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Question on CMOS small signal analysis??

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Anachip

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Guys,

Why when we do a small signal analysis for a transistor circuit, we ac ground any bias voltages and also why when we measure the input or output impedance (let say for a common source amplifier), we usually short the output when to find the input impedance and vise versa.

Thanks,
Anachip
 

v_c

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Anachip,

in ac small signal analysis we are only looking at linearized system about constant valued operating points. Since dc voltages have no ac signal they are grounded. You can think about this another way. Consider that any signal can have a dc and ac component. Now use superposition and take the dc and ac components separately to do the analysis and at the end add them up.

When we do bias component analysis (to get the dc operating points), we assume that the ac voltages are zero. We open circuit the capacitors, and then we do the analysis. For ac analysis, we short the capacitors, etc. The true amplifier waveforms contain both dc and ac.

As far as impedances are concerned, you have to go back to the definition of impedance. It is generally calculated by applying a signal at the port of interest and looking at the ratio of voltage/current. But for this definition to hold, the applied signal at the port of interest must be the *only* excitation applied to the circuit. All other independent sources must be turned off -- voltages sources set to zero (shorted) and current sources set to zero (opened).

I hope this helps you.

Best regards,
v_c
 

    Anachip

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Anachip

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Thx for the explanation V_C.

v_c, can u explain me as well on my posting about slew rate, settling time in last 2 days.
 

v_c

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Is it this topic ? I will try to answer it on that thread, if you don't mind.

Best regards,
v_c
 

beckchm

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ac ground reprents that voltage of the nodes does not change
 

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