real world versus calculated power and operating conditions

[Wilder Amplification]

Thanks again, still having a hard time wrapping my mind around this. Thanks for humoring me.

I don't like that there is no compensation for the fact that a tube isn't a perfect device. I understand the whole +- 20% tolerance thing but how do we account for the fact that a tube can't pull the plate all the way to ground?

That's a function of the transconductance characteristic. At the point where Vg1 = 0, any increase in grid voltage makes the grid positive and as such the grid will start to draw current. Negative electrons flow to the grid and offset the positive value of the signal voltage, which acts like a limiter and is called "grid limiting". This is why the plate current line starts to flatten out after the "knee" and as such why plate current doesn't increase very much if at all past the point where the load line crosses the Vg1 = 0 curve. If there's no increase in plate current, there won't be a decrease in plate voltage. This "grid limiting" characteristic is a big cause of that.

Where does peak power versus average power fit into that scheme?


jamie

It depends on whether you're using "peak swing voltage" or "RMS swing voltage" to calculate total power consumed by the load. If using peak voltage, you get peak power. If using RMS voltage, you get average power.

This is where it gets weird. You have "peak voltage", and then you have "peak to peak voltage". This means you technically have two different RMS voltages...the RMS value of the peak and the RMS value of the peak-peak voltage. In each scenario, RMS = 70.7% of either or.

In other words. RMS value of the peak-peak voltage is 70.7% of the peak-peak while RMS of the peak voltage is 70.7% of the peak voltage. If you want to calculate average power across the full winding of the primary, you square the RMS of the peak-peak (70.7% of the peak-peak voltage) and divide THAT by the load, not the RMS of the peak voltage.

[imaradiostar]

You guys are really not going to want to help me if I keep arguing, right? I hope you don't hate me for experimenting and sharing!

When I got back from class I spent some time in the shop trying to get a grip on these concepts. I pulled out my mini-plexi (or low voltage plexi, whatever you wanna call it) and started measuring things.

First off I replaced the first resistor in the power supply with a choke to minimize screen sag. I installed a pair of chinese EL34's and biased them pretty moderately, around 60% of 25 watts. I tried other settings too and it changed the gain a bit but the limits were about the same.

I connected meters to the plate and screen voltage, a regular old digital VOM to one tube's plate and the scope probe to the other. Output voltage was read with another VOM across the resistive 8 ohm load.

Output transformer was measured at 6635:8 ohms based on the unloaded voltage ratio- excessive for EL34's at low voltage but I did that on purpose for the sake of this experiment.

I used a sine wave at 500hz for most of the testing- I played around with other frequencies with little change.

I gradually increased the input voltage till the scope showed a sine wave that was 6 divisions from peak to peak. The scope was set at 5v/div and I was using a 10x probe. It's a tektronix 5110 and it's been professionally calibrated (as in I didn't do it!) in the last several years. 6*5*10=300 volts. 300 volts/2.8=107 volts. The reading on the voltmeter on the opposing plate was 106 volts. That's off a volt from what I expected, not too shabby.

I continued to increase the input voltage and checked along the way- yes, a sine wave continues to have a peak to peak voltage of about 2.8 times the reading on the voltmeter. At 8 divisions I was getting 400 peak to peak at one plate and 142 on the voltmeter- volt swing at on one side of the primary was now well in excess of B+ which had by now sagged to about 325 volts.

By the time I got to clipping the voltmeter read 199 and the scope was unable to keep up. By taking it out of calibrated input mode I could at least verify that the output was still clean. I messed with this uncal mode and was able to approximate that these ratios held true up till clipping. The scope was now showing a full 550 volts peak to peak. Plate voltage at full output had sagged to 320 volts and screen voltage was 318 volts.

Actual output to the load was 11.5 volts at that point. I verified with the scope that the output was similarly clean.

If the plates were swinging 550 peak to peak then the El34's were capable of pulling the plates down to about 45 volts based on 550/2=275, 320-275=45. So 275 is the "real world plate voltage" that we can use to calculate the output of the amp.

Using your formula- (275^2)/(6635/4)=45.6 watts peak instantaneous power, roughly 22.8 "average" power.

Using my screwy method, 550 volts/14.41=38.2 peak-to-peak at the secondary. 38.2/2.8=13.64 volts into an 8 ohm load. 13.64^2/8 ohms equals 23.3 watts- darn close to the 22.8 "average" power by your method.

So...they're not that different and to be clear the plates DO swing peak to peak well over the B+ voltage. Each plate can swing about as high over b+ (minus the drop of the idle current through the DCR of the winding) as the opposing plate swings negative with respect to the plate supply. Of course the tube that's working can only pull the plate to about 45 volts, not all the way to zero. This varies with different tubes, loads, voltages, etc.

For the record the actual output of the amp was 11.5 volts or 16.5 watts into 8 ohms. The output transformer had lots and lots or primary DCR and I'd imagine it's not the most efficient TF I've ever seen- plate to plate DCR was over 700 ohms. I'm probably losing a lot of output right there. This little transformer does have lots of bass and sounds nice though so I can't complain. It was originally in a Knight kit amp with ridiculously high voltages (around 475 volts!) on a pair of 6973 output tubes.

OK, flame on!

jamie

[Wilder Amplification]

So...they're not that different and to be clear the plates DO swing peak to peak well over the B+ voltage.

That's what I've been saying all along. Across the plates (i.e. full primary winding) the voltage peak-peak will swing over B+. In the beginning you made it sound like from one plate to the center tap it would swing over B+, and that's impossible to happen.

[martin manning]

Plate voltage on each side can swing higher than B+, along the class-A load line, because the transformer, being an inductor, will not allow an instantaneous change in current. This overage won't contribute to the amplitude of the output signal though, because it is in the overlap region and the signal peaks come from the other extreme. Head on over to Merlin B's Valve Wizard web site, and read both the push-pull and single-ended articles.

[imaradiostar]

I don't think it's impossible. Please try this before you dismiss what I have to say.

With your scope input set to the least sensitive position and a 10x or greater probe, connect your probe to one of the plates of a tube amp. I chose a lower voltage amp because my scope can't handle very high input voltages. Connect the ground clip or jumper or whatever to ground on the amplifier. Connect an adjustable sine source to the phase inverter input and some sort of dummy load to the output.

Set the scope to "DC" input mode and fire up the amp. Adjust the scope trace toward the bottom half of the screen- it should be a straight line representing ground at this point. When you flip the standby switch you should see the scope trace rise as the plate voltage comes up. This increase, assuming your scope as the range, should correspond to the plate voltage of the amplifier. If you turn the amp off the trace should return to the bottom of the screen as the supply voltage bleeds off.

Turn the amp back on (trace goes up) and make a mental note of the position of the trace. Apply a signal and you'll see the plate lead oscillate above and below B+. As the load increases and the power supply sags a bit the dc offset will change but I think you'll see that opposing ends of the OT primary do indeed go well above the supply voltage of the amp.

jamie

[imaradiostar]

Plate voltage on each side can swing higher than B+, along the class-A load line, because the transformer, being an inductor, will not allow an instantaneous change in current. This overage won't contribute to the amplitude of the output signal though, because it is in the overlap region and the signal peaks come from the other extreme. Head on over to Merlin B's Valve Wizard web site, and read both the push-pull and single-ended articles.

How about it's inherent to how a transformer works? Sure, if you put DC across a transformer and load it up with energy then suddenly release that energy it can greatly exceed the initial voltage. That's the concept behind a spark coil in an automobile.

It's like winding up a spring that can suddenly be released. For example- it might take you half a second or more to pull pack your arm before you take aim with a bow and arrow. When you let that arrow fly it's a tiny fraction of a second but all (most anyway) of the energy you exerted pulling on the string is released all at once.

This applies a little bit to tube amps but to a much larger extent it's just how a transformer works. Take any old push-pull OT you have laying around a connect a low-voltage AC wall-wart between the center tap and one of the plate leads. Assuming the transformer is decently made, the voltage measured from the center tap to either plate lead will be the same. Even if you start to put a load on the secondary or a load on the opposing side of the primary this will continue to be true. Of course if you put a 2 ohm resistor across the opposing side the resistance of the primary windings of the OT will bear the brunt of the load and start to get hot but hopefully you see where I'm going with this.

In much the same way- once a tube is biased off or mostly off it doesn't present a huge load to the opposing tube and its plate swings well above B+.

It's like a see-saw- if one side doesn't go up at the same time and rate as the other side comes down it's either already broken or you're both too fat (ie: too much current is flowing) and you're bending it so that it's no longer linear.

And I know Steve or someone else will say it's not that simple- you're right, it isn't, but it's more like that than it isn't!

jamie

[imaradiostar]

Another example- the voltage coming into your house from the transformer on the pole. If things didn't work this way we wouldn't be able to power our oven on 240v and our computers on 120v without different transformers.

jamie

[Wilder Amplification]

Another example- the voltage coming into your house from the transformer on the pole. If things didn't work this way we wouldn't be able to power our oven on 240v and our computers on 120v without different transformers.

jamie

That's because the transformer that powers your house is center tapped (the neutral is your CT). The turns ratio is DOUBLE on the CT windings than it is across the full winding so you get 1/2 the voltage on each rail relative to the neutral CT. Across the neutral (again the CT) and either outside rail you have 120V, but go across both rails with a double breaker and you get 240V.

[Steve Conner]

It's like a see-saw- if one side doesn't go up at the same time and rate as the other side comes down it's either already broken or you're both too fat
...
And I know Steve or someone else will say it's not that simple

No, it is that simple, in this context at least. The center tap of a PP OT is just like the fulcrum of a see-saw. When the tube that's conducting "jumps" on its end and shoves it down below B+, the opposite end rises above B+ by an equal amount.

That ignores leakage inductance, which lets the ends "spring" a bit further than they're supposed to. Imagine a bendy fiberglass see-saw with a couple of really fat American kids on it.

The push-pull output stage is just the FWCT rectifier turned backwards: the peak voltage across the whole PT secondary is twice B+.

[roughcut studios]

"Imagine a bendy fiberglass see-saw with a couple of really fat American kids on it."

I suppose that IS a better image than a couple of fat old Scots in kilts on a see-saw!

(sorry, now back to the serious discussion)

[Wilder Amplification]

"Imagine a bendy fiberglass see-saw with a couple of really fat American kids on it."

I suppose that IS a better image than a couple of fat old Scots in kilts on a see-saw!

(sorry, now back to the serious discussion)

Now THAT was funny...should've said "skirts" though. (sorry Steve...had to)

Maybe I took Jamie's original idea wrong. The way I understood it was that he was speaking as if the voltage from one plate to the center tap would equal B+ x 2 at max swing. Thinking that this is what he was saying, I came on and informed him that it's the plate-plate voltage that equals B+ x 2, not plate - center tap voltage. That's because assuming the plate can drop to zero (in the perfect ideal amplifier), this effectively straps one side of the primary across the power supply, and the power supply can't just manufacture voltage so this would make the plate-center tap voltage equal B+, not DOUBLE B+. But the plate-plate voltage would equal double B+ due to the push-pull/differential nature of the circuit.

Was I misinterpreting what Jamie originally was trying to say?

Maybe I misinterpreted what he was originally stating?

[Steve Conner]

Well, in the West of Scotland we have about the worst health in Europe, but we're still not as lardy on average as you doughballs.

I always think of voltages in a circuit with respect to ground. The two ends of the OT primary winding oscillate between ground and twice B+ (give or take a tube drop) while the center tap stays fixed at B+.

If you look at Tom Hornak's article on visualizing circuits:
Analog circuit design: art, science ... - Google Books

that more or less sums up the way I visualize them too. (I like to draw the OT primary winding horizontal, and I hate the old Fender schematics that draw one tube of a push-pull pair upside-down.)

[pat19c]

Sorry for the Interruption.

I'm new to the forum world and I was wondering if there might be a help menu somewhere that would go into depth on the proper way to post threads and so on. Any input would be greatly appreciated. Once again I'm sorry I barged into the middle of this thread but I think its going to be the best way to find my answer.