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In Reply to: RE: UTC output transformers LS-63 posted by Tre' on April 16, 2015 at 07:46:05:
Assume use of 2A3 tube which develops 3 watts of output power per published ideal design parameters.
pri impedance 2500 ohms
AC signal volts (full power) will be 86.6 vrms
AC signal current .0346 Arms
now let's calculate and then add in vectorially the currents wrought by the inductive reactance of the primary;using the 15 henries that you spec'd @ 31 hz and full power (86.6 vrms) we would have;
Iex .0296 Arms
Itot .0458 Arms (vector sum)
notice that our total current draw is 132% greater than the signal current.
Now let's keep all the same parameters but reduce the freq to 26.5 hz.
this is the -3db large signal point. It is where the inductive reactance equals the reflected impedance.Heres the resulting current draws;
Iex .0346 arms
I total .049 (142% greater than the signal current)
also note that the Iex equals the Iac signal current. This is because we are delivering half of our power into the shunt arm (the inductive reactance) hence the currents btwn the two "circuits" are the same value.
but now the 2A3 must work (from the current standpoint) 142% harder to drive the complex reactive loadline that the anode of the tube is looking into.
And, again, real world losses will be even greater than shown above since we are not accounting for any copper losses (remember we are now shoving 142% more current through the copper circuit) or core losses.
Not an entirely pretty picture.
MSL
Builder of MagneQuest & Peerless transformers since 1989
Edits: 04/16/15
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Follow Ups
- some current calculations using your example... - mqracing 11:40:38 04/16/15 (0)
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