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G0HZU_JMR · 21st Nov 2022, 1:51 am

This all assumes that I understand the circuit correctly, of course...

However, I think it's possible to predict at what DC voltage (at pin 7 of the second AD603) the circuit will start to misbehave. If the base of Q1 is at about 4.6V then for Q1 to conduct weakly there will be maybe 0.5V dropped across its Vbe junction. So the emitter would be at 4.1V DC in this case.

The problems will occur if 250uA is drawn by Q1 in the 'no-signal' condition. 250uA through your 820R resistor R12 drops another 205mV.

So, I think your circuit could have problems if the DC voltage at pin 7 of the second AD603 is close to (or lower) than about 4.6 - 0.5 - 0.205 = 3.895V DC in the 'no-signal' condition.

21st Nov 2022, 1:51 am	#4
G0HZU_JMR Dekatron Join Date: Sep 2010 Location: Cheltenham, Gloucestershire, UK. Posts: 3,077	Re: AD603 IF Amp. This all assumes that I understand the circuit correctly, of course... However, I think it's possible to predict at what DC voltage (at pin 7 of the second AD603) the circuit will start to misbehave. If the base of Q1 is at about 4.6V then for Q1 to conduct weakly there will be maybe 0.5V dropped across its Vbe junction. So the emitter would be at 4.1V DC in this case. The problems will occur if 250uA is drawn by Q1 in the 'no-signal' condition. 250uA through your 820R resistor R12 drops another 205mV. So, I think your circuit could have problems if the DC voltage at pin 7 of the second AD603 is close to (or lower) than about 4.6 - 0.5 - 0.205 = 3.895V DC in the 'no-signal' condition. __________________ Regards, Jeremy G0HZU