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A lot more than that.

IMO best practice is to assume the effective Zin of an A2 triode grid is around 1k. If you use this value then your 1.5 mA supply would fold once the A2 voltage on the grid got to around +3 volts each cycle, which is "why bother" territory. If you are going to swing the grid strongly positive you need to sink a fair bit of current.

You can calculate grid currently directly if you know a single parameter for the V/T, the proportionality constant for grid current, which is in turn dependent on the ratio of the gap between the anode and grid to that between the grid and cathode.

The calculation is then Ia/Ig = k.SQRT(Va/Vg), k is the proportionality constant mentioned above. The equation only holds if Vg < Va, but operating the grid positive to the anode is not recommended.

There's a calculation in Beck somewhere, IIRC values around 5 are common.

BTW Paul is right (as usual), once you overpower the bias supply the thing will self correct by sliding towards class C and sound completely terrible.




Mark Kelly



Edits: 09/08/14

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  • A lot more than that. - Mark Kelly 05:28:37 09/08/14 (0)

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