# Open circuit voltage gain of JFET transistor circuit

So this is one of the questions from my previous past papers, I worked out my Vgs and Vds value correctly (-1.25v and 10v). However, when calculating my open circuit voltage gain, I use the equation Av=-gm.Rd, and get the value -25. When I asked my tutor if this was correct, he said that the voltage gain was wrong. He said because there's no capacitor, I need to use the equation Av=-gm.rf/(1+gm.rs) (I assume he meant -gm.rs?) he hasn't elaborated further. Any help would be really appreciated :D (Also ignore the transfer and output characteristic)

## migrated from physics.stackexchange.comMay 23 at 18:44

This question came from our site for active researchers, academics and students of physics.

• Note that Rs provides negative feedback. As Id increases, Vgs decreases by Id*Rs. So what does that do to your gain? Can you write the equations taking that into account? – John D May 23 at 18:50
• Why is $A_v = -g_m R_d$? – TimWescott May 23 at 19:12
• The cap goes across Rs to allow full Vgs AC swing to get the full gain to the drain. When gm*Rs >>1 your gain without the cap is the R ratio = -Rd/Rs. Note that the text is old and we use capital S for Siemens now. so remember to use 10 ms instead of 10mS – Tony Stewart Sunnyskyguy EE75 May 23 at 19:20

If you look at the AC small signal model of a MOSFET, you will see this:

Therefore, you need to convert your MOSFET circuit into this form and analyze it. Beware of memorizing and thinking with formulas. This is circuit analysis, not circuit memorization.

First since this is AC, we can just short out the capacitors so $$\V_G=V_{in}\$$ and $$\V_D=V_{out}\$$ (Assuming capacitors are big enough). Then, we can just ignore the $$\R_G\$$ as it is huge making it essentially an open circuit in the system. We can also ground the $$\+V_{DD}\$$ DC source as we are looking at this in AC domain. So now gate of the MOSFET goes directly to $$\V_{in}\$$, drain goes to the ground through RD and source again goes to ground through RS as you can see below.

simulate this circuit – Schematic created using CircuitLab

Now, onto the analysis:

We are trying to find $$A_v=\frac{V_{out}}{V_{in}}=\frac{V_{GS}}{V_{in}}\frac{V_{out}}{V_{GS}}$$

For $$\\frac{V_{GS}}{V_{in}}\$$, $$\V_{in}=V_G\$$ and $$\V_S=g_mV_{GS}R_S\$$. So $$\V_{GS}=V_G-V_S=V_{in}-g_mV_{GS}R_S\$$. Therefore, $$\\frac{V_{GS}}{V_{in}}=\frac{1}{1+g_mR_S}\$$.

This is our first expression.

For $$\\frac{V_{out}}{V_{GS}}\$$, since all the current, $$\g_mV_{GS}\$$, is flowing through $$\R_D\$$, $$\V_{out}=-g_mV_{GS}R_D\$$. So $$\\frac{V_{out}}{V_{GS}}=-g_mR_D\$$

This is our second expression.

Finally, if we put both expressions into $$\A_v\$$,

$$A_v=\frac{V_{out}}{V_{in}}=\frac{-g_mR_D}{1+g_mR_S}$$.

When you plug in these, the answer you get is -4.17 V/V. As you can see with this, $$\R_S\$$ provides negative feedback into this amplifier, lowering the gain.