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Old 27th October 2010, 01:08 PM   #1
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Default Power triode drive current requirements

My big thing is direct coupled SE triodes. I currently have a DC Darling, with a mu follower type 6SN7 input running at 9.6 mA.

I've been designing a full DHT amp, with a 26 driving a 45, to see what all that DHT talk is about. The 26 can push a good 6-8 mA of current, which I presume should be fine for the 45. However I recently found a pretty good deal on some NOS 30 triodes: NJ7P Tube Database Search

It looks really tempting for a couple of reasons; 1) it would be a cheap way to get my feet wet on the DHT front 2) with that low filament current I could use some serious chokes to get it properly clean.

However, I began to wonder about the current drive capabilities of the 30. Is 3.1 mA really sufficient to drive a 45? What about the 1626? How much current is enough? Is there a way to calculate it from some parameters?
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Old 27th October 2010, 04:41 PM   #2
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You don't drive a class A amplifier with current. Well, at the high frequency end one does need to drive the load capacitance. Calculate the current needed for whatever high end you desire and make sure the tube draws at least that much.

What is DHT? Why do people insist on using acronyms to make themselves seem knowledgeable?

I am an experienced engineer and have, I think, a lot to offer technically. But I am not 'hip' to some of this jargon I read.
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Old 27th October 2010, 04:59 PM   #3
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DHT means Directly Heated Triode.

Also, very useful for this thread is acronym PIM.
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Old 27th October 2010, 11:40 PM   #4
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bob91343:

I have found that the driver stage's current pushing capability greatly affects a power triodes performance - if you don't have enough current, you loose dynamics and body.

"to drive the load capacitance. Calculate the current needed for whatever high end you desire"

How?

PIM? A personal information manager (often referred to as a PIM tool or, more simply, a PIM)?
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Old 27th October 2010, 11:43 PM   #5
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Also, PIM means Phase Inter-Modulation. In tube amps it is often dominated by a weak driver (like 12AX7 with 100K load) driving power triode with huge Miller capacitance.
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Old 28th October 2010, 08:31 AM   #6
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Quote:
Originally Posted by MrCurwen View Post
How?
Calculate the Miller capacitance of the triode, add a few pF for wiring stray capacitance, and then calculate the reactance at the highest frequency you wish to reproduce. This combined with the max output swing of the driver gives the max current that is needed.
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Old 28th October 2010, 10:25 AM   #7
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Quote:
You don't drive a class A amplifier with current.
Although not often done, there is no reason why a Class A output cannot in principle be used in the grid current regime. You would need a good driver, of course. The fact that grid current is usually associated with Class B (or AB1, AB2) does not mean that Class A cannot use grid current. The Class of an amplifier is defined by the output conduction as a proportion of the total cycle, not the input conduction. Otherwise, you would not be able to do Class A with BJTs!
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Old 28th October 2010, 11:01 AM   #8
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Eimac 100TH being a good example of a candidate for SE class A2 operation.
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Old 28th October 2010, 11:49 AM   #9
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Quote: The fact that grid current is usually associated with Class B (or AB1, AB2) does not mean that Class A cannot use grid current.

Wrong. Grid current is associated with class A2 and B2.
When, in a PP stage, one of the active elements is temporarily cut off during a full cycle, we have class B operation.

Pieter
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Old 28th October 2010, 11:55 AM   #10
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To be more precise: when one of the devices is always cut off during half a cycle, we have pure class B.
Temporarily cut off depends on biasing; we have class AB then.

Pieter
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