> So the way it's normally meant to operate is that grid current flows
> down through R3, and when the voltage drop exceeds about 0.6V it turns
> on Q2 which trips the protection relay. That would happen at about 200mA
> grid current, which makes pretty good sense (200mA x 3.3 ohms = about
> 0.6V).
That is correct Ian.
> Note that no grid current flows through Q2, and Q2 does not "interrupt"
> grid current. It simply operates a relay which switches the amp to
> standby.
Right again. In essence, Q2 "samples" the emitter current on the darlington
pair.
> What happens to Q2 in a major surge depends on the other things
> connected to R20. If the circuit is at all conventional, R20 is the grid
> current metering resistor, so there must be a meter connected across it,
Yes. I failed to mention that the R20 sample point also routes over to the
multi-meter circuit for grid I monitoring.
> and hopefully also a protection diode (anode to ground). Then the
> B-minus rail and anode current meter connect to the top of R3.
Could be...I didn't look this far.
> Q2 *will* survive OK if the base
> resistor is increased from 100 ohms to say 10k (and maybe change Q2 for
> a device with higher current gain... I don't know the 2N5184).
I agree...perhaps going from 100-ohms to 10K in order to increase input
isolation and changing Q2 to a darlington-pair may be added
insurance.
-Paul, W9AC
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