Why do we short the input when calculating the output impedance?

Question

I believe the title says everything about the question. My issue has to to do with the following: Consider a two-port given by:

$$V_1 = Z_{11}I_1 + Z_{12}I_2$$ $$V_2 = Z_{21}I_1 + Z_{22}I_2$$

Cleary by the equations above, one would expect that the measurement of the \$Z_{ij}\$ parameters to be performed while keeping one port at a time in open loop configuration to guarantee \$I = 0\$ for the "unwanted" current. However this seems not to be the case, as can be seen for instance in the derivation of the of the output resistance of a simple CE amplifier where the input is shorted to determine the output resistance. What am I missing here?

Answer 1

The two test configurations tell you two different things.

If you short the input, you're testing the output impedance of the circuit as it is normally used. This is because you typically drive the circuit with a (relatively) low impedance source.

If you open the input, you get the \$Z_{22}\$ as defined in circuit theory as a characteristic of the circuit itself (without effects from any driving circuit).

The first option is often more practically useful.

Answer 2

If you want to solve for any of the Z parameters it is clear that you must assume Iac=0, which means Idc=const.

However, remember that there are other sets of parameters for describing the BJT:

• Y-parameters are all "short-circuit"-parameters (Vac=0, Vdc=const)

• h-parameters are "short-circuit-parameters" for h11 and h21 and "open-circuit-parameters" for h12 and h22.

And remember: "Short ciruit" means: DC conditions constant (no ac signal).