Consider the source-follower circuit shown in Figure P4.43. The most negative output signal voltage occurs when the transistor just cuts off. Show that this output voltage \(v_{o}(\mathrm{~min})\) is given by

\(v_{o}(\min )=\frac{-I_{D Q} R_{S}}{1+\frac{R_{S}}{R_{L}}}\)

Show that the corresponding input voltage is given by

\(v_{i}(\) min \()=-\frac{I_{D Q}}{g_{m}}\left(1+g_{m}\left(R_{S} \| R_{L}\right)\right)\)

Transcribed Image Text:

CCT VDD CC2 RG Vi RL Rs Figure P4.43 - V