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Old 18th November 2008, 02:50 PM   #1
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Default a question about drive power

Hi there

First of all, I am clueless when it comes to tube audio amps, so bear with me.

I plan to use a simple triode as a pulse modulator in a project I'm doing, and since people here discuss tubes, I think this is a good place to ask something.

One thing I don't understand is driving power. I assume it is the product of grid voltage and grid current in any moment in time (if we are talking about DC power)
But if there are no grid "leaks", shouldn't there be no current on the grid?
In other words, isn't it true that a perfect triode has zero driving power in its entire operating range of grid voltages?

And is there a way to reduce this grid current to zero by any means outside the tube itself.
And if yes, how come it's not done always (is there a reason to desire grid current?)

thank you
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Old 18th November 2008, 06:16 PM   #2
m6tt is offline m6tt  United States
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Grid current will not really flow until the grid is trying to be driven + in relation to the cathode (with most common audio tubes)...when this happens, power is consumed in the grid circuit. This is done in A2, AB2, sometimes B and C types of operation. Often tubes designed for this will have a radiating fin welded to the grid wire supports at the top of the tube to help dissipate the power. Frequently this is used to obtain more output power, but generally requires the design to provide a very low-z output to the tube, such as a transformer coupling, a beefy cathode follower...etc...(hmm.. LM3886 driver anyone? ). This is because the grid circuit resistance starts to present a much lower input Z as the grid is driven more positive. Grid current is often limited, at least in power tubes, by series resistors to the grid which also use the miller cap. to limit the frequency response of the stage. Some tubes like 6N7 require grid current to get anywhere near their design rated max output power.
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Old 18th November 2008, 07:26 PM   #3
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Default Re: a question about drive power

Quote:
Originally posted by biasedED
One thing I don't understand is driving power. I assume it is the product of grid voltage and grid current in any moment in time (if we are talking about DC power)
But if there are no grid "leaks", shouldn't there be no current on the grid?
In other words, isn't it true that a perfect triode has zero driving power in its entire operating range of grid voltages?
If there is no grid leak, there will be no grid current, and there will be no plate current either. A floating grid will attract electrons until it builds up enough static charge to drive the tube into cutoff. Not very useful.

All VTs have a parasitic diode formed from the control grid and cathode. If you have an input signal that takes Vgk positive, that diode turns on and conducts. That costs you input power, which if not supplied, simply results in grid clipping and distortion since there won't be any more plate current. Going into Class *2 can get you more output, and with some types (1624, 814) is really the only way to get significant output for the operating voltage. Only way to get rid of the grid current is to use a MOSFET instead.

Quote:
And is there a way to reduce this grid current to zero by any means outside the tube itself.
And if yes, how come it's not done always (is there a reason to desire grid current?)
No, if they coulda they woulda. The one case where grid current is desireable is with Class C RF amps. These frequently derive some, if not all, their operating bias from clipping off the positive peaks and integrating the negative peaks into a steady DC bias voltage.
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