Magnatone Melodier OT imp

[shaun]

Heyo,
I think I know the answer to this one, but thought it might be of interest - these old amps are super fun and quirky, too.

This amp is designed without an inverter tube. But because it has two 6V6 power tubes, it has what is apparently called a self-split circuit, where the the screen grid of one 6V6 acts almost like a plate, in that it feeds the signal down to the second 6V6's input grid.

The Melodier 109 amp I have on my workbench is missing its OT - gone. I am assuming the new OT should suit an 8kohm input impedance due to its PP 6V6 configuration. However, I am slightly worried that, because of the self-split circuit, the output imp of the two 6V6's may be different than in a typical PP 6V6 amp that uses a 12AX7 or similar to invert the signal.

I believe the output impedance of a pentode PP circuit is set by the design of the tubes themselves during manufacture to some extent. But obviously, voltages on the pins and the circuit design effect where the tubes' impedance is expected to sort of "land". In any case, my assuming could be making an ass out of me. In looking at the schematic, can anyone see any reason why I couldn't use a typical 6V6 PP OT, rated around 12-15 watts?

Thanks as always.

[sluckey]

can anyone see any reason why I couldn't use a typical 6V6 PP OT, rated around 12-15 watts?

I can't

[PRR]

Pentodes are never "matched" to a load. The tube has limits on voltage, current, and power. This leads to a preferred load zone.

Stupid tricks like this one might shift the "perfect optimum" but the whole idea is such a compromise that I'm sure it won't matter. It won't make even 10 Watts clean, but most 2-6V6 OTs are rated at least that high so that is what you use.

[shaun]

Aha. Thank you.
And the knowledge that output might not reach 10watts is very helpful. It confirms my thinking regarding what must surely have been a tiny OT shoehorned onto the chassis - it's in a weird location bang up against the control pots. It must have been tiny to fit, so maybe 8watts or less.

[trobbins]

Note that self-split is default class A only, and the OPT can't be pushed in to class AB1 type power levels.  As such, the max power swing (assuming a resistive 'straight-line' load line) will depend on the mid-point idle bias setting, as well as the reflected load impedance causing the load line to target the knee region for one peak, and be sufficiently away from cut-off for the corresponding other peak excursion along the load-line.

[shaun]

Thanks Trobbins. Very interesting. Are you saying that, because the circuit is a default class A, the input impedance of the OT only needs to accommodate one 6V6, as if it were single-ended? ie 5kohms? That would inform OT choice.

The output bias setting is straight from the schematic, so I'm going to follow that. And from what seems to be available in OTs, Hammond offers a 10w 8k with multiple output taps, so that was where I was going.

[trobbins]

No, I'm saying it would be worthwhile drawing up a PP class A loadline for 6V6.  You would need to determine the available B+ voltage for the nominal output stage idle current, and the idle screen voltage for that amp.  Then calculate the idle Vak and screen-cathode voltage of a 6V6 and use that as the idle point on your loadline using a set of 6V6 pentode mode plate curves closest to your nominal screen voltage.  Add in the likely OPT impedance straight line and see if it targets the knee region and stays out of cut-off on the other end.

[tubeswell]

... Are you saying that, because the circuit is a default class A, the input impedance of the OT only needs to accommodate one 6V6, as if it were single-ended? ie 5kohms?

No. You still have 2 tubes working through the primary (so even though each plate is in opposing phase to the other plate, there is 2 x as much 'current' in the primary).

[shaun]

Thanks tubeswell. That's what I thought