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Old 19th November 2010, 12:26 AM   #1
ChrisA is offline ChrisA  United States
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Default How to compute current at control grid of power tube?

This should be an easy question, almost embarrassing to ask.

Let's say I have a power tube, Use a "standard" KT88 push pull audio amp as the example. I know what voltage I need to apply to the grids. It's in the graphs on the data sheet.

What I don't know is how to compute the current that must be supplied by the driver tube. What's in input impedance at the control grid? (I'm asking about the general case not for any one amp)

From experience and looking at various amp schematics I know what might work and I have Spice to simulate it, but that is not the same as knowing how to compute the values myself.
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Old 19th November 2010, 12:36 AM   #2
stoc005 is offline stoc005  United States
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Tube grids are voltage driven. They are very, very high impedance and a bit (small) of capacitance. Anything from 100K to 1 Meg will work. Greater than that , approaching 10-22Meg, you get into grid leak bias, which probably isn't what you want. All this changes if you drive the grid positive wrt the cathode and draw grid current.
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Old 19th November 2010, 12:48 AM   #3
ChrisA is offline ChrisA  United States
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Originally Posted by stoc005 View Post
...Anything from 100K to 1 Meg will work. Greater than that , approaching 10-22Meg, you get into grid leak bias...
This sounds like to are saying how to select a grid leak resistor.

Maybe I was not clear. What I don't understand is how much current goes into the tube. It would seem to me that a tube like a 6L6 or KT88's control grid would have almost infinite input impedance. But if that were true then I could drive a dozen power tubes with a single 12AX7. I know that can't be done. I'm needing to know how to compute the current that a driver tube must supply.
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Old 19th November 2010, 01:07 AM   #4
Ian444 is offline Ian444  Australia
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I think the limiting factor is the grid leak resistor.

If you were driving 10 x 6L6, each with a 500K grid resistor, the load on the driving tube would be around 50K.
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Old 19th November 2010, 01:10 AM   #5
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look here - a nice rundown (bit overly techo, but thorough!)
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Old 19th November 2010, 01:14 AM   #6
SY is offline SY  United States
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It depends on the input capacitance and the desired bandwidth. In the case of pentodes, which have very low input C, you can indeed drive a whole pile of them off a single small tube.

Calculate input capacitance. Choose an f3. Then you can get the required current. Let's say you have 20pF input capacitance and you want the f3 at 20kHz at full power. The tube is biased to -40V. We're running class A or AB1. f3 = 1/2piRC. R = V/I. So, with a little algebra, I = 2piVCf3 = (6.3)(40)(20E-12)(20,000) = about 0.1mA.
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Old 19th November 2010, 01:36 AM   #7
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Usually datasheets show grid currents when driven positive. They depend on the tube, and are different, you can't calculate them.
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Old 19th November 2010, 01:52 AM   #8
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RCA datasheets are good about showing grid current curves. Of course, some tubes were not designed for operation in positive grid region and they don't have grid current curves in the datasheet.
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Old 19th November 2010, 02:29 AM   #9
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Quote:
Originally Posted by ChrisA View Post
But if that were true then I could drive a dozen power tubes with a single 12AX7.
Nahh, of course you can.

You will loose the top edge of frequency response (which will be worse in triode mode), and the required grid leak resistor will cause significant loading, reducing gain and increasing distortion.

Most audiophiles choose a ridiculously powerful driver, because it "feels right". This results in bandwidth on the order of 200kHz, which means the output transformer and feedback network will dominate the overall response. That's better than having the phase shift of repeated poles.

Back in the day, 6L6s were regularly driven from 12AX7s, since it works for the specified distortion and frequency response (maybe 5% at full power, and ~20kHz upper limit, -3dB).

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Old 19th November 2010, 05:37 AM   #10
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Quote:
Originally Posted by ChrisA View Post
This should be an easy question, almost embarrassing to ask.

Let's say I have a power tube, Use a "standard" KT88 push pull audio amp as the example. I know what voltage I need to apply to the grids. It's in the graphs on the data sheet.

What I don't know is how to compute the current that must be supplied by the driver tube. What's in input impedance at the control grid? (I'm asking about the general case not for any one amp)

From experience and looking at various amp schematics I know what might work and I have Spice to simulate it, but that is not the same as knowing how to compute the values myself.
Hello,
I have trying to wrap my brain around similar thoughts the last several days. My thoughts have included grid current among other things perhaps unimportant detours. This is about getting the driver output into the power tube as cleanly as possible. What things need to be overcome to get there; the time constant of the coupling capacitor and drain resistor, the current flow into the drain resistor, the grid stopper resistor, the Miller capacitance of the power tube, and the current of the power tube grid. For A1 operation grid current is the least of our worries, it is all about voltage. For A2 or positive grid voltages that is where grid current becomes an important consideration.
Without an accurate mathematical model it is not possible to calculate A2 grid currents. Perhaps the best we have is empirical data. As noted in other posts that data for some tubes is on the tube data sheet.
SPICE models. If you want to make a project of modeling perhaps you can tweak the tube models to have improved performance in the low impedance positive grid voltage range. Even then it will be trying to get the model to match the empirical data.
Also look up tubelab.com PowerDrive. George has driven tubes beyond where tube data sheet has gone before.
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