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enter image description hereWhen biasing the common base as shown I was expecting the input impedance to be (1/gm1 + R3/Beta)//(R1//Rpi2). So we with Beta of 100, and biasing current of 250 uA I was expecting the impedance to be around 100+50 (ignoring R1 and Rpi2). But when I simulate the circuit it is coming up to be < 2 Ohms. Could someone please help me understand why?

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schematic

simulate this circuit – Schematic created using CircuitLab

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Your circuit is an amplifier (Q2) with high open-loop voltage gain and strong negative feedback back (via Q1) back to its input: -

enter image description here

This means that the input impedance will be very low because it now behaves like a virtual ground (as per classic op-amp theory). None of these transistors are in common-base by the way.

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  • \$\begingroup\$ Initially I was tying to create a common base topology with Q1, and installed Q2 for biasing purposes to generate a constant current across R1 with diode drop. It seems the circuit behavior different now. Could you please further explain the virtual ground, I am assuming that I am connecting to ground through Rpi2. Further explanation will be greatly appreciated. \$\endgroup\$
    – electronicah
    Commented Feb 9, 2021 at 12:07
  • \$\begingroup\$ Link to virtual ground. Regards this: I am assuming that I am connecting to ground through Rpi2 - I have no idea what this means @electronicah \$\endgroup\$
    – Andy aka
    Commented Feb 9, 2021 at 12:27
  • \$\begingroup\$ I am talking about the small signal model of both transistors, Doing KCL on Q1 emitter, shows impedance as follows 1) R1 2) Rpi of Q2 3) re of Q1+ reflected R3 into the emitter. all should be in parallel. Is my understanding is correct or not. \$\endgroup\$
    – electronicah
    Commented Feb 9, 2021 at 13:56
  • \$\begingroup\$ @electronicah did you forget something in your comment? Anyway, you can do KCL to your hearts content but it still won't alter the fact the the input is connected to a virtual earth and that has theoretically zero ohms input impedance. The emitter follower formed by Q1 will have an output impedance (at the emitter) of only a few tens of ohms so, I'm not sure what else you might expect. \$\endgroup\$
    – Andy aka
    Commented Feb 9, 2021 at 13:58
  • \$\begingroup\$ I do understand that it will be virtual ground, but I am trying to understand the reason why? actually simulation supports your opinion. I just did not see the feedback in my analysis and trying to understand how can I correlate small signal analysis "KCL" to the feedback loop. \$\endgroup\$
    – electronicah
    Commented Feb 9, 2021 at 14:13

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