Newsgroups: sci.electronics
Path: utzoo!utgpu!news-server.csri.toronto.edu!rpi!zaphod.mps.ohio-state.edu!maverick.ksu.ksu.edu!ux1.cso.uiuc.edu!kline
From: kline@ux1.cso.uiuc.edu (Charley Kline)
Subject: Re: Power supply design
Message-ID: <1991Jun20.182012.10847@ux1.cso.uiuc.edu>
Organization: University of Illinois at Urbana
References: <1991Jun18.175131.8374@ux1.cso.uiuc.edu> <2578@gold.gvg.tek.com>
Date: Thu, 20 Jun 1991 18:20:12 GMT
Lines: 51

Thanks for all the replies and email.

For what it's worth, one response shamed me into actually going
and researching this myself, and what I was missing in my
original analysis were the 1.4V drop across the full-wave
rectifier and the 10% slop required for low AC line voltage. So
I thought I would share my results with the net in the hopes
that it would be constructive.

My analysis goes something like:

We want 13.8 volts out at 3 amps. The regulator has a 2-3 volt dropout,
so we're conservatively up to 16.8. Let's set the ripple on the filter
capacitor to 2 volts at maximum load (more on this later), so now we're
up to 18.8V. The full wave rectifier will create a 1.4V drop, giving us
20.2V. That's peak-to-peak voltage, so divide by sqrt(2) to get 14.29V
RMS for the transformer secondary. But transformer secondaries are set
given the "average" primary voltage of 117V, and it could slip as low
as 105V and still be power company legal, so multiply 14.29 * 117/105
to get right about 16Vrms for the secondary. Well, no one stocks 16V
transformers, which is why they went up to 18V, I guess. This does not
include the resistive drop inside the transformer secondary, which I
suppose could be a volt or so.

Okay, what size filter capacitor to use?  The regulator can be
modeled as a constant-current source at maximum load, which is
3A. Capacitance is defined as Q = CV, so differentiating we get
I = C dV/dt, or dV/dt = I/C. Since current is constant, the
capacitor discharges in a linear ramp, so dV/dt is a constant,
the slope of the ramp. The peaks of the full-wave rectified AC
from the secondary are 1/120 second apart, so we want to set the
capacitor's value such that 1/120 * dV/dt is about our design
ripple voltage of 2V. So 2 = dV/dt * 1/120 = I/C * 1/120 =
3/C/120. This gives C=12,500 microfarads, and since
electrolytics have very high tolerance ranges, we'll round this
up by 20% or so to get 15,000 uF.

Many pointed out that running a regulator continuously at its
maximum rated current isn't good conservative design, so I've
upped the regulator to an LM338 which is rated at five amps.

Heat sinking? Well, at the maximum power company voltage of
125VAC, the secondary is now at 19.2V = 27.2V p-p. Power
dissipation in the regulator is then P=IV = 3 * (25.8 - 13.8) =
36 Watts. Clearly a largish heat sink will be required on the
regulator.

And no, this isn't for a car stereo in my house. :) It's for an
amateur television transmitter.

/cvk