Question 7.13: Determine the input-referred noise voltage and current for t...

To compute the input-referred noise voltage, we must short the input port. In this case, we can also short the output port as shown in Fig. 7.38(b), and find the output noise current due to $R_F  and  M_1.$ Since both terminals of $R_F$ are at ac ground, a KVL yields

$\overline{I_{n 1, \text { out }}^2}=\frac{4 k T}{R_F}+4 k T \gamma g_m$                (7.63)

The output impedance of the circuit with the input shorted is simply equal to $R_F$ , yielding

$\overline{V_{n 1, \text { out }}^2}=\left(\frac{4 k T}{R_F}+4 k T \gamma g_m\right) R_F^2$                          (7.64)

We can calculate the input-referred noise voltage by dividing (7.64) by the voltage gain or by dividing (7.63) by the transconductance, $G_m.$ Let us pursue the latter method. As depicted in Fig. 7.38(c),

Dividing (7.63) by $G^2 _m$ gives

$\overline{V_{n, i n}^2}=\frac{\frac{4 k T}{R_F}+4 k T \gamma g_m}{\left(g_m-\frac{1}{R_F}\right)^2}$                          (7.67)

For the input-referred noise current, we first compute the output noise current with the input left open [Fig. 7.38(d)]. Since $V_{n,RF}$ directly modulates the gate-source voltage of $M_1$, producing a drain current of $4kT R_F g^2_m$, we have

$\overline{I_{n 2, \text { out }}^2}=4 k T R_F g_m^2+4 k T \gamma g_m$                                  (7.68)

Next, we must determine the current gain of the circuit according to the arrangement shown in Fig. 7.38(c). Noting that $V_{G S}=I_{i n} R_F$, and hence $I_D=g_m I_{i n} R_F$ we obtain

Dividing (7.68) by the square of the current gain yields

$\overline{I_{n, i n}^2}=\frac{4 k T R_F g_m^2+4 k T \gamma g_m}{\left(g_m R_F-1\right)^2}$                          (7.71)

The reader is encouraged to repeat this analysis using the output noise voltage rather than the output noise current. The above circuit exemplifies cases where the output noise voltage is not the same for short-circuit and open-circuit input ports. The reader can prove that, if the input is left open, then

$\overline{V_{n 2, o u t}^2}=\frac{4 k T \gamma}{g_m}+4 k T R_F$                                                  (7.72)