What Happens If....?

[The Ballzz]

What happens in a four tube, push/pull power section if the input signal to the grid is turned off to two of the power tubes, one tube on each side of the push/pull? Does the required/expected primary impedance of the output transformer remain the same, or does it boil down to being basically the same as just pulling those two tubes out of circuit?
Likely A Dumb Question, I Know!
Gene

[shooter]

if the input signal

what happens to it when you quit playing 

the better question might be IF I lost my bias am I up sheet creek 

pull 'em all, verify you have bias in the socket
ohm the tranny should read other than open

[The Ballzz]

if the input signal

what happens to it when you quit playing 

the better question might be IF I lost my bias am I up sheet creek 

pull 'em all, verify you have bias in the socket
ohm the tranny should read other than open

I'm not troubleshooting here, simply exploring a possibly stupid and/or uninformed thought. I'm talking about keeping the bias voltage applied to all four tubes and simply removing the audio input signal to two of those power tubes while still driving the other two. Again, still leaving one P/P pair driven and operating while not driving the other P/P pair and still leaving their plates and screens connected. I'm guessing that each tube (or at least each P/P pair) would need to be biased separately? As I ask more questions and get them answered, it will become evident what I'm shooting for and trying to accomplish and why.
Thanks,
Gene

[The Ballzz]

For the purposes of discussion, we could even consider cathode bias, to remove the added bias voltage from the grids.
Just Sayin'
Gene

[shooter]

removing the audio input signal to two

I'm playing (**EDIT:** with crayons  )

they are driven in pairs, so guessing you'll hear it and it won't make the band members smile
If you break "inside" the pairs, guessing a good player will know

just signal shouldn't melt anything

[The Ballzz]

I'm trying to think of a simple way to switch off  the amplification power of two of the power tubes or even four, in an amp like, say a six- 6550 SVT, without having to compensate for the different impedance requirements of just removing the tubes! I'm thinking this may be simpler than it might seem and just that no one else has ever bothered to try it. I mean: "Why would anyone ever want to reduce the output power of an amp?" 

Just Sayin'
Gene

[sluckey]

I mean: "Why would anyone ever want to reduce the output power of an amp?" 

EXACTLY!

[kagliostro]

This matter was discussed time ago

Unfortunately I don't remember exactly the answer

May be HotBluePlates can say something about

Our friend Da Geezer (or it was Tubenit ??) proposed something like you say some years ago

--





Franco

[pdf64]

My understanding is that the undriven pair are effectively out of circuit, in regard of ac, they’re not there.

[The Ballzz]

My understanding is that the undriven pair are effectively out of circuit, in regard of ac, they’re not there.

Ahhh yes, but what impedance are they wanting to see at the output transformer primary, given that the screens, cathodes and plates of all four tubes are still connected?

Just Askin'?
Gene

[j_bruce]

Here's a thought from an utter novice: when you turn the volume knob down to zero the OT impedence does not change, so why not come up with a way to switch between all four on the signal and two on the signal and the other two on a zero signal?

[PRR]

> but what impedance are they wanting to see

Tubes do not have wants desires or needs.

If you mis-load a tube you get less power than you could have.

[KeithR]

Everything is still connected related to output. You are only dropping the inputs so as far as the transformer is concerned, all is well. What you will be changing is the loading of the LTP out. An extra resistor to ground might seem the way out but that resistance remains static and the resulting impedance won't react as driven tube would.

[jjasilli]

What happens in a four tube, push/pull power section if the input signal to the grid is turned off to two of the power tubes, one tube on each side of the push/pull? Does the required/expected primary impedance of the output transformer remain the same, or does it boil down to being basically the same as just pulling those two tubes out of circuit?
Likely A Dumb Question, I Know!
Gene

Back to the original question.  Impedance matching remains constant, because operationally all 4 tubes are still in the circuit. (But if 2 tubes were pulled, the plate-to-plate "output" impedance of the power tubes would be doubled, resulting in a mismatch with the "primary impedance" of the OT.)


The output of the amp in watts would be halved.  But in terms of volume (SPL), there is only a barely noticeable 3dB change if el power in watts is halved or doubled.

[tubeswell]

My understanding is that the undriven pair are effectively out of circuit, in regard of ac, they’re not there.

Ahhh yes, but what impedance are they wanting to see at the output transformer primary, given that the screens, cathodes and plates of all four tubes are still connected?

If you ‘neutralise’ the control grid voltage to a steady state DC (which means having no signal at the grid, so it’d need to be disconnected from the other grids), then the voltage swing on the plate (that is still connected to the OT Primary, whose voltage is changing as a result of changes in the other tube), will attempt to produce tube current that has to be sourced/sunk by the other electrodes. Only positive voltage at the plate and screen can attract surplus electrons from the space charge at the cathode (while the relatively-negative voltage at the signal grid would repel most electrons). Anything that got past the signal grid, would be further shielded by the fixed voltage at the screen, which would in turn mostly shield any voltage changes in the plate from current sourced from the cathode. (If you lowered the screen voltage enough, you could turn the tube off.)


Impedance is resistance to changes in current, so if there is minimal change in current at the plate (despite having large voltage swings on the plate), then there will be minimal impedance to that tube.


The other tube (that is powering the changes in current through the OT) now only sees 1/2 the load impedance from the OT secondary (all other things being equal) despite having the same voltage swing as before, so more current has to be sourced through the tube under signal conditions, and that tube works harder (for no additional benefit).


All this is academic.