Positive grids on 12ax7's do not amplify
,
the signal is rectified.
Neither statement should be understood as either cause or consequence of the other.
Is the signal rectified?
Well, what do you mean by rectified?
1) "the current trough the grid to cathode diode is higher when forward biased than when reverse biased", like on *any* diode in the universe. <-- agree, nothing new here.
2) "under those conditions (forward bias, "positive grid" to comply with this post's obsession) the signal dies/hits_a_brickwall/kaputt, etc. (take your pick) meaning it dissappears or takes a fixed DC value which means it's no "signal" anymore." <-- do *NOT* agree, signal simply is attenuated (very different from dissappearing as such).
Why is it attenuated?:
it comes from a relatively high impedance generator, driving a grid to cathode diode, which has variable impedance: very high while reverse biased and much lower when forward biased.
Does "much lower" mean "0"?
Not at all.
The grid will continue to take positive values which will be faithfully amplified by the next tube.
Will said signal hit a brickwall some day?
Yes, but it's not the positive grid's fault.
Simply the earlier stage (which, by the way, is negative biased) , will either hit its own rail, or run out of current, said current being a function of the plate resistor and +B value.
If , to throw some numbers in, the driving stage can furnish 1mA and the grid to cathode diode has an internal impedance of, say , 1K at 1 mA, 1 Volt will develop across it and will be duly amplified.
Just like that.
So, analysing the first half of sentence, it's *wrong*.
Positive grids on 12AX7 *do* amplify, as predicted by physics and confirmed by datasheets and actual experiments.
If the grid gets 1V positive, it doesn't care where it came from, it just does what grids do.
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