I am trying to do a small signal analysis for the following circuit enter image description here

I did not have any experience with BJT so I am having a hard time to analyze the circuit. From research, I know Q1 is common base and Q2 and Q3 are common emitter. To draw the small signal model, I would have to short all DC source and capacitors.

Then I ended up with the following circuit enter image description here

I am not sure what to do next. I tried to break the circuit into 3 parts to analyze but I am not sure what to do with the 47K, 56K,15K resistors that's connected to Q1.

Can anyone help with the small signal analysis?

Now I have this for the first part if I were to divided the circuit to 3 parts. enter image description here

To find the input impedance, applying KCL then it would be 12/47k + gmVpi + Vpi/rpi + Vin/10K?

I am a bit confused with Vpi and rpi. Do I need to calculate that or it's from the BC547 data sheet?

  • 1
    \$\begingroup\$ Why are you using a common base input? what's your Mic impedance? \$\endgroup\$
    – S.s.
    Commented Dec 4, 2020 at 0:28
  • 2
    \$\begingroup\$ @MathKeepsMeBusy no, its common emmiter with shunt feedback \$\endgroup\$
    – S.s.
    Commented Dec 4, 2020 at 0:33
  • \$\begingroup\$ The 15K and 56K are bias resistors for the base of Q1. In small signal analysis, the base of Q1 is shorted to ground by the 220uF capacitor. The 15K and 56K resistors then become redundant (each end is shorted to ground). It would help if you kept the wire to the base of Q1 to see that. (Short it to ground). \$\endgroup\$ Commented Dec 4, 2020 at 0:35
  • \$\begingroup\$ From my research, this amplifies dynamic microphones with 200 to 600 ohm output impedance, so I guess the mic impedance is between 200-600 ohms? \$\endgroup\$
    – lsi
    Commented Dec 4, 2020 at 0:36
  • 1
    \$\begingroup\$ @lsi The first thing I'd want to do is to perform a sanity check on the DC quiescent operating point. The next thing is to know exactly how the microphone itself works. (Too many kinds floating about.) Only then, would I dive into the small signal part. The common-base first stage is DC biased just as any common-emitter stage. In fact, you can see all of the usual culprits -- the base biasing pair, the emitter resistor, and the collector resistor. If you spend a little time working out the DC quiescent point, my interest may climb a bit. Just FYI. \$\endgroup\$
    – jonk
    Commented Dec 4, 2020 at 7:47

1 Answer 1


This common base operates with about 2 volts on the base. Thus Iemitter is 1.3/10K or 130 uA.

The 'reac' is 26 ohms/0.13mA == 150 ohms. Low yes, however.

The frequency response of 100uF (Cin) and 100 ohms is 0.01 second, or 16Hz.


Regarding 'reac'

When first delighting in the use of small_signal analysis, I soon realized I could predict the gain of bipolar devices (FETs were rare, expensive, and not stable in parameters) simply by

  • dividing the combined Rs on the collector, in parallel


  • the series emitter resistances;

And that series emitter resistance (if no discrete external R) was just

  • 26 ohms / Re in mA

thus a device operating at 1mA would have

'reac' of 26/1 = 26 ohms

And if that bipolar has a 260 ohm collector resistor, with only scope probe as the load, then the voltage gain was precisely

  • Gain = Rcollector / Remitter = Rcollector / 26 == 260/26 = 10X

And I've in the intervening decades used my personal lable 'reac' for that value, which is the slope of the emitter_base diode.

Another way to compute it is 1/tranconductance.

Again, this value is just the small_signal resistance of emitter_base diode.

  • \$\begingroup\$ What is reac? input reactance? How is it ohms/Amp ? \$\endgroup\$
    – lsi
    Commented Dec 4, 2020 at 3:01

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