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19th November 2010, 01:26 AM  #1 
diyAudio Member
Join Date: Jan 2008

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. 
19th November 2010, 01:36 AM  #2 
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Join Date: Feb 2007
Location: Midwest in the USA

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 1022Meg, 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.

19th November 2010, 01:48 AM  #3  
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Join Date: Jan 2008

Quote:
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. 

19th November 2010, 02:07 AM  #4 
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Join Date: Nov 2008
Location: Brisbane QLD

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. 
19th November 2010, 02:10 AM  #5 
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Join Date: Mar 2007
Location: Auckland, NZ

look here  a nice rundown (bit overly techo, but thorough!)
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19th November 2010, 02:14 AM  #6 
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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)(20E12)(20,000) = about 0.1mA.
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19th November 2010, 02:36 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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19th November 2010, 02:52 AM  #8 
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Join Date: Nov 2007

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.

19th November 2010, 03:29 AM  #9  
diyAudio Member

Quote:
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). Tim 

19th November 2010, 06:37 AM  #10  
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Join Date: Aug 2009
Location: Sacramento

Quote:
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. DT All Just for fun! 

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