Exercise 4.10 Basic amplifiers with an NPN and an NMOS transistor

a)

An answer:
$$\begin{array}{rcll}{a}_v& =-g_m{R}_C& & \\ g_m& =40\cdot I_C& & \\ I_C& =\frac{V_{R_C}}{R_C}& & \end{array}$$

from which it follows that the maximum achievable (absolute) voltage gain is

$$\begin{array}{rcll}& |a_v|& =40\cdot V_{\mathit{RC}}& \end{array}$$

which is directly determined by (and limited by) the supply voltage. Note that the $40=q∕\mathit{kT}\approx 40$ at room temperature; at low temperatures this factor is higher.

b)

An answer:
Using the square law: $$\begin{array}{rcll}{a}_v& =-g_m{R}_D& & \\ g_m& =\sqrt{2K{I}_D}& & \\ I_D& =\frac{V_{R_D}}{R_D}& & \end{array}$$

from which it follows that the maximum achievable (absolute) voltage gain is

$$\begin{array}{rcll}& \left|a_v\right|=\sqrt{2K\frac{V_{R_D}}{R_D}}\cdot R_D=\sqrt{2K{R}_D{V}_{R_D}}& & \end{array}$$

From which it follows that for maximum (absolute) voltage gain, the $K$ and the $R_D$ should be as large as possible and consequently the $I_D$ should be set very low. There seems to be no limit to these which would indicate that $|a_v|$ can be set arbitrarily high.

c)

An answer:
For the MOS question, you probably used the square law. This square low is not valid for $v_{\mathit{GS}}-V_T<\mathit{kT}∕q$ as then a different operation region is entered: moderate inversion (MI) or even weak inversion (WI). Weak inversion is ignored in this course: there the MOS transistor is modelled as being ”off”. Moderate inversion is not dealt with in this book as it is a transition region between strong inversion (square law) and the (here neglected) weak inversion region. This simplification for the MOS element equations makes derivation (results) for which $v_{\mathit{GS}}-V_T<\mathit{kT}∕q$ far from realistic. Including MI and WI equations would show that MOS amplifiers end up at lower gain than for the bipolar case.