Calculate Effective Capacitance

ZoOneR said:

Hi,
We have 4C connected from node Y to VDD, and 2C connected from node Y to gnd. I think we can all agree on this.
To calculate the output impedance of the circuit we connect to the output a source, let's say a voltage source, and we measure the current drawn from it.

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You are describing the Applied Source Method. I like to use a 1 amp AC current source instead of a voltage source, because then the calculated voltage at that point is also the impedance. The OP did not ask for the impedance, which is only defined for AC. He asked for "effective capacitance", whatever that is. The impedance of all the caps at DC is infinity.

But in this process all other voltage sources become shorted and current sources become open circuit. This means that the voltage source, VDD, is replaced by a short circuit to ground. Now the 4C connected previously to VDD are connected to ground. Since the 2C are also connected to ground we obtain a total of 6C connected to ground.

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That procedure only applies to independent voltage and current sources. In this case, there is only on independent source. Dependent sources keep their impedances.

From above discussion we can see that the 4C are not phisicaly connected to ground, but they modeled as so when we do a small signal analysis.

Hope this helps.

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