Re: calculating ripple

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Jim Lux by way of Terry Fritz <twftesla-at-qwest-dot-net>" <jimlux-at-earthlink-dot-net>

Ignoring such things as source impedance and assuming constant current
draw...

The voltage drop after the peak will be I/C * delta T, where delta T is 1/2
cycle time (8.6 mSec for 60 Hz)

So, for 1 Amp, and 1000 uF, 1/(1000E-6) * 8.6E-3 = 1E3 * 8.6E-3 = 8.6
Volts... Not too bad for a 1000 V power supply, terrible if it's a 5V
supply...

Since a lot of circuits are sensitive to percentage ripple, as opposed to
absolute ripple, you can see why you need a much bigger capacitor for a low
voltage supply.  Fortunately, as capacitors get lower voltage, getting more
uF gets cheaper.

----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Wednesday, August 21, 2002 9:39 PM
Subject: calculating ripple


> Original poster: "Crow Leader by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<tesla-at-lists.symmetric-dot-net>
>
> How do some people here calculate ripple on a capacitor being charged at
> 120Hz from a bridge rectifier and a load of x amps? So far every place
I've
> looked has a different formula, and none have consistent answers.
>
> - the Rado Shack "Building Power Supplies" book suggests
>
>    Vrms = 2.4 I/C
>
> where Vrms = ripple volts
> 2.4 comes from apparently nowhere
> I = current in mA
> C = capacitance in uF
>
> - this link on audio amplifier design says
>  http://sound.westhost-dot-com/power-supplies.htm#capacitor-value
>
> which comes to a different answer yet again.
>
> - another link has something similar
> http://www.mitedu.freeserve.co.uk/Circuits/Power/unreg.htm
>
> - and one more
> http://my.integritynet-dot-com.au/purdic/power1.html
>
> and the botton of the page here
> http://www.signaltransfer.freeuk-dot-com/powerout.htm
>
> So, what should I use, and where do these strange numbers like the "factor
> of 7" for 60Hz and "2.4" come from?
>
> KEN
>
>
>