# Working of differential pair using MOSFETs

I am currently learning about differential pair using MOSFET from this course. In this particular lecture, the teacher says that to have a differential pair we need to apply the second signal v2 at the source through a buffer as shown below because impedance at source of the MOSFET is $$\1/g_m\$$.

To implement a buffer he uses a source follower like this:

The source voltage of this source follower will be almost equal to v2 if the $$\g_mR_l >> 1\$$, where $$\R_l\$$ is the load connected at the source terminal of source follower because:

Output voltage of source follower: $$\frac{g_mR_l \times v_2}{1+g_mR_l}$$

But later he connects the voltage buffer to the initial amplifier as shown below:

Now the doubt that I have is that now the impedance at the output of source follower is $$\1/g_m\$$ which is much much less than 1 and therefore according to the formula above the output of voltage buffer is no longer equal to the input voltage v2 and therefore it is not a good differential pair anymore.

Is my logic correct or am I making a mistake?

• Your differential pair isn't biased properly, for one thing. Commented Jul 12 at 13:42
• I know that. The professor in the linked lecture is only talking about the small signal model. In the later lectures he tells how to bias. Commented Jul 12 at 14:29
• Yes, the voltage at the middle node (source) is vs = (v1+v2)/2 which can be obtained by superposition. Assume v2 = 0, vs= 1/gm1/(1/gm1+1/gm2)*v1 = 1/2*v1. Similarly you get 1/2*v2 so the total voltage is vs = (v1+v2)/2. Commented Jul 12 at 16:34