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In Reply to: RE: grid resistors first stage of amp posted by Tre' on June 25, 2012 at 14:34:57:
only in a terrible design. (or maybe a pentode driver)
If you had a 100k source impedance (without the grid resistor in place) and you used a 100K grid resistor.....the grid resistor being in parallel with the source....your drive impedance would be 50K and your voltage would be -6db.
If you had a 100k source impedance and you used a 10K resistor, your drive impedance would be 100k//10k=9.09k and you would have 1/10th of your drive voltage.
In either case the load line of the driver stage would be vertical (not good for triodes).
To my way of thinking the source impedance should be low enough to drive the Miller capacitance before the grid resistor value is considered. At that point the value of the grid resistor won't matter (except the lower it is, the more vertical the load line for the driver stage).
In any event, it's the source impedance that has to be low enough to fully drive the Miller and the grid resistor value is just a small subset of that.
I hope that is clear.
Tre'
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Follow Ups
- It dawns on me that the grid resistor does lower the source impedance but... - Tre' 14:55:16 06/25/12 (5)
- A triode driver example - Tre' 15:30:26 06/25/12 (4)
- RE: A triode driver example - Triode_Kingdom 16:26:14 06/26/12 (3)
- RE: A triode driver example - Tre' 20:01:05 06/26/12 (2)
- RE: A triode driver example - hennfarm 21:14:01 06/27/12 (1)
- RE: A triode driver example - Tre' 09:13:50 06/28/12 (0)
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