Maximum wattage, 2 6V6 in PP, cathode bias

[jhawk]

How many watts can you get from a 2x6V6 push-pull cathode biased output section, assuming your PT and OT aren't limiting factors (plenty of current), and your're throwing +350VDC at the plates of the 6V6s (which should put the bias at about 100% plate dissipation)?

 

[robrob]

With 350v plates you're going to be around 13 or 14 watts. You might be able to get to 16 watts with 420v on the plates.

 

[Old Tele man]

With 350v plates you're going to be around 13 or 14 watts. You might be able to get to 16 watts with 420v on the plates.

Agreed, although the 6V6 data sheet claims 14W with only 285Vp and 285Vs, but that's with fixed bias and using 8KΩ plate-to-plate OT, not 6KΩ ala' Fender Deluxe, etc.

https://frank.pocnet.net/sheets/127/6/6V6.pdf

 

[peteb]

What's the maximum power out for a pair of fixed bias 6V6s in Push pull?

 

[Dacious]

What's the maximum power out for a pair of fixed bias 6V6s in Push pull?

Depends how long you want them to live and how attenuated the bass.

Bad Cat claims 30. 20 is probably reasonable.

 

[printer2]

Depends on how much crossover distortion you want. More cooler bias,more crossover distortion, more power.

 

[Bendyha]

What's the maximum power out for a pair of fixed bias 6V6s in Push pull?

Jim Kelley amps workout at 60W for a quad of 6V6, TUT5 from Kevin O'Conner ( I spell his name wrong in the same way that he spelt Jim Kelley's name wrong) gives the math for it if you are interested. So if your voltages are right, and the OT matched as Old Tel man points out, then 30W is possible with a pair of 6V6 (good ones), and is said to sound very good.

Edited the wrong number out.

 

[printer2]

I would really like to see how 35W is achieved. I would think the amp manufacturers would be all over it.

 

[Bendyha]

Ah....my mistake...not 70W...memory ain't what it was...I think... 60W is the correct rating for a Kelley....so 30W for a pair. I've edited the mistake above.

 

[peteb]

Jhawk, Did not mean to sidetrack your thread at all, but what I am thinking is that the starting place is understanding the max power out from fixed bias and then the refigure it for cathode bias.


complicating factors: true power out is clean power by some standard. true power out is from the speaker.


How would one quantify the power produced by the power tube? I thought I had read it but cant remember. one would think the voltage of the signal times the (idle) current flowing thru the power tubes would be power out of the tube but this is wrong as the number would be way too small.




some ideas - probably wrong:


the plate voltage (much larger than the signal voltage) times the idle current could be considered the power input to the tube, and is also considered to be the idle plate dissipation. the max plate dissipation for a tube type is roughly equal to the max power out for the amp. example: the max plate dissipation for a 6v6 is 14 W per tube, and a pair can output 28-30 watts.
but the power out of the tube is based on the AC not the DC. as said above, if you multiply the AC signal by the idle current the result will be much less than the max plate dissipation or max power out.


Something doesn't make sense?

 

[Old Tele man]

1) First, let's start with the input drive signal (Vg(ac)) and estimate the output power:

Po = Zo*(gm*Vg(ac))^2
Po = (Zoo/4)*(gm*Vg(ac))^2
Po = (8KΩ/4)*(0.003gm*35Vg)^2 = 22.05 ≈ 22W (NOTE: this is ONLY a rough estimate!)

where:
Po = Power output in W(avg)
Zoo = OT plate-to-plate, 8KΩ
Zo = OT effective, (Zoo/4)
gm = tube transconductance, 0.003A/V
Vg = AC-voltage drive signal, 35Vg

Of course, this all assumes there is sufficient plate (Vp) and screen voltage (Vs) to provide the assumed peak plate current of about 100mA(*) across the 2KΩ effective load line.

2) Now, you'd turn to the Eb-Ib plate curves (which don't exist for Vp = 420Vdc) and draw in the representative lines. At this point -- as my professor would declare -- "We leave the solution as an exercise for the student to solve."


(*) 100mA ≈ 0.003A/V*35Vg = 105mA

 

[jhawk]

Lots of great replies! Thank you.

So I'm trying to read as much as possible to gain a better understanding of how circuits function. Still struggling with lots of stuff, but the picture is slowly getting clearer!

I've been piecing together a circuit that straddles a Princeton Reverb and a 5E3 Deluxe. I noticed a lot of power transformers provide way more voltage than necessary and not enough current. So I've been looking at just spec-ing my own transformers (from Edcor, most likely).

This got me wondering if you could get more watts from a 5E3 output section. It seems the transformers typically limit the output. Could you upgrade the transformers to get more output? A Deluxe Reverb is rated at 22ish watts and just has bigger iron and a LTPI, right? Or does a DR obtain this power rating mostly by running it's 6V6s hot? Would more current allow a 5E3 to have more output?

I played a PR at a local shop a couple days ago and it really has less headroom than I remember. Bass and treble maxed, vol at 5 and there was pretty good grind going on with just a Tele. I like the idea of a 5E3 output section (pretty similar to a PR), but I'd like just a little more headroom without running my tubes real hot.

 

[jhawk]

Also, the Milkman Creamer claims to be 18-22W with 2x6V6. False? It's cathode biased with a cathodyne PI (as far as I know).

 

[Old Tele man]

A cathodyne PI means ALL the small-signal voltage gain has to come BEFORE the PI (meaning tone controls between two tubes).

Cathode-bias will net slightly LESS power output than will fixed-bias, but the "playing" responses WILL be different: C-bias compresses & expands with volume changes (depending if Rk is by-passed or not and by how much capacitance); while, F-bias is clear & crisp and cleaner.

 

[peteb]

thanks old tele man!

1) First, let's start with the input drive signal (Vg(ac)) and estimate the output power:

Po = Zo*(gm*Vg(ac))^2
Po = (Zoo/4)*(gm*Vg(ac))^2
Po = (8KΩ/4)*(0.003gm*35Vg)^2 = 22.05 ≈ 22W (NOTE: this is ONLY a rough estimate!)

where:
Po = Power output in W(avg)
Zoo = OT plate-to-plate, 8KΩ
Zo = OT effective, (Zoo/4)
gm = tube transconductance, 0.003A/V
Vg = AC-voltage drive signal, 35Vg

Of course, this all assumes there is sufficient plate (Vp) and screen voltage (Vs) to provide the assumed peak plate current of about 100mA(*) across the 2KΩ effective load line.

2) Now, you'd turn to the Eb-Ib plate curves (which don't exist for Vp = 420Vdc) and draw in the representative lines. At this point -- as my professor would declare -- "We leave the solution as an exercise for the student to solve."


(*) 100mA ≈ 0.003A/V*35Vg = 105mA

Is this a form of: power = (V^2)/R or (V^2)Z

from combining Power = volts X current and ohms law.


I could be over simplifying it, but is V the signal and Z is the impedance of the tube?

 

[Old Tele man]

thanks old tele man!


Is this a form of: power = (V^2)/R or (V^2)/Z . . . .( you omitted the "/" )

from combining Power = volts X current and ohms law.

I could be over simplifying it, but is V the signal and Z is the impedance of the tube?

Neither, it's a form of: power = R*(I)^2 ...where Zo subs for " R " and gm*Vg subs for " I ".

Zoo is the OT plate-to-plate impedance and Zo ≈ (Zoo/4) is the effective plate impedance seen by each tube individually.

However, you are CORRECT in assuming that power = (V^2)/R can be used to estimate the minimum plate voltage (B+) needed from the PT:

1) estimate Vp(dc) voltage drop at idle: 5Vp(dc) ≈ 35mA*150Ω
2) estimate Vp(ac) from Po and Zo: Vp(ac) = SQRT[22W*2KΩ] = 210Vp(avg)
3) estimate Vp(pk): Vp(pk) = Vp(avg)*SQT[2] = 210Vp(avg)*1.414 = 297Vp(pk)
4) estimate "diode line" Vp(min) for tube: Vp(min) > 50Vp(min)...but 2X is preferred.
5) add the voltages: B+ ≥ 100V+297+5V = 402Vdc

Thus, the PT needs to produce at least 352Vdc and preferably 402Vdc (B+) to sustain that 22W output.


P.S. - when C-bias is used, then also add that cathode-bias voltage (Vk) to above numbers too.

 

[peteb]

thank you

 

[jhawk]

Cathode-bias will net slightly LESS power output than will fixed-bias

Needed to hear this. I'm not sure why I had it in my head that there was much of an output difference between the two schemes.

Agreed, although the 6V6 data sheet claims 14W with only 285Vp and 285Vs, but that's with fixed bias and using 8KΩ plate-to-plate OT, not 6KΩ ala' Fender Deluxe, etc.

So what will happen if I use the DR output transformer with the 6KΩ input impedance?

With 350v plates you're going to be around 13 or 14 watts. You might be able to get to 16 watts with 420v on the plates.

I've read several places that 350v is supposed give you 100% dissipation at idle and 70% with signal. Isn't 70% supposedly optimal? Please set me straight!

Depends on how much crossover distortion you want. More cooler bias,more crossover distortion, more power.

So if I bias too cold, I get crossover distortion. If I bias too hot, I get red-plating?

 

[Old Tele man]

With C-bias, the voltage on the cathode actually 'subtracts' from the voltage available on the plate, because the bottom of the cathode resistor is referenced/anchored at GROUND. A little less Vp means a little less plate voltage "swing" and thus a little less power output.

If, however, the B+ is increased exactly the same as the Vk voltage, then there would be no loss in power output, but there certainly would be a drop in power utilization efficiency (Po/Pi) because now MORE B+ had to be used to create the same power output...thus, efficiency went down ↓ .

Go back to the first equation, Po ≈ (Zoo/4)*(gm*Vg)^2, and plug-in 6KΩ and you'll see that reducing the OT from 8KΩ to 6KΩ will cut power to about 16-17W...a 25% reduction.

The closer the power tubes are biased to Class-B operation, the more "cross-over" distortion becomes a problem.

 

[printer2]

So if I bias too cold, I get crossover distortion. If I bias too hot, I get red-plating?

Yes, yes and no. If you bias it cold you are running in Class B (one device on at a time), a little warmer and you are in Class AB (no signal and low levels both output devices are on at the same time while at larger signals the nonconducting device is off part of the time). Now for example the 18 Watt amp can have an amount of crossover distortion, it gives the amp some of its grit. Bias it warmer and it smooths out, some prefer a little more edgy sound. Now you can bias the amp fully on where both tubes are on all the time (Class A). If the amp is designed to get the most out of the tubes you probably will get red plating.

But if the output tubes are not dissipating above their limits with either the amp limited in voltage or current (or high impedance but that is another kettle of fish) then cranking the bias will only add more heating. I built an amp where I adjusted the bias to -1V, not much room for a signal but the tubes were dissipating many times more watts than the signal getting through. The amp had a VVR voltage reducer and had about 50V on the plates. I guess you could say the bias was not too hot as the tubes could not red plate under those conditions. Just one of those 'yeah but... ' conditions.