# Find the gain in mosfet

I am trying to find the gain for vd/vi and vs/vi. My initial attempt as to find vd in terms of vdd,Rd, and Id. What I obtained was: Vd=Vdd-(RdId) I attempted the same thing with Vs and got: Vs=(RsId)-Vss

My problem is how do I relate them to the gate voltage, vi? Something tells me I did this wrong(the setup) as I think Id has no purpose here.

Note: The problem doesn't tell me what region of the graph the MOSFET is in and doesn't give me numerical values.

Thanks for the help!

• You need to understand what gain means in such transistor/mosfet based amplifiers. Such circuits dont have, what is called large signal gain. They have small signal gains related to small signal parameters. You have calculated the value of Vs when Vi = 0. That is correct but doesnt tell you to gain, rather the operating point. To proceed further, draw the small signal model, extract the small signal parameters and determine the gain formula. – Adil Malik Apr 17 '17 at 13:03
• Mosfet amplifiers are almost always biased in the saturation region. So a good bet would be to assume the mosfet is saturated. – Adil Malik Apr 17 '17 at 13:05
• look for value of $g_m$ – Tony Stewart Sunnyskyguy EE75 Apr 17 '17 at 13:08
• So, if I converted this to a small signal, would vgs be equal to vi-Vs(the one i found above)? – John G. Apr 17 '17 at 13:21

You could plot or write the expression for Vout vs Vin and then take the derivative of it to get the large signal gain $A_V= \frac{dVout}{dVin}$ The simulation result is:

For $R_D=1k\Omega$,$R_S=100\Omega$,$V_{DD}=10V$,$V_{SS}=-10V$

And notice the maximum gain is equal to $\frac{R_D}{R_S} \approx 10$

And this is not the coincident.

Or you can do small signal analysis and solve for the gain.

From inspection we can write:

$$V_{IN} = V_{GS}+I_D*R_S$$

And because $I_D = g_mV_{GS}$

We have

$$V_{IN} = V_{GS}+g_mV_{GS}*R_S = V_{gs}\left (1+g_mR_S \right )$$

The output voltage can also be find by inspection

$$V_{OUT} = -I_D*R_D$$

This minus sign comes from the fact that the $I_D$ current is flowing from GND into drain terminal in our small signal equivalent circuit.

In reality this minus sign only inform us that our amplifier output voltage is 180° out of phase with respect to the input voltage.

Any increases in $Vin$ will increase $I_D$ current also, the voltage across $R_D$ increases too, therefore $V_D$ voltage drops due to large voltage drop across$R_D$ resistor, $V_d = V_{DD} - I_D*R_D$

Hence the voltage gain is equal to: $$\frac{V_{OUT}}{V_{IN}}= \frac{-g_mV_{gs}R_D}{V_{gs}\left (1+g_mR_S \right )}= - \frac{g_mR_D}{1+g_mR_S} = - \frac{R_D}{\frac{1}{g_m} + R_S}$$

• Thank you so much! A couple of questions. 1. Lets say we move vout to be at Vs. What does vout become. Is it like my equation for Vs above, only with Vss becoming 0? – John G. Apr 17 '17 at 15:52
• @JohnG. From "large signal" perspective $V_S = Vin-Vgs$. And from the "small signal" perspective we will have $V_S = I_D*R_S=g_m V_{gs}*R_S$ so the voltage gain is ?? – G36 Apr 17 '17 at 16:27
• And yes, for "small signal" all DC voltage sources become a short circuits. And this is why VSS becomes 0V (ground). – G36 Apr 17 '17 at 16:30

MOSFETs (enhanced ) are rated as follows;

$g_{fs} = 3 S$ (S=Siemens = 1/Ohms=A/V)

$g_{fs}$= Forward transconductance $V_{DS}$ = 15 V, $I_{D}$ =1.5A 3 S (1)

1. Pulsed: pulse duration=300µs, duty cycle 1.5%

From this you choose a Source Follower design (unity gain) or Common Source design with drain load or Drain/(Rsource+RdsOn) impedance ratio for small signal ratio gain.