Please see images of question and solution. Please see my working out using the small-signal model

I'm really confused about how they went about solving the input impedance of this circuit? Did they use DC biasing or the small-signal model to work out the input impedance?

Question Solution Small-Signal working out

  • 1
    \$\begingroup\$ The input impedance is found from the small-signal model. Why don't you draw out the small signal model of the circuit and add it to your question so we can discuss it? There's a schematic editor in the post editor. \$\endgroup\$
    – uint128_t
    Commented May 1, 2016 at 16:46
  • \$\begingroup\$ en.wikipedia.org/wiki/Common_base (scroll down to low frequency characteristics. \$\endgroup\$
    – Andy aka
    Commented May 1, 2016 at 17:00
  • \$\begingroup\$ Still have a couple of questions. 1)when finding input/output resistances do you ground all nodes in the small signal model? 2) Is there other method in working out input/output impedances without having to resort to the small-signal model? \$\endgroup\$
    – Arsenal123
    Commented May 1, 2016 at 17:32
  • \$\begingroup\$ 1) no, only those that connect to a small-signal ground. 2) No there isn't. The in and output impedance are by definition small signal properties so you need a small signal equivalent circuit to determine their values. \$\endgroup\$ Commented May 1, 2016 at 18:42

1 Answer 1


Small signal is all about change. Input impedance merely means the ratio of the change in input voltage at the point by some source and the change in current obtained by it. You could wiggle a current through the node too and measure the change in voltage at the node.

If you pass a small current di into the node (causing the total current to become I+di), since di = gm.dvgs (gm is transconductance), the change in voltage developed is 1/gm times di. Thus, the answer would be dv/di = 1/gm. The resistor appears as a resistor to ground (a small current change would occur through the resistor since other voltages are fixed giving the impression of resistance to the current source that caused the di) and thus zin = 1/gm || r.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.