Figure shows a MOSFET diff-amp with an active load. Transistors M1 and M2 are n-channel devices and form the differential pair biased with Iq. The load circuit consists of transistors M3, and M4, both p-channel devices, connected in a current mirror configuration. A one-sided output is taken from the common drains of M2 and M4. When a common-mode voltage of v1 = v2 = von is applied, the current IQ splits evenly between M1 and M2 and id1 = id2 = Iq/2. There are no gate currents; therefore, id3 = id1 and id4 = id2·

If a small differential-mode input voltage vd = v1 - v2 is applied, then from equation, we can write

$ i_{D1} = \frac {I_Q}{2} +i_d$

&

$ i_{D2} = \frac {I_Q}{2} -i_d$

where id is the signal current. For small values of vd, we have id = (gm.vd)/2. Since M1 and M3 are in series, we see that $ i_{D3}=i_{D1} = \frac {I_Q}{2} +i_d$

Finally, the current mirror consisting of M3 and M4 produces

$ i_{D4}=i_{D3} = \frac {I_Q}{2} +i_d$

Figure is the ac equivalent circuit of the diff-amp with active load, showing the signal currents. The negative sign for id2 in equation shows up as a change in current direction in M2, as indicated in the figure.

Figure (a) shows the small-signal equivalent circuit at the drain node of M2 and M4. If the output is connected to the gate of another MOSFET, which is equivalent to an infinite impedance at low frequency. the output terminal is effectively an open circuit. The circuit can be rearranged by combining the signal grounds at a common point, as shown in figure (b).

Then,

$v_0=2(\frac{g_mv_d}{2})(r_{02}||r_{04})$

and the small-signal differential-mode voltage gain is

$A_d = \frac{v_0}{v_d} = g_m(r_{02}||r_{04})$

Equation can be rewritten as

$A_d = \frac{g_m}{\frac{1}{r_{02}}+\frac{1}{r_{04}}}=\frac{g_m}{g_{02}+g_{04}}$

If we recall that,

$g_m = 2 \sqrt{K_nI_D}=\sqrt{2K_nI_Q}$

$g_{02} = \lambda_2I_{DQ2}=\frac{\lambda_2I_Q}{2}$

$g_{04}= \lambda_4I_{DQ4}=\frac{\lambda_4I_Q}{2}$

then the equation becomes,

$A_d = \frac{2\sqrt{2K_nI_Q}}{I_Q(\lambda_2+\lambda_4)}=2\sqrt{\frac{2K_n}{I_Q}}.\frac{1}{\lambda_2+\lambda_4}$