 1 through n inputs will latch the >corresponding output until another input is pulsed. Rather easy: o +Vdd | | o Data-latch |<= switch --- o | | |-----------------------------|D O|--o out | | | | E | | | --- |100nF --- 100| | --- | | K | | | 1N4148 100nF | ___ ___ -|>|----------||--------| /// /// | | | | | | | | | | - | from other -|>|-- | | | | ^ --> to E other switch 1N4148 | | | latches ___ ___ ___ /// /// /// 10K 100K 1N4148 How it works: when pushing the switch, a High level is applied to the D-input of the associated data-latch. Any switch pushed will also cause a High-level pulse on the E input of all latches (the diodes function as a wired-or circuit). This way all data-latches will transfer their input level to their output; if only one switch is pushed, only one input will be high (and therefore the output will go high too), the others will stay low. The rightmost 1N4148 diode prevents the E-input from receiving a voltage below 0V, when the switch is released. So what you need per switch is a 100K resistor, a 100nF capacitor (to debounce the switch contacts), a 1N4148 diode and one data-latch unit. The pulse-generating circuit (10K/100nF/100K/1N4148) is only needed once. When you use the CMOS 40373 or the 74HC373 you get 8 latches per device. Success, Richard Richard Rasker Calslaan 54-11 7522 MG Enschede Holland tel. +31-(0)53-4350834
Article Reference: 4B992E43BE0E6F67
ramp converter
Date: Fri, 19 Jul 1996 16:49:30 GMT
Original Subject: CHALLENGE QUESTION ONE
>HERE IS A CHALLENGE DESIGN QUESTION. >ANYONE WITH A GOOD WORKABLE SOLUTION WILL BE WELL REWARDED. >PLEASE RESPOND DIRECTLY TO ME VIA EMAIL, AND PERHAPS ALSO WITH A POST > >I have a potentiometric bridge type sensor (** see definition below), >a microprocessor, and possibly some 555 type timers. How can I >use RC timing techniques to successfully measure the sensor value. > >Obviously, the combination of a microprocessor and a 555 timer may >be used to determine an unknown R when in a circuit with a known C. >For example, the 555 can be wired with the R and C into a multivibrator >circuit and the microprocessor can be used to time the circuit's >period. From there, the unknown R may be calculated. As a result, >a resistive temperature sensor such as a thermistor may be used in >combination with the microprocessor, 555, and known C in order to >measure temperature. > >With a potentiometric bridge type pressure sensor, however, you can't >(and don't want to have to) get at the resistive elements independently. >I don't necessarily need to know the exact value of all four resistors. >I only need to know enough to calculate the pressure that's exerted >on the sensor. To do this, the appropriate ratios of the left and >right half bridge is sufficient. > >In case you're not familiar with this >field, such a sensor is modeled by 4 resistors connected as if they >were the four sides of a diamond. Typically, the sensor is wired as >follows. The bottom point (common to R3 and R4) is connected >to ground. The top point (common to R1 and R2) is connected >to a known voltage reference (the excitation voltage, VI). Meanwhile, >the left point VL (common to R1 and R3) and the right point VR (common >to >R2 and R4) are the outputs. The difference in voltage between the >outputs is indicative of the pressure, e.g. P=k(VR-VL). Obviously, >VL=VI(R3/(R1+R3)) and VR=VI(R4/(R2+R4)). >Thus, (VR-VL)=VI( R4/(R2+R4) - R3/(R1+R3) ) > >For the purpose of this question, note that it's only necessary to know >particular ratios involving the resistors, not the actual values of the >resistors themselves. Why bother even trying to find out the resistor values? Your equation P=k(VR-VL) is sufficient. Simply run these voltages to an op-amp then compare with a ramp generated by another op-amp. The time it takes the comparitor to switch from the beginning of the cycle is proprotional to the input voltage. Monitor this time with your processor. Like this: R |-----~~-| | | Comparitor | |\ | R | | \ | |\ VR -~~-*-|- \ | | \ | >--*-------|- \ VL -~~---|+ / | >------ To processor R | / *---|+ / |/ | | / Op-amp | |/ | |----------------| | Ramp Generator | |----------------| What's my reward? Brian Rose
Article Reference: D049C8A52790542D
remove negative RS-232 transitions
Date: Mon, 28 Sep 1998 10:13:38 +0100
Original Subject: Re: rs232 rectify diode
In article <6uhlia$77a@sjx-ixn6.ix.netcom.com>, alex c peper <xxalexx@ix.netcom.com> wrote: > I would like to know best design to allow only positive > rs232 transmit signals to switch a transistor. > Place diode in series or parallel. > Most designs I have seen used parallel, for some reason > this does not work for me. > I am using a IN4007 in series. Would this give better port protection? A series-resistor between the Txd output and the base of the transistor and a shunt diode between base-emitter would seem the better way to go. That keeps the load on the Txd output symmetrical, plus it allows the negative swing of Txd to smack the transistor OFF, but clamping the max negative Vbe to -0.7v. |/ c Txd-------[10k-ish]--+---| | |\ e --- | / \ | --- | | | 0v------------------+-----+ -- Tony Williams.
Article Reference: EB32155B95E74CDF
reverse DC motor
Date: 24 Nov 1996 18:44:26 GMT
Original Subject: Re: reverse current to motor through switch
Guy Duplain (gduplain@sympatico.ca) wrote: : I'm looking for a three way switch (off in center) or diagram of a circuit : to reverse the direction to a DC motor through switch selection. Only : about 6 volts from batteries to be switched. ---------------------------------------------- Guy, Try this: O----------------+ BATTERY | | O-----+----------------------+ | | | | NC NO | NC NO | +-->| <---+---->| <--+ | | | | C O-- MOTOR -- O C The switch is a DPDT Center-Off (ON-OFF-ON) switch. John Fields
Article Reference: 3F7B6E6AA8303093
Sallen-key circuit
Date: Sun, 15 Sep 1996 15:22:24 -0700
Original Subject: Re: Active filter design
************ I fixed your drawing ********************** > >4. Assuming I cascade two second-order VCVS low-pass filters to make a > >fourth-order filter: Sallen-key circuit > > > > +--------------|C1(--------+ > > | | > > + ---~~R1--+--R2--+-------+|\ | > > | |op-amp----+----- + > > | | / | > > Vin C2 | |--|/ | Vout > > --- | | > > --- +-------------+ > > | > > - ----------------|------------------------- - > > Normalized values for a sallen key circuit with gain = 1 are Stage 1 Stage 2 --------------------------- R1=1 R1=1 R2=1 R2=1 R3=OPEN R3=OPEN R4=0 R4=0 C1= 2Q C1= 2Q C2= 1/2Q C2 1/2Q Q values 4 th order: Q(stage1) = .54 ; Q(stage2) = 1.31 Choose a crossover frequency in Hz and multiply it by 6.28 to convert it to rad/sec. Now divide all the caps in both stages by this frequency (in rads/sec). This is called frequency scaling. All the resisters should be 1 ohm at this point. Choose a desired resister size (say 10K) and multiply all resisters by that number (lets call it Km). Then divide all cap values by Km. This yields the same exact frequency response (cutoff freq) as the normalized circuit except the componant values are much more reasonable. This is call magnitude sacling. Magnitude in this case refering the the componant size. And thats it. play with the Km until the resistors and caps a reasonably close to standard sizes and pick the nearest one. I suggest using polypro or metal film caps in the signal path. You can magnitude scale for small cap values (.1uF) which are cheap. As for opamps I suggest the LF347 from National Semiconductor. This on was designed from the ground up for audio purposes. You can use an 741 from radio shack to see if it all worked out but the sound quality won't be outstanding. Should sound OK though. Gain can be whatever you want. The higher the gain, the higher the noise floor on the input side though. If your radio drives you amp to full output now, Let k=1. That should do it Dan Cousin
Article Reference: 870F358C198B778C
Schmitt Trigger Oscillator
Date: 29 Feb 1996 21:23:41 -0500
Original Subject: Re: Schmitt Trigger Oscillator help!
From: davidr@pobox.com (Dave Rahardja) > Okay. I've made a self-oscillating Schmitt Trigger rig like so: [deleted] > Because of the inherent capacitances of the transistors, this > circuit naturally generates a nice 1.4V+ pulse wave output at almost > exactly 1MHz. Now, is there any way that I can scale the frequency > down to about 1kHz? I've tried caps and inductors, etc, etc, but no > success! Any suggestions? > Is there a way to delay the feedback from the output transistor? >Inductors don't work! (at least they didn't for me) You need some more resistors to set the upper and lower trigger levels. In this circuit if R1=R2=R3 the trigger levels will be about 1/3 and 2/3 of the supply. The other resistors need to be about 1/10 or less of R1. The frequency is about Rf*C and the symmetry will depend on the gain of the NPN pair (capacitor supplies base current on discharge but not charge). You can adjust the symmetry with another resistor and diode in parallel with Rf if you need an exact square wave. +V +--------+------+-----------+ | | | | | \ \ | | / / R1 | | \ \ E | | | |/ | +--------------B| PNP | | | |\ C C | C |/ \| | R2 | +-----+--B| NPN |B- +---\/\/\---+-----+---> out | | |\ /| | | | --- | E E \ \ | C --- | ----+--- / / | | | | \ R3 \ | | | \ | | | | | / | | | | | \ | | | | | | | | | +---- | ----------+----------+-----------+ | | | Rf | Gnd +--------------------\/\/\---------------+ Bill
Article Reference: 6816D8002A5F51EB
simple amplifier
Date: 6 Jul 1996 17:18:22 GMT
Original Subject: Re: Q: 100gain AMP without OP-AMP ??
Sam Goldwasser (sam@stdavids.picker.com) wrote: : In article <4qtau3$9l@jagalchi.cc.pusan.ac.kr> ysaum@hyowon.cc.pusan.ac.kr (Youngsub Aum) writes: : > Thank you 4 read this : > I want 2 have Amplifier which have 100 gain. : > Just use NPN transistor. : > Theoritical design is not a Perfact Answer. : > What make it deficult? : > Some help...... PLEASE Which gain -- Av, Ai, or Ap?? I assume we're talking about Av. : How close to 100 do you want it? The nice thing about op amps is that with : very high open loop gain, controlling the closed loop behavior is for many : applications very accurate and the theory works. : With a single NPN of high enough Hfe, you can get 100 gain easily - you will : need to use an emitter resistor as feedback to stabilize the collector : current vs. input voltage. The gain will be roughly the ratio of the collector : to emitter resistances. However, since Hfe is not infinite and can vary : quite a lot from transistor to transistor - or with temperature, etc. - this : will not be nearly as constant as an op amp. : o : | : / : \ : / Rc : \ : | : +-----||-----o Output : | : |/ C : Input o----------| NPN : |\ E : | : / : \ Re : / : \ : _|_ : - : (Biasing not shown). : Vout (small signal) ~= Vin * Rc/Re. The Re results in the transistor : operating as an emitter follower which means that Vin appears on Re. : Since nearly the same current (actually, reduced by Hfe/(Hfe+1)) flows : through Rc, the voltage across Rc will be Rc/Re times the voltage on Re. The voltage gain is high only as long as the Rl on the output is >> Rc. : With multiple transistors, however, you can design a very stable high gain : amplifier. The usual preamp stage has two transistors in a shunt series connection. They can have nice flat gain with wide bandwidth. The negative feedback works wonders. See H&H's Art of Electronics for more info, as he states below. : See any introductory book on transistor circuit design or Horowitz and Hill.. : --- sam -- #======P=G=P==k=e=y==a=v=a=i=l=a=b=l=e==u=p=o=n==r=e=q=u=e=s=t======# | John Lundgren - Elec Tech - Info Tech Svcs. | jlundgre@ | | Rancho Santiago Community College District | deltanet.com | | 17th St at Bristol \ Santa Ana, CA 92706 | "http://www.rancho"| | My opinions are my own, and not my employer's. | .cc.ca.us | | Most FAQs are available through Thomas Fine's WWW FAQ archive: | |"http://www.cis.ohio-state.edu:80/hypertext/faq/usenet/FAQ-List.html"| | "You can flame your brains out -- it won't take long." |
Article Reference: 3EB614C160C3825E
simple FM transmitter
Date: 26 Oct 1996 21:54:00 +0100
Original Subject: Re: How to build a radio transmitter
On 26 Oct 96 (12:15), t96uro@student.hk-r.se wrote: > Is there any one who knows how build a radio > transmitter for 70-110 MHz FM? Yes, me. It's very simple, when you are willing to use a xtal. Modulation can be done via two BB405B's or similar devices. VCC | | +-+ | | R1 +-+ | | | | +----------+--------| |------------->> | | | | XTAL | | |--+ C1 +-------| T1 | |--+ +-+ | | | R2 | +-+ | | | _|_ _|_ GND GND R1 = 470R...2k2 (depending on VCC) R2 = 1M0 C1 = 4.7pF...68pF T1 = either BF256A, BF256B or BF245A (other n-channel jfets may be used as well) -- Stephan A. Maciej, stephanm@muc.de, "http://www.muc.de/~stephanm/"
Article Reference: 1D89C2DF99C928FA
simple thermostat using LM34
Date: Fri, 19 Apr 1996 20:10:48 GMT
Original Subject: Re: Simple Thermostat
ravalent@liii.com (Bob Valentine) wrote: >I'm looking for a simple circuit to trigger a 12-volt relay at a >certain temperature, and then turn it off at another pre-set >temperature, or at at x degrees less than turn-on. This is not a very difficult task. The first answer is to use an LM75 temperature sensor, which has a programmable over-temperature set point and a hysteresis set point. When the device temperature exceeds the over-temperature set point, the overtemp shutdown (O.S.) output goes low. When the temperature comes back below the hysteresis set point, the O.S. output goes back high. The part has an I2C (I-squared C) interface and has resolution to 1/2 degree C. The other thing is to use a temp sensor, like the LM34 or LM35, with a simple comparator: V+ ____ | | | |\ | In this circuit, the output will go |LM34|------|+ \ high when the LM34 output goes above |____| | \_____ Output the value at the comparator (-) | | / | input, which is found to be: __|__ -|- / | (V+) * (Req1) / (Req1 + Ri), where ___ | |/ | Req1 = Rs * Rf / (Rs + Rf). When the _ | | output from the LM34 goes high, it ______|___ Rf __| will then remain high until the LM34 | | output falls to a value of: Ri Rs (V+) * Rs / (Rs + Req2), where Req2 = | | Ri * Rf / (Ri + Rf). | __|__ V+ ___ _ I will leave it to you to calculate the needed resistor values. The LM34 puts out 10mV per degree Fahrenheit. -- Nick Gray Applications Engineer National Semiconductor Corporation National Data Sheets at "http://www.national.com"
Article Reference: C2C7C1412DC621BD
spark coil driver
Date: 1 Dec 1995 16:57:31 -0500
Original Subject: Just one spark!
> I need to produce a single spark at a very precise time > My problem: The FET heats up beyond reason. > How would I trigger a single spark from a coil, using a 555 timer, > based on a repeating trigger (The trigger is located on a rotating > shaft which houses the sparkplug... but I only need one spark). Try a 25-30 amp FET. The on state resistance should be about 0.06 ohms and will dissipate only 2 watts at 6 amps. Or, you can use a CDI arrangement, but you need a high voltage supply which could be made from a 555 and a buffer to drive a transformer. The supply impedance (R) needs to be high enough so the SCR turns off when the coil discharges. Otherwise it will trigger once and leave a short on the supply. R 1-2 uf 400 volt non polorized +300 volts DC <--------\/\/\---+-----+----| |----+ V | --- ) Trigger pulse <--------\/\/\-------/ | 10 amp ) Ign coil (+12) 510 ohms | 400 ) primary | volt ) | SCR | | | | | common negative <------------------+-----------+ Bill
Article Reference: 9EB0026910276207
speaker out microphone in
Date: Tue, 13 Aug 96 08:42:00 GMT
Original Subject: Re: 8 ohm connection to mike input
In article <32100F5C.CAD@cdc3.net> jj@cdc3.net "Jerard Johnston" writes: > Does anyone know how to take a connection for an 8 ohm speaker (such > as an earphone jack) and directly connect it to microphone input on > something else? Any help is appreciated Well, for a *direct* connection you just need wire ;-) But... ---R1---+---C1---+ ---- | | Spkr R2 R3 Mic Input | | --------+---C2---+----- Say you have a 10V speaker output and a 100mV mic input and you want to load the speaker output to simulate an earphone (say 100R), and the mic input is high impedance. So... R1 = 100R R2 = 1R approx 100:1 divider R3 = 10k load and bleeder for Cs C1 = C2 = 330nF DC blocking, approx 50Hz -> 10k These values are very approx and don't forget that R1 could dissipate up to 1W using the assumptions above. Jenny -- Mayes uk
Article Reference: F268FBC5944AD40A
stereo to mono connection:001
Date: 15 Oct 1998 07:08:24 GMT
Original Subject: Re: Will this ruin my Discman?
In article <362564A2.650B1B32@wcnet.org>, Joe Konecny <gmi@wcnet.org> wrote: >I built an FM transmitter from a kit and it seems to work well. Can I >connect both channels >from my Sony Discman directly together so as to create a mono channel to >run through the >transmitter? > >Thanks for any info! >Joe --------------------------- You will want to isolate Left and Right from each other with either huge resistor values and/or use an op-amp as an adder, or mixer, to mix the two together. This is done with an op-amp: 10K pot +---/\/\/\/\--+ | ccw ^ cw | | | | 4.7uF 10K | +------+ left --|(--/\/\/\----+ . 7 | CD-player 4.7uF 10K | 3|\-- +V | right --|(--/\/\/\----+----|-\ 6 | 10uF CA3130 2| >-----+----|(---- Xmttr +--|+/4 | |/-- -V | . \ / 4.7K \ / _|_ - Make the feedback pot bigger to get more gain. -Steve -- -Steve Walz rstevew@armory.com "ftp://ftp.armory.com:/pub/user/rstevew" -Electronics Site!! 1000 Files/50 Dirs!! "http://www.armory.com/~rstevew" Europe:(Italy) "http://ftp.cised.unina.it/pub/electronics/ftp.armory.com"
Article Reference: CEC96875F192332C
stereo to mono connection:002
Date: Thu, 15 Oct 1998 06:43:39 GMT
Original Subject: Re: Will this ruin my Discman?
On Wed, 14 Oct 1998 22:57:38 -0400, Joe Konecny <gmi@wcnet.org> wrote: >I built an FM transmitter from a kit and it seems to work well. Can I >connect both channels >from my Sony Discman directly together so as to create a mono channel to >run through the >transmitter? > >Thanks for any info! >Joe No, do this: 10k L ----/\/\/\----+ |---- L+R (to transmitter) R ----/\/\/----+ 10k GND --------------- Make sure to tie all the grounds together. If you want one channel louder than the other, simply reduce the resistance with respect to the other channel.
Article Reference: ED1F4A388302B29D
stereo to mono connection:003
Date: Sat, 21 Nov 1998 11:14:39 GMT
Original Subject: Re: Question: Mixing 2 sound sources to one set of amplified speakers? (TIA)
On Sat, 21 Nov 1998 09:50:10 GMT, ddaniels@deletethis.com wrote: >I have two computers systems with sound and one set of speakers. > >I don't want to purchase speakers for both computers so how could I go >about feeding the two separate audio sources into these amplified >speakers? That is, of course, without resorting to purchasing a >stereo mixer (over-kill). You could prabably get by with making a simple passive mixer for each channel using 2 resistors like this: 10k-56k PC A >----/\/\/\/------+------> AMP | 10k-56k | PC B >----/\/\/\/------+ The optimal resistor values will depend on the output impedence of the sound cards and the input impedence of the speaker amps. Tiny 1/8th watts could be soldered to the terminals of a metal-covered phono plug. Of course there'll be some loss in volume, but encreasing the PC output volume may compensate enough. If not, you could try lowering the resistor values. If the volume ranges of each PC are highly mismatched, you can lower the resistor value of the weaker one. In any case, try to avoid going lower than ~1k for audio outputs and ~100 for speaker outputs.
Article Reference: 77393CB38A5D240D
switch debounce
Date: 18 May 1996 19:01:49 +1200
Original Subject: Re: toggle switch debounce
In article <4n7q30$i4r@miner.usbm.gov>, Dubaniewicz <dubanith@ptbma.usbm.gov> wrote: > I need to manually switch a 5 v, 1-10 mA variable current signal on-off. > I would like to have a clean signal within 1 ms of throwing the switch. > Board power supply is 6 V. Any suggestions on how to do this would be > greatly appreciated. The simplest way is to ensure you have a break-before-make switch, ground the changeover contact and use the NC and NO contacts to drive a S-R flip-flop. There are lots of variations depending on your requirements and preferences. A simple way is like this. (off) |\ o----------------| >o------------------- Q-bar -------o----- | |/ | | o | | GND (on) | ----/\/\/\/------------- | | | | ----/\/\/\/---------- | | | | | | |\ | ----------------| >o------------------- Q |/ This circuit is a simple S-R latch made from two cross-connected inverters. The switch holds the latch in one or other state. During switching, the switch goes open circuit to both contacts, and the latch holds its state. As soon as one of the contacts closes, the latch flips to that state. As long as you use a switch that breaks before it makes (also called non-shorting), you get a bounce-free output. You can use CMOS gates (4049, 4069, 40106, 4584, 74HC04, 74HC14 etc) with 10K or 100K resistors. For LS or TTL you should use lower resistor values e.g. 1K because they draw more current. You can even short the resistors out entirely, if you use open collector TTL or LS parts (7405 or 74LS05 I think from memory), so you get three debouncers per chip with no external components apart from the switch. Kris -- Kris Heidenstrom Time flies like the wind kheidens@actrix.gen.nz Fruit flies like bananas Wellington, New Zealand -- someone else's .sig ----------------------------------------------------------------
Article Reference: 9AF8768FAF8E867D
switch debounce circuit:001
Date: 23 Jan 1996 16:30:19 GMT
Original Subject: Re: Switch Debouncing
Chanawala SC <df192@city.ac.uk> writes: >Can anybody help me in devicing a way of switch debouncing. The only >componants available are: >D-type flip flops; Invertors; quad input NAND gates. Debouncer: +5 +5 __________ | | | ______ | SPDT R R -| | | Switch | | | Nand |-|------- output o--|-------|______| | | --o--- | ______ | | | o----------| | | | | | Nand |- | Gnd -|______| | |____________| So a quad nand chip is a dual debouncer. Doug Jones
Article Reference: BD53BF73D654EF61
switch debounce circuit:002
Date: 01 Sep 1998 12:13:35 GMT
Original Subject: Re: [Q] How to make a switch debouncer
In article <cwatson-3108982112530001@dialup-611.hip.cam.org> cwatson@cam.org (Sean McBride) writes: > I'm trying to make a debouncer for a SPST switch. It is a pushbutton > really, normally open, and closed when pushed. I'd like each push to > generate a pulse (hi) to increment a counter. Vcc o | / \ 10K / \ | |\ 74xx14 +----+-----o| >-----> To clock input. | | |/ 2uF _|_ \ --- | _|_ _|_ - - --- sam : Sci.Electronics.Repair FAQ: "http://www.repairfaq.org/" Latest Sam stuff: "http://www.repairfaq.org/sam/" Lasers: "http://www.repairfaq.org/sam/lasersam.htm" "http://www.misty.com/~don/lasersam.html"
Article Reference: 8559DB660EE3179B
switch interface to CPU
Date: Wed, 16 Dec 1998 00:12:51 -0600
Original Subject: Re: RS232 Control - Almost Solved...Thanx!!
Use a reed switch, and either pull a pin high or low. As many switches can be connected in series as needed in either circuit. The output will change only when all are CLOSED, or even one is OPEN. Either way, a cut wire to the switch will sound an alarm. Poll the port connected to the switch for 100 mS if you see a transition, watching for the same state, and reset the 100 mS timer if the state changes between polls, to avoid triggering on noise. Only after the same state has been polled for 100 mS do you consider that state valid. You will wake up in the middle of the night less frequently. The 12V battery should be float charged to 13.8V max. (lead acid example). This will give a long stand-by time, and drive a siren if desired. In long signal runs, a small cap across the PIC input to ground (1 uF, 25V, positive to the PIC pin, negative to ground) will help eradicate noise. ckt 'a' This makes a logic 1 when reed sw is closed Note: put 22K and Zener near PIC pin. / +12V >--39K----o/ o----+----+--> to PIC pin reed sw | | 22 K _|_/ | /_\ 1N4733 | | 0V >-----------------+----+--> to PIC ground ckt 'b' This makes a logic 0 when reed sw is closed Note: put 22K and Zener near PIC pin. +12V >--39K------+----+-----+---> to PIC pin | | | o | _|_/ reed sw \ 22 K /_\ 1N4733 o \ | | | | | 0V >-----------+----+-----+---> to PIC ground Kool Breeze wrote in message <3676155c.36973116@news.mindspring.com>... >Looks like I am gonna get a basic stamp for my needs. >I now have two. I wish to detect when a door/window is >open. The basic stamp inputs 5v == 1 or 0.6v (or less) == 0. > >What's the best way to run wiring from windows/doors >(using a 12v battery) to do this? > >
Article Reference: B6783EC6C873F459
telephone partyline blocker
Date: 23 Feb 1996 02:59:39 -0500
Original Subject: Re: Telephone blocking circuit?
From: jlundgre@delta1.deltanet.com (John Lundgren) > The way the SCR and zener works is that the first phone that goes > off-hook gets the full 48 volts DC which makes the 21 V zener conduct > and fires the SCR for the rest of the time the phone is off hook. > Any other phone then can't get the 21V to fire the SCR, and they all > remain open. Simple and effective. Some circuits add another zener in series with the SCR allowing an unprotected line to disconnect the protected lines when it is picked up, since the voltage will fall below 8.2 volts and turn off the SCR's. The other diode passes the negative ring voltage. Anode Cathode + <---+------------- SCR ---------|<----------+-------> | Gate 8.5 volt | To | 12 volt | zener | line +------|<--------+ | To phone | | +------|<--------------------------------+ - <----------------------------------------------------> Bill
Article Reference: A005BBC5FC4C020D
telephone to radio patch
Date: Mon, 18 Dec 1995 10:23 EST
Original Subject: Re: Phone Patch
In Article <4b35uo$egm@fountain.mindlink.net> "andrew_taylor@mindlink.bc.ca (Andrew Taylor)" says: > Just wondering if anybody has made a phone Patch (with > DTMF encode/decoder!)...If you can point me in the right > direction Id appreciate it!...Im going to be using it > with VHF radio!. > > Andrew T. > andrew_taylor@mindlink.bc.ca > > PHONE PATCH FOR HANDHELD RADIOS This circuit was published in _73_ magazine ~10 years ago. I've built lots of them. Push-to-talk circuit as shown works with most handhelds, which key their transmitters when a dc path through the microphone is completed. Base/mobile rigs use a separate push-to-talk line. to radio earphone jack ^ ^ | | | / T1 | \ R2 (optional) C1 | / _____ o-----||---)| | | |( | )| ______|__.______ |( \ R1 | PTT phone 1k )|(8 ohm | 8 ohm)|( 1k / --- line )|(______._________)|( \<------o o---o to radio mike input )| |( / (use shielded wire.) o----------)| |(_____|_____________o C1 1 uf 200 volt non-electrolytic. R1 ~5k screwdriver-adjust pot to set microphone input level. R2 resistance to equalize speaker and patch levels. T1, T2 Audio-output transformer: Radio Shack part # 273-1330 (center taps not used). -- Frank reid@indiana.edu W9MKV
Article Reference: 5738AA7F72FB4DB5
tester makes scope patterns
Date: 6 Oct 1996 03:17:48 GMT
Original Subject: curve trace
I found the old text to go with the diagram; this device has worked for me for over 20 years and I hope it will be useful for you................ In the August 1975 issue of Popular Electronics author Lou Garner wrote in a story called "A simple On-Board Tester" about this fairly simple piece of test equipment. The device can be used with any type of oscilloscope and consists of a 6 volt filament transformer, three 1/4 watt resistors and two test probes. Half the filament voltage is applied to a voltage divider consisting of 220 ohm and 100 ohm, yielding 1 volt ac on top of resistor 1000 ohm. This voltage can be applied to any component or combination of components across which the test leads are placed. The current is limited to one milliampere by the 1K resistor. The voltage across the probes is connected to the horizontal input of a scope while the voltage across the 1K resistor as a result of the current through it is connected to the vertical input. What we see on the scope is a voltage across a component under test versus the current through the component. Resistors: open horizontal line 10000 10 degree 1000 45 degree 0 vertical line Capacitor .1uF shallow ellipse 2.6uF circle 50uf narrow vertical Transformer ellipse depending on impedance Diode germanium right angle display silicon right angle one side longer (any leakage showing less sharp angle) Transistor test as two diodes (base to emitter and base to collector) IC's input for gates and counters show a certain signature display outputs display a different signature a short will show a vertical line an open will show a horizontal line In circuit testing is done with NO power applied to the equipment under test. With some experience one is able to test components in and out of circuit and troubleshoot without danger of a damage to components. --- 200 Ohm---------o---------o-------------- Vert. Oscilloscope | | | 1 K Ohm to the 3V | | winding of a | o-------------- Ground 6 volt | | filament 100 Ohm | transformer | V RED component test lead | --------------------o----------------------- Hor. Oscilloscope | | V BLACK component test lead -- W.Thiel
Article Reference: A7D4CAD6832974F7
transistor noise source
Date:
Original Subject:
Nick Bryant (bryant@mpr.ca) wrote: : I am now looking for a simple circuit which will generate a white : noise signal, suitable for the AUX input of a spare amplifier I own. : Does anyone have a _proven_ circuit they would care to share? Yes, I have a _proven_ circuit. An easy way to realize white noise ================================== There is a very easy way to realize a white-noise-circuit. Take a bipolar npn-transistor and a resistor. That's all. The trick is, you must need the inverse basis-emitter- breakdownvoltage (about 7V) as following circuit shows: |-----------------------------------------o +15V | R1 = 470 k-Ohm | |------------------> U-noise = 5 - 20 mV | e \| To amplify U-noise take a highimpedance-opamp, b |---- BC107 like a bi-fet-type is, as example LF356, TL071. c /| | | | |----------------------------------o GND With other transistor-type it can be you must change the resistor R1 between about 100k-Ohm and 1M-Ohm for best results. So I hope I can help you. If you have more questions write to my mailbox please Thomas ============================================================= | Tel + FAX: * * * | | +41 01/632'6583 * * * | | (2768) * * * E-Mail: | | +41 01/256'0823 (FAX) * * * schaerer@isi.ee.ethz.ch | | * * * ======================= | | * * * | | * * * | | Thomas Schaerer | | Organ.: Swiss Federal Institute of Technology Zurich, ETHZ | | Depart.: Signal and Information Processing Laboratory ISI | | Address: Sternwartstr. 7 CH-8092 Zurich, Switzerland | =============================================================
Article Reference: 84C25FE5A381DF01
transistorized fluid sensor
Date: Sat, 10 Feb 96 14:53:44 GMT
Original Subject: Re: Air or fluid detector?
In article <4fb2op$ldh@newsbf02.news.aol.com> ifordir@aol.com "IFOR dir" writes: > Anyone heard of any OEM suppliers of a resistor heater sitting next to a thermistor > (in, say, a small tube package) used to detect whether the pair were > sitting in a gas or fluid? I don't need to detect flow, just whether or > not (say) air or water is present. > Any leads appreciated, Rob Peterson Liquid detector: ----+------+------+ A Few Volts | | | | R | Zener bias and base current resistor | 1 | | | | C | C B----+----B Both NPN low power E | E | | | (O1)----+ | +----(02) | + | R Z R 2 D 3 | | | ----+------+------+ 0V Select ZD zener voltage and resistors R2 and R3 such that the transistors are dissipating unequal amounts of heat, say a few milliwatts for one and maybe a hundred or so for t'other. Most of this heat will be generated in the Collector areas and should be chosen to result in one transistor running fairly warm in air. The Vbe drop varies from its nomunal 0.7V by about 2mV/K and thus there will be some small voltage measured between O1 and O2. If both transistors (or just the hot one if you prefer) are immersed in some liquid, heat will be conducted away, reducing the junction temperature of (particularly) the hot transistor, and the Vbe difference voltage will change accordingly. The trick is essentially that the Collector regions do the heating while the Base-Emitter junction does the temperature meaurement. These regions are all, of course, in intimate thermal contact. This works quite well. In practice device variations mean that there will be some offset to be nulled out. Ferranti ZTX300 type transistors work well as the collector plate is pretty close to the outside world. The Zener may be replaced by a couple of normal diodes in series connected the other way round giving about 1.4V drop. This circuit detects just about any liquid with a time constant of about a second upon immersion. On drying out, the time constant depends greatly on the liquid used: a light organic solvent will dry out in a couple of seconds, while water is about the slowest to go having such a large specific heat capacity, and may take ten seconds or so. --
Article Reference: 67A9562DD0497212
tri-state detection
Date: Wed, 21 Feb 96 22:55:19 GMT
Original Subject: Re: logic probe
In article <4fvjak$ut@beyla.ifi.uio.no> olee@ifi.uio.no "Ole Christian Kollen Evensen" writes: > > I'm going to build a probe to sense logical "O", "1" or high-impedance state. > I will have three LED's of different color and some logical gate's etc. > My problem is how to sense the high-impedance state. Anyone have an idea ? > > Maybe some kind of 'current-detecting' or something. > > Thanks in advance ! > > -- > Ole Christian Evensen > e-mail: olee@ifi.uio.no You can't just measure the node voltage as others have suggested because Hi-Z outputs are often connected to pull ups or pull downs. +--/\/\/\--5V 1 L +-------------) 2 E | )XOR>----) D o-----+ +--) )XOR>--+ | | +--) L | | | E +--/\/\/\--+----+---+ D | +--/\/\/\--0V OSC o---0V I think this will work. When both sides of the input resistor do the same, that is, probe connected to Hi-Z (or unconnected) then XOR 1 output is low and XOR 2 output is clocking. Both LEDs light. When probe is connected to a Lo-Z logic high, XOR 1 output is clocking out of phase with OSC (XOR 1 inverts the clock) so XOR 2 output is High and lower LED lights. When probe is connected to Lo-Z logic low, XOR 1 output is clocking in phase with OSC (XOR 1 passes the clock) so XOR 2 output is Low and the upper LED lights. OSC may be made using spare XORs from a quad package and a couple of passives. Don't forget the OV connection to the circuit under test. If the oscillator runs at around 10Hz, the lights will flash when Hi-Z which will remind you to switch it off. Beware that this probe will drive a Hi-Z node up and down and may thus have some effect on the following circuit. -- Keith Wootten
Article Reference: C7E6CBFA0975DB08
Triangle Generator, 250kHz
Date: 3 Sep 1996 19:52:24 GMT
Original Subject: Re: Harris Class D Amplifier
Frank Boylan (boy@iol.ie) wrote: : Please view this in a fixed width font such as Monaco or Courier 9 or : 10. : I'd like to thank you all for your answers to my question. I found the : Class D evaluation board on the Harris site. It's very interesting but : there are a couple of things that have me puzzled. : This is the triangular wave generator. It uses three inverters.I always : thought that such inserters were logic in nature and therefore their : outputs would always be either high or low, in which case I can't figure : out how the output could be a triangular wave. : However ,if they are analog inverters, then what kind of characteristics : do they have? : In other words would this circuit produce a linear triangular waveform ? : Tri-Wave Generator 250kHz CD4069UB : ____ : | | : |\ |\ | _ |\ : __| \o____| \o___/\/\/\__|_/\/\/\__| \o_______o : | | / | / | | | / | : | |/ |/ | | |/ | : | | |---||---| : |_____/\/\/\/\__| | : |__________________/\/\/\__________________| A CMOS inverter such as the 4069 can be biased to operate in the linear mode. That is what the left-most resistor does. But they make lousy amplifiers, and no one would seriously use this circuit, however, as there are many better ones available (and in fact, it makes far more sense to use a proper PWM chip for modulating a class D amp). : My second question is: Is this an N channel or P channel MOSFET ? : RFP22N10s : D + : | : |--' : G ----||<-. : |--| : | : | : S - : In the circuit diagram, the output stage uses four of these Mosfets. : They are described as N channel Mosfets, but if so shouldn't the arrow : be facing the other way? That is an N channel FET. P channel FETs have the arrow pointing the other way. As you have drawn it, it makes perfect sense; the parasitic diode is reverse biased. If it were drawn the other way around (and connected with the same polarity) the diode would continuously be "on" and short out the drain to the source.
Article Reference: 084A71ED0AA1F149
ultrasound impulse drive
Date: Thu, 25 Jan 1996 07:07:03 GMT
Original Subject: Re: impulse generating for Ultrasound Transducer
Heath Bebout <dhb@engr.uark.edu> wrote: >I am working on using a ultrasound transducer to measure a distance to a >surgical pin in the human body. I am going to use a microcontroller for >the control and analysis of the signal. The question I have is how do I >genrate a 150 to 250 V spike to intiate a signal from the transducer. >The spike must be of a very short duration, approximately 20 - 40 ns. >Thanx for any help you can give. Standard method for generating an impulse drive is charge a capacitor to desired voltage thru a current limiting resistor (R1), discharge capacitor (C1) with a SCR (SCR1) or other fast switching device. This is the standard method used in driving transducers for ultrasonic non-destructive testing (these are 1 to 5 MHz units with really low Q). R1 C1 D1 D2 B+ o---/\/\---+---||---+---|<|---+--|<|---+---o output to xdcr | | | | - / / / SCR1 V \ R2 \ R3 \ R4 - / / / C2 /| | | | in o--||--+-/ | | | | | gnd gnd gnd gnd / \ R5 / | gnd R1: 220k, use 1 watt carbon composition, not film if possible R2: 1.0k, 1/2 watt carbon composition, not film if possible R3: 2.2k, 1/2 watt carbon composition, not film if possible R4: 4.7k, 1/2 watt carbon composition, not film if possible R5: 100 ohm, 1/4 watt C1: 1.0 nF, many volts, polyester (mylar) or polycarbonate C2: 10 nF SCR1: Something that can handle 50 to 100 amps at voltage you require. Motorola MCR729 is a hefty part (expensive) which is used in these types of amplifiers. You might try a MCR265 and see if that works. D1: 1N4608 (you can try a 1N4148) D2: 1N4608 (you can try a 1N4148) You want to use a big size capacitor for C1, preferably made for high current pulse applications. Wima has some very nice units. You might consider a 2000 volt unit so you can carry the current pulse. Best to use carbon composition resistors for R1-R4. This is a pulse circuit which means that you will be applying very short, high stress pulses to the parts. Carbon film resistors don't handle high voltage pulse well. Carbon film resistors work fine for experimenting, but don't use them for the final design. You might have to adjust the value of R1 to adjust the charge rate of C1. Depends on your pulse rep rate. SCR1 must be able to handle high current. You will be dumping close to 100 amps thru the SCR for a few nanoseconds. You might also try using a power FET. I have a funny feeling that a FET won't work as good as a big SCR. You may have to adjust the value of R4 to make sure that you do not load your receive signal from the transducer too much. Be sure that your T/R switch can handle the high voltage pulse. You will have to build some sort of preamplifier which needs to be protected from the high voltage pulse. D1 and D2, I like using the 1N4608 (handles lots more current than the 1N4148), but it is hard to find sometimes. 1N4148 will do if you can't find the 1N4608. The input (marked "in") will require a positive going pulse of at least 5 volts. The circuit will fire on the positive edge of the trigger pulse. Be sure to use a somewhat healthy drive to the gate. The "B+" input is connected to your high voltage power supply. You can use anything up to about 800 volts (make sure your SCR and C1 can handle the voltage). If your high voltage supply is noisey (you typically will be using a small DC-DC converter from Endicott Research Group), then you can put an RC filter at the output of the HV power supply (10k and a 47nF as a starting point). You want to keep all leads really short. Be sure to use a copper plane for ground. You need to keep inductance to a minimum to keep the rise time fast. Remember that you are switching 100 amperes (about 10 toasters), thus keep everything tight. Above all, beware of HIGH VOLTAGE. 250 volts DC is enough to hurt you really bad, if not from the direct shock, then from hitting yourself against some sharp object as you fly across the room. Trust me, I've banged myself up using less voltage. - Mark Chun | mark@reson.com | Santa Barbara, CA
Article Reference: 3EC067E3E562573D
Variac schematic
Date: Thu, 13 Apr 2000 18:15:18 GMT
Original Subject: Re: What is a Variac?
The renowned lharts@unm.edu wrote: > NEGATIVE!!!! > First off Variac is a trade mark of ( I forget but think it was General > Radio) AND MOST IMPORTANT! are not isolated but are a single winding > on the core with a brush riding on a track for the output. So one terminal > is common to input and output, this should always be the "neutral" line. There's a few manufacturers, Staco and Superior (I think the latter is the original.. could be wrong tho) The schematic looks like this (often, though some don't have the extended winding from A to B): A o----- ) B ( o----- ) ( )<----------o Out L ( ) C ( o----------------o Out N C goes to the incoming neutral. If you connect incoming Line to B you get something like 0~140 VAC out (with 120V in), if you connect it to A, you get 0-120VAC out with 120V in. As others have said, there is no isolation, the winding is shared between primary and secondary (an autotransformer), and as has already been said, the wiper is a weak point. The Out L line should be fused because it's quite possible to draw excessive current through the wiper and burn the winding at that point. Best regards, =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= Spehro Pefhany --"it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: "http://www.trexon.com" Embedded software/hardware/analog Info for designers: "http://www.speff.com" Contributions invited->The AVR-gcc FAQ is at: "http://www.BlueCollarLinux.com" =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Article Reference: 657C7DA5A02946CB
Wilson constant current source
Date: 13 Jun 96 01:01:05 GMT
Original Subject: Re: Constant Current Source (1-5 mA)
Geoff Raynak <graynak@u.washington.edu> writes: >Sorry for the confusion. I meant to say that I tried the circuits >provided by H&H but they don't seem to be _precision_ in the sense that >they cannot maintain the load against the resistance. I am trying to >sense geometry changes in a tube filled with saline. Using a resistance >method, i need a constant current source (very constant) to measure >resistance-induced voltage changes without any current-induced voltage >changes. >thanks again >geoff A couple of thoughts come to mind: Transistor based: Cascode or modified Wilson source: Gnd--R--+ | | \_/ Iin +-----+ | | | | \ | / |---+--------| / \ | | | | | +-----+ | | | \ | / |--------+---| / \ | | | | -V -V All these attempted transistors are npn's (this is the modified Wilson source). It should keep compliance to within about 1V of the -V rail which is well within your request. You can set the output current with the input resistor (approximate current is (V-2Vbe)/R. You need transistors with a VCEmax rating high enough to meet whatever voltages you should expect to see. This accuracy of this will require that the temperature of the lower pair remain pretty constant (try touching one of the lower transistors and watch the output current change...). The output resistance of this current source should be pretty large O(beta*ro) where beta is the beta of the upper transistor on the right and ro is the output resistance of the lower transistor on the right (2N3904 at 1mA is about 100k, I think). monolithic approach: modified Howland Source V2--R1-+---R1-----+ | | | |\ | | | \ R2 +-|- \ | | >----+ +-|+ / | | | / R2 | |/ | | | V1--R1-+---R1-----+ | | \_/ Iout | This looks like a modified instrumentation amplifier. There are 2 resistors in series going from the output to the input. Your current source output is actually between the 2 resistors. With a difference of voltage applied to the input, you will get an output of Iout=(V1-V2)(1/R1 + 1/R2). The output impedance depends on the CMRR of the difference amplifier. Burr-Brown (amongst others, I'm sure) has an application note on current sources/receivers which includes this circuit. This circuit might be a bit overkill... It is nice though in the fact that you can get a nice bidirectional current programmed by a voltage. You can also boost the current by placing a power buffer at the output of the opamp, if necessary. There are plenty of other circuits... Art Zirger.
Article Reference: E0BE023A604EE4BD
wireless bug
Date: 19 Jul 1996 17:06:49 GMT
Original Subject: Re: Electronic 'bugs', 'recievers', detectors
For a simple wireless FM bug you could use this circuit. / 100k 1/2 W ----------o/ o--------/\/\/\/\/\------+ _______ | | ________ | \ | / - | / \ | \|/ --- C \/ \ | | 6" or so - / \ _ \ | | --- | \| | | B | | | PNP transistor | | |------|---+ | | | _/|_| | | | | \ \| / --- 100 pf | _ E \/ / --- | >|_) carbon mic /\________/ | | | or other | | | | analog input | L | | +---------------+-()()()()()()-----------+----------------+ | _ | | ||/| | +-------||---------------+ /|| 10 -> 365 pf or fixed To make L: |-36t #28 C.T.| _________________ _ | ||||||||||||||| | | | ||||||||||||||| | 1/4" -+------+------+- - | | | good luck.
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