To measure a resistance respecticely a complex impedance in a simulator, I'll inject a current into the node and measure the voltage.
But all other methods would work in AC analysis as well. The only important point is to achieve correct DC bias without introducing AC feedback or circuit loading.
as long as you biased your circuit with dc voltage source and the only ac source is at the output then your method is fine
If this means two AC sources active at the same time, it's clearly wrong.in the same time I connected the normal signal voltage source that we use already when we simulate the frequency response of the amplifier.
If this means two AC sources active at the same time, it's clearly wrong.
What should be the meaning of this expression? Vi hasn't to do with output impedance, Zo = Vo/Iobecause the simulator will calculate it as Vo/Vi/I
What should be the meaning of this expression? Vi hasn't to do with output impedance, Zo = Vo/Io
Io is injected in the test, Vo is the measured voltage.
adding to FvM's response. if you check textbook of how output impedance is calculated in small signal model, you basically short out the input voltage source, so you don't need any other ac source to get Rout. i recommend you read about how to calculate output impedance from ss model then you would understand what you are doing in the simulation exactly.
Yes I am now totally agree with you but my question was what should be the status of the two input terminals of the op-amp, ok one is at VCM and what about the other ??I dont have an inductor to use in my simulator as you show me in the image you have applied before.
The required connection can be derived from the definition of output impedance, or reviewd in an analog design text book, I presume.
- connect a DC input voltage that places the amplifier in linear operation range, e.g. output voltage at about mid supply
- no AC input
This is usually ahieved by DC feedback as in my example. I presume that a usable circuit simulator has some means to setup a DC bias circuit.
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