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..."
--
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|_| |_| |_| / |_| | | where Engineering
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Date: Fri, 23 Aug 1996 05:22:49 GMT
Original Subject: Re: Help my rusty math please