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Consider the following circuit:

circuit

I want to find the gain of this configuration, with and without the bypass capacitor (the 47uF capacitor).

Vx will be taken as the input voltage and Vout as the output.

Part 1: using the bypass capacitor

Using the small signal pi model, we obtain the following circuit.

pi model equ circuit

Therefore:

Vout = -(ro || 4.7k || 10k) * gm * Vin

Av = Vout / Vin = -(ro || 4.7k || 10k) * gm

I built this circuit and measured and found that the current through the drain is 1.8074 mA. This would give ro = VA/ID = 25/1.8074 = 13.86 ohms. If we consider gm to be equal to 0.5 * 10^(-3) Siemens and we calculate Av, we get -6.90 * 10^(-3) V/V.

Question:

Why is the gain so low?

Part 2: without the bypass capacitor

Using the T-model, we get the following circuit.

t-model circuit

Analyzing this circuit, we get:

Vout = - gm * Vgs * (4.7k || 10k)

Vin = (1/gm + 1k) * gm * Vgs

Therefore, Av = Vout/Vin = -1.067 after plugging in the values.

Question: This gain makes more sense. Is this because ro was ignored in the T-model?

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1 Answer 1

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In your calculation, you have made an error. It is correct that the gain is determined by ro||RD||Rload - however, in this expression the quantity ro is the small-signal output resistance of the FET (equivalent to the inverse slope of the Id=f(VDS) curves, normally in the higher kOhm range). Hence it is wrong to use the DC values for calculating ro (VA/ID, resulting in a very small value).

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  • \$\begingroup\$ "equivalent to the inverse slope of the Id=f(VDS) curves, normally in the higher kOhm range" can you please elaborate on this? how would i calculate the output resistance? thanks so much! edit: is it by doing ro = vds / id = 4.2933 V / 1.8074 mA = 2.38 kOhms (these are the experimental values i obtained)? \$\endgroup\$ Commented Apr 4, 2016 at 16:06
  • \$\begingroup\$ If I ignore ro in the pi-model, I obtain a gain of -1.60 . Is this the correct gain? \$\endgroup\$ Commented Apr 4, 2016 at 16:28
  • \$\begingroup\$ The small-signal output resistance of the FET is ro=d(Vds)/d(Id). But this resistance should be listed in the data sheet. \$\endgroup\$ Commented Apr 4, 2016 at 16:46
  • \$\begingroup\$ Without bypassing the source resistor the gain is G=-3.2k/(2k+1k)=-1.066 (assuming gm=0.5mS and ignoring ro). \$\endgroup\$ Commented Apr 4, 2016 at 16:50
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    \$\begingroup\$ Yes , -1.6 is corect assuming gm=0.5mS and RD||Rload=3.3k. Perhaps the transconductance gm was larger in your experiment? By the way- up to now we didn`t consider the series resistor at the input (10k). We have a voltage division between 10k and the bias voltage divider. \$\endgroup\$ Commented Apr 4, 2016 at 17:34

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