Here're some measurements I did long ago: you can see that, unsurprisingly, the Leak ST60 (blue) is not really optimized, whereas the ARC D70 is optimized for minimum THD @ rated impedance. The X axis is log Ohms... I think.

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Here're some measurements I did long ago: you can see that, unsurprisingly, the Leak ST60 (blue) is not really optimized, whereas the ARC D70 is optimized for minimum THD @ rated impedance. The X axis is log Ohms... I think.

Yup. 24 watts from a pair of cheap triode wired pentodesThat's for triode-wired, no?

OK, excellent. I guess we are on the same page and I have now no confusions!Yes, in class A both tubes (current sources for AC analysis) are delivering power simultaneously to the same load. Not veryefficientlybecause of all the standing current. But both are delivering power. At the same delivered voltage, the current goes to 1/2 for each.

I completely agree with the statement, although I don't think I got the graphs right. Basically you are saying that the speaker impedance changes over the frequency. So the primary impedance of the OPT to the tubes also changes as the frequency changes (so called "playing music"), hence 99+% of the time when we are listening music, the tubes are off the load line that you think the tubes are on.Interesting to note that you don't really have to get the load line super right because the impedance of the speakers will certainly mess it up bad.

Here're some measurements I did long ago: you can see that, unsurprisingly, the Leak ST60 (blue) is not really optimized, whereas the ARC D70 is optimized for minimum THD @ rated impedance. The X axis is log Ohms... I think.

View attachment 1063531

I guess that is why you just can't say anything about the quality of the amplifier from the datasheet or even the measurements using a dummy load or a fixed frequency sinusoidal signal. Probably this is a true statement "you just listen to it and you decide if you like it or not".

No.plate resistance specification. Doesn't that give you a good clue as to what load impedance might work well for it?

A pentode is a 2-slope device. The usual plate resistance is the horizontal part, very high impedance. Just as important for Power is the vertical part, a few hundred ohms. Ignoring limits, you can run almost anywhere

Think of the load line as a target that you want to be statistically close to. The real “dynamic” load line is elliptical anyway, because of speaker reactance. But in practice that doesn’t hurt things as badly as it does with transistor amps (transistor amps tend to blow up or go into very nasty sounding limiting modes unless you oversize the hell out of the output stage). If you plot it with a music signal containing a wide spectrum it looks like a hair ball, stretched out as sloped through where you would expect it to be. The more widely your speakers vary in impedance, the more “extra” current capacity you need to design for. The hair ball gets fatter and slopes up a bit more. Extra current capacity in tube amps is not free - it costs heater power and screen dissipation which must be managed.I completely agree with the statement, although I don't think I got the graphs right. Basically you are saying that the speaker impedance changes over the frequency. So the primary impedance of the OPT to the tubes also changes as the frequency changes (so called "playing music"), hence 99+% of the time when we are listening music, the tubes are off the load line that you think the tubes are on.

I guess that is why you just can't say anything about the quality of the amplifier from the datasheet or even the measurements using a dummy load or a fixed frequency sinusoidal signal. Probably this is a true statement "you just listen to it and you decide if you like it or not".

There are actually quite a few knobs to turn with push pull output stages, especially pentodes. Within limits, you can dial in the output power you require, given the power supply voltage and load impedance you

I'm interested to understand what this means. I don't think I'm following you...No.

A pentode is a 2-slope device. The usual plate resistance is the horizontal part, very high impedance. Just as important for Power is the vertical part, a few hundred ohms. Ignoring limits, you can run almost anywherebetweenthese two slopes; getting different powers.

You mention the "horizontal part". Of what, please? Of the pentode plate curves? So the "horizontal part" is plate voltage (x axis), and the "vertical part" is plate current, the y axis?

slopes of the curves, 5500 ohms is (270-40)/(.2-.15) = 230/0.05 = 4600 ohms , missed?

slopes of the curves, 5500 ohms is (270-40)/(.2-.15) = 230/0.05 = 4600 ohms , missed?

???

Sorry, I'm not educated about this. I don't understand.

Push Pull 4k plate to plate and with a center tap, to an 8 Ohm tap that is loaded by an 8 Ohm non-inductive resistor:

1/2 winding = 1k Ohm (one plate connection to the center tap). The other 1/2 winding is the same, 1k Ohm.

But . . .

In Class A, each tube is On, and each tube is Aiding each other. So each tube sees a 2k load.

In Class B, only one tube is on, so they do Not aid each other. So each tube sees a 1k load.

In Class AB1:

Sometimes both tubes are On (each tube sees a 2k load).

Sometimes only one tube is On (each tube sees a 1k load).

When the tubes are in transition from one state to the other, the load impedance is between 2k and 1k.

Simple, straightforward, yet quite complex.

I hope that makes it clear, and makes it memorable.

Have fun deciding what operating Class you want to use.

I didn't realize that was what TonyTecson was referring to. Thanks for clarifying that.

Yes, I'm familiar with the change in apparent load impedance with class of operation, but I have to admit I don't have a full understanding of impedances, reactive loads and electro-magnetics in general.

I do get that:

1) In class A push-pull operation, the output tube sees an impedance that is half of the total plate-to-plate primary impedance of the OPT. (The rp of an EL84-triode will be about 2k ohms. so for push-pull class-A operation choose an 8k ohm plate-plate primary impedance relative to the load impedance on the secondary.)

2) In class B PP operation, because one tube is in cutoff (not drawing current) for each half-cycle, each output tube now sees only one-quarter of the total plate-to-plate primary impedance of the OPT. (For a push-pull pair of EL84-triodes running class-B, that 8k ohm primary impedance will look like 4k ohms.)

3) In class AB PP operation, the load each output tube sees varies with how close to cutoff each side of the push-pull pair goes into on each cycle.

Deciding what class of operation I want to use?

Many years ago, I had built up a crazy fixed-bias push-pull 300B amp which had this mammoth tube series regulator (giant 8H choke input power supply, 360mA rated power transformer, 6336 pass tube) with which I could adjust the output voltage. I could also adjust bias voltage for the output tubes, so I could play with operating points pretty easily. For listening to recorded music at home, I found that I liked the sound of the push-pull output stage with lower voltage/higher current operating points, so the output stage stayed in class A as much as possible, even though that meant less output power. When I adjusted the operating points for higher plate voltage and lower plate current, I didn't like that sound as much. I recall that it sounded 'drier' and less 'full' the more I adjusted it to high voltage/low current operation.

So I can understand what's going on with triodes. But what about choosing the primary impedance for a pair of pentodes?

What I'm having trouble with is what PRR posted in post #25, after I'd asked whether the plate resistance (rp) specification of a*pentode* would give you a good indication of what OPT primary impedance to choose for it.

PRR posted this graph to explain:

When I posted that I didn't understand what the graph is showing,

I see the vertical slope alongside the y axis is labeled with 125 ohms and the horizontal slope alongside the x axis is labeled 5.500 ohms. But I don't understand the significance of that. Is that 125 ohm line a second kind of load line?

Yes, I'm familiar with the change in apparent load impedance with class of operation, but I have to admit I don't have a full understanding of impedances, reactive loads and electro-magnetics in general.

I do get that:

1) In class A push-pull operation, the output tube sees an impedance that is half of the total plate-to-plate primary impedance of the OPT. (The rp of an EL84-triode will be about 2k ohms. so for push-pull class-A operation choose an 8k ohm plate-plate primary impedance relative to the load impedance on the secondary.)

2) In class B PP operation, because one tube is in cutoff (not drawing current) for each half-cycle, each output tube now sees only one-quarter of the total plate-to-plate primary impedance of the OPT. (For a push-pull pair of EL84-triodes running class-B, that 8k ohm primary impedance will look like 4k ohms.)

3) In class AB PP operation, the load each output tube sees varies with how close to cutoff each side of the push-pull pair goes into on each cycle.

Deciding what class of operation I want to use?

Many years ago, I had built up a crazy fixed-bias push-pull 300B amp which had this mammoth tube series regulator (giant 8H choke input power supply, 360mA rated power transformer, 6336 pass tube) with which I could adjust the output voltage. I could also adjust bias voltage for the output tubes, so I could play with operating points pretty easily. For listening to recorded music at home, I found that I liked the sound of the push-pull output stage with lower voltage/higher current operating points, so the output stage stayed in class A as much as possible, even though that meant less output power. When I adjusted the operating points for higher plate voltage and lower plate current, I didn't like that sound as much. I recall that it sounded 'drier' and less 'full' the more I adjusted it to high voltage/low current operation.

So I can understand what's going on with triodes. But what about choosing the primary impedance for a pair of pentodes?

What I'm having trouble with is what PRR posted in post #25, after I'd asked whether the plate resistance (rp) specification of a

PRR wrote:

No.

A pentode is a 2-slope device. The usual plate resistance is the horizontal part, very high impedance. Just as important for Power is the vertical part, a few hundred ohms. Ignoring limits, you can run almost anywherebetweenthese two slopes; getting different powers.

PRR posted this graph to explain:

When I posted that I didn't understand what the graph is showing,

TonyTecson replied:

slopes of the curves, 5500 ohms is (270-40)/(.2-.15) = 230/0.05 = 4600 ohms , missed?

I see the vertical slope alongside the y axis is labeled with 125 ohms and the horizontal slope alongside the x axis is labeled 5.500 ohms. But I don't understand the significance of that. Is that 125 ohm line a second kind of load line?

Last edited:

4k is the reflected impedance seen by two tube plates when an 8 ohm resistor load is connected to the secondary. this an impedance ratio of 4000/8 or 500 and a turns ratio of 22.4,Hello,

I have a simple question about a push pull output transformer.

For example, when I go to Hammond, say one of their PP OPTs has "4000 Ohm ct" impedance on the primary side and 8 Ohm on the secondary side. My question is exactly where you get this 4 kOhm impedance on the primary side (assuming the secondary load is an 8 Ohm speaker). Is it plate to plate impedance (so that plate to center tap becomes 1 kOhm) or is it plate to center tap (so that plate to plate becomes 16 kOhm). Since plate-ct is the effective load to each tube, I am guessing that the transformer is 4 kOhm plate-ct (so you can directly use this value on a tube's load line) such that plate-plate is actually 16 kOhm, but I could not find a definite description...

Thanks!

now if you connect a 4 ohm resistor to secondary, then this becomes 2000 ohms, now when you connect 16 ohms, the reflected impedance becomes 8000 ohms....

take note that once an OPT is built, the turns ration is fixed, now a resistor is much different than a real world speaker in that the speaker impedance is never constant for all frequencies..

lastly, take note that sine waves that was used as basis for design calculations is much different than actual music that we hear in our speakers...

there are procedures for dawning plate load lines.. like this one... https://groupdiy.com/threads/designing-push-pull-power-amp-stages.73279/ and this too... https://www.diyaudio.com/community/threads/pentode-loadlines-pp.324990/

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