simulate this circuit – Schematic created using CircuitLab

N.B. I_D=KVov^2 without the 2

I’m really confused with this excercise, my problem is to calculate the A_v=v_out/vd and the cutting high frequency fH neglecting the parasites capacitors of the transistors (there is only C1) . I am confused because with the small signal circuit I can’t calculate the amplification of the differential amplifier because is asymmetrical and the source in not grounded.

I have Vo=-R1gmV_gs2 and I can’t find V_gs2

single ended differential

Here it is my solution

solution page 1

I have tried to use sovrapposition effect and obtained that Vgs_2= -Vd/2 but with a lot of effort obtaining a so clear result means that i'm not getting something about it. btw in the end i obtained A_v=-19.8 and fH=2.5MHz but i'm not sure they are correct


  • \$\begingroup\$ How did you obtain FH = 16MHz? As for the gain, the Vgs is equal to Vd/2. Also, how did you get the A = 112.5? gm1 = gm2 = ?? and gm3 = ?? \$\endgroup\$
    – G36
    Sep 9, 2020 at 14:10
  • \$\begingroup\$ gm1=gm2=4.4 10^-4 gm3= 8 10^-3 I tought there was no point ti show a mistaken calc but im going to add it as soon as possible \$\endgroup\$ Sep 9, 2020 at 14:19
  • \$\begingroup\$ Hmm, why I've got a different result? \$ g_{m1} = \sqrt{2KI_D} = \sqrt{2 * 1m * 50 \mu A} = 316\mu S \$ and \$ g_{m3} = \sqrt{2KI_D} = \sqrt{2 * 10m * 100 \mu A} = 1.41mS \$ \$\endgroup\$
    – G36
    Sep 9, 2020 at 14:26
  • \$\begingroup\$ Why the 2 Is inside the Square root \$\endgroup\$ Sep 9, 2020 at 14:45
  • \$\begingroup\$ Btw im going to sent my solution, wait fifteen minutes \$\endgroup\$ Sep 9, 2020 at 14:46

1 Answer 1


If the equation for a drain current in the saturation region (active region) looks like this:

$$I_D = K(V_{GS} - V_T)^2$$

Then the MOSFET transconductance is equal to:

$$g_m = \frac{dI_d}{dV_{GS}} = 2K(V_{GS} - V_T)$$

But if we substitute \$(V_{GS} - V_T) = \sqrt{\frac{I_D}{K}}\$ we will end with:

$$g_m = 2K\sqrt{\frac{I_D}{K}} = 2\sqrt{KI_D}$$

Now we can calculate the transconductance for your amplifier.

\$g_{m1} = g_{m2} =2\sqrt{1\text{m} [\frac{A}{V^2}]\times 50\mu A} = 447.2 \mu S \$

\$g_{m3} =2\sqrt{10\text{m} [\frac{A}{V^2}]\times 1.6\text{mA}} = 8 \text{mS}\$

Now the voltage gain. The \$M_3\$ voltage gain is:

$$A_{V3} = -g_{m3}R_2 = 8\text{mS} \times 1.25\text{k}\Omega = -10 V/V $$

And the voltage gain of a single-ended differential amplifier we can obtain using this small-signal equivalent circuit:


simulate this circuit – Schematic created using CircuitLab

$$V_O = -I_D * R_D = -g_{m2}*v_{gs2}*R_d$$

For this circuit \$v_{gs2} = -\frac{V_D}{2}\$ why?

Notice that becouse \$1/g_{m1} = 1/g_{m_2}\$ we have a voltage divider therefore.

KVL \$V_D = v_{gs1} + v_{sg2}\$ and notice that , \$v_{sg2} = - v_{gs2} \$)

$$\frac{V_O}{V_D} = \frac{g_{m2}*R_D}{2} = \frac{447.2\mu S * 18\text{k}\Omega}{2} = 4 V/V$$

So, the overall voltage gain is \$-10 * 4 = -40 V/V\$

As for the capacitor effect on the circuit, we need to find the pole frequency. and because \$M_3\$ is working as a common source amplifier we have this equivalent circuit:


simulate this circuit

We can see that because the drain of a MOSFET in the saturation region has a property of a current source we see that the pole frequency is

$$F_H = \frac{1}{2\pi RC} \approx\frac{0.16}{R_DC_D}\approx\frac{0.16}{1.25\text{k}\Omega * 100\text{pF}} = 1.28\text{MHz}$$

  • \$\begingroup\$ But ID3 Is not 100uA isnt ? And i didnt get why in this circuit vgs=Vd/2 i cant see it \$\endgroup\$ Sep 9, 2020 at 16:28
  • \$\begingroup\$ I update my answer. Do you see is why vgs=Vd/2 is true? \$\endgroup\$
    – G36
    Sep 9, 2020 at 16:47
  • \$\begingroup\$ i tought that the current generator cant able me to use the voltage divider formula \$\endgroup\$ Sep 9, 2020 at 16:51
  • \$\begingroup\$ Ok now its clear thank you! \$\endgroup\$ Sep 9, 2020 at 16:55
  • 1
    \$\begingroup\$ @SergioPiccione notice how I mark the voltage polarity Vd and Vgs1 and Vsg2. \$\endgroup\$
    – G36
    Sep 10, 2020 at 13:54

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