RMS vs Average volts with DC offset

[DuaneOh]

I'm playing with some circuits in LTspice, which will give values for both RMS and Average volts for a sign wave with a DC offset. The circuit I'm looking at has a local feedback loop from the plate of the output tube to the plate of the phase inverter, like a Bassman AB165. I'm trying to determine the power dissipated by the feedback resistor at full power output. I'm not sure if I should be looking at RMS or average voltage to do my calculations? I'm thinking that it should be RMS?

Thanks for your help!

[acheld]

I'm not sure if I should be looking at RMS or average voltage to do my calculations? I'm thinking that it should be RMS?

With the caveat that I'm not the expert around here (the southern boys are laughing their butts off with that), I believe the answer is neither.

You're talking power dissipation, so it's really the current flowing and the voltage drop across the resistor that count, and the DC offset has no effect on this.

[DuaneOh]

With the Spice I can see the voltage waveform on both sides of the resistor but they are out of phase. And I get the current waveform through the feedback resistor. Spice will give me the RMS and average of these over the time that I have set it up to record. So I'm wondering which would give an accurate dissipation.

[PRR]

What exactly is a "Bassman AB165"?

If the one I found, then it is not a practical concern.

> for a sign wave with a DC offset.

There is no "DC offset" here; in fact there is a DC-block cap.

Yes, you want RMS.

> at full power output.

Is this a test lab? No, a stage amp. "Full output" is Gross Clipping, nearly square waves. (It is very embarassing to design a part to stand unclipped sines, and smell it smoke in real life.)

And.... make a wild guess. I guessed 14V output (50W in 4r), so 15V against a ~1V input. 15V in 47,000 Ohms is milliWatts. Doubled for square and doubled again for reliability is 0.02W.

Can you even buy a part that small?

[DuaneOh]

The Bassman AB165 has local feedback on the 6L6's, the 220k resistor highlighted in the attached schematic. On the 6L6 side of the feedback resistor is 425v and on the 12AT7 side is 225v at idle.

RMS voltage with a DC offset is something I'm unfamiliar with. I'm trying to understand the power dissipated by the local feedback resistor when the amp is clipping.

I plan on building a version of this without the global negative feedback to to hear it's effects, trying different levels of local feedback, and want to size the feedback resistor correctly.

I built an audio amp design by Steve Bench in the early 2000's using local feedback similar to this, using EL34's in triode mode:
http://diyaudioprojects.com/mirror/members.aol.com/sbench102/PowerAmps/e34lamp.gif
In an email with Steve he said "There seems to be no sonic penalty for local feedback, vs global feedback which tends to produce a more sterile but less satisfying sound quality."

[sluckey]

The Bassman AB165 has local feedback on the 6L6's, the 220k resistor highlighted in the attached schematic. On the 6L6 side of the feedback resistor is 425v and on the 12AT7 side is 225v at idle.

RMS voltage with a DC offset is something I'm unfamiliar with. I'm trying to understand the power dissipated by the local feedback resistor when the amp is clipping.

I don't see this as something to get hung about. Just calculate the dc wattage dissipated by that resistor. You have 200Vdc across a 220K resistor so power dissipated is...

     P = E² / R = 200*200 / 220,000 = 0.18 watts. Fender used a half watt resistor which is plenty big enough, even with a wee bit of AC power thrown in.

[acheld]

I'm trying to understand the power dissipated by the local feedback resistor when the amp is clipping. . .

OK, there is DC in that local feedback circuit. But, the answer to the power dissipation still lies in Ohm's law.  PRR's back of the napkin calculation remains relevant.  You need to measure the voltage drop across the feedback resistor.  RMS voltage taken from each side of the resistor is what you want.  The DC offset does not matter, as you have taken it into account by measuring voltage on each side of the resistor.  You know the resistance, and can then calculate power dissipation.

Of course while writing this tome, sluckey has chimed in with the actual practical answer.  The AC voltage will be on the order of < 5% of the DC voltage, hence can be ignored.  I still think that RMS voltage drop would work fine, but it would be interesting to work that out.

, and want to size the feedback resistor correctly. 

Practically, no-one does this since typical resistors in this position (whether global or local) do not dissipate much power.   Personally, I use 1W resistors because that is what my default value is in my amps, knowing that it is waaaay over spec'd.   In a true production environment, with production software and simulation, I'd know what the power dissipation is and would buy my parts as appropriate.

PRR's and sluckey's back of the napkin approaches are practical and spot on.   

[DuaneOh]

I don't see this as something to get hung about. Just calculate the dc wattage dissipated by that resistor. You have 200Vdc across a 220K resistor so power dissipated is...

     P = E² / R = 200*200 / 220,000 = 0.18 watts. Fender used a half watt resistor which is plenty big enough, even with a wee bit of AC power thrown in.

Practically, no-one does this since typical resistors in this position (whether global or local) do not dissipate much power.   Personally, I use 1W resistors because that is what my default value is in my amps, knowing that it is waaaay over spec'd.   In a true production environment, with production software and simulation, I'd know what the power dissipation is and would buy my parts as appropriate.

PRR's and sluckey's back of the napkin approaches are practical and spot on.

Your right, I know that it's not much wattage. I'm fairly new to using spice on these circuits and I can get bogged down on the amount of data, trying to understand it. For instance, on the amp I'm building when EL34's are clipping the plate's average volts and current can be 355v and 92ma (30 watts) but the RMS can be 421v and 140ma (54 watts), according to spice. Obviously, the EL34's can handle this but I'm wondering what is really going on, are the tubes seeing 30 or 54 watts?

Thanks for everyone's help.

Edit, Cathode biased, forgot to subtract 32v on the cathode.

[pdf64]

I strongly suggest not to probe an overdriven output valve anode, there will be back emf spikes above the meter’s ratings.
I zapped my first Fluke that way 
The dissipations you’re calculating are a combination of dissipation by the anode and power delivered to the load.

[tubeswell]

… using spice … trying to understand it. … I'm wondering what is really going on, are the tubes seeing 30 or 54 watts?

Are you asking about what’s happening to output tube dissipation in a push-pull amp when it changes from Class A to Class B? Or?

[PRR]

OIC. You did not specify which resistor.

....The AC voltage will be on the order of < 5% of the DC voltage....

The 6L6 plates will pull-down to like 75V and up past 425V the same amount. So 247V sine RMS. This is comparable to the 200V of DC, more actually; though only when loud.

IMHO, you would be better to yank all that 1960s cruft off and wire as a 5F6a. But I see no harm in trying.

[DuaneOh]

… using spice … trying to understand it. … I'm wondering what is really going on, are the tubes seeing 30 or 54 watts?

Are you asking about what’s happening to output tube dissipation in a push-pull amp when it changes from Class A to Class B? Or?

It is a EL34 cathode biased class AB amp and I am clipping it pretty hard, so class B. On the spice simulation, I'm looking at the plate voltage and current waveforms, spice will tell me both the average and RMS values for both. The 30 and 54 watts are from calculating the wattage for both average and RMS values. So that's what I'm asking, is which is real, the average or the RMS?

This may have no value at all other than curiosity, I just trying to have a better understanding. This amp will produce about 30 clean watts to the speaker, I'm just suprised to see 1 EL34 dissipating 54 watts RMS when producing about 40 distorted watts to the speaker.

I've attached a spice screenshot of the circuit and waveform.

[pdf64]

…
The dissipations you’re calculating are a combination of dissipation by the anode and power delivered to the load.

And also includes dissipation from the cathode bias resistor.

[shooter]

fwiw;
I like tuning your waveform into 8-10 on the drive pot.  1-7 + guitar 1-10 + pedals, waveform's created from strings to the PA driver tube, 8-10 on the drive knob adds flavor.

[PRR]

...... suprised to see 1 EL34 dissipating 54 watts RMS when producing about 40 distorted watts to the speaker....

The tube dissipation usually goes DOWN when clipped all to heck.

Are you sure you don't have 54W input, 40W output, just 14W lost in the tube?

[DuaneOh]

The tube dissipation usually goes DOWN when clipped all to heck.

Are you sure you don't have 54W input, 40W output, just 14W lost in the tube?

I am measuring 10ms of the amp producing about 40w at the speaker. The current through the plate of the EL34 is 140ma RMS, the voltage at the plate is 420v RMS, the cathode voltage is 31.5v RMS:
(420-31.5)x.14=54.4

If I measure 1/2 of the output transformer primary I'm seeing 139ma.
Ig2 is 33.5ma
Ik is 173.5ma
And the feedback resistor current is .62ma
All RMS values

All added up it seems to come out close with some rounding errors.

The plate voltage is a kind of a square wave, peak to peak 550v, with about 275v DC offset.

All of this is from the LTspice simulation, I'm not sure if I'm doing something wrong, the software is lying, or it is really dissipating 54 watts?

[DuaneOh]

At max clean output (30 watts), RMS vs. Average, same measurements as above:

Ia=124ma RMS
Va= 410v RMS
Vk= 27.8v
Wa= 47.4w

Or:
Ia=81ma Average
Va=359v Average
Vk=27.8v
Wa= 26.8w Average

[acheld]

Well, I've got Kuehnel's book on my desk right now, have made it to p3 only.

But I do know that LTSpice has poor modeling for tubes, especially at edge cases, so I'd be inclined to suspect your tube model.   

[DuaneOh]

Well, I've got Kuehnel's book on my desk right now, have made it to p3 only.

But I do know that LTSpice has poor modeling for tubes, especially at edge cases, so I'd be inclined to suspect your tube model.

I think that you are probably correct, I've gotten tube models from different sources and some are definitely better than others.

I did not know that Kuehnel had a book on simulation, I'll have to pick that up. Thanks

[PRR]

> the software is lying

Usually it is giving the correct answer to the wrong question. You seem to be getting INPUT power and thinking it is Dissipation. LTspice has an alt-rt-click tool to read dissipation but as acheld says, this may not work on sub-models.

It may be interesting to get a long-term DC power number at the power supply. (Put in a 1 Ohm resistor, multiply B+ voltage times current, for several cycles of the audio.)

[DuaneOh]

> the software is lying

Usually it is giving the correct answer to the wrong question. You seem to be getting INPUT power and thinking it is Dissipation. LTspice has an alt-rt-click tool to read dissipation but as acheld says, this may not work on sub-models.

I had to look up input power (power consumption?), from what I read it's the total power consumed by the entire circuit?

I did not know about the dissipation function in LTspice, and it looks like it worked. I've attached waveforms for both 40 watt (clipping) and 30 watt (clean) outputs using:
V(A1)*Ix(U3:Anode)+V(S1)*Ix(U3:Screen)+V(N004)*Ix(U3:Grid)+V(k1)*Ix(U3:Cathode)

and got: 40w = 13.1 watts, 30w = 14.3 watts    Lower dissipation for clipping like PRR predicted earlier. But I'm still a bit confused, 13.1w /(420v - 31.5v) = 33.7ma for measured RMS plate voltage. For average plate voltage,  13.1w /(355v - 31.5v) = 40.4ma

It may be interesting to get a long-term DC power number at the power supply. (Put in a 1 Ohm resistor, multiply B+ voltage times current, for several cycles of the audio.)

I put a 1 ohm resistor after the first smoothing capacitor,
at 40w output, B+ = 358v, Ir = 247ma, W = 88.4w
at 30w output, B+ = 362v, Ir = 216ma, W = 78.2w
at 0w output,  B+ = 366v, Ir = 174ma, W = 63.7w

Thanks, this has been very helpful.

[pdf64]

You don’t seem to be taking on board that power delivered to the load, despite it being fed to and passing via the output valves, is not dissipated by the output valves?

[DuaneOh]

You don’t seem to be taking on board that power delivered to the load, despite it being fed to and passing via the output valves, is not dissipated by the output valves?

Are you asking about the power dissipated in things like the output transformer, cathode resistor, etc? When LTspice measured the EL34 dissipation, it looked at the anode, cathode, g1 and g2. I'm assuming it doesn't measure the heater dissipation. The previous wattage measurements were made when the amp was in class B operation, when it is setting idle the EL34's are dissipating 28 watts, and with 6w (Class A) to the speaker they're at 25 watts.

[HotBluePlates]

... correct answer to the wrong question. You seem to be getting INPUT power and thinking it is Dissipation. ...

I had to look up input power (power consumption?), from what I read it's the total power consumed by the entire circuit?

... got: 40w = 13.1 watts, 30w = 14.3 watts    Lower dissipation for clipping like PRR predicted earlier. But I'm still a bit confused

pdf64 is saying this a different way, so here's another try:

Make note of the numbers you're seeing at the output tube plate.  But now also use Spice to measure the power across the "Speaker."

Power Input - Power Output (across the speaker) = Tube Dissipation

Earlier, you were stuck measuring values that described the Power-Input from the power supply, through the output transformer primary, to the output tube plate.  But there is an audio component that is being transferred from the OT primary to the secondary, and out to the speaker.  That transferred-power does not heat the output tube, because it has been transferred away to the speaker.

So only the remainder (of Power Input - Power Output, minus an additional small loss in the transformer) is tube plate dissipation.

This is why Class A amps dissipate the most power in their tubes when idling, and are actually cooler when played loud.  For Class AB, the maximum tube dissipation happens somewhere between minimum and maximum power output (but not at either extreme).

[DuaneOh]

pdf64 is saying this a different way, so here's another try:

Make note of the numbers you're seeing at the output tube plate.  But now also use Spice to measure the power across the "Speaker."

Power Input - Power Output (across the speaker) = Tube Dissipation

So I measured again and did some math:
(Va - Vk) x Ia - (speaker @ 30w) - (W @ OT / 2) - (W @ Rk / 2) = 11.9 watts plate dissipation
Comes out close to the 14.6w @ 30w that LTspice gives for the total dissipation for the EL34 which includes g2 watts. Does this look correct?

Earlier, you were stuck measuring values that described the Power-Input from the power supply, through the output transformer primary, to the output tube plate.  But there is an audio component that is being transferred from the OT primary to the secondary, and out to the speaker.  That transferred-power does not heat the output tube, because it has been transferred away to the speaker.

So only the remainder (of Power Input - Power Output, minus an additional small loss in the transformer) is tube plate dissipation.

This is why Class A amps dissipate the most power in their tubes when idling, and are actually cooler when played loud.  For Class AB, the maximum tube dissipation happens somewhere between minimum and maximum power output (but not at either extreme).

The A-Ha moment for me from Aiken's web page: "Actually, a full-power square wave results in the least plate dissipation. This is because the plate voltage and plate current are out of phase, i.e., when the plate voltage is at a maximum, the plate current is at a minimum." It's so obvious looking at the traces.    After reading the linked Aiken article, it seems like "Power Input - Power Output = Tube Dissipation" is more of a rule of thumb than a precise measurement?

I think that this has put me on the right track, thanks to everyone for your help!

[PRR]

> it seems like "Power Input - Power Output = Tube Dissipation" is more of a rule of thumb than a precise measurement?

No. It is exact and self-enforcing. There is no free power. All power can be accounted. But it can be tough with side-paths like screen current and non-trivial waveforms like not-sine not-square roundy-cornered.