OT Plate Load

[Wilder Amplification]

OK, I think I got the power output formula wrong, but I've posted the right one before.

It's basically P=IV or P=V^2/R. But because you're dealing with a sine wave, and V and I are peak values, you have to multiply each one by 0.707 to get the RMS. That's the same as multiplying the product VI by 0.5.

I've never seen peak cathode current ratings for most tubes either. The DC cathode current rating is understood to be measured by a moving-coil meter or DMM, which gives the average.

To a certain extent the peak current is under your control as the designer. You choose the screen voltage to limit it to a sensible value, which is what your dual rail philosophy is all about, not letting any more electrons out to play than you need.

Gotcha...OK so basically what I'm finding (and I probably should've already known this) is that amplifiers are rated for AVERAGE output power (not the typical "RMS Power" that everyone thinks which I'm starting to see means absolutely nothing), and the formula for such is -

Pavg = Vrms x Irms

Pavg = Vrms^2 / Z

So if I took my plate voltage of 600V and place that across my plate-center tap load of 1250 ohms -

600V / 1250 = 480mA

480mA is my PEAK current. I would then multiply that by 0.707 to get Irms -

480mA x 0.707 = 340mA (rounded up)

600V is my peak so I would multiply that by 0.707 to get Vrms -

600 x 0.707 = 424Vrms

Then multiply the two to get Pavg -

424Vrms x 0.34 (340mA)rms = 144 Watts Average

With 144 Watts average across the primary load this would theoretically give me the 70% of just over 100 watts average out.

Is this correct?

[Steve Conner]

What you call average power there, is the same as what the industry calls "RMS" power. That phrase is technically meaningless, and it really should be called average power, or just power. But it gets called "RMS" because you calculate it by multiplying Vrms and Irms together.

Unlike PMPO where you multiply Vpeak, Ipeak, and, say, the national debt of Iceland, to arrive at a stupendous marketing figure.

The 144W is what my method predicts, but it ends up nearer 100 because it gets nibbled away by imperfections that my method doesn't take into account. Tube drop, losses in the OT, and so on.

[Wilder Amplification]

What you call average power there, is the same as what the industry calls "RMS" power. That phrase is technically meaningless, and it really should be called average power, or just power. But it gets called "RMS" because you calculate it by multiplying Vrms and Irms together.

Unlike PMPO where you multiply Vpeak, Ipeak, and, say, the national debt of Iceland, to arrive at a stupendous marketing figure.

The 144W is what my method predicts, but it ends up nearer 100 because it gets nibbled away by imperfections that my method doesn't take into account. Tube drop, losses in the OT, and so on.

Which pretty much means that the supply I'm using for this thing has no chance of sagging even it if wanted to. I spec'ed the HT winding at 500mA and it looks like peak current falls right below that.

This should be one stiff bitch...makes me wonder what it will be like to play through.

[Joey Voltage]

This should be one stiff bitch...makes me wonder what it will be like to play through.

Probably not very fun to be honest, as you now don't have any of that elastic, somewhat forgiving nature on the pick attack that most people desire for lead playing. but who knows... It's subjective anyway

[tubeswell]

Except insofar as the maximum theoretical efficiency for a Class-B amp, with unclipped sinewave output, negligible tube drop and negligible idle current, is 70%. This figure is derived from the mathematical properties of a sine wave.

Yeah I did wonder about chucking a .707 in to it somehow

I've previously posted the formula I like to use to estimate load impedance:

Impedance is roughly equal to the HT voltage you have available, times 4, then divided by the maximum peak cathode current rating of your tubes.

How do you get the peak cathode current rating? Is it the same thing as maximum tube current (for that amp)? Or is it something else? I looked on the Tung Sol 6V6GT datasheet and saw a 'maximum peak cathode current rating' of 105mA given for "Vertical Deflection Amplifier - Triode Connection AB". Is that the right peak cathode current rating? Please forgive my stupendous ignorance. Cheers

[imaradiostar]

How do you get the peak cathode current rating?

Isn't that the point just before the tube welds a screen to another element and starts taking other components with it?

I'd wager there is no cut and dry way to know what a given tube is capable of because so much of what we have today bares little resemblance to the tubes available when the datasheets were made. EH and JJ 6v6's, JJ and EH 7591's and pretty much any russian 6L6/5881 come to mind. Sure, they sound great but they're not really what they had back in the day when guys in white coats generated curves on graph paper to help engineers design the next Bogen PA amp.

Someone else here once told me that the datasheets are ideal scenarios for happy usage in your parent's TV set or car radio. I think all of those conditions get tossed out the window when we put them in a guitar amp! No wonder guitar amps eat tubes so quickly- there are no mentions of safe operating conditions for "class AB exceed dissipation by 50% operation." I think guitar players, not engineers, invented that operating mode.

jamie

[imaradiostar]

So...it's been over a year. I just pulled the trigger on a large Edcor transformer since the order deadline for Christmas is tomorrow and now I'm curious again. How did your split supply turn out, Jon?

jamie