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Old December 26th 07, 11:40 PM posted to alt.binaries.pictures.radio
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Default Westinghouse H-104 power supply

The part of the schematic that I didn't post does indeed show 117 volts as
the expected input.

So one final question, and then I think I've got it. I've gone online and
found a few sources for what should be a compatible transformer. 270-0-270,
plus 5V and a 6.3V. I also looked up the current draw of a 5U4 and found
that I need at least 3A across the filmanent, so I'll be sure to buy one
that is appropriate. But about that 270V - do I need to take into account
any kind of voltage drop across the plate? Obviously the schematic is
showing resulting voltages, not transformer output voltages as I originally
though. So is 270 the right spec? The 5U4 has about a 50V drop across the
plate.

Jeff



"Carter-k8vt" wrote in message
. net...
Unrevealed Source wrote:
"Carter-k8vt" wrote in message
t...

The 268 volts you are seeing on each plate is 268 volts -AC- to ground;
to put it another way, you are looking at a transformer with a 268-0-268
volt secondary (or a transformer with a 536 volt *center-tapped*
secondary).


I've got that part - thanks for confirming.


Let me toss you a minor curve ball: If you were repairing this radio in
its era, you would indeed be looking for a replacement transformer with
approximately a 270-0-270 high voltage secondary winding; this is
because it was designed for 110 volts AC coming out of the socket on
your wall. Today, that voltage is 115 to 120 volts or even a bit higher
and using -today's- line voltage with a 270-0-270 winding would give you
a bit more than the 265 volts DC B+ that you are looking for. So, what I
am saying is that you could (should?) look for a transformer with a
220-0-220 or 250-0-250 high voltage winding; i.e. less (than 270-0-270)
is better.

The 265 volts you refer to is +265 volts -DC-, measured from the
filament (cathode) to ground. The voltage -across- the 5U4 filament
is 5 volts AC, the filament voltage of a 5U4.


So I need to apply 5V AC across the filament (pins 2 and 8). That's
consistent with what my understanding would have been.


Yes. That is why your replacement transformer needs a 5 volt secondary
winding (with enough current for a 5U4 filament).

And the 265V is generated by the tube, correct?


Yes, the tube changes (or 'rectifies') the 268 volts AC to 265 volts DC.

This would be a pulsed DC, would it not, coming directly off the tube?


Yes.

In circuit, it is constant DC due to the filter capacitors?


Yes (and also due to a filter choke if one is present (rest of the
schematic is cut off).

Do I understand this correctly?


Yes.

By the very nature of your question, it indicates that you are a
beginner -- which is OK; we were all beginners at one time. Just be
careful; these voltages, if not fatal, can at least be harmful.


Yes, relative to almost all of you, I am indeed a relative beginner. And
this is my first experience with a transformer/rectifier-based radio, so
I apologize if my questions are somewhat stupid.

Jeff


Nope, not stupid. Again, we were ALL beginners at one time. Asking
questions is how you learn...



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Old December 27th 07, 01:16 AM posted to alt.binaries.pictures.radio
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Default Westinghouse H-104 power supply


"Unrevealed Source" wrote in message
...
The part of the schematic that I didn't post does indeed show 117 volts as
the expected input.

So one final question, and then I think I've got it. I've gone online and
found a few sources for what should be a compatible transformer.
270-0-270, plus 5V and a 6.3V. I also looked up the current draw of a 5U4
and found that I need at least 3A across the filmanent, so I'll be sure to
buy one that is appropriate. But about that 270V - do I need to take into
account any kind of voltage drop across the plate? Obviously the
schematic is showing resulting voltages, not transformer output voltages
as I originally though. So is 270 the right spec? The 5U4 has about a
50V drop across the plate.


That 50V drop is figured into the resulting DC output voltage, once
filtered. (DC filtered voltage = AC voltage (of one half of the transformer
at any peak) x 1.414)


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