Article Reference: 1265A97FE4958B33
Date: Sun, 31 Jan 1999 19:26:42 -0000
Original Subject: Re: what is D.F. in capacitor?
A few equations... DF = dissipation factor (%) = Tan d = ESR / Xc Xc = 1 / (2pi f C) Power Factor (%) = ESR / total impedance = sin (loss angle) = cos (phase angle) Q = quality factor = cotan (loss angle) = 1 / DF Power loss = 2pi f C V˛ DF = I˛ ESR See "http://www.novacap.com" ! Paul. ratman@execpc.com wrote in message <790cm2$649@newsops.execpc.com>... > Probably better to say dielectric absorption is low when >insualtion resistance is high. > > > >On Tue, 19 Jan 1999 13:14:30 +0900, jin <mindblue@hotmail.com> wrote: > >>hello >> >>In the spec of capacitor, what does D.F(dissipation factor) mean? >> and how can I do modeling D.F as ideal R,C,L? >>If D.F is low, what is good ? >>I just read dielectric absorption is low if D.F. is low, am I right? >> >>thanks in advance. >> >> > >e<nospam>ndlr@execpc.com
Article Reference: 75E9F1FFE29A1EBA
Date: 14 Sep 1996 11:39:02 -0700
Original Subject: Re: Zero dB Phase Advance Cct - Possible?
It's called an ALL-PASS network. Two RC's: one time constant = sqrt(2)-1
times 1/(2*pi*f), other = sqrt(2)+1 times 1/(2*pi*f). (f = center of
operating range.) Drive R end of networks with "ein", C end of networks
with "-ein". Tap points are the two outputs. Phase difference between
the two is 90d, good over close to an octave. Amplitude is essentailly
0db over the octave.
...Jim Thompson
% James E.Thompson, P.E. Consulting Engineer % mens %
% Analog/Mixed-Signal ASIC's and Discrete Systems % et %
% Phoenix, Arizona Voice: (602)460-2350 % manus %
% analog@primenet.com Fax: (602)460-2142 % Brass Rat 1962 %
Chris. Garnett <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. I have a circuit, using a passive low
: pass filter in the negative feedback path of an Op-amp that gives phase
: advance but also has 12dB/octave gain. I also have a Cct that gives a 0-180,
: degree phase lag with 0dB gain.
: Does anyone know of a phase advance Cct with 0dB gain or am I chasing the
: holy grail?
: Please reply by E-mail as I'm using a mates account at the moment!
: Iain Hunter
Article Reference: 52B7E08258724B36
Date: 13 Dec 1996 00:28:20 GMT
Original Subject: Re: who defines american wire guage?
User622230 (user622230@aol.com) wrote: : need to know who, and how the guage corresponds to the actual size.. Dunno who. Diameter in inches = 10^(-0.05035 * AWG_number - 0.4884) Accurate to better than 0.1%. Use -2 in formula for 000AWG, for example. (You could say that the AWG gauge number _is_ the size; it's just in inconvenient units for some things.) (Also, resistance, ohms/1000 feet, at 20C, is very nearly 10^(0.1*AWG - 1), though this is not as accurate as the diameter formula above...just easy to remember.) -- Cheers, Tom
Article Reference: E663C6AA23B53D4D
Date: Tue, 29 Oct 1996 16:33:16 GMT
Original Subject: Re: Chebyshev Poles ??
On 29 Oct 1996 06:56:06 GMT, abarak@iil.intel.com (Amit Barak) wrote:
>Does anyone have the formula for poles location of a
>chebyshev and/or butterworth lowpass filter of order n ??
>Either S plane (continuous time) or Z plane (discrete
>time) is ok.
>
>thanks,
*** Butterworth poles location ***
Pi = -sin {(2i-1)pi / 2n} + j cos {(2i-1)pi / 2n}
where :
Pi : pole location number i
n : filter order
i : 1...n
pi = 3.14...
angles in radians
The Chebyshev poles location is a bit too complicated to give the
description here. So if you don't find it, let me know your fax number
and I'll send it to you.
jacques.fournier@dmt.epfl.ch
Article Reference: 6F19719A0F2CB7D5
Date: Sat, 07 Sep 1996 23:22:24 -0700
Original Subject: Re: Butterworth digital filter
Micheal,
One reference I have for you is "Digital Signal Processing"
by Stanley, Dougherty and Dougherty. Published in 1984 by
Reston Publishing Co. Look in the section discussing
"Infinite Impulse Response Digital Filter Design".
The equation for a 2nd order analog Butterworth response
is:
1
H(s) = -----------------------
s^2 + 1.4142136*s + 1
The -3 db frequency is at s = 1 radian/sec.
Hope this helps.
phil kenny
Michel Ruel wrote:
>
> Hi!
> I want to implement a Butterworth second order filter in a program
> (Delphi). Where can I find the equations (Z-transform or analog
> equivalent)?
>
> Thank you for your help!
> --
> <<<<<<<<<<<<<<<<<<<<<<----------------------->>>>>>>>>>>>>>>>>>>
> |Réglages MIRE Inc | TOP Control Inc |
> |Michel Ruel, ing. | Michel Ruel, eng. |
> |Directeur du support technique | Technical Support Director|
> |RÉGLAGE EN USINE-FORMATION-ANALYSE| TUNING-TRAINING-ANALYSIS |
> |223 du Pont, St-Nicolas, Qc, Canada G0S 2Z0 |
> | Tel.:(418)836-6666 Fax:(418)836-6062 |
> <<<<<<<<<<<<<<<<<<<<<<----------------------->>>>>>>>>>>>>>>>>>>
Article Reference: 9FA002E643AB25CC
Date: Wed, 25 Aug 1999 23:13:40 GMT
Original Subject: Re: another capacitor question...
Justin Headley <jheadley@centralva.net> wrote:
>so they say that a capacitor blocks dc in a circut, does that mean it
>doesn't block ac? can someone tell me about what a capacitor does in an
>ac circut please, i want to make a very small transistor amplifier, this
>will be my final project for a while, since school starts, i won't be on
>for a while, maybe 15 minutes a day, tops. i know some of you are going
>to be jumping for joy though....
>
The amount of ac that gets through a capacitor depends on where it is
going...in series, to the load, or in parallel, to ground...the
reactance (ac resistance) of a capacitor follows the formula
X=1/[2*pi*fC]
so, as the frequency or capacitance, or both, increase, the AC
resistance decreases. It's voltage divider stuff, with ac...This can
easily be graphed...we'll jump for joy when you graduate, get a job,
and cut us in for all of the excellent advice you've received...<g>
Tom
>
>--
>Factorials were someone's attempt to make math *look* exciting.
>
>--- Stephen Wright
>
>
Article Reference: 3D1CD340F0D010D7
Date: 23 Aug 1996 00:17:12 GMT
Original Subject: Re: Help my rusty math please
JWISNIA (jwisnia@aol.com) wrote:
: 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
: could be 90% efficient, but didn't go into much detail on how they work.
: I assumed that resistive charging would "waste" half the power, and
: similarly for the dischargin action. But, perhaps I'm not looking at this
: correctly, maybe someone could straighten me out on this too?
Someone else answered the integration part of the question, but for
this part, consider just how they work. You waste some energy getting the
caps charged initially, but once charged it's very efficient. If you
have two caps, each 1 farad, one charged to .99 volts and one to 1.01
volts, and connect them together, charge is preserved so the final
voltage is 1.00 on each. The starting energy is (0.99^2 + 1.01^2)/2,
and the ending energy is 1.00^2. But the starting energy can be
expressed as
((1-delta)^2+(1+delta)^2)/2
= ( 1-2*delta+delta^2 + 1+2*delta+delta^2) / 2
= ( 2 + 2*delta^2 ) / 2
= 1 + delta^2
so if you make delta small, the energy lost is vanishingly small on
each transfer. For the 0.01 volt delta suggested above, the loss is
only .0001 joules loss per cycle. The energy transferred to the output
cap was (1.00^2)/2 - (.99^2)/2 or going back to the deltas...
(1^2)/2 - ((1-delta)^2)/2
= 1/2 - (1-2*delta+delta^2)/2
= delta - delta^2/2
So energy transferred is essentially linearly proportional to delta
for small delta, but energy lost is proportional to the square of delta.
(Yes, in a practical charge pump you will have two _or_more_ such
transfers going on in each cycle, so the above isn't a complete analysis,
but the point remains the same... additional transfers just double/triple/
etc the amount lost.)
--
Cheers,
Tom
tomb@lsid.hp.com
Article Reference: 361E1E204CDA415B
Date: 10 Jan 1996 17:45:12 GMT
Original Subject: Re: What makes 75ohm coax 75ohms?
The characteristic impedance of coax is determined not only by C, but also L, the
inductance. Z = sqrt( L/C ). This is only valid at RF, of course. L and C are
defined by the geometry of the cable and the dielectric. If A is the radius of the
inner conductor and B is the radius of the outer conductor,
L = mu/(2*pi)*ln(B/A)
C = ( 2*pi*epsilon ) / ln(B/A)
An excellent book on this, and many other related subjects, is
_Microwave_Engineering_ by David Pozar, 1990, Addison-Wesley.
Regards,
---
/\ Weston Beal Signal Integrity is the
\\ \ Signal Integrity Engineer application of analog
\ \\ / SUN Microsystems principles to digital
/ \/ / / 2550 Garcia Ave. MS MTV03-01 problems.
/ / \//\ Mountain View, CA 94043-1100
\//\ / /
/ / /\ / (415) 336-1450
/ \\ \ (415) 336-5469 fax
\ \\ weston.beal@sun.com w.beal@ieee.org
\/
Article Reference: 603DCC60DDE83F41
Date: 6 Sep 1998 12:39:03 GMT
Original Subject: Re: Beginner's Questions About Diodes
Alorac Jones wrote: > > Question 1. A circuit described as a clipper or limiter has a pair of > silicon diodes of opposite polarites connected across an audio signal. > According to the explanation, silicon diodes do not conduct until they > are forward-biased by at least 0.7 volts, and as a result, the pair of > diodes has no effect on signals less than 1.4 volts peak-to-peak, but > clip larger signals. The text also states that germanium diodes begin > to conduct when forward biased at 0.3 volts. > > On the other hand, a circuit for a "crystal" radio uses a 1N34A > germanium diode as a detector. Surely the voltage across this diode > is in the microvolt range, and in any event, much less than 0.3 volts. > If the assertions in the first paragraph are correct, how can this > diode pass any signal and how can it rectify? > > Question 2. Diodes such as the 1N914 and 1N4148 are described as > silicon signal diodes. Does the signal have to be at least 0.7 volts > for these diodes to work? Often hear folks say that diodes "don't conduct below 0.7 volts (0.3 volts) forward." Said it myself! But, ain't quite true. Diodes, silicon or germanium, have a V vs I curve that is a more-or-less exponential function, IOW: I ~ e^(V/Vth) where Vth is a "threshold" voltage dependant upon temperature and diode material properties. For silocon diodes at room temp, Vth ~ 25 mV. What this all means is that diode current increases _very_ rapidly with forward voltage. Over the range of currents commonly seen in run of the mill analog circuits (say, 50 uA to 50 mA) forward voltage will change only a few hundred mV. So, statement about diode threshold voltage might be more accurately stated, "diodes don't conduct enough current to be interesting for most solid state analog circuitry below a threshold voltage of about 0.7 volts (0.3 volts)." As far as XTAL radio is concerned, rely to some extent on XTAL passing a bit of current, down in uA range, even at signal levels below nominal threshold voltage. In point of fact, though, XTAL RCVR's don't really have enough sensitivity to pick up signal in uV range. Built one when I was a kid, up in White Mountains of NH, not an AM MTR within 50 miles. Connected long wire antenna, plugged in phones, and heard.... "Hsssssss," all across band. Took RCVR with me, next time family went to visit relations in Boston area; RCVR worked fine there, with coat hanger for antenna! W Letendre Dir Eng NEAT
Article Reference: 92724571CBBAC70C
Date: 7 May 1996 23:38:03 GMT
Original Subject: Re: making capacitors
Robert Forman (rsf1@ix.netcom.com) wrote:
: Does anyone have experience making capacitors for laminate used to make
: printed circuit boards.. I'd like to construct a large (approx .5
: farad) 20,000 v cap using this method. Any ideas out there??
(I assume you mean "from laminate," not "for laminate"...)
Lesseeee... E^2*C/2 = 10E8 joules = 27.7 kWh
Lesseeee... C = .0885 er*A*(N-1)/t
where
C = capacitance in pF
er is dielectric constant relative to air
A = plate area in sq cm
t = thickness in centimeters
Assuming t = .0625" = 0.1588cm
A = 34" x 46" = 1E4 cm^2
er = 4
then to get C = 0.5E12 pF -- you need a pretty big N.
In fact, it looks like the thing would take up quite
a bit of room. Maybe you would do better with a
higher dielectric constant material?
Article Reference: 0AC525BC6C7F173F
Date: 21 Jul 1996 20:05:10 -0400
Original Subject: Re: Wire Resistances
Martyn Watts <100722.3014@CompuServe.COM> writes:
>Does anyone know where I could find a list of resistances for
>wires of different materials at different gauges, in lengths of
>1m? An FTP or WWW address would be nice.
I have a few figures for copper wire. Note that these figures are for a
round trip (out and back) circuit.
Copper Wire
-----------
AWG Feet/Ohm Ohms/100ft Ampacity* mm Meters/Ohm Ohms/100M
10 490.2 .204 30 2.588 149.5 .669
12 308.7 .324 20 2.053 94.1 1.06
14 193.8 .516 15 1.628 59.1 1.69
16 122.3 .818 10 1.291 37.3 2.68
18 76.8 1.30 5 1.024 23.4 4.27
20 48.1 2.08 3.3 0.812 14.7 6.82
22 30.3 3.30 2.1 0.644 9.24 10.8
24 19.1 5.24 1.3 0.511 5.82 17.2
26 12.0 8.32 0.8 0.405 3.66 27.3
28 7.55 13.2 0.5 0.321 2.30 43.4
These Ohms / Distance figures are for a round trip circuit.
Specifications are for copper wire at 77 degrees Fahrenheit
or 25 degrees Celsius.
- murr -
--
Article Reference: 20AA29CBEC08DDF4
Date: Mon, 4 Mar 1996 10:40:00 GMT
Original Subject: Re: Transformer question
MK>mcovingt@ai.uga.edu (Michael Covington) wrote:
MK>>Is it OK to run a transformer at higher than the rated voltage if the
MK>>frequency is also higher, so as to prevent saturation?
MK>>The application is a 700-volt power supply for a Geiger counter.
MK>>I'm wondering if a small power transformer (6.3V to 120V) could be
MK>>run at a higher frequency (a few kHz) and could safely deliver
MK>>700V p-p on the 120V side. (Note that the normal voltage, 120V
MK>>rms, is 338V peak to peak so this would only be slightly more
MK>>than a factor of 2 overvoltage.)
You didn't give a lot of information, like the rating of the
transformer, or how much power you need at 700 Volts.( almost no
current???) Is it a toroid, or a cheapy iron Class A type that runs
hot with no load, etc.
Most power cores are designed to have half their power in the windings
(at full rating) and half in their core loss [this *widely* varies, but
you have to start somewhere]
Let's assume your transformer is a little 2 A at 6.3 V filament
transformer. So full load is 12.6W and maybe this is a cheap 75%
efficient transformer so 4.3W is dissipated in the thing. Ok, that'll
be hot let's make it 80% so 3.2W gets dissipated in the transformer.
More like it.
1.6 W in the windings and 1.6 W of core loss. The power loss in the
windings is usually half in each, too. .8W in the secondary means that
the secondary winding has around .2 Ohms and the primary has around 72
Ohms.
Now we've established that no load is 1.6 W and full load is 3.2W.
Normal cores have a characteristic called volt-sec saturation. That's
pretty much a constant.
The higher voltage the faster you saturate, or the higher the frequency
the higher the voltage it can take.
So twice the voltage only needs twice the frequency.
Ah, but now losses.....
One comes from eddy current in the laminations and another comes from
the hysteresis of the core.
Ph = Kh*f*Bm^x*V hysteresis loss
Kh is a constant which depends on the chemical analysis of the
material and the heat treatment and mechanical treatment to which
it has been subjected
f is frequency in Hz
Bm is maximum flux density in webers / m^2 (Teslas)
V is the volume of the material in cubic meters
Pe = Ke*f^2*c^2*Bm^2*V eddy current loss
Ke is a constant which depends on the resistivity of the material
f is frequency in Hz
c is lamination thickness in meters
Bm is maximum flus density in webers / m^2 (Teslas)
V is volume of the material in cubic inches.
Let's first guess at the Bm....
Assume that the inductance of the core is constant as a function of
frequency (poor guess for iron but a start) Then Bm will be inversely
proportional to the frequency and proportional to the voltage.
At 60 Hz you have Bm at the transformer rating of 120V, If you ran this
transformer at 120 Hz you could put 240V across it, or 680 Vpp So with
10% margin you had better count on putting 270 Vac across it. So we had
better up that frequency to 135 Hz. Note that Bm has just slightly
increase.
So we can change the above equations to
Ph = Kh'*f hysteresis loss
Pe = Ke'*f^2 eddy current loss
since everything else is constant.
Pc = Ph*135/60 + Pe*(135/60)^2 total core loss
If *ALL* the power were in the hysteresis, you'd get a loss of 3.6 W,
which is more than the whole transformer was dissipating under full
load. Probably get warm.
Now let's *GUESS* that half the power went into hysteresis and half into
eddy current losses, then you would have....
Pc = .8*135/60 + .8*(135/60)^2 total core loss
= 5.9 W !!!
Now that to me looks a little bit too warm for the little transformer.
Even with no load.
However, *if* your transformer was a torroid and not the normal PS core
material (which starts running out of gas at around 2KHz) and is heavily
overdesigned (very efficient, running very cool compared to capability)
you could probably get away with it.
However, if your transformer is a standard little E I core; I doubt it.
The conclusion is that you should get a 240 Volt filament transformer
and run it at a slightly higher frequency ( 68 Hz ) Or put a step up
after the one you have, but don't forget you'll have to supply the
step up transformer's core loss power from your little filament
transformer.
- Robert -
* OLX 2.1 TD * Give a gift that keeps on giving - a female kitten.
Article Reference: 89CFFBAEF464BED1
Date: 28 Mar 1996 23:29:19 GMT
Original Subject: Re: Switching Power Supply
Bob Wilson (bwilson@newshost) wrote:
: [........]
:
: In fact, the basic equation is:
: V = k [FNAB]
: where V is the maximum input voltage
: k is a constant that depends on the type of excitation
: (e.g. square wave, sine wave, etc.)
: F is the frequency of excitation
: N is the switching frequency
: A is the effective core area
: B is the core flux density
: As you can see, if we keep all other things unchanged, increasing N
: (frequency) causes a direct and equal reduction in A (core cross
: sectional area). The reduced A obviously translates to a reduced core
: size.
OOPS!! I type faster than I think. And I don't type very fast either!
The above should read:
In fact, the basic equation is:
V = k [FNAB]
where V is the maximum input voltage
k is a constant that depends on the type of excitation
(e.g. square wave, sine wave, etc.)
F is the frequency of excitation
N is the number of primary turns
A is the effective core area
B is the core flux density
As you can see, if we keep all other things unchanged, increasing F
(frequency) causes a direct and equal reduction in A (core cross
sectional area). The reduced A obviously translates to a reduced core
size.
That's better...
Bob.
Article Reference: 14616C8B320E0D4A
Date: Fri, 22 Mar 1996 03:05:33 GMT
Original Subject: Re: [Q] winding transformers for dc-dc conv.
bwilson@newshost (Bob Wilson) wrote: >"winding portion" (preferably only one or two layers), but to give you >an idea, the skin depth at 100 KHz is about 0.008". Obviously, from >this, it would be pretty stupid to wind a 100 KHz transformer with, say, >#18 or #20 wire!. I think you need to re-think who is stupid and who isn't. As you say, the skin depth in copper at 100 KHz is about 8 thousandths of an inch. If you used a wire that was 16 thousandths of an inch in diameter, say a #26 gage, then that would be the bigest wire that would make sense. Right? Wrong! Wire at 26 gage size has a cross section of about 254 circular mils and a resistance of about 42 ohms per thousand feet. A 18 gage wire has a diameter of 40 mils and a cross section of 1624 circular mils. If you subtract the skin depth from the diameter of a 18 gage wire you get 40 -16 or 24 mils. That 24 mils is the part of the wire that *doesn't* have any appreciable current in it. That's also about 600 circular mils. Now subtract 600 circular mils from 1624 circular mils and you get about 1024 circular mils. That's the cross section of the 18 gage wire that *does* have the 100 KHz current in it. A cross section of 1024 circular mils is about what a 20 gage wire has. A 20 gage wire has about 10 ohms per thousand feet. So you can see that by going from a 26 gage wire to a 18 gage wire is *far* from a stupid thing to do at 100 KHz. You drop the resistance of the winding by nearly three quarters despite that silly old skin effect. Jim "With a wire table and a calculator." Meyer
Article Reference: 735DCF40EFBCDBE7
Date: Mon, 2 Nov 1998 21:23:31 -0600
Original Subject: Re: determining decibels of audio signals
Hi Tim, First, decibels are not absolute units. They are relative, and only give information about how much greater or smaller a voltage (or current, or electrical power, or acoustic pressure, ...) is, as compared to another voltage (or current, or electrical power, or acoustic pressure, ...). When comparing two voltages V1 and V2, the expression to compute how many decibels are they apart, is: Ratio [dB] = 20 * log10( V1/V2 ) where log10( x ) means base 10 logarithm of x. V1 and V2 may be in V, in mV, in kV or whatever, as long as both are expressed in the same units, and are positive and nonzero. For instance, a voltage of 50 V is 20 dB above a voltage of 5 V. If you compare powers, instead of voltages, then the expression is: Ratio [dB] = 10 * log10( P1/P2 ) For instance, a power of 50 W is 10 dB above a power of 5 W. Knowing a voltage at a certain point in a circuit does not give you any absolute information about what is the acoustic pressure that signal would represent when "played" through a certain amplifier and a certain speaker. The actual acoustic level will depend on the gain of the amplifier, and on the "gain" of the speaker. It is up to you to choose what voltage you will call "0 dB." If you call Vref to that "reference" voltage, then you can immediately express any other voltage in dB by using this expression: Level in dB of voltage Vx = 20 * log10( Vx / Vref ) How to choose Vref? One suggestion is to choose it as your maximum voltage that you will ever have at that point. For instance, the (analog) output of a CD player is constrained between -2.5 V and +2.5 V. You can forget about the sign and think only about absolute values. They go from 0 to 2.5 V. In your case, your voltages go from 0 to 5 V. If you choose Vref=5 V, then you are calling 0 dB to the maximum voltage you can have at that point, and all the other possible voltages (of course, smaller than 5 V) will be negative, when expressed in decibels. For instance: 0.1 V will be called " -33.98 dB" 0.5 V will be called " -20 dB " 1 V will be called " -13.98 dB " 5 V will be called " 0 dB " Every time you multiply the voltage by 10, you add 20 dB. Remember that calling " 0 dB " to 5 V will be just one possible choice. You cannot choose Vref as zero, for obvious reasons (you will have 1 / 0). The usual way to choose Vref is the _recommended_ voltage at that point. Above that voltage you will have high nonlinearity, and below it you will have bad signal to noise ratio. Regards, Cesar. Tim Curry wrote in message <363E59CF.164CB88E@traveller.com>... >I have a peak detection circuit that measures and holds the maximum >amplitude of an audio signal for a given time interval. I want to tell >the user what the decibel level of the peak sound in decibels. I >currently have a voltage between 0.0 and 5.0 VDC. > >How do I convert this sound level to decibels? I know that decibels are: >10.0*log(out)/log(in). > >Is a person talking normal around 35 decibels? >
Article Reference: 00B05CD283375600
Date: Tue, 27 Aug 1996 08:46:18 -0700
Original Subject: Re: Using Diode (IN4148) or Transistor as Temperature Sensor.
queksp@singnet.com.sg wrote:
>
> Hi!. I wants to use a general purpose diode as a temperature sensor
> (-2.1mV/deg celsius). Sensing range from 0 to 60 deg C. The question
> is:
> How closely match were diode or transistor follows this physics
> equation??
> Also what is the possible spread in temperature characteristic from
> device to device??
> Would a thermistor be better off???
>
> LM35 is out as it cost US$ 2.90 each for 1Kpcs.
>
> Regards
> queksp@singnet.com.sg
Diodes are not a good idea due to the heavy doping.
Base-emitter voltage of a small signal transistor
Vbe = kT ln( Ic )
-- --
q ( Is )
Is depends on various factors.
If you switch currents between two levels and measure the _change_ in
Vbe, using the same transistor the Is term cancels (as long as no
saturation effects occur):
deltaVbe = kT * ln( Ica ) - kT * ln( Icb )
-- --- -- ---
q ( Is ) q ( Is )
= kT * R
--
q
where R is the log of the ratio of the collector currents.
This method gives very good accuracy (claimed within 1 degree absolute
over >150 degree range) with off-the-shelf transistors.
Cheers
Ian
Article Reference: 2290996DEB43844B
Date: 27 Nov 1995 23:14:30 GMT
Original Subject: Re: Burning Capacitor
Thor A. Johansen (thorj@aimnet.com) wrote:
> I find this strange, because an ideal capacitor does not dissipate energy.
An ideal capacitor does dissipate energy (it gets rid of it somehow).
Most college physics courses (in the electronics section) give a problem which
looks something like this:
/
*----* *-----*
| |
----- -----
----- C1 ----- C2
| |
*-------------*
You have two ideal capacitors, C1 and C2, with equal capacitance. C1 is
charged to 10V before the switch is closed; C2 is charged to 0 volts. Close
the switch. What happens?
There are two rules operating here, conservation of charge (CV before =
CV after) and conservation of energy. But they both can't operate here,
because they lead to conflicting results. Conservation of charge gives the
correct answer, which is V(C1)=V(C2)=5v. However, the energy in the system
before the switch was closed was:
1 2 1 2
- CV = - * (C1 in farads) * 10v = 50 * (C1 in farads) joules
2 2
and afterwards
1 2 1 2
- CV = - * (C1 in farads) * 5v = 12.5 * (C1 in farads) +
2 2
1 2
- * (C2 in farads) * 5v = 12.5 * (C2 in farads) =
2
25 * (C1 in farads) joules
So, half the energy in C1 is dissipated (the professor said "it goes somewhere
outside the circuit") when the switch is closed. CMOS IC designers know that
half the energy disappears (it usually turns into heat on the IC) every time
an on-chip capacitive line is charged/discharged -- that's where most of the
power is dissipated in low-power CMOS circuits. A practical circuit and
capacitor have series resistance -- in practice, the resistance turns most of
the extra energy into heat. In most power supply circuits, the capacitor has
some ripple voltage across it which usually turns into heat inside the
capacitor.
Back to the original question, though, your friend's claim may have
something to do with inrush current. When a capacitor is fully discharged and
the first quarter cycle of AC charges it, a lot of current rushes into/thru
the cap. Unless some sort of an inrush current limiter is used (most circuits
use thermistors or thermistors shorted by time-delay relays), the inrush
current might blow a fuse, a circuit breaker or the capacitor itself.
Thanks,
Carl Wuebker * clw@f.rose.hp.com * (916) 785-4296
"My opinions only -- not those of my employer"
Article Reference: 5BD91B08EB0052B9
Date: Mon, 07 Oct 1996 12:30:42 -0400
Original Subject: Re: Help with calculation
Jim Drew wrote: > Being a TTL/microprocessor kind of guy, I know enough about analog to be > dangerous (one a good day). I have a question that I just can't figure > out. > If I charge a 500,000 ufd capacitor at ~90 volts (via a 2 amp 24vct A/C > transformer and voltage quadroupler), I get 68 volts across the cap when > full charged... it stays at that value and will not rise/fall. Now, the > question I have is: how much energy (amperage) is stored in this > capacitor? I know that I can blast a 1/2" steel rod in half if I short it > across the capacitor polls, so there is some serious power here. :-) > I'm trying to build something to control my wife when she gets out of > line... just kidding! ;-) Well, for your energy E = QV/2, but (charge) Q = CV. So, E = (1/2)CV^2 In your case, Q = (500,000)*(10^-6)*(68) = 34 coulombs. This is getting serious. This is in the relm of the amount of coulombs transfered during a moderate lightning bolt exchange. Severe bolts can transfer up to 50 coulombs. Energy (or work) = (34*68)/2 = 1156 joules. Very, very serious stuff. Your current is a different story. You are discharging down from 68 volts. So, discharging through 1 ohm gives you 68 amps, initially. Discharging through the metal rod (i.e. 1/10 ohm let's say) gives you 680 amps initially. You're at the high end of the initial current running through a car starter. This is before the motoring effect. I'm going to nailed for this one but, Watts = Volts*Amps = 46,240 watts impusively. I'm not surprised the rod exploded. I'm surprised you haven't been hurt yet. Be very careful. Your RC time constant will decide the time it takes to discharge through one time constant (tc). You want to go basically more than 5 tcs. The R represents the bleeder resistor. In the past, I used an R that was 1/10 the load seen by the power source. Too high a repetitive discharge current through the cap will damage it. The plates in the cap have a resistance and they can heat to dangerous levels. Let's assume you have a 680 ohm type (imagine that). I = E/R = 1 amp @ Watts = RI^2 = 680 watt rating for resistor at 100%. This means you now have a time constant of 340 seconds. This means you can say the cap is discharged at 5*340 = 1700 seconds = 28 minutes. One final note, just when you think you've discharged a cap as large as yours, the voltage will patially come back. What the hell is this? It's called "dielectric absorption". Briefly, it takes work to align the atoms of the dielectric when imposed by a voltage. Well, it takes work to 'disalign' them. Be very careful of this effect. I've gone on much too long, but I think others will agree this is way serious stuff. Be very careful. As a final note, I would contact the mfr of the cap or any cap mfr for that matter and see what they recommend. While I was in the Army, a newbie was discharging input filter caps to a shed supplied with AC from a substation thinking it was a safe way to handle caps quickly. He did this with a screwdriver. An automotive mechanics screwdriver - a biggie. Not only did he vaporized the screwdriver, he suffered severe injuries to include burns from flying molten metal. He also suffered partial blindness because the arcing caused such severe ultraviolet radiation, it partially burned part of his retina. Several hundred times the arc of a normal electric welding torch. ******************************************************* Doug McKean doug_mckean@paragon-networks.com ------------------------------------------------------- The comments and opinions stated herein are mine alone, and do not reflect those of my employer. ------------------------------------------------------- *******************************************************
Article Reference: E28521E3256D7834
Date: 10 Jan 1996 03:57:09 GMT
Original Subject: Re: MOSFET Noise modeling
Gregg R Castellucci (greggc) <4ctr80$ltt@mdnews.btv.ibm.com> wrote:
>
>Can someone offer some advice or pointer on how to best model the noise due to
>gate resistance in a MOSFET? Currently I calculate the gate resistance and add
>it as a lumped element in series with the gate. I don't think this method
>accurately models the device.
>
Hi, Gregg:
Even though I am quite new for noise analysis, however I just studied
the noise model for MOSFET. As far as I recalled, two major noise
sources are usually considered in MOSFET:
1. Thermal noise
thermal noise current density = Si(thermal)
= 4kT/Gds if in ohmic region
= 8KT*gm/3 if in saturation region
2. Flicker noise (1/f noise)
Flicker noise current density = Si(flicker)
= 2*Kf*K'*Id/(Cox*L*L*f)
where
Kf is the flicker noise coeffecient, a constant
K', Cox, L are all device parameters.
Id is the bias current
f is the frequency.
Be careful about the region of operation (satuation or ohmic) of the
MOS. It is simple but sometimes confusing.
I think the next two references have been helpful to me a lot, so you
might want to check them out:
1. Analysis and Design of Analog Integrated Circuits, by Paul Gray and
Robert G Meyer, Wiley, (third edition). See Chapter 11, this give
a good overview of Noise Analysis.
2. VLSI Design Techniques For Analog and Digital Circuits, By Randall
Geiger and ... , McGraw-Hill, See: section 3.1.8. This gives a good
and brief review on MOSFET noise model as well as SPICE noise model.
Two more recent papers may help to get to the detail:
3. JSSC Feb. 1993, p.184. for a complete trace of noise.
4. JSSC July 1994, p.833 for MOSFET thermal noise
(where JSSC = IEEE Journal of Solid State Circuit.)
I don't know if it is "accurate" to model the MOSFET noise source as
a current source between Drain and Source (or a voltage source in series
with Gate), however, it seems to me that a lots of people using it
and using it as a mean to judge the noise performance of a circuit.
I wish to learn more on noise, more comments on this topic
are welcomed !!
--
-------------------------------
Mr. Chiajen Lee
c532237@showme.missouri.edu
km.bbs@bbs.ee.nctu.edu.tw
km.bbs@bbs.cis.nctu.edu.tw
Article Reference: 2AFA5B53833E71CE
Date: 23 Feb 1996 19:25:45 -0800
Original Subject: Re: digital filtering
In article <4gfmg6$4am@erinews.ericsson.se>,
Tomas Lundborg ERA/AR/BU <eralob@eras70.ericsson.se> wrote:
>Two common filter structures are used:
>
>FIR- Finite Impulse Response
> The input samples are put in a shift register with limited length.
> The output is a function of what is stored in the shift register
> for the time being.
>
This above is true, while the stuff below has some errors/typos:
> ex. y(k)=a*y(k)+b*y(k-1)+c*y(k-2)
The above difference equation is recursive, which makes it IIR.
(Actually it's not even a filter, since there is no input)
I guess it was a typo :)
It should read:
y(k)=a*x(k)+b*x(k-1)+c*x(k-2)
Transfer function: H(z) = Y(z)/X(z) = a + b*z^-1 + c*z^-2
H(z) has only zeros, no poles.
> coefficients a, b, c may be determined by studying the
> response of a dirac pulse (spike).
> a=0, b=0.5, c=0.25 would probably give a nice low pass
> filter, to make an unscientific guess. To be scientific,
> use the z-transform.
>
>IIR- Infinite Impulse Response
> A recursive algorithm, where the filter value is determined by
> the input value and the current value.
I think this is more accurate:
A recursive algorithm where the filter output value is determined
by the current and past input values, as well as past output values.
>
> ex. first order low pass filter y(k)=(d-1)y(k)-d*y(k-1)
> where the rise time is determined by d and the sample rate
>
Again, this equation does not have any input. (Where do you put the
'stuff' you want to filter?)
Example of IIR: y(k) = b0*x(k) + b1*x(k-1) - a1*y(k-1) - a2*y(k-2)
Transfer function of same filter:
H(z) = Y(z)/X(z) = (b0 + b1*z^-1) / (1 + a1*z^-1 + a2*z^-2)
H(z) has both zeros and poles.
>
>This is the general idea, off the top of my head.
>Study a book on digital filters to get the details.
>
Good advice.
>Hope this helps.
>
>/Tomas
>
Best regards,
Thor Arne Johansen
Article Reference: CC0C303A7E773EEB
Date: 21 Feb 1996 17:58:30 GMT
Original Subject: Re: digital filtering
Two common filter structures are used:
FIR- Finite Impulse Response
The input samples are put in a shift register with limited length.
The output is a function of what is stored in the shift register
for the time being.
ex. y(k)=a*y(k)+b*y(k-1)+c*y(k-2)
coefficients a, b, c may be determined by studying the
response of a dirac pulse (spike).
a=0, b=0.5, c=0.25 would probably give a nice low pass
filter, to make an unscientific guess. To be scientific,
use the z-transform.
IIR- Infinite Impulse Response
A recursive algorithm, where the filter value is determined by
the input value and the current value.
ex. first order low pass filter y(k)=(d-1)y(k)-d*y(k-1)
where the rise time is determined by d and the sample rate
This is the general idea, off the top of my head.
Study a book on digital filters to get the details.
Hope this helps.
/Tomas
Article Reference: 62E607C478C00B8B
Date: Wed, 27 Nov 1996 00:44:20 GMT
Original Subject: Re: How do you offset and amplify a voltage using single ended op amp
> I've looked in the H&H "Art of Electronics" and in there is a sample > circuit that will offset a voltage in the positive or negative > direction, but it uses an op amp whose output can swing both positive > and negative. How do I offset a voltage using a single ended supply > op amp? Hello Dave Op-Amps have *two* inputs. One is marked "+", and the other "-". You can offset a positive input by putting a positive voltage up the *inverting* input. Without getting into diagrams...let us try and describe. Resistor from output to "-" input is Ro, the feedback resistor. Resistor from inverting input "-" to ground is Ri, a gain setter. There *may* be a resistor from the sensor to the "+" input, but it does not have a gain role (equalises bias currents). The gain of this non-inverting arrangement is: Vout = Vsensor*( 1 + Ro/Ri) Now if you want to subtract 4.2V, then make a positive voltage from a potentiometer, or potential divider using two resistors..whatever, and offer it *between* ground (0V) and the end of the resistor Ri that leads to the "-" input. In other words, your input goes up a non- inverting route, while your offset goes up a inverting scheme. The gain up the inverting route is different: Vout = Voffset*(Ro/Ri) If Ro was the same value as Ri, and large compared to your sensor source resistance, then the Vsensor gets gain (1 + Ro/Ri) = 2. The 0.5V change becomes 1V at the output. Without an offset input, the 4.2V causes 8.4V The offset voltage sees gain Ro/Ri = 1. So you need +8.4 volts to cancel it. Of course you can play tunes with the values to make more convenient voltages, but take care. Op-Amps outputs often cannot work right up to the supply rails, and even those that claim they do, cannot do it while supplying realistic currents. You would have to accept that trying to cancell *all* of that offset in a single supply scene might be beyond reason for the op-amp you chose. Hope this helps -- Graham
Article Reference: B96A70F7D10B9467
Date: Fri, 12 Jan 1996 21:11:51 UNDEFINED
Original Subject: Re: What makes 75ohm coax 75ohms?
In article <4d6ee9$cnv@driene.student.utwente.nl> vincent@mat002102.student.utwente.nl (Vincent J. Arkesteijn) writes:
>From: vincent@mat002102.student.utwente.nl (Vincent J. Arkesteijn)
>Subject: Re: What makes 75ohm coax 75ohms?
>Date: 12 Jan 1996 19:56:57 GMT
>Adam Seychell (s921880@minyos.its.rmit.EDU.AU) wrote:
>: Could someone explain what physical properties of a coaxial
>: cable give it an ohm rating? When you look at a 50ohm
>: Another wierd thing I see about coaxial cables is that how can
>: they carry radio frequencies when they have such a high
>: capacitance between the center cable and its shielding.
>: Adam
>Howdy Adam,
>A coaxial cable indeed has a certain capacitance per unit length. However, it
>also has a certain self-inductance per unit length. These series-inductance
>and shunt-capacitance thus form a circuit as shown below.
> ---+---^^^^^---+---^^^^^---+---^^^^^---+---
> | L | L | L |
> = C = C = C = C
> | | | |
> ---+-----------+-----------+-----------+---
>The C's and L's aren't real, lumped components, but distributed ones.
>Now, a bit of math can show you that a cable of infinit length yields a
>resistive impedance at the input. This impedance (the characteristic impedance
>of the cable) depends on the ratio of self-inductance to capacitance (both
>per unit length).
>It is also related to the ratio of the inner diameter of the outer conductor
>and the outer diameter of the inner conductor. By looking at that ratio (and
>a little experience) you could tell the difference.
>A rather simple formula to calculate the impedance from these two diameters
>exists, but I can't remember it right now and I can't find it either.
The formula is roughly:
138
Z= -------- * log(D/d)
sqrt(er)
where
Z = characteristic impedance,
er= relative dieelectric constant of insulating material (air = 1)
D = inner diameter of outer conductor
d = outer diameter of inner conductor
Regards,
Richard
Richard Rasker
Calslaan 54-11
7522 MG Enschede
Holland
Article Reference: B208E3B9ED466325
Date: 10 Jan 1996 15:17:48 GMT
Original Subject: Re: What makes 75ohm coax 75ohms?
In article <4cvkc9$59o@aggedor.rmit.EDU.AU>, s921880@minyos.its.rmit.EDU.AU (Adam Seychell) wrote: > Could someone explain what physical properties of a coaxial > cable give it an ohm rating? When you look at a 50ohm > cable (say RG-58) and 75ohm (RG-59) I can't see any differences? > Can it be measured? What would be the ohms of a standard audio RCA lead? > > Another wierd thing I see about coaxial cables is that how can > they carry radio frequencies when they have such a high > capacitance between the center cable and its shielding. > for example, the RG-58A/U has 101pF/meter capacitance. > so for 20 meters of cable at a frequency of 477Mhz then > the reactance (Xc) is equal to > C = 20 * 101pF = 2.0nF > > Xc = 1/ (2* 3.1416 * 477e+6Hz * 2.0nF) > = 0.165 ohms ! > > This is *much* lowwer than the 50ohm resistance from the > source. wouldn't his cause major attenuation in the signal? > A quick calculation will show that the above example will reduce the > signal by 303 times (or a gain of -49dB) > how can signals at this frequency ever travel down these cables, how ??? > > Adam The impedance of a line consists not only of capacitance, but also of inductance and resistance. Excluding the case with resistance, calculating the impedance Z of a line by dividing the distributed reactance into small segments and add these properly, is possible. Think of the line as an endlessly long chain of segments. Each segment consists of an inductor connecting the segment before to the one after, and a capacitor between the segment and the common earth. The impedance seen at the end of this line will be either: Z = Xl/2 + Squareroot( Xl*Xl + 4*Xl*Xc)/2 or Z = Xl/2 - Squareroot( Xl*Xl + 4*Xl*Xc)/2 The reactance of the inductor and capacitor is Xl resp. Xc To make this infinitely long line, finitely long, the number of segments must increase. The increase of segments makes the inductance and capacitance of each segment smaller and smaller. By making the capacitance ( C ) and inductance ( L ) approaching zero, Xl will go towards zero and Xc towards infinity. Z = Squareroot( Xl*Xc ) = Squareroot( J*2*pi*f*L / J*2*pi*f*C ) = Squareroot( L / C ) Which is independent of frequency and length but depends on inductivity and capacitivity or as in the case of vacuum, permeability and permittivity.
Article Reference: 234E80B533169885
Date: Thu, 03 Oct 1996 07:47:51 -0700
Original Subject: Re: Be Warned: Mutant Capacitor !
Jack Climent wrote:
...[clip]...
> Since the Mutant Capacitor is perfectly realizable, I think the question
> is valid. I am aware that the *Instantaneous* value of I & V in a circuit
> with such a component is measurable and stantionary. I am also aware that
> the capacitance would reach a steady-state and a single value on a meter,
> so that it would have, as you say: ("the same properties at either
> terminal").
>
> But the question is: What happens in a *Non-Instantaneous* (AC) circuit
> where the I & V are not at a steady-state and the capacitor becomes
> reactive (Frequency-Domain)..... ("Would filter circuits filter out
> different frequencies on the negative side of the waveform than the
> positive side of the waveform?").... and what happens in the charging and
> discharging sequences..... ("Would one side of the capacitor get filled
> up and start to discharge before the other side filled up?").....and if
> so,..("what would happen to those nice, tidy, trig relationships that we
> depend on to figure out what's going on in a circuit?") etc.. etc...
The question has already been answered...the instantaneous value of
charge on the capacitor will be given by the equation:
q=int[C(V)dV] + q_0; where q_0 is initial charge
Simply use this value of q in the differential equations you use to
describe the circuit under consideration, and all other tools normally
used in the circuit analysis apply.
> I realize that this is a theoretical question, however I don't see why it
> can't be answered. Who knows, maybe it might turn out to be something
> useful...............It wouldn't be the first time !!
...it has been answered, many times in print, and a few times in this
thread (we aren't talking about *new_science* here!!)
> Inquiring Minds Want To Know..
Ok...use the equation above in a simple series RLC circuit, write
down the differential equation, then try to solve it. If you are
truly an inquiring mind, this should provide hours of entertainment ;)
Steve
Article Reference: 23AA9398DCAABA10
Date: 18 Aug 1998 19:02:21 GMT
Original Subject: Re: Help, thermistor conversion
In article <6r7bp8$7qq$1@phobos.brunnet.net>, "Ryan" <swilson@brunnet.net> wrote: > I have a thermistor and my multi meter reads the the thermistor in K. > Is there IC or conversion number to convert K to ferhiet or celuis? C = K - 273.15 F = (9 / 5) * C + 32 TGB \\ The opinions expressed herein are my own. //
Article Reference: 58BFCFEAEBFAA658
Date: 17 Aug 1998 14:26:55 -0400
Original Subject: Re: Help, thermistor conversion
Ryan <swilson@brunnet.net> wrote: >I have a thermistor and my multi meter reads the the thermistor in K. >Is there IC or conversion number to convert K to ferhiet or celuis? Subtract 273 from K to get Celsius. Then multiply Celsius by 9/5 and add 32 to get Fahrenheit. M Kinsler -- ............................................................................. Interpretation and instruction of physical science and technology Athens, Ohio, USA. "http://www.frognet.net/~kinsler"
Article Reference: D37B47EE41573457
Date: Wed, 27 Mar 1996 23:41:18 -0800
Original Subject: Lietz wire - the horses mouth
There has been some discussion of Litz wire and why it works. This topic is treated in great detail in: Radio Engineers Handbook by Frederick Emmons Terman, Sc.D. First Edition Tenth Empression McGraw-Hill Book Company, Inc. New York and London 1943 [I tried used book stores and advertized in the local free ad paper] On page 30 is information relating the ratio of AC to DC resistance as a function of the factor x (defined in eq. 3) which we call today skin depth. This is studied for not only circular cross sections but also for rectangular and hollow tubes. Starting on page 36 is a discussion of Proximity effect. This relates to the current in one conductor effecting the current distribution in an adjacent conductor because of it's magnetic field. Starting on page 37 is a discussion of Litz (Litzendraht) wire including a formula for the AC to DC resistance ratio. equation (8a): Rac/Rdc = H + k * (n*Ds/Do)^2 * G where H= resistance ratio of individual strand when isolated (plain Cu wire) G= constant taking into account the proximity of neighboring wires n= number of strands in cable (Litz cable) Ds= diameter of individual strand Do= diameter of cable k= constant depending on n (varies from 1.55 for n=3 to 2.0 for n=infinity) There are many versions of this equation depending on the geometry and wire spacing given in section 19. the values of the parameters are given in tables in section 18 Losses in Air-cored Coils at Radio Frequencies on pages 74 through 77 and in section 19 Calculation of Coper Loss in Coils on pages 77 through 83. Section 19 has some information on toroidal coils. Section 20 talks about Distributed Capacity of Coils with diagrams of different windings and the order of the turns. Starting with section 23-Coils with Magnetic Cores, is information on transformer design for power and audio frequencies. If you are working with Litz wire this is the place to get a lot of information. Have Fun,
Article Reference: 4290D50E991206ED
Date: Mon, 16 Aug 1999 12:53:36 +1000
Original Subject: Re: Math function for log potmeter
I have actually seen a plot of log pot resistance years ago, and I
think it might have been in a philips book.
The plot was proper 100:1 exponential.
Res = Rlow x 10^(2 x Rotdeg/270) for a 270 degrees pot.
HarryD wrote:
>
> Does anybody actually *know* the answer ?
>
> Again, the question was: what is f(a) = R for a log potmeter, where:
> - a is the rotation (from 0 .. 1)
> - R is the max resistance
>
> Barry Lennox <barryl@ntplx.net> wrote in message
> news:37b572e6.9184969@news.ntplx.net...
> > On Tue, 10 Aug 1999 16:10:00 +0200, "HarryD" <keep@spam.out> wrote:
> >
> > >For linear tapered pots, the transfer function would be something like
> > >Rp=a*Rmax, where a = 0 .. 1.
> > >What is the function for an audio tapered (log) potmeter?
> >
> > The other posters have answered many of the points, but there was a
> > very good article in "Electronics and Wireless World" about 3-4 years
> > ago on this very topic, and some tricks you can play to get weird
> > functions. It's a UK magazine.
> >
> > Barry Lennox
--
Regards,
Russell
Article Reference: 9AFF3BCAB47FB30E
Date: Fri, 22 Nov 1996 13:42:10 -0800
Original Subject: Re: Parallal caps & conservation of energy...
Thomas A Maier wrote: > > > It is easiest to see if you have 2 identical caps and also easier to > follow if they are relatively large (>=100uf or so). Clip a voltmeter > across capacitor A and make sure that cap B is completely discharged. > Charge cap A by placing a battery or supply across it for a second or > two. Read the voltmeter. Connect cap B across the voltmeter and cap A. > > At this point the total capacitance is 2*C since the two caps are now in > parallel. Check the voltmeter. Lo and behold it will read exactly 1/2 > of the original voltage. Keeping equ (1) happy and balanced. > > For one capacitor the stored energy is: W1=(1/2)(C)(V^2) Now connect a second identical capacitor, uncharged. The total C is 2C and the voltage is V/2 (charge is conserved). The energy is now: W2=(1/2)(2C)(V/2)^2=(1/4)(C)(V^2)=(1/2)W1 The stored energy has been reduced to 1/2. The missing energy has been either lost by electromagnetic radiation or by wire resistance heating or both.
Article Reference: 62DB077C8CE56252
Date: 3 Oct 1996 16:56:14 GMT
Original Subject: Re: Winding Inductors
Bruce Sutherland (bsutherl@commpass.awinc.com) wrote:
: I would like to know what the equation for winding my own inductors.
: Does anybody know it or know where I can find it.
There is no equation. At least not a one-size-fits-all equation.
If you are winding them on a core (Iron, ferrite, permalloy or
whatever), then you need to find the AL value (written as "A" with a
subscript "L"). This is the number of turns per nanohenry *squared*. For
example if the AL value is 100, then 1 turn will give you a 100 nH
inductor; 2 turns will result in 10000 nH (10 uH), and so on. The
formula is:
AL = L/N^2
DC saturation is an important criteria as well. For any inductor wound
on a permeable core (that is not an air core), there is a maximum field
strength beyond which the inductor will magnetically saturate, and act
as an ordinary piece of wire. Magnetic field strength can be determined
according to:
H = (I x N)/Ie
where: H is the DC field strength in Amps/Meter
I is the DC current in the windings
N is the number of turns
Ie is the effective magnetic length if the core (from the
core data book.
The idea is to determine the maximum possible field strength that can be
tolerated by the particular core, at the effective permiability you are
using (determined by the core material, geometry, and air gap), and
design the inductor so the H value above does not exceed the maximum.
Adding an air gap is generally done with an inductor, as a means of
increasing the maximum value of current before saturation, but at the
cost of reducing permiability. Without an air gap, the inductor will
saturate at very low DC currents. You as the designer have to trade off
increasing air gap with the need for more turns (as a result of reduced
permiability caused by adding the air gap.
There are some of the basic, simple concepts to get you going.
So, as you can see, there are no magic formulas. Like anything else, you
need to understand the subject before you can proceed.
Article Reference: 281D25F96C9B2FE4
Date: Wed, 16 Oct 1996 04:45:49 GMT
Original Subject: Re: Help finding the characteristic impedance of a PC trace
jwisnia@aol.com (JWISNIA) wrote:
>I can't seem to find a reference in my office giving the derivation or
>formula for the characteristic impedance of a PC trace. Specifically, if
>I have a 1 oz copper trace on layer 3 of a 4 layer .062" thick G10 board,
>with layer 4 a ground plane, how does the characteristic impedance of a
>straight trace change with its width, and what is it, for widths of .050"
>to .125"? (Nothing much else near the trace on its layer or the other
>layers either.)
The following information is from an excellent unknown reference
source (somebody didn't copy the title page!).
For a microstrip line (trace above a ground plane):
Z = [87 / sqrt(er + 1.41)] * ln[5.98 * h / (0.8 * w + t)]
where: t = copper thickness in inches.
t = 0.0014" for 1 oz copper
t = 0.0028" for 2 oz copper
w = width of trace in inches.
h = height of dielectric (spacing between the
trace and ground plane) in inches.
er = dielectric constant of the circuit board.
er is approximately 4.8 for G10 and FR4
epoxy-glass boards (value is an effective
value that takes into consideration the
non-ideal nature of the material which
is full of holes and components).
Z = impedance in ohms.
ln = natural logarithm (base e).
sqrt = sqare root operator.
The value of Z is most accurate for 0.1 < w/h < 3.
You best ask the pcb manufacturing house what all the cross sectional
dimensions of the circuit board are. When making multilayer boards,
the layers could be spaced at irregular intervals.
-
Mark Chun | mark@reson.com | Santa Barbara, CA
Article Reference: 9AEDCCDAF3FACAE5
Date: Fri, 22 Nov 1996 12:44:40 -0800 (PST)
Original Subject: Re: Parallal caps & conservation of energy...
jeffa@ix.netcom.com(Jeff Anderson) wrote: > In thinking about this, I believe it's because the energy has been >"radiated" away as electro-magnetic radiation. To move charges from >cap C1 to C2 the charges must move from 'resting' at C1 to C2, then >stop. That is, they first accelerate, then decelerate. And it is this >process of acceleration and deceleration which results in radiation. This question came up a few weeks back. In retrospect, I think the following is the best explaination: If we postulate zero resistance then there will be a Heavyside impulse (or Dirac delta function, if you prefer) of current in the loop comprising C1-C2. This will produce a broadband EM pulse which contains all the "missing" energy. Of course any such loop has finite inductance, so the thing would actually ring until all the extra energy is radiated over some finite time. The inductance will mean that there is not an impulse, but a damped siusoidal current flowing in the loop. -KF-
Article Reference: C6F72210A0817534
Date: Sun, 24 Nov 1996 11:56:33 -0600
Original Subject: Re: What's the maximum voltage gain from a MOSFET transistor?
Todd K. Moyer wrote:
>
> > While stumbling around here, I'd still like to hear from some real
> > MOSFET experts! For example, what's a good way to think about VMOS
> > g_oss r_o g_ds Va u_rs or whatever you want to call it! In device
> > design is there a tradeoff, like the beta-Early voltage tradeoff for
> > BJTs?
>
> I doubt that I qualify as a "real MOSFET expert", but I can speak in a
> limited way of device design tradeoffs. I've designed some analog
> circuitry in a primarily digital CMOS process where I had complete freedom
> to choose the gate width and length (within the design rules of course).
> If I wanted an op-amp to have very high DC gain, I could "stretch the
> channels" of the transistors, i.e. make the gate lengths very long (for
> instance make it 10u in a process that could go down to 0.8u). This had
> the effect of increasing the output resistance of the devices (or
> decreasing g_ds). Of course it also decreased g_m and therefore
> necessitated an increase in gate width to compensate. So as you can
> imagine, it eventually becomes impractical to make the device because it
> becomes too large and the undesirable gate capacitance gets so high that
> the amplifier is too slow. But I was able to easily achieve 80-90 dB of
> gain in what was essentially a 2-stage amplifier (diff amp followed by
> common source amp) of reasonable size and bandwidth. Unfortunately, I
> don't completely remember/understand why longer channels mean higher output
> resistance; maybe someone else can provide an explanation.
>
> Todd Moyer
> In Focus Systems
> todd.moyer@infocus.com
I'm no "real MOSFET expert" either, but I can post a (rather verbose)
reply as to why longer channel lengths mean higher output resistance.
First, consider the resistance of an n-channel mosfet. We know the
mobility is un, and the conductivity is mobility*charge on one
electron*number of electrons, which is un*q*n. Well, what is q*n? For
a mosfet in inversion, lets assume that it is, to a first order, all of
the gate voltage "left over" after we've inverted the channel times the
gate capacitance ( Q=C*V, right?). Well, the left over voltage is
(VG-VT -Vy), since VT is what we've defined to be the voltage needed to
invert the channel, and Vy is the voltage at some distance y from the
source (the Voltage in the channel will vary from source to drain
because we have current flow due to drift in the channel). So now we
have
conductivity in the channel = un * Cox * (VGS-VT-Vy)
Ok, we have conductivity. But notice that it varies along the length of
the channel. Also, notice that we would integrate chunks of resistivity
(1/conductivity) to get the total channel resistance. If we have a long
channel, we have a lot of resistance. If we have a shorter channel, we
have less resistance. So, total channel resistance depends directly on
L (channel length).
Now enter the drain. The drain, in saturation, creates a depletion
region (because it is always reverse-biased w.r.t. the body of the
device). It creates a depletion region everywhere, including the
channel. What does this depletion region do? To a first order (again)
it decreases the channel length. Why? Well, in the channel, electronic
conduction is limited by the mobile carrier charge and the lateral
electric field (refer to the conductivity equation above). In the drain
depletion region, however, there are no mobile carriers at all (again,
to a first order) but there are HUGE lateral electric fields, and the
transport of carriers is no longer limited by anything except carrier
scattering.
If you increase the drain voltage, you increase the depletion region.
This is the simple depletion width calculation we did in senior ee
devices class, where the votlage dependence was square root for step
junctions and cube root for graded juntions. But wait, no one likes
square roots. And besides, if we start out with a depletion region
already and increase the reverse bias, the increase in depletion width
can look kind of linear (I can't draw the graph here, check any device
physics text for the juntion capacitance versus reverse bias voltage to
see what I mean, noting that depletion capacitance per unit area equals
epsilon of semiconductor/depletion width).
Enter the parameter lambda. The gd of a mosfet (which is the inverse of
output resistance) is ID/lambda, where ID is the dc bias current and
lambda is the channel length modulation factor which you see in spice
models. The classical square-law mosfet equation in saturation is:
ID= un*Cox ((VGS-VT)^2) * (1 + (lambda*VD))
-------
2
The parameter represents the effect of the drain depletion region
spreading into the channel. For long channel devices in saturation, as
you increase the drain bias (and hence, the dc bias current) you
increase the drain depletion region and, to a first order, decrease the
channel resistance by a corresponding amount. For a shorter device, to
a first order, you also decrease the resistance by the same absolute
amount, but the percentage change in channel resistance is much greater,
because you had a smaller resistance (due to having a shorter channel)
in the first place.
In this explanation, I've repeatedly used the phrase "to a first
order". I've also used the term "shorter channel" rather than "short
channel". There are long, involved reasons for this. The above
arguments apply to long channel devices which fit the "gradual channel
approximation" only. They break down for short channel devices, where
drains and substrates have some control over channel charge. The best
thorough treatment I've seen in less than 200 pages is John Brews
"Physics of the MOS Transistor" published in Applied Solid State
Science, Supplement 2A, Academic Press, 1981 ISBN 0-12-002954-5. The
best treatment I've seen in more that 200 pages is Tsvidis' Operation
and Modeling of the MOS Transistor, McGraw HIll, 1987, ISBN
0-07-065381-X.
Hope this helps.
--
+-----------------------------------------------------------+
| Edward R. Deak | Email: erdeak@ti.com |
|--------------------------------| |
| Texas Instruments, Inc. | |
| Mixed Signal Design Department | |
| P.O. Box 660199, MS 8729 | |
| Dallas, Texas 75266 | |
Article Reference: 8CC284728FF45A4E
Date: Mon, 05 Feb 96 01:05:16 GMT
Original Subject: Re: Motor field strength??
In article <4f2sb9$2j5@news.esslink.com>
paulc@esslink.com "Paul A. Cianciolo" writes:
>Can anyone tell me the mechanism by which weakening the field in a
>shunt wound motor there by increases the speed.??
>
>This does not make sense to me... I am also looking this up in some
>motor books I have but the concept just doesn't jive. I have a motor
>that will almost explode if I dont excite the field. It seem like
>this would be a easy place to adjust the speed of a motor .. I have
>heard that in a shunt wound motor that the field current is only a
>fraction of the armature current .
>
The motor armeture a current = ((Vin - Vb)/Za) where Vin is the applied
(terminal) voltage, Vb is the back EMF produced by the windings moving in
the field from the field coils. Vb is proportional to The field strength
and the speed of the motor. If the field current is reduced the current will
increase (Due to lower Vb) increasing the speed until equalibrium is restored.
Hope this helps.
--
********************************************************
* And on the first day the lord said...... * *
*......Lx1, Go! and there was light! * Dan Mills *
Article Reference: AF282078ECB1C865
Date: Thu, 29 Feb 1996 00:13:41 GMT
Original Subject: Re: Looking for temperature sensor under $ 1.00
On Tue, 27 Feb 96 00:29:40 PST, 1RFD1642@MtSACvm.MtSAC.edu wrote: >I looked at the two motorola ones, don't remmember the part number, but >i could looke at the master selecction guide so that's not a problem. >Anyone out there knows of anything else...and an aplication circuit ? >please :-) . BTW something that will measure between 0 - 50in degrees >celsius would do, and accuracy of + or - 5 degrees is ok. An NTC with proper calibrating will measure temp within .1 deg C. Hook it up to an Ohm-meter and measure the resistance at 0-10-20-30-40-50 C by taping it to a thermometer with Al tape. List the measured resistances in Excel (or any other spreadsheet) and match them with the formula T=a-b*ln(k) where k is the resistance in (kilo)Ohms. Let Excel "solve" a & b by minimizing the sum of the squared errors in predicted temperature. I have calibrated over 100 NTC's this way and can measure temperature to .1 deg C accuracy. NTC are only $0.20 and SMD NTC's are very small. I can post (or mail) spreadsheets, formulas and extensive setups. Personally I made a setup with a DMM with RS232 output and a relay board so I select up to 16 NTC's with the parallel port and read the values with the serial port. A Pascal program handles the values and draws the graphs on screen. Voila: a sixteen-channel PC-controlled temperature data sampler for about $150. The DMM is $100.
Article Reference: 9D0551F88317AF74
Date:
Original Subject:
In article, John Lundgren jlundgre@news.kn.PacBell.COM wrote:
>
>Subas (subas@intnet.pc.my) wrote:
>: hi there everyone!!
>: is there a way simple and easy way to construct a parabolic
>: reflectors(dish).i have not bought the downconverter yet
>: and also the amplifier to receive this signal.
>: can anyone please give me the ful and simple way and
>: formulas to contruct the parabolic dish and also recommand
>: a good book in the construction of a parabolic dish.
>
>
>The formula for a parabola is a simple algebraic equation, something like
>Y = AX^2 + BX + C. (Where A not equal to 0)
>So with A=1 and B and C = 0, Y = X^2.
>Check on the newsgroup rec.radio.amateur.antennas for more info.
---------------------------------
The neat thing about a parabola is that if you choose to make a square
matrix of pieces that form the inner shape of a parabola, the pieces all
have parallel top edge curvatures, thus can be cut out of one piece of,
say, plywood. Then you can half-slot the pieces so that they fit together
like a 3D jigsaw puzzle, or kid's construction set. Then you can set corner
pieces of wood with screws at strategic spots to give it rigidity.
A parabola is a "locus" (set) of points equidistant from a point, the
focus, and a line on the convex side of it, the directrix, which is, from
observing that the closest point on the parabola is the apex, or center,
also at a same distance from the apex as is the focus, and is obviously
also both perpendicular to a line drawn from the focus and through the apex
to it, and symmetrical on either side of that drawn line.
The formula for a given parabola of desired focal length comes simply from
the fact that the distance between any point on the parabola and the focus
point is equal to the distance from that same point on the parabola to a
plane (or line in 2-D) in back of the reflector shape which is the same
distance from the apex, or center, of the parabola as is the focus. This
plane(line) is called the directrix, and it is perpencidular to a line
drawn through the apex of the parabola to the focus. Note that the distance
from the point on the parabola to the directrix is the closest distance, or
the perpendicular distance to the plane(line). The reason I use the words
plane(line) is that each point on a paraboloid, a 3-D parabola of rotation,
is given by the formula which gives only a vertical slice through the focus
and apex of the paraboloid and the directrix plane, the slice thus yielding
a line directrix for each rotation "slice" of the "spun" parabola in 2-D.
Now for this simple case it is good to use the origin of an x-y Cartesian,
(named after DeCartes, don't freak), to graph these points so that you have
a big paper pattern when you make one the size you want to mark the
material with for cutting. And it is most useful to call the apex of the
intended parabola the origin, or point (0,0), (x=0, y=0). Now, since the
focus is "above" the apex, then its x-coordinate = 0, so the focus is just
the point (0,f), where f is the focal length to the focus in the positive y
direction. The directrix is clearly the line "below" the apex, so that it
is the line on which all y = -f. Thus the apex is equidistant from the
focus and the directrix and is on the parabola! The simple distance
formulas come from old Pythagoras and algebra, and they are simply the
square root of [the vertical distance squared plus the horizontal distance
squared]. We need first the distance formula from the focus to a point on
the parabola, and we'll call it (x,y), since that's what we're trying to
find, (SQRT means square root and we'll use the ^2 symbol to mean squared)
and so:
Dfocus = SQRT [(x-0)^2 + (y-f)^2]
We use the difference in coordinates to get rid of the coordinates
themselves, since (y-f) would leave only their difference, which IS their
vertical distance! Likewise for (x-0) or just x in this case. The focus is
already AT the horizontal zero distance.
Now, the distance from a point, which we do not know yet, to a line, sounds
like a crazy thing to try to get, since we don't know what part of the
parabola it lies on or how close that part is to the directrix line! But we
DO! We know the vertical distance above the x axis is just its
y-coordinate, y, and the distance from the x axis to all points on the
directrix to be f, the focal length, but below the x-axis, and so in the
negative direction, that is, all points on the directrix have y-coordinate
= -f . Thus the distance from the directrix to any point on the parabola
is simply:
Ddirectrix = SQRT [(x-x)^2 + (y-(-f))^2]
And so now, to satisfy the definition of a parabola, these two distances
for all points (x,y) on the parabola must be equal, and so we simply set
one equal to the other and solve for a simple equation!:
Dfocus = Ddirectrix
Or: SQRT [ (x-0)^2 + (y-f)^2 ] = SQRT [ (x-x)^2 + (y-(-f))^2 ]
And now, since we have two things equal that are both square roots, we know
that if we squared them they would still be equal, thus we can dispense
with that and:
(x-0)^2 + (y-f)^2 = (x-x)^2 + (y-(-f))^2
also: x^2 + (y^2 -yf + f^2) = 0 + (y + f)^2
and so: x^2 + y^2 -2yf + f^2 = y^2 +2yf + f^2
And now eliminating terms common to both sides: x^2 - 2yf = 2yf
Or, adding 2yf to both sides: x^2 = 4yf
And so finally dividing both sides by 4f, we get: y = (x^2)/4f
And this is the equation for any parabola on a paper pattern using any
measuring unit, mm's cm's inches, feet, miles, etc. As long as you use the
same measure unit throughout and for the f or focal length you desire.
Lay it out with shelf-paper or whatever, by graphing enough points to the
accuracy you wish.
If anyone wants proof that all pieces of a square matrix of slats that one
wishes to use must be cut to the same parallel curvature, and so can be cut
from the same piece of material, I can supply the proof, although the
diagram accompanying the equations would be hard to render here! But I DO
guarantee it!! Try it empirically with cardboard! I did sit down and prove
it for a school project once and I have looked at it recently. I would not
have guessed it worked out so easily and practically stood up and proved
itself! You know, those great proofs that look horrific, and then
everything falls out leaving a simple elegant term? It is like that!
This surface will act as a perfect concentrator, whether for sound, light,
radio waves, or even tennis balls if it's strong enough. Enjoy!
-Steve Walz rstevew@armory.com
Article Reference: E5C4D8C8FBE98082
Date: 23 May 1996 17:11:17 GMT
Original Subject: Re: Calculate KVA
The original poster might not know some of the
subtleties so I'd like to add some clarification ...
John Freitag (Jfreitag@gsosun1.gso.uri.edu) wrote:
: Kilovolt-amps: VA=volts X Amps 1000VA = 1 kVA
Don't forget the units for the Volts and Amps:
Volt-Amps = Volts (rms) X Amps (rms)
^^^ ^^^
: Watts (AC) = volts(rms) X amps X power factor
Don't forget the units for the Amps:
Watts = Volts(rms) X Amps(rms) X Power Factor
^^^ ^^^
Power factor is defined as:
Power Factor = True Power / Apparent Power
Apparent Power = Volts (rms) X Amps (rms) (Volt-Amps)
_ T
/
|
True Power = (1/T) | V(t) I(t) dt (Watts)
|
/
0 -
_ T
/
|
Volts (rms) = [ (1/T) | [ V(t) ]^2 dt ]^(1/2)
|
/
0 -
and similarly for Amps (rms)
For LINEAR loads (motors etc.) it is true that:
: Power factor = Cosine of the phase angle between voltage and amperage
For NONLINEAR loads (like computers and other electronic devices),
Power Factor is only defined by calculating the Apparent Power
and True Power via the equations above. No simple relation (like
the Cosine) exists.
Of course, this is probably superflous information since the narrow
answer to the posters original question (restating what's already
been posted) is simply:
KVA = ( Volts (rms) * Amps (rms) ) / 1000
Brian
Article Reference: 0C6BF1B36D874BE7
Date: Thu, 26 Sep 1996 15:40:09 -0400
Original Subject: Re: Speed of an electric impulse over a circuit board trace....
Chris DiBona wrote: > > Okay, the subject says it all, I know the speed of light is 300,000 km > per second, but I seem to remember that the speed of electricity was > slightly slower. Hence my question.... > > Thanks in advance.... > > Chris DiBona The propagation delay of a circuit board trace is: t(pd) = 1.017 * sqrt( 0.475 * er + 0.67) nanoseconds/foot where er is dielectric constant of PCB material. For fiberglass-epoxy er is about 5.0 giving 1.77 ns/ft or 1.72 * 10^8 m/s (172,000 km/sec). Steve Mercer smercer@ositech.com
Article Reference: B221C50A6ADF1083
Date: Thu, 23 Jul 1998 12:28:55 GMT
Original Subject: Re: PT100 Signal Conditioning.
On Thu, 02 Jul 1998 15:23:31 GMT, "Ian Davies"
<Ian@vsltec.demon.co.uk> wrote:
>Hi,
>
>I am currently working on a thermal control system for one of my clients.
>The temperature sensing element has been specified as a PT100 (Probably Class A)
<SNIP>
We recently designed a PT100 based data logger for steam steriliser.
I decided the best approach for us was to go straight into a MAX132
ADC. This has ridiculous resolution (18 bits from memory), but has
the great advantage of a ratiometric front end. This means we only
need one single precision component: a 0.1% 15ppm resistor.
If we tried conditioning the signal before the ADC we'd need low
drift or chopper op-amps, plus a high stability reference voltage.
The PT100 we had to work with is a 3-wire one. We came up with a
circuit configuration where one lead appear in series with the
reference resistor and one in series with the RT100 element. With a
reference resistor value equal to the RT100 at the temperature of
prime interest, lead resistances become second order errors.
The RT100 is only mildly non-linear. For our application we software
calibrate at a reference thermocouple at a 130deg nominal operating
point. To get reasonable readings down to room temp I empirically
found a second order equation that linearises to 0.05deg. My
equation is
T' = T + w(T-130)^2
where w=1.33E-4
It works well for us, given our 130deg operating point.
What will work for you depends on how much accuracy you need over
what range of temperatures.
David Gibson, "http://www.microconsultants.com"
Makers of SPLat easy to use programmable controller
(Simply Programmed Logic automation tool)
and GaugeLink digital gauge to computer interface.
Free software/tutorials for machine control and SPC.
Article Reference: 089B84886808F6F7
Date: 10 May 1996 20:43:10 GMT
Original Subject: Re: Q:Formula for resistance due to skin effect
The formula is Skin Depth) = 1/sqrt(Pi*f*mu*sigma)
where
mu = pemeabillity of the metal
and
sigma = conductivity of the metal
This was found on page 113 of Introduction to Electromagnetic
Compatibility by Clayton R. Paul. published by Wiley Intersience
Article Reference: 5BD58D21ADCAF4BB
Date: Mon, 22 Jul 96 23:47:32 GMT
Original Subject: Re: How to calculate the turns of transformer
> : --->I think of building a transformer myself but I don't know the > : formula > : --->to calculate the nos. of turns for the primery & secondary. > : --->Can anyone help and tell me how to calculate the nos. of turns > : of a > : ---> (E I) plate transformer with the given value of the voltages > : and > : --->ampere First, measure the cross section of the leg of the core that goes through the winding. Use 7 turns per volt per square inch of core cross section. (60 hz) i.e. if the core has 4 square inches of cross section, use 1.75 turns per volt. Use 7.5 instead of 7 if you have space. Neil.
Article Reference: B9D080697C1FBF8F
Date: 5 Jun 1996 20:34:45 GMT
Original Subject: Re: skin effect question
utes (utes@fnalv.fnal.gov) wrote:
: Does anyone have a formula for the resistance due to
: skin effect as a function of frequency, over and above the
: DC resistance? All of our reference books have the formula for
: skin depth, of course, and one even gives a rule of thumb that
: virtually all of the current exists within two skin depths.
: I'd like a more specific formula. Thanks.
Ramo, Whinnery and Van Duzer in "Fields and Waves in Communication
Electronics", section 5.13 to 5.20 discuss skin effect.
In section 5.18 they present a general formula for impedance of
a round wire as a function of skin depth.
In section 5.17 they present approximations for low and high
frequencies as follows:
With ro = radius of the round wire (meters)
del = skin depth (meters)
= 1/(pi*f*mu*sigma)^.5
f = frequency
mu = material permeability (henrys/meter)
(4*pi*10^-7 for good conductors)
sigma = material conductivity (mhos/meter)
(5.8*10^7 for copper, for example)
Ro = 1/(pi*sigma*ro^2) (DC resistance)
O ro/del < 2 (low frequencies)
R/Ro = 1 + (1/48)*(ro/del)^4
(approx. 6% error at ro/del = 2)
O ro/del > 2 (high frequencies)
R/Ro = ro/(2*del)
By the way, in section 5.16 they present an equation for current
as:
| Iz/Io | = e^[-(ro-r)/del]
where r is the distance measured from the center of the wire
(So it looks to me that at 2*del from the conductor edge,
the current has decreased to e^(-2) of its value at the edge.)
Maybe this will help? Hit the bookstore/library for more
information!
Article Reference: CBCFC3E30323F4D3
Date: Wed, 8 May 1996 16:23:39 -0400
Original Subject: Re: Q:Formula for resistance due to skin effect
On Wed, 8 May 1996, Martin Wiseman wrote: > Does anyone know a formula for for calculating the AC resistance of a > piece of wire due to the skin effect at high frequencies. I have two For copper wire... divide 66 by the square root of the frequency in Hertz. That will give you the skin depth in milimeters. For 100 KHz, that works out to about 0.2 mm. Since the wire is already smaller than the skin depth, its resistance will change little if at all. Jim "Skin-man" Meyer
Article Reference: 3FBDF2D6C3BAF912
Date: Wed, 05 Jun 1996 18:34:27 -0700
Original Subject: Re: skin effect question
utes wrote:
>
> Does anyone have a formula for the resistance due to
> skin effect as a function of frequency, over and above the
> DC resistance? All of our reference books have the formula for
> skin depth, of course, and one even gives a rule of thumb that
> virtually all of the current exists within two skin depths.
> I'd like a more specific formula. Thanks.
For a round copper wire, no plating,
R=(1/(2a))*sqrt(rho*f*mu/pi) Ohms /meter length
where a =radius of wire in meters
rho= copper resistivity Ohm-meter = 1.74*10^-8
f =freq Hz
mu=permeability of copper 4*pi*10^-7 Henries per meter
R = 4.17*10^-6 *1/a * sqrt(f) for copper
Condition: radius a >> skin depth
Reference W.C. Johnson "Transmission Lines and Networks" 1950
Article Reference: C8F61C7033169885
Date: 6 Nov 1998 10:52:02 -0800
Original Subject: Re: Qts from Qes and Qms ?, any speaker buffs out there?
In article <1337CE389002D211A46FAA0004002A057B88BF@comet.bre.co.uk>, Morrison, Peter <MorrisonP@bre.co.uk> wrote: >I would like to know how the Qts parameter for a speaker is derived from >its Qes and Qms figures. Qts = (Qes * Qms) / (Qes + Qms) -- Bill McFadden Tektronix, Inc. P.O. Box 500 MS 50-350 Beaverton, OR 97077 bill.mcfadden@tek.com "http://www.rdrop.com/users/billmc" CAUTION: Don't look into laser beam with remaining eye. Any opinions expressed above are not necessarily those of Tektronix, Inc.
Article Reference: 0E19D978276CC927
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: B7AAC3AF49A814EC
Date: 5 Apr 1996 05:16:55 GMT
Original Subject: Re: AMPACITY OF PCB TRACES
In <4APR199610511068@vms1.tamu.edu>, bac4773@vms1.tamu.edu (COBO, BRADLEY A.) writes:
>I am currently building a low to medium quantity of small circuit boards.
>They are for an automotive (12v) application and involve switching DC loads
>around 5A. I am currently using traces on the board that are .125 or 1/8"
>wide. Is that enough surface area to adequately handle 5A? The finished
>product will be sealed in epoxy, which means it can't be repaired, so i need
>to make sure they will work properly. Thanks for any help.
>
Brad,
The PCB manufacturer should be able to supply you the plating spec.
for your boards in OZ/ft^2. Use this information to compute the
resistivity of your critical traces.
EXAMPLE:
3OZ plating yields a thickness of appx 4mils.
Use the the following formula:
l
R = --------- l = trace length
(Sigma) A Sigma = copper conductivity
A = trace x-sectional area
Regards,
Mike
Article Reference: 02656065E28193FC
Date: Wed, 22 May 1996 21:44:41 -0400
Original Subject: Re: Calculate KVA
In article <4nvrv8$6q1@news-f.iadfw.net>, Dean Humphus <"dhumphus@airmail.net"@mail.airmail.net> wrote: > Seeking formula(s) to calculate kilovolt amps from amps, volts, and/or > watts. > > Any help appreciated. > > DEH Kilovolt-amps: VA=volts X Amps 1000VA = 1 kVA Watts (AC) = volts(rms) X amps X power factor Power factor = Cosine of the phase angle between voltage and amperage kilovolt-amps = kilowatts when the power factor = 1, that is when the load in question is totally resistive JF -- THE REPLY, OF COURSE, DOES NOT REPRESENT ANY OFFICIAL POSITION OF THE UNIVERSITY, only my personal opinion.
Article Reference: 3DA66DB83DAD32F3
Date: 24 Aug 1996 21:43:19 GMT
Original Subject: Re: Using Diode (IN4148) or Transistor as Temperature Sensor.
queksp@singnet.com.sg, <queksp@singnet.com.sg> said... > > I want to use a general purpose diode as a temperature sensor > (-2.1mV/deg celsius). Sensing range from 0 to 60 deg C. ... > How closely [do] diodes or transistors follow the physics equation?? In my experience, you'll get surprisingly good Ebers-Moll equation following with even cheap transistors, if you use them at a low enough current. >Also what is the possible spread in temperature characteristic from >device to device?? In the equation Vbe = (kT/q) ln (1 + Ic/Is), the parameter Is varies from part to part, and it's physics tempco swamps the kT term, yeilding roughly -2mV per degree. However, a current proportional to absolute temperature T (in degrees Kelvin) can be easily obtained and calibrated with a single room temp adjustment: Connect the base of an npn transistor to a 1.25V temperature-independant voltage source (the silicon band-gap voltage, use a voltage divider or an LM385-1.2V etc), and the emitter to ground through a fixed plus adjustable resistor. Voila, the collector has your absolute-temperature current. >Would a thermistor be better off??? Despite the above transistor-boosting vote, when I needed 0.1 degree accurate temperature sensing in an oceanographic thermometer product, I used $12 thermistors. -- Winfield Hill hill@rowland.org The Rowland Institute for Science 100 Edwin Land Blvd. Cambridge, MA USA 02142-1297
Article Reference: 2FEF15CBF09B16E1
Date:
Original Subject:
In article, rdell@cbnewsf.cb.att.com (richard.b.dell) writes: >From: rdell@cbnewsf.cb.att.com (richard.b.dell) >Subject: Resistance of Wires -- simple formula >Date: Sat, 23 Jul 1994 15:36:38 GMT >I got tired of digging up a wire table every time I wanted specific >resistance values, and decided to type a wire table into my palmtop. >So, having the data in a spreadsheet, I did a bit of number crunching and >came up with this formula (which perhaps I'll remember). >Ohms per thousand feet = 10^(0.1 * Gauge Number - 1) > where Gauge Number is American Wire Gauge. >less than 1 percent error from 1 Gauge to 31 Gauge, and maximum error of 3 >percent all the way to 44 gauge. My mentor (now 72 years old, and still working with me) does this routinely in his head. It used to amaze me, but it is really quite simple. All you have to remember is two facts. One, the resistance of 40 AWG is about 1 ohm per foot. Two, the resistance doubles (or halves) every three gauges. So, the resistance of 37 gauge is 0.5 ohms/foot, 34 gauge is 0.25 ohms/foot.... So for any old number he finds the difference to 40, divides by three, and halves (or doubles) the number 1 that many times. Of course he never takes more than 5 seconds to come up with the answer! I can do it too, but it takes me a little longer . The accuracy is not perfect, 40 AWG is really 1.046 ohms/foot, but for a quick and dirty, nothing beats it. Andy >Now if memory will serve, all I need is a calculator. >-- >Richard Dell
Article Reference: 3A3C0FFF189B770F
Date:
Original Subject:
In article, Andy Cooper coopera@gvg47.gvg.tek.com wrote: ->In article rdell@cbnewsf.cb.att.com (richard.b.dell) writes: ->> ->>Ohms per thousand feet = 10^(0.1 * Gauge Number - 1) ->> where Gauge Number is American Wire Gauge. -> ->have to remember is two facts. One, the resistance of 40 AWG is about 1 ohm ->per foot. Two, the resistance doubles (or halves) every three gauges. So, ->the resistance of 37 gauge is 0.5 ohms/foot, 34 gauge is 0.25 ohms/foot.... Or, stated as a formula: ohms/1000ft = 0.1 * 2^(AWG/3) I haven't checked to see which of the two formulas is more accurate. -- Bill McFadden Tektronix, Inc. P.O. Box 500 MS 58-639 Beaverton, OR 97077 bill@tv.tv.tek.com, ...!tektronix!tv.tv.tek.com!bill Phone: (503) 627-6920 How can I prove I am not crazy to people who are?
Article Reference: 7D6B25C95C7E7792
Date:
Original Subject:
In article coopera@gvg47.gvg.tek.com (Andy Cooper) writes: >If you really want it more accurate, remember that 40 AWG = 1.046 ohms/foot, >so just multiply the result by 1.046. For those who prefer more civilized units :-), 35AWG = 1.079 ohms/meter. -- SMASH! "Sayy... I *liked* that window."| Henry Spencer @ U of Toronto Zoology "I enjoyed it too!" "Hmph! Some hero!"| henry@zoo.toronto.edu utzoo!henry
Article Reference: D6637D90814D5877
Date:
Original Subject:
In article bill@thd.tv.tek.com (William K. McFadden) writes: >From: bill@thd.tv.tek.com (William K. McFadden) >Subject: Re: Resistance of Wires -- simple formula >Date: 26 Jul 1994 17:54:28 GMT >In article, Andy Cooper coopera@gvg47.gvg.tek.com wrote: >->have to remember is two facts. One, the resistance of 40 AWG is about 1 ohm >->per foot. Two, the resistance doubles (or halves) every three gauges. So, >->the resistance of 37 gauge is 0.5 ohms/foot, 34 gauge is 0.25 ohms/foot.... >Or, stated as a formula: >ohms/1000ft = 0.1 * 2^(AWG/3) Actually the correct formula is: ohms/ft = 2^((AWG-40)/3) or: ohms/1000 ft = 0.001 * 2^((AWG-40)/3) Funny, looking at it this way, I can't do it in my head any more . >I haven't checked to see which of the two formulas is more accurate. If you really want it more accurate, remember that 40 AWG = 1.046 ohms/foot, so just multiply the result by 1.046. >-- >Bill McFadden Tektronix, Inc. P.O. Box 500 MS 58-639 Beaverton, OR 97077 >bill@tv.tv.tek.com, ...!tektronix!tv.tv.tek.com!bill Phone: (503) 627-6920 >How can I prove I am not crazy to people who are?
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