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Tube DIY Asylum Do It Yourself (DIY) paradise for tube and SET project builders. |
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Tube DIY Asylum Do It Yourself (DIY) paradise for tube and SET project builders. |
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In Reply to: Re: keep going...sooner or later, you'll get it. posted by mqracing on December 1, 2004 at 08:50:24:
We provided you many references, but you never commented them, just keep going Raa/4. We gave you many examples, why you never calculate this example of yours (4 Ohms /16 Ohms /2 x 8 Ohms)? You just use some sentences from various sources that saying about Raa/4.
No problem for me, any more. For the end, just the reference from the RDH4, I wrote it 100 times...Page 571 & 572, "Fundamental principles of push-pull operation":
"The load resistance R2 is connected across the secondary, and the reflected resistance across the whole primary is:
Rl=4R2(N1^2/N2^2)
and the reflected resistance across half the primary is
Rl``= 1/2 Rl = 2R2(N1^2/N2^2)
where N1=turns in half primary winding
and N2=turns in secondary windingIf valve V2 in fig. 13.31 were removed from its socket, the load resistance effective on V1 would then be
Rl` = 1/4 Rl = R2(N1^2/N2^2)
which is half the load resistance on V1 under push-pull conditions. This is the condition which occurs when one of the valves reaches plate current cut-off."
Well, this is it - it can`t be expressed more clearly.
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Follow Ups
- References... - Damir _ the real one 12/1/0410:51:27 12/1/04 (1)
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