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From: crs@lanl.ARPA
Newsgroups: net.analog
Subject: Re: Adding Video Input to TV.  Summary
Message-ID: <34212@lanl.ARPA>
Date: Tue, 3-Dec-85 11:09:55 EST
Article-I.D.: lanl.34212
Posted: Tue Dec  3 11:09:55 1985
Date-Received: Fri, 6-Dec-85 06:37:04 EST
References: <6770@amdcad.UUCP> <3500001@Clio> <610@ttrdc.UUCP>
Organization: Los Alamos National Laboratory
Lines: 60

> More to the point, yes it's cheaper.  And less efficient and it has less power
> handling capability, perhaps on the order of a 10-20% loss in voltage
> (due to resistive voltage losses in the windings which are not made up in
> an increased secondary turns count since the two transformers are obviously 
> symmetrically ratioed)...

This is not to disagree with Dan's article but merely to add an idea
for those applications where one can stand all of the disadvantages
except the voltage loss.	[See caveat at end of article.]

A filament transformer should provide its *rated* output voltage at
its *rated* current.  Thus a 6.3 volt, 10 ampere transformer should
supply 6.3 volts when it's load is 10 amps.  To compensate for IR
drops in the windings, I assume that the turns ratio is juggled
slightly.  It is here that trouble occurs because the "juggling"
that compensates when the transformer is used normally will work
against you when used backwards.  Since in the output transformer
both the IR drop and the compensation are *decreasing* "output"
voltage, at rated load, I'd guess the "system" will act as though
neither transformer is compensated and so there will be a voltage
loss as suggested.

Suppose that, instead of making the configuration symmetrical, you
make the input transformer (operated normally, as a step down
transformer) a 6.3 volt filament transformer.  Now, making suitable
allowances in power rating, make the output transformer (connected
"backwards" as a step up transformer) a 5 volt filament transformer.

If all goes well, under load the 6.3 volts out of the input
transformer will be about 26 percent more than the 5 volts "expected"
by the output transformer.  Assuming the 20% figure suggested by Dan
and assuming that half of it is lost in the turns ratio compensation
and the other half in IR losses:

			117/5 = 23.4

			0.9 * 23.4 = 21.06

			21.06 * 6.3 = 132.7

			0.9 * 132.7 = 119.4 volts out

Naturally, this output transformer must be *over* *rated* because with
6.3 volts across a 5 volt winding dissipation will be higher than
normal.

Naturally we aren't limited to filament transformers.  In these days
of solid state circuits, power transformers are also, typically, step
down transformers.  A *first* *approximation* of output current rating
for the combination would be the rated secondary current (when the
transformer is used normally) divided by the turns (or voltage) ratio.

[Note:	I haven't tried this -- it's just an idea.]

-- 
All opinions are mine alone...

Charlie Sorsby
...!{cmcl2,ihnp4,...}!lanl!crs
crs@lanl.arpa