Higher AM Detector

DC analysis:

Q1 and Q2 form a current mirror at DC.  Given a 6V supply voltage and a 0.5 volt drop across Q1 the current flowing into Q1 is about 0.275 uA. Assuming Q1 and Q2 are approximately matched a current of approximately 0.275 uA should flow through the collector of Q2. That is then multiplied by the Hfe of Q3 (eg. Hfe=300, Ic Q3=300*0.275=82.5 uA).  Anyway choose R2 such that there is a voltage drop of about 1 to 2 volts across it.

The current into the base Ib of Q2 is 0.275 uA/Hfe, approximately 1 nA.  Its DC input impedance is then about 26 mega-ohms. 

Since BJT transistors are current controlled devices the high input impedance should make the circuit insensitive since relatively relatively large voltage changes are needed to cause small current changes across 26 mega-ohms.

AC analysis:

The "diode connected transistor" Q1 is a short at RF,  even if C1 were absent it still presents a relatively low impedance. 

C1 also causes soft start up behavior.

The reason Q2 can allow high sensitivity AM detection is due to its intrinsic base emitter capacitance which lowers the input impedance at RF. The capacitance pulls charge into the base as it charges and supplies charge as it discharges.

Because the DC impedance is so high the transistor effectively changes from a current controlled device to a voltage controlled device (with a small parallel input capacitance.)

The non-linear behavior of Q2 results in AM detection. 

Since the Ic of Q2 is so low the various transistor junctions it is connected to filter away RF leaving only an audio signal going into the base of Q3 where it amplified.