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I understand this has been asked before, but I've never been able to find a satisfactory answer to the question I have, which is: in general, how do we treat the dependent sources in the small signal models when calculating the short-circuit time constants?

Case in point: I ended up with the following circuit (in part) when doing the low-frequency analysis of a CS-CB circuit, for one of the capacitors:

Part of the Circuit

From the example given in this question, I guessed that the equivalent resistance would be R1||R2. Or, as my lecturer said about a similar circuit, the right-half of the circuit is "completely separate" from the left-half.

However, I thought the way to compute equivalent resistance was to replace the capacitor with a voltage source, and to calculate the current? In that case, G would have a non-zero potential, and then wouldn't the dependent current source add current from the right-half of the circuit to the left-half...?

Am I understanding "equivalent resistance" incorrectly? When can we "ignore" the dependent source when finding the resistance "seen" by the capacitor?

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  • \$\begingroup\$ The left and right halves are only connected at one point. There's no return path for anything to flow from one side to the other. \$\endgroup\$ – Roger Rowland Sep 18 '16 at 6:30
  • \$\begingroup\$ @RogerRowland Thanks for the comment Mr. Rowland. In general, though, is it correct, as I described, to replace the capacitor with a test voltage and to find the resulting current? \$\endgroup\$ – user3109672 Sep 18 '16 at 6:40
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Yes generally speaking using an indipendent voltage generator as a stimulus to find equivalent resistance is more than correct, it is applying its definition.

Just a few add-ons:

It doesen't have to be a voltage generator, can be either voltage or current, very often one of the two choices gives much simpler maths.
So before starting reckless calculations it is definitevely worthwile having a short survey to find out which is best in that case.

When we say parallel resistance being \$R_\text{p}=R_1R_2/(R_1+R_2)\$ we have actually connected a voltage generator, done the maths and finally remember the result by heart.
This is usefull since it's such a common problem, it's no use re-invent the wheel every time.

The same applies to series resistance (but with a current generator though).

A few other configurations may be worthly learned by heart, e.g. common collector basic output resitance is \$1/g_\text{m}\$ parallel something else, basic BJT input resitance with emitter degeneration is \$(1+g_\text{m}r_\pi) R_\text{E}\$ series something else, and so on depending or what you need/ are studying.

Recapping, use equivalent resistance defintion, hence voltage or current stimulus, is always correct but usually longer, so should be reserved to cases not covered above.

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