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From: grege@gold.gvg.tek.com (Greg Ebert)
Newsgroups: sci.electronics
Subject: Re: Power supply design
Message-ID: <2578@gold.gvg.tek.com>
Date: 19 Jun 91 21:08:42 GMT
References: <1991Jun18.175131.8374@ux1.cso.uiuc.edu>
Organization: Grass Valley Group, Grass Valley, CA
Lines: 34

In article <1991Jun18.175131.8374@ux1.cso.uiuc.edu> kline@ux1.cso.uiuc.edu (Charley Kline) writes:
>Hi, we're trying to build a power supply for a device that requires 3A
>at 13.8 volts. The voltage regulator will be an LM350T, a 3A-rated
>variable voltage regulator.
>
>The problem is the power transformer. I can't find anywhere a 3A
>transformer at 18 volts.
>
Radio Snack used to (and maybe still does) sell an 18V center-tapped
xformer rated at 4 amps.

>My question is, am I guessing the secondary voltage wrong? Would a 14V
>secondary be okay since that is the RMS value and the
>rectified/filtered DC out of that would be sqrt(2) * 14 = 20V?

Pretty close. You need to include diode voltage drops (they are about
0.8 volts for currents around 1-3 amps. A full-wave bridge rectifier
will bite you twice and gobble up 1.6 volts. You will also experience
voltage 'sag' between AC peaks as your filter capacitor drains. If
you assume it's linear (a very good approximation), I*t=C*V. For
a 3 amp load, 60Hz in, you get a 1volt dip if your filter capacitor
is 25,000 uF.

Then the voltage regulator will take another 2.5 volts. Let's guess that
voltage drops in the Xformer are 0.5 volts.

So, 13.8 (load) + 2.5 (LM350) + 1.6 (Rectifier) + 0.5 (Xfmr) = 18.4 volts.

That leaves you with 1.6 volts of ripple coming from the filter cap.
You will need at least 15,000 uF of filtering to prevent 'dips'
below 13.8 volts out.

If you use the 18 volt xformer, be aware that your regulator will be
taking a larger voltage drop, hence it will need more heat-sinking.