Question 19.8: Differential pairs are often used as “current switches.” As ...
Differential pairs are often used as “current switches.” As shown in Fig. 19.69, the circuit routes its tail current to either of the outputs according to the large swings controlling the gates of M_1 \text { and } M_2 . Explain what happens at node X during switching. If the tail currents of a large number of differential pairs feed from node X, should this voltage be provided externally?
Recall from Chapter 4 that for the differential pair to experience complete switching, the differential swing \left|V_2-V_1\right| must exceed \sqrt{2}\left(V_{G S}-V_{T H}\right)_{e q}, where \left(V_{G S}-V_{T H}\right)_{e q} is the overdrive of M_1 and M_2 in equilibrium, i.e., if I_{D 1}=I_{D 2}. We denote the voltage at node P when the pair is completely switched by V_{P 1}, and in equilibrium by V_{P 2}. Thus,
V_{P 1}=V_2-\sqrt{2}\left(V_{G S}-V_{T H}\right)_{e q} (19.25)
In equilibrium,
V_{P 2}=\frac{V_1+V_2}{2}-\left(V_{G S}-V_{T H}\right)_{e q} (19.26)
Assuming that V_2-V_1=\sqrt{2}\left(V_{G S}-V_{T H}\right)_{e q}, \text { and hence } V_1=V_2-\sqrt{2}\left(V_{G S}-V_{T H}\right)_{e q} , we have
V_{P 2}=V_2-\left(1+\frac{\sqrt{2}}{2}\right)\left(V_{G S}-V_{T H}\right)_{e q} (19.27)
Thus, V_{P 2} is lower than V_{P 1} by (1-\sqrt{2} / 2)\left(V_{G S}-V_{T H}\right)_{e q}, indicating that during switching, V_P drops by this amount. This voltage change is coupled to node X through the gate-drain overlap capacitance of M_3, disturbing I_{D 3} and hence I_{\text {out } 1} or I_{\text {out } 2}.