real world versus calculated power and operating conditions

[imaradiostar]

I'm having trouble grasping how a power output section works...some of the printed values on datasheets don't seem to align with real world data.

As an example-

Assume plate voltage is 500 volts. Assume the pentodes can pull the plate within a few volts of the cathode and assume that they're "ideal" pentodes with the ability to sink whatever current we demand of them. Also assume we're using a 5K:8 ohm output transformer which amounts to a 25:1 turns ratio.

Each plate can swing +/- 500 volts about the center tap so that's 1000 volts peak-to-peak into one half of our 25:1 transformer. Right?

So the peak to peak voltage at the secondary should be 1000/12.5=80 volts. Assuming a sine wave, peak to peak voltage/2.8=28.57 volts RMS.

28.57 squared = 816.2449. That divided by 8 ohms load gives an ideal perfect world max clean sine output of 102 watts...which doesn't see too far off from reality for many amps out there. This all assumes that you can use all of the available voltage which probably wouldn't happen in reality.

Using the another formula I've seen, plate voltage squared/output z=power, I get only 50 watts output. That's half of the theoretical max!

My real-world observations using a number of different tubes but the above plate voltage and impedance show outputs between 45 and about 70 watts, but none reaching as high as 100. I assumed I was reaching grid limiting but checking with a scope showed a clean sine wave on the grids and voltages still "between the rails" and not exceeding the bias voltage. I'm driving the grids using caps and there was no bias shift at the onset of clipping.

So maybe the tubes can't supply the current...but lightening the reflected load by a factor of 2x scarcely has an effect on the output voltage. Of course it shows a drop in power because the reflected load is doubled.

So I checked this against a few other amps...and I'm left scratching my head.

One of my favorite amps is a little plexi-ish thing with a pretty solid supply (350 to 325 volts depending on load and bias) and 1k screen grid stoppers. I've run 6v6's, 6l6's, el34's, 7591's and KT88's in it. It only ever does 15 watts into 8 ohms, usually less with lower power tubes. I know the grid stoppers could be smaller and it's probably dropping voltage across the "choke" resistor but only 15 watts seems really lame! I've tried it at both 3300 and 6600 ohm loads with little change in output voltage. I suppose my screen grids are sagging and preventing me from reaching full output- the 6l6 made the least power so that makes sense to me. Even then, I would think it could do at least 25 watts.

Doing the original math with the values on the EL34 datasheet- They claim 35 watts at 5% distortion from a cathode biased pair of EL34's into a 3400 ohm load.

315 plate volts- 630 peak to peak
3400:8 ohms means 20.6:1 voltage ratio.

630/(20.6/2)=61.165 volts peak to peak.

61.165/2.8=21.84 volts rms...or 60 watts. OK, fair enough, it's theoretical.

315 squared /3400=29.18 watts. That's less than the datasheet! What?

Anyway, I spent several hours reading RDH4 trying wrap my mind around all this and though I know it doesn't matter much for guitar, I have a high gain metal amp and a bass amp project in the works and I'd like for each to make max power. I'm really frustrated trying to understand why the numbers don't always seem to match up and I couldn't find straightforward math that explains what I've been seeing. I didn't take the time to plot load lines for any of them so I probably deserve a hand slap for even posting this but I was hoping someone could say something that would help me wrap my mind around this.

I hate it when I think I understand something...but I don't.

jamie

[Bruce / Mission Amps]

Without all the crazy input power math crap... what is the actual voltage across a fixed non reactive load at the speaker jack?
If you get something like 13v - 14v across the 8 ohm load and it looks like it is clean as viewed by your scope probes on the dummy load then you have a very simple average power formula.
14^2/8 ohms = 24.5 watts etc.

[imaradiostar]

Yeah, crap is exactly how I feel about it. My mini plexi sounds great so who cares, right?

I use the method you mention, Bruce. I even have a "true rms" meter that's accurate within 5% compared to some much higher end stuff. I run a sine into the amp and crank it up till it flattens the waveform a bit then back it off till it's fully rounded again- often I'll use the dual trace feature to compare the two.

I'm just trying to understand what to expect from different designs. I guess that's why you use a load line, right?

jamie

[Alex/Tubewonder]

Assuming plate swinging double B+ peak-to-peak is were you depart from reality and enter utopia. In your last example take a real world EL34 being able to swing close to 70V plate-cathode saturation voltage i.e. 2(315-70) peak-peak and you'll get a Pout value very close to the one quoted in data sheet.

[imaradiostar]

I guess the math wasn't as far off as I'd thought. I guess the volts squared/load equation is a good ballpark number to make sure an output section is working as it should be.

Thanks again. As usual, I just need to build some stuff and make some noise!

jamie

[imaradiostar]

Assuming plate swinging double B+ peak-to-peak is were you depart from reality and enter utopia. In your last example take a real world EL34 being able to swing close to 70V plate-cathode saturation voltage i.e. 2(315-70) peak-peak and you'll get a Pout value very close to the one quoted in data sheet.

And you're totally right- I got 36 watts by that math- close to the datasheet value.

I guess this also explains why a high-voltage colder biased output can be more efficient than a similar output power using lower voltages and impedances- if you're always going to lose 50-75 volts then increasing plate voltage and impedance makes that loss less consequential.

jamie

[Old Tele man]

...you can't expect 100% accurate mathematical answers from ±20% devices (tubes, capacitors, etc.).

...remember, all the data book "numbers" represent a "boggie" tube, a theoretical tube that seldom was produced that had all its 'average' characteristics exactly equal to the published "average" characteristics, which were values derived from lot samples of 100 production tubes.

[Wilder Amplification]

Jaime, peak-peak swing = double the B+ across the full winding from plate-plate, not 1/2 the winding from plate-center tap. From plate to center tap it can only be equal to B+.

In your theoretical amp example, at the peak of 1/2 the swing, the tube is sinking full current. This essentially straps 1/2 the primary from plate #1 to center tap across the B+ supply as at that point the tube looks like a short to ground. This means that the peak swing from plate-center tap can only equal the B+ voltage.

Since they operate out of phase and are pulling current in opposite directions (primary coil winding direction relative to the center tap is backwards on 1/2 the primary, which amounts to a differential input), in your "theoretical amp" you end up with +/- 500V relative to the center tap. Since it is a differential input, (+)500V - (-)500V = 1000Vpk-pk from plate-plate across the full winding.

Here's a composite load line for my 2 x KT88 amp. Each graph represents 1 tube and 1/2 the OT primary. The "dotted arc" is max dissipation while the center dot is the idle bias. You can see that it idles at 600V @ 50mA, right at 70%. The load is 5K Zpl-pl (1250 ohms plate-center tap). If you look at the complete load line which is the plate-plate load, you can see that it extends from zero to 1200V (double the 600V B+), but across 1 graph, which represents 1 tube and 1/2 the primary, it only extends from zero to B+.

[imaradiostar]

Jaime, peak-peak swing = double the B+ across the full winding from plate-plate, not 1/2 the winding from plate-center tap. From plate to center tap it can only be equal to B+.

I don't understand that aspect- double check my logic here, help me understand what I'm doing wrong.

Assume a class B or really cold biased class AB amp.

Let's say the plate voltage is 650 volts and each pentode is capable of pulling the plate down to 50 volts. This means each plate can swing from 650v down to 600v, a change of 600 volts.

When the opposing output tube pulls the opposing half of the primary down to 50 volts, the plate on the other side should, in theory, swing positive by the same amount, right? (I know from practical experience with mosfets in my day job that this should be the case, assuming mosfets are the SS analog of tubes.)

This would mean that if the end of the primary was resting at roughly the B+ voltage, each plate would swing positive or negative depending on if it was working or resting on the given half cycle.

Since that voltage change is one plate referenced to the center tap (plus minus 600 volts relative to the center tap) we end up with 1200 peak-to-peak on each half of the primary, measured from the center tap to either end.

I used turns ratio rather than impedance ratio because it removes power from the equation and turns ratio=voltage ratio.

We know that the RMS voltage of a sine wave is the peak to peak voltage divided by 2.8.

As Bruce said, (RMS voltage into the load) squared/ speaker load impedance = output power.

So...the above seems to make sense to me but I could be way off.

How do we explain a 100 watt 2xEL34 output with 800 volts on the plates?

Keep in mind the the overlapped section of your load line would be puny with an output section like this and each tube is roughly working on its own.

800^2/11000= 58 watts out...that's not a hundred by a good margin!

Assume the El34's are able to pull the plates down to 70 volts. That's 730 volts of "usable" plate swing. Double that for the full peak-to-peak voltage, 1460.

An impedance ratio of 11,000 into 8 ohms makes for a turns ratio of 37:1. Half that is 18.5:1. 1460 volts/18.5=78.9 volts peak to peak at the secondary.

Divide that by 2.8 and we get 28.2 volts. Squared and divided by 8 ohms yields 99.4 watts, much closer to the quoted 100 watt number.

I'm not saying this applies to your average 60-70% biased class AB output- I'm looking for the limits of what a tube is physically capable of. For the average fender twin or marshall plexi volts^2/output z works perfectly- 450 volts squared/2000 ohms = 101 watts. Or perhaps a Fender Deluxe Reverb- 400 volts^2/6600=24 watts. Pretty damn close for the average class AB amp.

Have you tested the output power of your KT88 output section? I have a PT and OT that equal the values you list and I've not gotten as much output as I had hoped for. I get about 60 watts dead clean with KT88's and the screens at about 350 and with a pair of Russian 6L6's puking their guts out and the screens glowing red with 500 screen volts I get about 80 watts- but they won't do that for long!

I didn't have a chance to try higher screen voltages before melting one of my KT88's screens with an overload condition into a too-high impedance. In a week or two when the semester and finals are over I'll get some more. I'd also like to get some KT90's to try at 500 volts into a 2.5k load. I downloaded datasheets for the KT90EH but haven't had time to draw a load line and see how it would turn out.

Thanks for replying, I appreciate your input even if I disagree or don't understand!

jamie

[Alex/Tubewonder]

Here's a composite load line for my 2 x KT88 amp. Each graph represents 1 tube and 1/2 the OT primary. The "dotted arc" is max dissipation while the center dot is the idle bias. You can see that it idles at 600V @ 50mA, right at 70%. The load is 5K Zpl-pl (1250 ohms plate-center tap). If you look at the complete load line which is the plate-plate load, you can see that it extends from zero to 1200V (double the 600V B+), but across 1 graph, which represents 1 tube and 1/2 the primary, it only extends from zero to B+.

Nice graph of a class AB1 load line for an ideal tube. In reality the maximum swing you can expect in class AB1 is from B+ down to where the load line crosses Eg=0, here ca 125V.
So max peak swing is (B+)-125 which will be 475V across Ra=1250 Ohm for one tube, thus, assuming both tube swing the same peak amplitude relative to B+ you get no more than Pout= (475*.7)^2/1250=88W which de facto is reasonably close to measured 78W @ 5%THD as at Eg=0 distortion is higher than 5%.
An ideal tube swinging full B+ down to 0V would give you ca 140W - a rather unrealistic figure at the illustrated operating conditions.

All these figures and graphs are useful only as a tool for checking that your design targets are reasonable under chosen operating conditions.
After all, even with 1% resistors and capacitors and regulated B+ supplies it still is a +/-20% world of tube parameter variations as OTM pointed out.

[Wilder Amplification]

I don't understand that aspect- double check my logic here, help me understand what I'm doing wrong.

Assume a class B or really cold biased class AB amp.

Let's say the plate voltage is 650 volts and each pentode is capable of pulling the plate down to 50 volts. This means each plate can swing from 650v down to 600v, a change of 600 volts.

When the opposing output tube pulls the opposing half of the primary down to 50 volts, the plate on the other side should, in theory, swing positive by the same amount, right? (I know from practical experience with mosfets in my day job that this should be the case, assuming mosfets are the SS analog of tubes.)

This would mean that if the end of the primary was resting at roughly the B+ voltage, each plate would swing positive or negative depending on if it was working or resting on the given half cycle.

Since that voltage change is one plate referenced to the center tap (plus minus 600 volts relative to the center tap) we end up with 1200 peak-to-peak on each half of the primary, measured from the center tap to either end.

I used turns ratio rather than impedance ratio because it removes power from the equation and turns ratio=voltage ratio.

We know that the RMS voltage of a sine wave is the peak to peak voltage divided by 2.8.

As Bruce said, (RMS voltage into the load) squared/ speaker load impedance = output power.

So...the above seems to make sense to me but I could be way off.

Think about it. While one tube is in cutoff, the other is being driven to saturation. The "cutoff tube" is out of the circuit since it's in cutoff and therefore that side of the OT isn't drawing current, so we'll ignore that side of the OT.

Let's use your theoretical amp for example and assume that each tube can sink full current and that at peak swing plate voltage = 0.

Once that happens, 1/2 the OT primary is effectively strapped across the power supply since the center tap is connected to B+ and one side of 1/2 the winding is grounded since the tube is basically a short circuit at that point and therefore sinking full current. Well power supply voltage can't just all of a sudden increase to double...you can't just make voltage.

Now...the cycle reverses. The cutoff tube now becomes the saturation tube while the saturation tube in the first example is now in cutoff. Same thing happens.

They're both passing positive current. BUT...1/2 of the OT primary reverses it because relative to the center tap, 1/2 the OT primary appears to be wound BACKWARDS relative to the other 1/2, so you get the polarity reversal. This effectively makes it +600V on one side of the OT and (-)600V on the other side because current is being pulled in two different directions. Differentiate the two and you have a TOTAL of 1200V pk-pk.

How do we explain a 100 watt 2xEL34 output with 800 volts on the plates?

Keep in mind the the overlapped section of your load line would be puny with an output section like this and each tube is roughly working on its own.

800^2/11000= 58 watts out...that's not a hundred by a good margin!

Assume the El34's are able to pull the plates down to 70 volts. That's 730 volts of "usable" plate swing. Double that for the full peak-to-peak voltage, 1460.

An impedance ratio of 11,000 into 8 ohms makes for a turns ratio of 37:1. Half that is 18.5:1. 1460 volts/18.5=78.9 volts peak to peak at the secondary.

Divide that by 2.8 and we get 28.2 volts. Squared and divided by 8 ohms yields 99.4 watts, much closer to the quoted 100 watt number.

Your math here is incorrect. This proves my point...when using the total plate-plate impedance to calculate output power, you must use DOUBLE the B+ to calculate it because the peak-peak voltage across the full winding = double the B+.

So we double 800V, square it, then divide it by the plate-plate impedance -

1600V^2 / 11,000 = 232.7 Watts Peak

Average power is 1/2 of this -

232.7/2 = 116.35 watts average

Take about 70% of that and you have 81 watts.

Notice when they state "100 watts with an 800V B+ and an 11K plate-plate load"...look at the %THD. It's like 5% THD @ 100 watts output.


Now...let's use 1/2 the OT primary to calculate it. In this instance, you will use the B+ voltage because voltage across 1/2 the OT = B+ at full swing -

Zplate-center tap = 1/4 Zplate-plate = 2,750 Ohms

800^2 / 2750 = 232.7 Watts Peak

232.7 / 2 = 116.3 Watts Average

As you can see in both examples, the math matches up perfectly.

In regards to the KT88 amp, I just barely got the iron in for it last week and I've got a couple of 2210s in the shop that I'm doing a conversion on so I haven't had time to build it yet. Once I've tested and measured output on it, I'll be revamping the dual rail thread to post the results.

@Alex/Tubewonder...yes I'm aware of the fact that the max pk-pk swing = the Vg1=0 crossing point voltage x 2. I was merely using an "ideal/perfect world" scenario to explain what's going on on the load line. Glad you liked that!

[Steve Conner]

I guess this also explains why a high-voltage colder biased output can be more efficient than a similar output power using lower voltages and impedances- if you're always going to lose 50-75 volts then increasing plate voltage and impedance makes that loss less consequential.

You got it!

when using the total plate-plate impedance to calculate output power, you must use DOUBLE the B+ to calculate it because the peak-peak voltage across the full winding = double the B+.

This is mathematically correct, but I prefer to use the B+ value and the impedance of one half of the primary, which is one-quarter the plate-plate impedance. It gives the same answer, but it feels more in touch with the physical reality.

[Wilder Amplification]

This is mathematically correct, but I prefer to use the B+ value and the impedance of one half of the primary, which is one-quarter the plate-plate impedance. It gives the same answer, but it feels more in touch with the physical reality.

You and me both! Don't know why....6 in one, 1/2 dozen in the other but using 1/2 the primary and the B+ does feel more in touch with the physical reality.

[imaradiostar]

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?

You mention peak power and average power, both of which seem to be unclear to me. I know that in a perfect world I can look at my scope and count the peak to peak voltage based on the divisions and a 1 volt RMS sine wave will be 2.8 volts from peak to peak. At work we use amplifiers that are capable of many thousands of volts of output and this is a basic part of what we do so I hadn't questioned it.

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

It would seem that for a real-world class AB1 amp, plate volts^2/primary Z gets really dang close to the actual output power. I have to double-check against some measured values in my notebook but I am trying to understand what sets high voltage dual rail class B stuff apart...other than crossover distortion!

jamie

[Steve Conner]

I don't like that there is no compensation for the fact that a tube isn't a perfect device.

There is, it's called a MOSFET.


Wilder has some weird thing going on about peak power, for which I'm partly responsible, I fear. My method gives you the instantaneous power at the peak of the waveform. If the amp is to amplify a sine wave (which is the standard test condition for output power) then the average power is one-half of this, because the RMS values of a sine wave are 1/sqrt(2) of the peak values, and 1/sqrt(2) times 1/sqrt(2) is 1/2.

And this average power is what the audio industry calls the "RMS" power. The peak power is what they call "music" power.

In my method, I account for "tube drop" by guesstimating it and subtracting it from the B+.