Impedance

[lowell]

In my amp I have a switch that cancels the input signal to 2 of 4 power tubes. So I'm left w/ 2 power tubes in push-pull. I'm curious if impedance is an issue here. It sounds fine, but wondering if operation will be better using the 4ohm tap. I'm using an 8ohm speaker cab. Since 2 tubes doubles the primary load, I'm thinking it may be best to double the secondary load. So in my case, using the 4ohm tap for my 8ohm speaker. Being that I'm cancelling the input signal and not lifting the cathode of the 2 cancelled tubes I'm wondering if this is necessary. Just a thought.

[Robert M. Martinelli]

Hi Lowell,
Those two tubes appear as they're "not there" to AC, the impedance "seen" by the OT's primary is doubled, so it's OK to use an 8 Ohm load connected to the 4 Ohm tap as you correctly stated; OTOH, from a DC standpoint, you still have quiescent (bias) DC current (unnecessarily) flowing through the two tubes you "lifted" the grids on, so, IMHO, it's indeed a much better solution to disconnect the two cathodes instead. This way you'll have less DC current flowing in the OT's primary at all times, taking some strain away from the OT. It would also become more difficult for the OT to saturate.

JM2CW

Hope this helps

Best regards

Bob

[lowell]

ok thanks, that's what I thought. I'm cancelling the signal instead of lifting cathodes so I don't have to deal w/ fluctuating B+ in the 2 different settings, and potential bias offsets caused by that.

[pdf64]

'Those two tubes appear as they're "not there" to AC, the impedance "seen" by the OT's primary is doubled'
Surely, as the tubes are still in circuit (but not responding to any input signal), their plate impedance will form a parallel load path to the OT primary?
I'm not sure how that would affect what the best OT primary load impedance might be.
Try it both ways and see which puts out the most clean power - that's the best test. Peter.

[Old Tele man]

'Those two tubes appear as they're "not there" to AC, the impedance "seen" by the OT's primary is doubled'
Surely, as the tubes are still in circuit (but not responding to any input signal), their plate impedance will form a parallel load path to the OT primary?
I'm not sure how that would affect what the best OT primary load impedance might be.
Try it both ways and see which puts out the most clean power - that's the best test. Peter.

...impedance is opposition to AC, since the two "idling" tubes are no longer responding to inputs, they're not producing an AC-output into the OT, therefore AC-wise they're not present. DC-wise, however, they're still there and idling away...moving the OT closer towards eventual saturation.

...ie: what Bob Martinelli said.

[pdf64]

It's making my head hurt, but I think I get that, thanks for clarifying that OTM.
However, even though they are out of it ac-wise, won't their dc resistance still be in circuit and still present an additional load (ie 400 plate voltage/30mA plate current = 13k, or whatever), as described?

[Old Tele man]

...at idle, the DC-current 'follows' the DC-loadline; it does not 'follow' the AC-loadline unless it's an AC-signal.

...and 13K is nothing compared to the 100-150 ohm resistance of the primary winding...remember the old "10-times" rule for parallel loads, ie: "...if resistance X is 10-times more than resistance Y, then you can effectively ignore X."

[pdf64]

I was thinking that the inactive 6L6s would be equivilant to 13k resistors for dc AND ac. dc being 0Hz, and there being no reason for them to disappear as the freq increments above 0 - wrongheaded again?
If above is correct, then 13k either side of the OT primary would affect the 6V6 operation?

[Old Tele man]

I was thinking that the inactive 6L6s would be equivilant to 13k resistors for dc AND ac. dc being 0Hz, and there being no reason for them to disappear as the freq increments above 0 - wrongheaded again?

...sounds like you're thinking Fourier transforms, where the zero'ith harmonic is DC, ie: H0 = DC, H1 = fundamental (ƒ), H2 = first harmonic (2ƒ), etc.. Put an AC-signal onto a DC-signal and you have a "modulated" DC, which is only a "level-shifted AC. But, without an AC-signal, only the DC-characteristic exists, and that's the DC-resistance (R.pri) of the primary winding of the OT, which is the DC-loadline.

...the output transformer doesn't "pass" DC, only AC, thus providing the "blocking" effect between tubes and speaker(s)...an inductive "isolation" equivalent of capacitive "blocking."


QUESTION--has anyone invented/developed a "capacitive"-equivalent of the output transformer, where there's a "scaling" ratio between the input and output? I know I've seen "split" capacitors where the two "inputs" shared a common ground, but what about a 'combined' capacitor where internally the two capacitors not only provided isolation but also scaling or ratio'ing of signal levels (and thus impedances)?