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Old 6th May 2007, 08:37 PM   #1
nhuwar is offline nhuwar  United States
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Default Triode strapped tetrode

When running a tetrode in triode strapped configuration deos it has an effect on drive power requirements and how much usally?

You know like percent increase of drive power.

I'm just looking for like a ball park rough kind of approximation because I know it changes with the tube.


Thanks for the insight

Nick
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Old 6th May 2007, 11:32 PM   #2
ilimzn is offline ilimzn  Croatia
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Very difficult to say because it indeed depends very much on the tube in question, but in general, YES the drive requirements are higher, both in voltage and especially in current. Unlike tetrode/pentode mode, in triode mode your driver sees a (sometimes considerable) miller capacitance, which it has to charge/discharge. In order to do this quickly enough, it needs sufficient output current. The fact that the screen grid screens the control grid from the voltage swing of the plate in tetrode/pentode mode, makes the input capacitance of the tube more or less the one you read out from the datasheet, usually Cg1-k + Cg1-g2, Cg1-p being very small. In triode, Cg1-p is more visible (but still relatively small), but the real problem is that the capacitance the driver sees is multiplied by the effective amplification factor.
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Old 7th May 2007, 12:10 AM   #3
nhuwar is offline nhuwar  United States
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Thats what I was kind of figuring.

Just figured I'd ask though

Nick
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Old 7th May 2007, 09:14 AM   #4
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As Ilimzn says, you can expect a significant increase in input capacitance. What this means in practice is that the driver has to provide more current. For example, in the Mullard 5-20 the 12AX7/ECC83 (just) manages to drive the EL34s in ultralinear but would not be so good if the EL34s were triode-strapped.

The maximum voltage swing from the driver probably wouldn't be any greater in triode mode than in UL mode, for the same plate voltage and idle cathode current . The bias would be about the same, because the tube in quiescent state doesn't know that the screen has been strapped to the plate. This means the maxumum drive voltage will also be the same, if you are to avoid driving the control grid positive.

Where difference occurs is when a signal is fed in. Then the maximum power generated by the triode-strapped tube will be only about half that from UL. So, you put the same in but only get 50% out.
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Old 7th May 2007, 11:29 AM   #5
nhuwar is offline nhuwar  United States
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Thanks that I didn't know

Nick
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Old 7th May 2007, 01:08 PM   #6
ilimzn is offline ilimzn  Croatia
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Ray is of course correct, but i can add some details:
The least inpit capacitance and least drive power is used in pentode mode. UL already requires more because G2 is not on a fixed potential. Triode requires the most.
Converting from UL to triode will result in pretty much the same bias and drive voltage requirements, as will conversion from pentode to triode, if the G2 voltage is the same as the plate voltage. In most Ul designs G2 is also at the same DC potential as the plates.
Where things can change, is in the case of a pentode mode amp which has a Vg2 lower than the plate power supply. Assuming these can be triode strapped to begin with (Vg2 limitations must be checked!), they will need to be rebiassed and the drive voltage requirement will usually rise. This is because in such amps Vg2 is usually lower than the plate voltage, so triode strapping will increase it, requiring a greater negative voltage on the grids to get the same plate current at idle.
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