11-09-2019, 05:08 PM
(09-09-2019, 04:14 PM)Amie Wrote: In a pushpull output, does the transformer need to present an Impedance on its entire primary to match the individual R.a. of the valves used, or does it need to match the R.a. in each half of the winding?
The texts I've so far read on the subject have been written with ambiguity.
Easy part first - in Class A push-pull, you need the transformer to match double what each individual valve would require.
So if a valve single-ended requires 6kΩ, then two valves in push-pull would require 12kΩ anode-anode. This is because you have the same current as a single valve, but double the voltage swing between the two anodes. (If you had connected them in parallel, you'd need half the individual or 3kΩ, makes sense because you'd have the same voltage swing as a single, but double the current).
(09-09-2019, 06:55 PM)ppppenguin Wrote: The 12AU7/ECC82 seems a rather unlikely output valve. Single ended or pp. But then the EF80 has been used for audio output in some TVs and that's even less likely.
It is unlikely, but it nevertheless works - I have done it!
Now, the harder part - what IS the load that a single-ended valve needs (so you know what to double for P-P)? Triodes are fairly uncritical so easy to match, but surprisingly hard to match properly.
If your signal voltage is fixed and very small, then the load for maximum power is simply the ra (anode resistance) of the valve. But that is an unusual case - generally, the signal is not fixed and if changing the load changes the power, you just twiddle the volume control to get back to where you were. The question then arises, what is the optimum load if each time you alter it and readjust the volume control up to the onset of distortion?
Triodes start to distort if the signal voltage is large enough that the grid voltage reaches zero in the +ve direction (because grid current flows and clips the input signal). They also distort in the -ve direction if the anode current reaches cut-off. Ideally, you want these two things to kick in at the same signal level.
It turns out that for a particular valve and a particular HT voltage, there is an optimum Ia (standing anode current) for maximum power, equal to Va / (4 x ra). More than this and the bias is so close to zero that you get grid current at small signal levels, so power is curtailed. Less than this and you are closer to cutoff before you even start applying a signal. At the optimum standing Ia, the optimum load is Rl = 2 x ra, for a single valve.
But that is not the whole story either. If you bias your triode for this optimum Ia, you may find yourself exceeding the anode dissipation. So you have to bias it farther back anyway (obviously giving Ia = Pa / Va where Pa = max power dissipation the valve can withstand). In that case, maximum power is reduced, but so is the standing current, and it turns out that efficiency is improved a bit. The optimum load is given by (Va/Ia - 2ra).
Once you have the single valve sorted, just double the load for push-pull.
One of the best textbooks for this is F Langford-Smith's Radio Designers' Handbook.
Once you have your optimum load, then matching to a speaker can be achieved with a small mains transformer if it has primary windings for 0-120V-240V. You use the 120V tap for HT and the 0 and 240V terminals for the anodes. If you need 12kΩ anode-anode and want to match to a 8Ω loudspeaker, then the turns ratio will be √(12,000 / 8) = 38.7:1 so your mains transformer will need this ratio. It is easy to see that a 240V in, 6V out transformer has a ratio of 40:1 so it will get you going and while not hi-fi, you will get decent results.
