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What I've read in microelectronic texts like Prof's Razavi's Fundamentals of Microelectronics is that for calculating output impedance, one has to make all independent sources 0, i.e., short all independent voltage sources and open all independent current sources.

Sighting a simple example of output impedance calculation in common source amplifier given in the picture, if we go by the rules, we'd short Vin and Vdd as shown in the picture. Now, from the small signal model, the output impedance is said to be Rd||ro. But my question is that if gate and source both are shorted making Vgs = 0, i.e., lesser than Vth, this implies there is no channel at all for current to flow from drain to source. So, why do we take into account ro in output impedance calculations when no current can even flow through the MOSFET due to Vgs = 0?

enter image description here

P.S.:You may either refer to Fundamentals of Microelectronics by Dr.Behzad Razavi or even his lectures on youtube for the prescribed method my question is based on, the links given below.

The book Fundamentals of Microelectronics :

Razavi electronics lecture 37 - at 45:54

https://www.youtube.com/watch?v=UeXtyA42ElU&index=37&list=PL7qUW0KPfsIIOPOKL84wK_Qj9N7gvJX6v

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You simply forget that you are dealing here with a small-signal equivalent model (AC signal only).

So, in this case, shorting the transistor gate to ground means that any AC signal present at the gate is shorted to ground. Hence, no AC voltage is present at the transistor gate. But the DC condition (quiescent currents and voltages ) should stay unchanged.

This is why we also "short-out" the DC voltage source in a small-signal analysis.

The ideal DC voltage source has 0Ω internal resistance. And that's why AC-signals are shorted by DC voltage source. DC voltage is always constant so for any change in current there is no change in the voltage. So there is 0 Ohm internal resistance. Additional in real life circuit we always use a bypass capacitor connect parallel to DC Voltage. And this capacitor will short all AC-signal to the ground.

For example, if we have a 9V DC voltage source, now if we smoothly change the current that is drawn from this DC voltage source, from 40mA to 20mV (we change the resistance from 225Ω to 450Ω). We create AC-current. But the DC voltage does not change (0Ω internal resistance). So the dynamic resistance of DC-voltage source is equal to rd = 0V/20mA = 0Ω And this is why we say that DC-Voltage is short for an AC-signals.

Here you have CS amplifier

schematic

simulate this circuit – Schematic created using CircuitLab

And this is how his equivalent small-signal circuit looks like

schematic

simulate this circuit

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  • \$\begingroup\$ Thanks for answering but in your answer, you told why DC sources don't show up in small signal model. That is not my question. I want to know that while calculating output impedance, the procedure that I used needs me to short the biasing voltage too. If you saw the pdf whose link I've mentioned or the video I mentioned, the common procedure told there is :As with the impedance of two-terminal devices such as resistors and capacitors, the input (output) impedance is measured between the input (output) nodes of the circuit while all independent sources in the circuit are set to zero. \$\endgroup\$ – Rohan Aug 19 '18 at 15:28
  • \$\begingroup\$ Because the biasing voltage is the DC voltage also. \$\endgroup\$ – G36 Aug 19 '18 at 15:31
  • \$\begingroup\$ Just tell me one single thing, if in the diagram above the small signal model i've made in the picture, when there is no Vgs, how would the MOSFET conduct? No channel means no Id and so, because ro = 1/(Id*lambda), ro should go to infinity and not show up in the output impedance but it has shown up wherever I've checked. I want to know if the method being employed is incomplete or should the independent sources be shorted only in the small signal model. \$\endgroup\$ – Rohan Aug 19 '18 at 15:43
  • \$\begingroup\$ Do you understand what small-signal model is? \$\endgroup\$ – G36 Aug 19 '18 at 15:44
  • \$\begingroup\$ Small-signal model is an equivalent circuit for an AC signal (from AC single point of view only). Shorting in AC small signal model vgs = 0V means the there is no AC voltage present between the gate and the source. And no AC Id current is flowing (no change in Id current). \$\endgroup\$ – G36 Aug 19 '18 at 15:50

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