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In the lesson on BJT Amplifiers, we were told the we can find the AC equivalent circuit by replacing all capacitors by short circuits, inductors (if any) by open circuits, dc voltage sources by ground connections and dc current sources by open circuits. What I don't understand is WHY should we replace dc voltage sources by ground connections and dc current sources by open circuits ?

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  • \$\begingroup\$ I'm not sure why you would replace the caps. by shorts, etc., unless you are looking at the high frequency behaviour. Remember that you are looking at the small signal behaviour around an operating point. If a node has a constant voltage, then the deviation from the operating point is always zero, hence it behaves as an AC ground. \$\endgroup\$ – copper.hat Mar 1 '15 at 4:55
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In linear circuits with both AC and DC components, one can split the circuit, using the Superposition Theorem, into a DC equivalent circuit to model the effect of the DC components and an AC equivalent circuit to model the effect of AC components. By "components" I mean voltage sources and current sources.

For AC analysis, you are only concerned with time-varying voltages and currents. Therefore, any DC voltage sources are considered shorted (0 V of voltage drop across them), while DC current sources are considered open circuits (0 A of current through them). Of course, the DC sources are included in DC analysis while all the AC sources are made 0. It is exactly the same as applying the Superposition Theorem to circuits with multiple DC sources - you consider the effect of one DC source at a time, and negate all the others.

This is the best link I found for explaining AC analysis so far.

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    \$\begingroup\$ Just to nitpick, it is not an application of superimposition theorem. Superimposition theorem is only applicable if the circuit is linear, and when you are dealing with BJT you have a nonlinear circuit. Although the practical application is almost the same, the reason why you can do this is because you are linearizing the circuit and this means that you only consider small variations of electrical quantities around their DC operating point. The mathematical details involve a multivariable Taylor expansion on the nonlinear equations of the circuit. \$\endgroup\$ – Lorenzo Donati -- Codidact.org Mar 1 '15 at 1:36
  • \$\begingroup\$ That's a great point. Small signal analysis makes it possible to apply these rules to these nonlinear transistor circuits. \$\endgroup\$ – FullmetalEngineer Mar 1 '15 at 3:02
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A perfect voltage source has zero output impedance. A perfect current source has infinite output impedance.

Don't replace a DC voltage source with a ground connection, replace it with a short circuit. There is often a difference.

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