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Electrical Engineering student here! (AKA beginner)

Suppose we have a general circuit (ie bjt cascode amplifier, differential amplifier etc)

To perform theoretical DC analysis we treat capacitors as open circuits. To take lab measurements we DO NOT add capacitors to our breadboard.

To perform theoretical AC analysis we treat capacitors as short circuits. To take lab measurements we DO add capacitors to our breadboard.

Why this difference? Thanks in advance.

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  • \$\begingroup\$ Counter question : do you understand how capacitors work differently in DC and AC circuits? What's the capacitor impedance at DC and AC? \$\endgroup\$
    – Justme
    Nov 25 at 11:10
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To perform theoretical DC analysis we treat capacitors as open circuits. To take lab measurements we DO NOT add capacitors to our breadboard.

To perform DC circuit analysis, it's true that you can simplify by treating the capacitor as an open circuit. This is because the effective resistance of a capacitor (i.e. its impedance) is 1/(2πfC). As f -> 0 the value of this expression goes to infinity (i.e. an open circuit).

If you have a breadboard of your circuit and are taking DC measurements you COULD leave the capacitors out and it would still perform the same way at DC. But I don't ever recall doing this in the real world.

To perform theoretical AC analysis we treat capacitors as short circuits. To take lab measurements we DO add capacitors to our breadboard

It's not clear where you got this idea but it's wrong. Capacitors only behave as short circuits when the frequency is high enough so that their impedance is much less than surrounding circuit elements. Again the formula for the impedance of the capacitor, 1/(2πfC), is used to calculate the effective resistance of a capacitor in the circuit at a particular frequency.

So whether or not you can treat the capacitor as a short circuit depends on the AC frequency you are using. There is no general rule that says you can always treat capacitors as short circuits under AC circuit analysis.

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