Chapter 11, Problem D11.105DP

The transistor parameters for the circuit in Figure $\text{P}11.105$ are: $K_n=$ $0.2\text{mA}/{\text{V}}^2,V_{TN}=0.8\text{V},$ and $\lambda =0.$ The output resistance of the constant-current source is $R_o=100\text{k}\Omega .$ (a) For $v_1=v_2=0,$ design the circuit such that: $v_{o2}=2\text{V},v_{o3}=3\text{V},v_o=0,I_{DQ3}=0.25\text{mA},$ and $I_{D{Q}^4}=2\text{mA}$ and $A_d=v_o/v_d.$ (c) Determine the common-mode voltage gains $A_{cm1}=$ $v_{o2}/v_{cm}$ and $A_{cm}=v_o/v_{cm},$ and the overall CMRR ${}_{dB}$ .

To determine

To design: The circuit for the given parameters.

Answer to Problem D11.105DP

The design of the given circuit is shown in Figure 2.

Explanation of Solution

Given:

The given circuit is shown below.

  

Figure 1

  $\begin{array}{l}{K}_n=0.2{\text{mA/V}}^{\text{2}},V_{TN}=0.8\text{V}\text{}\text{},R_o=100k\Omega ,\text{}\text{}\lambda =0\\ v_1=v_2=0,v_{o2}=2\text{V},v_{o3}=3\text{V},v_o=0,I_{DQ3}=0.25\text{mA},I_{DQ4}=2\text{mA}\end{array}$

Calculation:

Consider the differential circuit part,

  

The circuit is symmetrical because currents divides between both MOSFETs. Now consider one by one part of the circuit.

  $\begin{array}{l}{I}_Q=\frac{5-2}{R}\\ R=\frac{5-2}{I_Q}=\frac{5-2}{0.5\text{mA}}=\frac{3}{0.5}=6k\Omega \end{array}$

Now find $V_S$ for $M_3$ ,

  $\begin{array}{l}{V}_S=I_{DQ3}\times R_{S1}\\ I_{DQ3}=K_n{\left(V_{GS}-V_T\right)}^2\\ 0.25\text{mA}=0.2{\left(V_{GS}-V_{TN}\right)}^2\\ {\left(V_{GS}-V_{TN}\right)}^2=\frac{0.25}{0.2}\\ {\left(V_{GS}-V_{TN}\right)}^2=1.25\\ \left(V_{GS}-V_{TN}\right)=\sqrt{1.25}\\ \left(V_{GS}-V_{TN}\right)=1.1180\\ V_{GS}=1.118-0.8\left[\because V_{TN}=0.8\text{V}\right]\\ V_{GS}=1.918\text{V}\\ V_{GS}=V_s=1.918\text{V}\\ V_G=V_{o2}=2\text{V}\\ V_{o2}-V_s=1.918\\ V_S=2-1.918=0.082\text{V}\end{array}$

Now find $R_{S1}$ by substituting the calculating values in above equation,

  $\begin{array}{l}{V}_S=I_{DQ3}{R}_S\\ 0.082=\left(0.25\text{mA}\right)R_S\\ R_S=0.082/0.25=328\Omega \end{array}$

Now calculate $V_{S4}$ ,

  $\begin{array}{l}{V}_{S4}=I_{DQ4}{R}_{S2}-5\\ I_{DQ4}=k_N{\left(V_{GS4}-V_{TN}\right)}^2\\ 2\text{mA}=0.2{\left(V_{GS4}-V_{TN}\right)}^2\\ {\left(V_{GS4}-V_{TN}\right)}^2=2\text{mA}/0.2=0.1\\ \left(V_{GS4}-V_{TN}\right)=\sqrt{0.1}=0.3162\\ V_{GS4}=0.3162-0.8\\ V_{GS4}=1.1162\\ V_{G4}-V_{S4}=1.1162\\ V_{S4}=-1.1162+3\left[V_{G4}=V_{o3}=3\text{V}\right]\\ V_{S4}=1.8838\text{V}\\ \text{Now,}\\ R_{S2}=\frac{1.8838+5}{2\times {10}^{-3}}\\ R_{S2}=3441.9\Omega \end{array}$

Now the design is shown below.

  

Figure 2

To determine

The differential-mode gains.

Answer to Problem D11.105DP

  $A_{d1}=1.8973\times {10}^{-3}$

  $A_d=-3.282\times {10}^{-6}$

Explanation of Solution

Given:

The given circuit is shown below.

  

  $\begin{array}{l}{K}_n=0.2{\text{mA/V}}^{\text{2}},V_{TN}=0.8\text{V}\text{}\text{},R_o=100k\Omega ,\text{}\text{}\lambda =0\\ v_1=v_2=0,v_{o2}=2\text{V},v_{o3}=3\text{V},v_o=0,I_{DQ3}=0.25\text{mA},I_{DQ4}=2\text{mA}\end{array}$

  $A_{d1}=v_{o2}/v_d\text{and}{A}_d=v_o/v_d$

Calculation:

Consider the $A_{d1}=v_{o2}/v_d\text{and}{A}_d=v_o/v_d$

  $\begin{array}{l}{A}_d=v_o/v_d=\frac{v_o}{v_{o3}}\times \frac{v_{o3}}{v_{o2}}\times \frac{v_{o2}}{v_d}\\ \text{Find}\frac{v_o}{v_{o3}}\\ \Rightarrow v_o\left(1+g_{m3}{R}_{s2}\right)=g_{m3}{R}_{s2}{v}_{o3}\\ \frac{v_o}{v_{o3}}=\frac{g_{m3}{R}_{s2}}{\left(1+g_{m3}{R}_{s2}\right)}\end{array}$

Put the calculated values in above equation,

  $\begin{array}{l}\frac{v_o}{v_{o3}}=\frac{8.944\times {10}^{-4}\times \mathrm{3.441.9}}{\left(1+8.944\times {10}^{-4}\times \mathrm{3.441.9}\right)}\left[g_{m3}=8.944\times {10}^{-4}\right]\\ \frac{v_o}{v_{o3}}=0.7548\end{array}$

Now,

  $\begin{array}{l}\frac{v_{o2}}{v_d}=\frac{g_{m2}\times 12}{2}\left[g_{m2}=3.1622\times {10}^{-4}\right]\\ \frac{v_{o2}}{v_d}=\frac{3.1622\times {10}^{-4}\times 12}{2}\\ \frac{v_{o2}}{v_d}=1.8973\times {10}^{-3}\\ A_{d1}=1.8973\times {10}^{-3}\end{array}$

Now find,

  $\begin{array}{l}\frac{v_{o3}}{v_{o2}}=\frac{-g_{m3}\times R_D}{1+g_{m3}{R}_{s1}}\left[g_{m3}=3.162\times {10}^{-4}\right]\\ \frac{v_{o3}}{v_{o2}}=\frac{-3.1622\times {10}^{-4}\times 8}{1+3.1622\times {10}^{-4}\times 328}\\ \frac{v_{o3}}{v_{o2}}=-2.292\times {10}^{-3}\end{array}$

Put the values in $A_d=v_o/v_d=\frac{v_o}{v_{o3}}\times \frac{v_{o3}}{v_{o2}}\times \frac{v_{o2}}{v_d}$

  $\begin{array}{l}{A}_d=0.75480\times -2.2920\times {10}^{-3}\times 1.8973\times {10}^{-3}\\ A_d=-3.282\times {10}^{-6}\end{array}$

To determine

The common-mode voltage gains and $CMR{R}_{dB}$ .

Answer to Problem D11.105DP

  $A_{cm1}=0.0548$ , $A_{cm}=0.548$ , $CMR{R}_{dB}=26.24dB$ .

Explanation of Solution

Given:

The given circuit is shown below.

  

  $\begin{array}{l}{K}_n=0.2{\text{mA/V}}^2,V_{TN}=0.8\text{V}\text{}\text{},R_o=100k\Omega ,\text{}\text{}\lambda =0\\ v_1=v_2=0,v_{o2}=2\text{V},v_{o3}=3\text{V},v_o=0,I_{DQ3}=0.25\text{mA},I_{DQ4}=2\text{mA}\end{array}$

  $A_{cm1}=v_{o2}/v_{cm}\text{and}{A}_{cm}=v_o/v_{cm}$

Calculation:

Consider the $A_{cm1}=v_{o2}/v_{cm}\text{and}{A}_{cm}=v_o/v_{cm}$

  $A_{cm}=v_o/v_{cm}$

  $\begin{array}{l}{A}_{cm1}=\frac{v_o}{v_{1cm}}=\frac{v_o}{v_{o3}}\times \frac{v_{o3}}{v_{o2}}\times \frac{v_{o2}}{v_{1cm}}\\ =\left(-0.058\right)\left(-1.71\right)\left(0.548\right)\\ A_{cm1}=0.0548\\ CMRR=\frac{A_d}{A_{cm1}}=\frac{-3.282\times {10}^{-6}}{0.0548}\\ CMR{R}_{dB}=20\log \left(20.527\right)\\ =26.24dB\end{array}$

  $\begin{array}{l}{A}_{cm}=\frac{v_o}{v_{cm}}\\ A_{cm}=\frac{v_o}{v_{cm}}=\frac{g_{m4}\left(1.36k\Omega \right)}{1+g_{m4}\left(1.36k\Omega \right)}\\ =\frac{0.895mA/V\left(1.36k\Omega \right)}{1+0.895mA/V\left(1.36k\Omega \right)}\\ A_{cm}=0.548\end{array}$

Hence the $A_{cm1}=0.0548$ , $A_{cm}=0.548$ , $CMR{R}_{dB}=26.24dB$ .