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Old 25th April 2013, 11:47 PM   #1
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Default calculating signal current for power dissipation

Hello everyone,
and thanks for all your help so far.

I have been having trouble finding concrete information on calculating singal current in a tube amp. I read in several places that in general it is very small, and I don't see resistors in the signal path being rated at 1w or more in various schematics.

However, I suspect that my application may involve enough current to fry an LDR or two. There are some Silonex photocells that can dissipate 200mw each (TO-5 series) and 500mw each (TO-8 series), but that's at 25 degrees C and I'm thinking it may well get to be 50 degrees C in that amp. So I'm derating this (linearly to 75 degrees C) according to spec to 100mw and 500mw. I need to vary the resistance between 50k and 330k.

The resistor goes in a local feedback circuit, out the anode of a 12au7 and back in the grid. There is a capacitor keeping DC out of the LDR in question. The anode voltage is about 12k and thus is drawing nearly 5ma. I am not clear how this relates to the signal current, or if it is even relevant.

How would I go about calculating exactly what the signal current passing through this local feedback loop is? Or, for that matter, just in general, how does one calculate the signal current leaving a valve preamp stage?

Thanks again,
Ted
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Old 26th April 2013, 04:24 AM   #2
DUG is offline DUG  Canada
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V * V / R = P

V = volts rms
R = resistance in ohms
P = power in watts

If that is not the answer, then I don't understand the question.
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Old 26th April 2013, 05:42 AM   #3
Enzo is offline Enzo  United States
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Quote:
The anode voltage is about 12k
Either I am dense tonight or this is a typo? Plate voltage on your 12AU7 is 12,000 volts? Or the plate RESISTOR is 12k ohms?


In any case, you want to know signal? What is the signal voltage at the one end of the part, and what is it at the other end? The difference is the voltage across the part. Consider the highest and lowest resistances in the cell, and plug them into Ohm's Law with the voltages as DUG suggested, and find the highest and lowest currents. Or solve directly for watts if you prefer.
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Old 26th April 2013, 10:26 AM   #4
DF96 is offline DF96  England
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Determine signal voltage (depends on input signal and gain). Use Ohm's Law to find current.

Alternatively, get a worst case estimate by assuming peak signal current equals quiescent DC current, so RMS signal current equals 0.707*DC current.
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Old 27th April 2013, 12:32 AM   #5
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[QUOTE=Enzo;3468695]Either I am dense tonight or this is a typo? Plate voltage on your 12AU7 is 12,000 volts? Or the plate RESISTOR is 12k ohms?[QUOTE]

Typo- and some misinformation as well. I meant 12k ohms, and yes it's a plate resistor. Actually the DC current is more like 24ma.

I can see that it's quite enough to know the signal voltage and the resistance.

"Alternatively, get a worst case estimate by assuming peak signal current equals quiescent DC current, so RMS signal current equals 0.707*DC current."

Thank you!
That's going to be a lot of watts for this little LDR...
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Old 27th April 2013, 03:52 AM   #6
DUG is offline DUG  Canada
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12AU7:

http://www.wooaudio.com/docs/tube_data/12AU7.pdf

24 mA on a 12AU7...seems a little high...5mA sounds better

Have you plotted the load line on the plate characteristics graph? (page 4)

Calculated plate dissipation? (12K resistor dissipation is close to 7W at 24mA)

What is your cathode resistor value?

What is the supply voltage?
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Old 27th April 2013, 02:04 PM   #7
JMFahey is offline JMFahey  Argentina
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[QUOTE=tapehead ted;3469685][QUOTE=Enzo;3468695]Either I am dense tonight or this is a typo? Plate voltage on your 12AU7 is 12,000 volts? Or the plate RESISTOR is 12k ohms?
Quote:

Typo- and some misinformation as well. I meant 12k ohms, and yes it's a plate resistor. Actually the DC current is more like 24ma.

I can see that it's quite enough to know the signal voltage and the resistance.

"Alternatively, get a worst case estimate by assuming peak signal current equals quiescent DC current, so RMS signal current equals 0.707*DC current."

Thank you!
That's going to be a lot of watts for this little LDR...
Can you please post the actual schematic of what you are building?
Otherwise at best you'll get "general purpose" answers such as "use Ohm's Law" , which is technically perfect, of course, but hard to put into numbers for lack of data.

We don't know what's on "each side" of your switching LDR.

Side note: I can't imagine much need for a 12AT7 fed, say, 250 or 300V and at the same time passing 24 mA (or even 12mA) unless it's, say, driving a reverb tank or a Class AB2 power stage , and anyway I would not ground its plate to mute it but its grid, infinitetly less power involved, so please letīs see what you are talking about.
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Old 1st May 2013, 12:39 AM   #8
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Hi all,
sorry again, I'm unable to post a schematic. Here at the library computer, no scanner, and very little time- that's why I'm only here every few days.

Yes, 24ma is off the chart- but I am somewhere around 19 ma maximum at 240 V B+. I have charted this and am using that as an absolute maximum value with 125 VAC as the mains voltage.

"Calculated plate dissipation? (12K resistor dissipation is close to 7W at 24mA)"

I'm right up against the plate dissipation curve somewhere well above vg=0 at my latest charted value of anode resistor =4.7k ohms, at 240v, which again is what I'm using for a worst case maximum.

Sorry for all the scrambled figures- I'm doing this from memory at the library in the few minutes I have after searching the archives. I will try to post more lucid questions in the future!

Thanks all, I probably seem pretty erratic just checking in here and there with obtuse questions, but I am gaining in comprehension all the time and I certain appreciate DUG's byline.

I'm actually astonished that I'm capable of learning this sort of material at all, and it's opening up some pretty thrilling horizons for this musician... it's hard to get excited about store bought amps anymore...
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Old 1st May 2013, 02:16 AM   #9
DUG is offline DUG  Canada
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Running tubes or resistors at max rating shortens life...sometimes very significantly.

Why run above Vg=0 ?

I learned by looking at schematics...I still do.

Look at stuff that works and figure out why...then start tinkering.

Old pilots don't learn from their mistakes...they learn from other peoples mistakes.

Good policy to adapt.
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Old 2nd May 2013, 11:25 PM   #10
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[QUOTE=DUG;3474410]Running tubes or resistors at max rating shortens life...sometimes very significantly.

Why run above Vg=0 ?
QUOTE]

It's a guitar amp- certain valves get pushed way into grid current limiting. I figure I ought to at least consider what life is like for the tubes when that happens.

Thanks for the thoughts- I've been studying some schematics and interestingly enough some of the amps that I really like in serious overdrive were never intended to be overdriven- Ampeg V2 and Leslie amps come to mind- seesaw phase inverters in freak-out mode, power tubes standing up to sing, healthy headroom up front. So I'm thinking that since I know I will be pushing the amps that way, I ought to keep the tubes as happy as possible as they sing.

I'm actually trying to design things much more conservatively than is common practice (at least in guitar amps) as far as the tube operating conditions are concerned. For one thing, I'm not interested in a maximum of clean power, so I have the power tubes well away from the plate dissipation curve and I'm doing my best to make life safe for the screens as well with some healthy sized screen grid stoppers.

However I see a lot of common established practice as "mistakes" when it comes to audio gear. A few exceptional pieces and designers have opened my eyes to what life is like without potentiometers in the signal path, and with a very minimum of phase shift in the audible band, for instance.

I feel I owe some amends to the tubes for the way guitar players have been blowing them up and scorching them all these years- I do plan to take good care of my little friends.
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