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In Reply to: power tranny question posted by Vinnie on April 4, 2007 at 17:52:24:
Vinnie, the transformer is likely to be capable of tremendous amounts of current. It should be more than adequate for powering a pair of 845 SE, or even a pair of parallel push-pull 845. It has been suggested that you measure the resistance of the windings. I'll repeat that suggestion.Concerning the question of how you can get over 1000V DC from it: traditional power supplies for tube equipment use a center tapped transformer for the simple reason that the AC can be rectified with just two diodes. Both diodes might be contained in a single tube rectifier...
With a center tapped transformer and two diodes configured as a full wave center tapped (FWCT) rectifier, the effective AC voltage that gets rectified is that between the center tap and either end. In your case that would be 425 volts. With the 1500VCT transformer that Grover specifies it would be 750 volts.
However, if you use four diodes (tube or SS) connected as a full wave bridge (FWB) rectifier then you can use the entire 850 volts present across the whole secondary (taps 5 & 7.) The center tap (tap 6) would be left unconnected. That's actually 100 volts more than what Grover specs.
Fundamentally, tube rectifiers and SS rectifiers perform the same basic function. You might choose to use either or even a combination of the two. The high voltage that we are talking about here limits your choices of tube rectifiers; most of the types used in more common tube amps will not survive.
There are diagrams of both FWCT and FWB rectifiers on Wikipedia.
-- Dave
Follow Ups:
Thanks for the explanation Dave. I think I understand about the voltage now. I measured the resistance of the windings and got the following:
pri - 1&2 5 ohms, 3&4 - 3.5 ohmssec 5 & 6 15 ohms, 6 & 7 15 ohms, 5 & 7 30 ohms.
So what does that tell us?
To be honest, I was expecting the resistances to be a little lower based on the weight, but I was just making a wild guess.Anyway, in the end all that really matters is the temperature rise. For the wire that's determined entirely by i 2 R heating. I'm going to make another wild guess and say that in transformer this big can handle total copper heating of about 10W.
Normally, the primary heating would be about equal to the total of all the secondaries (just one secondary in this case.) I'll assume the designer intended that to be the case with a FWCT rectifier. Using a bridge rectifier across the entire secondary will tilt things towards the secondary. Maybe 60% of the heating will be in the secondary. (I suppose I could break out the textbooks and figure it out more precisely, but I'm making so many guesses and estimations already I don't see that it would buy us anything.)
OK, so if we make the conservative estimate that 10 watts of heating is safe for all windings combined, then the secondary can dissipate 6 watts. That means about 450mA through 30 ohms. Again, 10W is probably conservative with a transformer this big. You might be able to draw more than that. The simple test is to see how hot the transformer gets. At 450mA it will probably just begin to get warm to the touch.
If you happened to have a 2000 ohm 1000 watt resistor you could put it across the secondary and let it run for a while to see if the transformer got hot. Doubt you've got that resistor handy though. :)
i doubt he has that resistor around, but 6 light bulbs in series and a variac will do the same thing.just set the voltage across the string to 750V and choose the wattage for the desired current draw. Just size the lightbulbs to the expected current draw and see how hot the sucker gets. 60W would give you a total current draw of around half an amp.
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