Chapter 4, Problem 4.66P

The ac equivalent circuit of a CMOS common−gate circuit is shown in Figure P4.66. The parameters of the NMOS and PMOS transistors are the same as given in Problem 4.61. Determine the (a) small−signal parameters of the transistors, (b) small−signal voltage gain $A_{\upsilon }={\upsilon }_o/{\upsilon }_i$ , (c) input resistance $R_i$ , and (d) output resistance $R_o$ .

Figure P4.66

To determine

The small-signal parameters for the given circuit.

Answer to Problem 4.66P

  $r_{o1}=200k\Omega$ , $r_{o2,3}=133.33\text{ k}\Omega$ , $g_{m1}=0.922\text{mA}/\text{V}$ , $g_{m2,3}=0.632\text{mA}/\text{V}$

Explanation of Solution

Given Information:

The given values are:

  $\begin{array}{l}{V}_{TN}=0.5\text{ V,}{V}_{TP}=-0.5\text{ V,}{\lambda }_n=0.05{\text{ V}}^{-1},{\lambda }_p=0.075{\text{ V}}^{-1}\text{,}{\left(\frac{W}{L}\right)}_1=50\\ {\left(\frac{W}{L}\right)}_{2,3}=50,k_n^{\text{'}}=85\mu \text{A}/{\text{V}}^2,k_p^{\text{'}}=40\mu \text{A}/{\text{V}}^2\text{,}{I}_{DQ}=0.5\text{ mA }\end{array}$

The given circuit is shown below.

  

Calculation:

The small-signal output resistance of transistor M₁

  $r_{o1}=\frac{1}{{\lambda }_n{I}_{DQ}}=\frac{1}{0.05\times 0.1}\text{ k}\Omega$

  $r_{o1}=200k\Omega$

Similarly, the small-signal output resistance of transistors M₂ and M₃

  $\begin{array}{c}{r}_{o2,3}=\frac{1}{{\lambda }_p{I}_{DQ}}\\ =\frac{1}{0.075\times 0.1}\end{array}$

  $r_{o2,3}=133.33\text{ k}\Omega$

Then the transconductance the transistor M₁ ,

  $\begin{array}{l}{g}_{m1}=2\sqrt{\frac{k_n^{\text{'}}}{2}{\left(\frac{W}{L}\right)}_1{I}_{DQ}}\\ g_{m1}=2\sqrt{\frac{0.085}{2}\left(50\right)\left(0.1\right)}\end{array}$

  $g_{m1}=0.922\text{mA}/\text{V}$

Similarly,

  $\begin{array}{l}{g}_{m2}=2\sqrt{\frac{k_p^{\text{'}}}{2}{\left(\frac{W}{L}\right)}_2{I}_{DQ}}\\ g_{m2}=2\sqrt{\frac{0.04}{2}\left(50\right)\left(0.1\right)}\end{array}$

  $g_{m2,3}=0.632\text{mA}/\text{V}$

To determine

The small-signal voltage gain.

Answer to Problem 4.66P

  $A_v=70.51$

Explanation of Solution

Given Information:

The given values are:

  $\begin{array}{l}{V}_{TN}=0.5\text{ V,}{V}_{TP}=-0.5\text{ V,}{\lambda }_n=0.05{\text{ V}}^{-1},{\lambda }_p=0.075{\text{ V}}^{-1}\text{,}{\left(\frac{W}{L}\right)}_1=50\\ {\left(\frac{W}{L}\right)}_{2,3}=50,k_n^{\text{'}}=85\mu \text{A}/{\text{V}}^2,k_p^{\text{'}}=40\mu \text{A}/{\text{V}}^2\text{,}{I}_{DQ}=0.5\text{ mA }\end{array}$

The given circuit is shown below.

  $g_{m1}=0.922\text{mA}/\text{V}$ , $r_{o1}=200k\Omega$ , $r_{o2,3}=133.33\text{ k}\Omega$

Calculation:

Input resistance looking into the source of transistor M₁is ,

  $R_{i1}=\frac{1}{g_{m1}}\text{=}\frac{1}{0.922}\text{=1}\text{.085 }k\Omega$

Then,

  $\begin{array}{l}{V}_{gs1}=-\left(\frac{R_{i1}}{R_{i1}+0.05}\right)V_{in}\\ V_{gs1}=-\left(\frac{1.085}{1.085+0.05}\right)V_{in}\\ V_{gs1}=-0.956V_{in}\end{array}$

The output voltage

  $V_o=-g_{m1}{V}_{gs1}\left(r_{o1}𑨀r_{o2}\right)$

then

  $V_o=-g_{m1}\left(-0.956V_{in}\right)\left(r_{o1}𑨀r_{o2}\right)$

  $A_v=-0.922\left(-0.956\right)\left(200𑨀133.33\right)$

  $A_v=70.51$

To determine

The input resistance of the circuit.

Answer to Problem 4.66P

  $R_i=1.135\text{ k}\Omega$

Explanation of Solution

Given Information:

The given values are:

  $\begin{array}{l}{V}_{TN}=0.5\text{ V,}{V}_{TP}=-0.5\text{ V,}{\lambda }_n=0.05{\text{ V}}^{-1},{\lambda }_p=0.075{\text{ V}}^{-1}\text{,}{\left(\frac{W}{L}\right)}_1=50\\ {\left(\frac{W}{L}\right)}_{2,3}=50,k_n^{\text{'}}=85\mu \text{A}/{\text{V}}^2,k_p^{\text{'}}=40\mu \text{A}/{\text{V}}^2\text{,}{I}_{DQ}=0.5\text{ mA }\end{array}$

The given circuit is shown below.

  $g_{m1}=0.922\text{mA}/\text{V}$ , $r_{o1}=200k\Omega$ , $r_{o2,3}=133.33\text{ k}\Omega$

Calculation:

Input resistance is,

  $R_{i1}=\frac{1}{g_{m1}}\text{=}\frac{1}{0.922}\text{=1}\text{.085 }k\Omega$

Then,

Input resistance of the circuit,

  $R_i=R_{i1}\text{+0}\text{.05=0}\text{.05+1}\text{.085 }k\Omega$

Then $R_i=1.135\text{ k}\Omega$

To determine

The output resistance of the circuit.

Answer to Problem 4.66P

  $R_o=80\text{ k}\Omega$

Explanation of Solution

Given Information:

The given values are:

  $\begin{array}{l}{V}_{TN}=0.5\text{ V,}{V}_{TP}=-0.5\text{ V,}{\lambda }_n=0.05{\text{ V}}^{-1},{\lambda }_p=0.075{\text{ V}}^{-1}\text{,}{\left(\frac{W}{L}\right)}_1=50\\ {\left(\frac{W}{L}\right)}_{2,3}=50,k_n^{\text{'}}=85\mu \text{A}/{\text{V}}^2,k_p^{\text{'}}=40\mu \text{A}/{\text{V}}^2\text{,}{I}_{DQ}=0.5\text{ mA }\end{array}$

The given circuit is shown below.

  

From part (a).

  $g_{m1}=0.922\text{mA}/\text{V}$ , $r_{o1}=200k\Omega$ , $r_{o2,3}=133.33\text{ k}\Omega$

Calculation:

The output resistance of the circuit

  $R_o=r_{o1}𑨀r_{o2}\text{=}\frac{\text{200}\times \text{133}\text{.33}}{333.33}\text{ k}\Omega$

Then,

  $R_o=80\text{ k}\Omega$ .