Question 7.21: A differential pair with current-source loads can be configu...
A differential pair with current-source loads can be configured to act as a large “floating” resistor [7]. Illustrated in Fig. 7.58(a), the idea is to bias M_1 and M_2 at a very small current so as to obtain a high incremental resistance between A and B, approximately equal to 1/g_{m1} +1/g_{m2}. Determine the noise associated with this resistor. Neglect channel-length modulation.
Viewing A and B as the outputs and modeling the circuit by its Thevenin equivalent, we must determine the noise voltage that appears between these nodes. To this end, we construct the half circuit shown in Fig. 7.58(b) and write the noise voltage at A as
\overline{V_{n, A}^2}=\left(4 k T \gamma g_{m 1}+4 k T \gamma g_{m 3}\right) \frac{1}{g_{m 1}^2}+\frac{K}{(W L)_1 C_{o x}} \frac{1}{f}+\frac{K}{(W L)_3 C_{o x}} \frac{1}{f}\left(\frac{g_{m 3}}{g_{m 1}}\right)^2 (7.127)
The noise measured between A and B is thus equal to
\overline{V_{n, A}^2}=\left(4 k T \gamma g_{m 1}+4 k T \gamma g_{m 3}\right) \frac{1}{g_{m 1}^2}+\frac{K}{(W L)_1 C_{o x}} \frac{1}{f}+\frac{K}{(W L)_3 C_{o x}} \frac{1}{f}\left(\frac{g_{m 3}}{g_{m 1}}\right)^2 (7.128)
We recognize that this resistor is noisier than a simple ohmic resistor of the same value (≈ 2/g_{m1}). It is also much less linear (why?).