I am teaching myself electronics in my spare time and I am trying to look at circuits and understand how they work. I am currently studying amplification and transistor basics, and I have started analysing the audio amplfication circuit below. audio amplifier

In particular, I am trying to figure out how exactly the darlington pair Q7 and Q9 associated with Q5 work: this is all quite confusing to me at the moment. The circuit description reads: 'the predriver stage (Q7, Q9) is a Darlington connection the load of which employs a constant current source (D5, Q5), resulting in a high voltage gain.'

Am I correct with the following?

  • constant current source: Q5 acts as a constant current source for the darlington pair. As such, if there were an increase in current at the collector of Q5, the emitter-base forward bias of Q5 would reduce (from the higher voltage drop across the emitter resistor of Q5), and the resulting lower Ib would reduce Ic and Ie. D5 maintains forward bias between the emitter and the base.
  • Darlington pair: this darlington pair is configured as a common emitter amplifier with Q5 as its load. With an AC input signal at its base, the positive portion of the signal increases the forward bias of Q7 and Q9, which would increase the collector currents. Since current is constant, what happens here, a smaller Vce drop for Q9 as the collector of Q9 gets closer to -B2? If that is the case, assuming that with constant current the voltage drops across R31 and R29 should remain constant, is also Vc of Q5 getting closer to -B2? Am I correct to deduce that this is a phase-inverting configuration?

I would appreciate if you could help me clarify these questions. Thanks.

  • \$\begingroup\$ Well, you are correct. The main purpose of using a constant current source as a load (active load) for the CE amplifier is to increases the stage voltage gain. Because in this amplifier topology almost all voltage gain is provided by a VAS stage. The gain will be high because the smallest difference in collector current between the active load (Q5) and CE stage will give a large voltage swing. A transistor which tries to make the lowest current will win the fight. \$\endgroup\$ – G36 Oct 24 '20 at 14:56

Q5 and its associated components form a constant current source and therefore the collector current of Q5 is held at a constant current which nominally doesn't vary.

The signal at the base of Q7 is a varying current signal, the voltage at the base of Q7 is held pretty much constant at 1.4V.

As the injected base current of Q7 varies, its collector current will also vary. If Q7/Q9s' collector currents increase then there is less current available from Q5 to drive the base of Q23 and therefore the amplifier's output will fall. (Remember there is a constant current out of Q5). If the base current of Q7 reduces, its collector current will also reduce. This will force more of Q5's constant current output into the base of Q23 and also less current will be pulled from the base of Q25. This results in the amplifier's output rising.


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