Push/Pull Pair Removal Theory & Questions

[Jazz P Bass]

"There is a reason it's called Ohm's LAW, not Ohm's Vague Suggestion."

Hah!
Spilled my coffee on that one.

[nickb]

There is a reason it's called Ohm's LAW, not Ohm's Vague Suggestion. If the current is half, then the Voltage can't be the same,,, unless somehow the load is doubled. Like Kevin, I suspect you don't have the 100X probe needed to look at the plate Voltage in a power amp stage.

The current is NOT halved. That's where this is coming off the rails.

[Mick Bailey]

On the subject of impedance mismatch I'll raise the Orange OR80 and OR120 as an example of how tolerant an amp can be to mismatched impedance. I'm talking here about the 70s amps. Both models use the same output transformer and the same tappings, 4/8/16 ohms set by a rotary plug. The OR80 has 2xEL34, the 120 4xEL34. I get plenty of OR80s to repair and most have been running for decades with a mismatch. I've only had two that I recall required replacement OTs. There are large transformers in these amps and maybe they're built well enough to last. The tubes certainly last under these conditions - for some reason older Orange amps always show up with really tired tubes and the owners just keep running them until something breaks.

[pdf64]

What power output do they actually manage?
I guess the HT is up around 500Vdc and the OT ~1k7?
I think that would push the EL34 plate dissipation rather high with signal, so best biased as cold as feasible?
But at least the mismatch might take some dissipation off the screen grids. Or is it the OR120 that's mismatched, ie the OT primary is up ~3k4?

[bob p]

Hey Bob, you've mixed up turns ratio with impedance ratio - the turns ratio for the example given would be 26.46:1.

Aw crap.

Sorry for the error. I should have known better, but I was rolling in high speed typing mode. Oops.

[Mick Bailey]

The transformer is designed for the OR120, as is the mains transformer. The last OR80 I had did run just over 500v. I recall that one making about 45W clean at 1Khz and biased very much on the cool side.

[J M Fahey]

And with a primary inductance in that range, it doesn't seem that the OT primary impedance would support the cab's bass resonance?
I'm likely missing something, but wouldn't that mean that the OT would end up dissipating the amp's output at the cab's bass resonance?

I take your point but think it seems low as I'm used to seeing values x10 that, in Hammond's and other info.
And with a primary inductance in that range, it doesn't seem that the OT primary impedance would support the cab's bass resonance?
I'm likely missing something, but wouldn't that mean that the OT would end up dissipating the amp's output at the cab's bass resonance?

Sort of.

The OT inductance is in parallel with reflected impedance.

As any inductor, it is not dissipative, so the transformer by itself does not *dissipate* more from that, it won´t heat up more (might marginally, through wire DC resistance, but resistance is , say, 1/10th or less than reflected impedance unless it´s a *very* cheesy and underrated OT, not the case here)

The full deal is (and I´m probably missing extra details) that:
1) going down with load impedance does not create *extra* power as in a standard SS amplifier, simply because tubes are current limited by design.
Just look at datasheet current curves, limit is reached with grid voltage = 0 , and above a certain voltage (100 something Volts) it becomes quite horizontal, meaning current does not go beyond that level.

This EL84 shows (I use the -1V grid curve, because the 0V one is incomplete) that going from 100V plate to 600V , a HUGE 6:1 swing, current varies only from 160mA to 200mA , a meager 25% increase.


If that´s not current limiting, then I don´t know what the label might apply to.

So in a normal Tube amp, pulling 2 tubes will *basically* halve current, and power will go to 1/4 as before :O

Why?
We live in a mostly SS World , so we think *voltage* and don´t care much about *current* because transistors will easily give you all you want (that many die trying to do so is an unfortunate side effect) so for us Power=V squared/Z

But in the Tube World, we´d better use "the other" equation: Power =I squared * Z.

So half available I means 1/4 available power. (-6 dB)

Only way to recover (somewhat) from that loss is doubling Z , by connecting load to next lower tap (if available). (-3dB)

Now this is not done in a Lab but onstage, to lower annoying loudness, and we all know that -3dB is the bare noticeable minimum, so many amps pull 2 tubes but do NOT correct impedance.

Which as counterintuitive as it looks, is the *useful* thing to do.

So when I see amps who just pull 2 tubes I think: "stage experienced guys".
When I see 2 tubes pulled *and* impedance switched I think: "Lab guys, not much stage experience"

[Mike Sulzer]

So the blue case has a peak power (vg = 0) of (335 - 10)*.085 = 27.625 watts, and if I halve the load and put two tubes in parallel, each does this and I get 55.25 watts peak. The red case is one tube into the half load. The peak power is (335 - 40)*.145 = 42.775. That is not that much less, but if I did that, the tube would melt. I do not think that current limiting is a good description what is happening.

Sort of.

The OT inductance is in parallel with reflected impedance.

As any inductor, it is not dissipative, so the transformer by itself does not *dissipate* more from that, it won´t heat up more (might marginally, through wire DC resistance, but resistance is , say, 1/10th or less than reflected impedance unless it´s a *very* cheesy and underrated OT, not the case here)

The full deal is (and I´m probably missing extra details) that:
1) going down with load impedance does not create *extra* power as in a standard SS amplifier, simply because tubes are current limited by design.
Just look at datasheet current curves, limit is reached with grid voltage = 0 , and above a certain voltage (100 something Volts) it becomes quite horizontal, meaning current does not go beyond that level.

This EL84 shows (I use the -1V grid curve, because the 0V one is incomplete) that going from 100V plate to 600V , a HUGE 6:1 swing, current varies only from 160mA to 200mA , a meager 25% increase.


If that´s not current limiting, then I don´t know what the label might apply to.

So in a normal Tube amp, pulling 2 tubes will *basically* halve current, and power will go to 1/4 as before :O

Why?
We live in a mostly SS World , so we think *voltage* and don´t care much about *current* because transistors will easily give you all you want (that many die trying to do so is an unfortunate side effect) so for us Power=V squared/Z

But in the Tube World, we´d better use "the other" equation: Power =I squared * Z.

So half available I means 1/4 available power. (-6 dB)

Only way to recover (somewhat) from that loss is doubling Z , by connecting load to next lower tap (if available). (-3dB)

Now this is not done in a Lab but onstage, to lower annoying loudness, and we all know that -3dB is the bare noticeable minimum, so many amps pull 2 tubes but do NOT correct impedance.

Which as counterintuitive as it looks, is the *useful* thing to do.

So when I see amps who just pull 2 tubes I think: "stage experienced guys".
When I see 2 tubes pulled *and* impedance switched I think: "Lab guys, not much stage experience"

[Malcolm Irving]

...
So when I see amps who just pull 2 tubes I think: "stage experienced guys".
When I see 2 tubes pulled *and* impedance switched I think: "Lab guys, not much stage experience"
...

I like this, but can I add one:

When I see 3 tubes pulled I think "harmonica player".

[Chuck H]

I like this, but can I add one:

When I see 3 tubes pulled I think "harmonica player".

Or 1 tube

[Mike Sulzer]

The tube curves above are for a constant 300 volt screen supply. But, as has been discussed here many times, the screen voltage in a guitar amp is not held constant, but sags because of the screen resistor. Thus, the plate current cannot be as high with low plate voltage as those curves imply. This is a large part of the reason why you can pull out one tube on each side and the remaining tubes do not melt when the impedance is not matched properly. Current limiting, yes, but caused by the screen resistor, not by the flat curves of the tube characteristics for a fixed screen supply.

[SoulFetish]

P=R*I^2. Half the current means ONE FOURTH the power. KOC doesn't understand this and gets it wrong every time!

Now half the current into twice the load impedance, that's half the power.

YES!! THIS^^^
I think this can cause confusion because because the first formula we encounter for calculating "power" is it's most simple form - volts*amps=watts. So, we're used to having the measurements of volts and current to work with and approaching it from this direction. This is still true in the power equations LT shows above, but it allows us use ohms law to make calculations based on the relationship of power, voltage, current, and resistance.
(Please forgive me engineers. I'm not trying to condescending, just offering this to those who may be confused why it is 1/4th the power)

This makes total sense if one takes a moment to work this out and bring it back to our simple truth of volts*amps=watts.
LT says "P=R*I^2. Half the current means ONE FOURTH the power". Lets see if he's correct:
lets say we have 2 tubes conducting 56mA into a 4k load at idle. Ohm's law tells us that .056(Amps) into 4000Ω produces a voltage of 224V across it.
so: 224*.056=12.544(Watts)

Now, lets cut the current in half:
one tube conducts 28mA into a 4k load at idle. Ohm's law tells us that .028(Amps) into 4000Ω produces a voltage of 112V across it.
(if LT is correct, we should get 12.544/4=3.136 Watts)
so: 112*.028=3.136W=there might be more to old LT besides his 100x probes and dual beam scopes.

[nickb]

YES!! THIS^^^
I think this can cause confusion because because the first formula we encounter for calculating "power" is it's most simple form - volts*amps=watts. So, we're used to having the measurements of volts and current to work with and approaching it from this direction. This is still true in the power equations LT shows above, but it allows us use ohms law to make calculations based on the relationship of power, voltage, current, and resistance.
(Please forgive me engineers. I'm not trying to condescending, just offering this to those who may be confused why it is 1/4th the power)

This makes total sense if one takes a moment to work this out and bring it back to our simple truth of volts*amps=watts.
LT says "P=R*I^2. Half the current means ONE FOURTH the power". Lets see if he's correct:
lets say we have 2 tubes conducting 56mA into a 4k load at idle. Ohm's law tells us that .056(Amps) into 4000Ω produces a voltage of 224V across it.
so: 224*.056=12.544(Watts)

Now, lets cut the current in half:
one tube conducts 28mA into a 4k load at idle. Ohm's law tells us that .028(Amps) into 4000Ω produces a voltage of 112V across it.
(if LT is correct, we should get 12.544/4=3.136 Watts)
so: 112*.028=3.136W=there might be more to old LT besides his 100x probes and dual beam scopes.

No that is not it at all. It's too simplistic. The current is NOT halved. Even ignoring the valid and significant effect of screen grid resistors that Mike S mentioned, take a look at these load lines for 2 vs 4 x EL34 using the loading and voltages right out of the data sheet. Without changing the load impedance the output power is only about halved:

[SoulFetish]

No that is not it at all. It's too simplistic. The current is NOT halved. Even ignoring the valid and significant effect of screen grid resistors that Mike S mentioned, take a look at these load lines for 2 vs 4 x EL34 using the loading and voltages right out of the data sheet. Without changing the load impedance the output power is only about halved:

[ATTACH=CONFIG]47420[/ATTACH]

I'm sorry, nick. I should have been clear. I was expanding on the general point LT was making. I understand that there are specific nuance to the operating conditions you are talking about

[bob p]

we're discussing a P-P configuration, so where are the composite load lines?

those charts show 400V B+ in both conditions, and all of the calculations are derived from an assumption that B+ never changes. With a real world transformer B+ will increase as you move from a full-load to a half-load condition.

granted, if the transformer has a good regulation spec then the difference may not be much, but if it's got a poor regulation spec (like most MI transformers) that change could be significant. it's probably worth taking that into account and using a new tube characteristic with the proper B+ when doing the comparison. and while we're at it, we should also be looking at composite load lines.