Chapter 11, Problem 11.84P

a.

To determine

The small-signal voltage gain $A_v=v_o/v_i$ and the output resistance $R_o$ for the given circuit.

Answer to Problem 11.84P

  $\begin{array}{l}{A}_v=-30053\\ R_o=192.2M\Omega \end{array}$

Explanation of Solution

Given:

The given circuit is,

  

  $I_Q=25\mu A,\beta =100V_A=50V,V_{TN}=0.8V,K_n=0.25mA/V^2,\lambda =0.02V^{-1}$

The two amplifying transistors $M_1\text{and}{Q}_1$ are biased in the saturation region and forward-active region.

Calculation:

Consider the given figure,

  $I_c=I_Q=25\mu A,$

Now find trans-conductance,

  $\begin{array}{l}{g}_{m2}=\frac{I_c}{V_T}=\frac{25\times {10}^{-6}}{26\times {10}^{-3}}\\ g_{m2}=0.9615mA/V\\ \text{Now}\text{find}{r}_{\pi },\\ r_{\pi }=\frac{\beta V_T}{I_c}=\frac{\left(100\right)26\times {10}^{-3}}{25\times {10}^{-6}}\\ r_{\pi }=104k\Omega \\ \text{now}{g}_{m1}\\ g_{m1}=2\sqrt{k_p{I}_Q}\\ g_{m1}=2\sqrt{0.25\times {10}^{-3}\left(25\times {10}^{-6}\right)}\\ g_{m1}=0.158mA/V\end{array}$

Now calculate $r_{01},r_{02}$ ,

  $\begin{array}{l}{r}_{01}=\frac{1}{\lambda I_Q}=\frac{1}{\left(0.02\right)25\mu A}\\ r_{01}=2M\Omega \\ r_{02}=\frac{V_A}{I_Q}=\frac{50}{25\mu A}\\ r_{02}=2M\Omega \end{array}$

Now consider small-signal equivalent circuit,

  

Apply nodal method at node $V_{\pi }$ ,

  $\begin{array}{l}-\frac{V_{\pi }}{r_{\pi }}-\frac{V_{\pi }}{r_{01}}+g_{m1}{V}_i=0\\ V_{\pi }=\left(\frac{r_{\pi }{r}_{01}{g}_{m1}}{r_{\pi }+r_{01}}\right)V_i\\ \text{Put}\text{the}\text{values},\\ V_{\pi }=\left(\frac{\left(104\times {10}^3\right)\left(2\times {10}^6\right)\left(0.158\times {10}^{-3}\right)}{\left(104\times {10}^3\right)+\left(2\times {10}^6\right)}\right)V_i\\ V_{\pi }=15.62V_i\end{array}$

Now apply nodal method at node $V_o$ ,

  $\begin{array}{l}\frac{V_o-\left(-V_{\pi }\right)}{r_{02}}{g}_{m2}{V}_{\pi }=0\\ V_o\left(\frac{1}{r_{02}}\right)+V_{\pi }\left(\frac{1}{r_{02}}+g_{m2}\right)=0\\ \text{Put}\text{the}\text{values}\text{,}\\ V_o\left(\frac{1}{2\times {10}^6}\right)+15.62V_i\left(\frac{1}{2\times {10}^6}+0.9615\times {10}^{-3}\right)=0\\ V_o=-30053V_i\\ A_v=-30053\end{array}$

Now consider,

  

Find $V_{\pi }$ ,

  $\begin{array}{l}{V}_{\pi }=-I_x\left(r_{01}\left|\right|r_{\pi }\right)\\ V_{\pi }=-I_x\left(\frac{r_{01}{r}_{\pi }}{r_{01}+r_{\pi }}\right)\\ V_{\pi }=-I_x\left(\frac{\left(2\times {10}^6\right)\left(104\times {10}^3\right)}{\left(2\times {10}^6\right)+\left(104\times {10}^3\right)}\right)\\ V_{\pi }=-9868\times {10}^3{I}_x\end{array}$

Apply nodal analysis at node $V_x$ ,

  $\begin{array}{l}{I}_x=g_{m2}{V}_{\pi }+\frac{V_x-\left(-V_{\pi }\right)}{r_{02}}\\ I_x=g_{m2}{V}_{\pi }+\frac{V_x+V_{\pi }}{r_{02}}\\ I_x=-\left(0.9615\times {10}^{-3}\right)\left(98.86\times {10}^3\right)I_x+\frac{V_x-\left(98.86\times {10}^3\right)I_x}{\left(2\times {10}^6\right)}\\ \frac{V_x}{I_x}=R_o=96.1\left(2\times {10}^6\right)\\ R_o=192.2M\Omega \end{array}$

Hence,

  $\begin{array}{l}{A}_v=-30053\\ R_o=192.2M\Omega \end{array}$

b.

To determine

The small-signal voltage gain $A_v=v_o/v_i$ and the output resistance $R_o$ for the given circuit.

Answer to Problem 11.84P

  $\begin{array}{l}{A}_v=-5.80\times {10}^7\\ R_o=192.2M\Omega \end{array}$

Explanation of Solution

Given:

The given circuit is,

  

  $I_Q=25\mu A,\beta =100V_A=50V,V_{TN}=0.8V,K_n=0.25mA/V^2,\lambda =0.02V^{-1}$

The two amplifying transistors $M_{1}and{Q}_1$ are biased in the saturation region and forward-active region.

Calculation:

Consider the given figure,

  $I_c=I_Q=25\mu A,$

Now find trans-conductance,

  $\begin{array}{l}{g}_{m1}=\frac{I_c}{V_T}=\frac{25\times {10}^{-6}}{26\times {10}^{-3}}\\ g_{m1}=0.9615mA/V\\ g_{m1}=g_{m2}=0.9615mA/V\\ \text{Now}\text{find}{r}_{\pi },\text{for}{Q}_1\\ r_{\pi 1}=\frac{\beta V_T}{I_c}=\frac{\left(100\right)26\times {10}^{-3}}{25\times {10}^{-6}}\\ r_{\pi 1}=104k\Omega \\ r_{\pi 1}=r_{\pi 2}=104k\Omega \end{array}$

Now,

  $\begin{array}{l}{g}_{m3}=2\sqrt{k_p{I}_Q}\\ g_{m3}=2\sqrt{0.25\times {10}^{-3}\left(25\times {10}^{-6}\right)}\\ g_{m3}=0.158mA/V\end{array}$

  $\begin{array}{l}{r}_{03}=\frac{1}{\lambda I_Q}\\ r_{03}=\frac{1}{0.02\left(25\times {10}^{-6}\right)}\\ r_{03}=2M\Omega \\ r_{01}=\frac{V_A}{I_Q}\\ r_{01}=\frac{50}{\left(25\times {10}^{-6}\right)}\\ r_{01}=2M\Omega \\ r_{01}=r_{02}=2M\Omega \end{array}$

Now consider circuit,

  

Apply nodal at top part of the circuit,

  $\begin{array}{l}\frac{V_o-\left(-V_{gs}\right)}{r_{03}}{g}_{m3}{V}_{gs}=0\\ V_o\left(\frac{1}{r_{03}}\right)=+V_{gs}\left(\frac{1}{r_{03}}+g_{m3}\right)\\ V_{gs}=-\left(\frac{1}{1+g_{m3}{r}_{03}}\right)V_o\\ \text{put}\text{the}\text{values}\text{,}\\ V_{gs}=-\left(\frac{1}{1+\left(0.158\times {10}^{-3}\right)2\times {10}^6}\right)V_o\\ V_{gs}=-V_o/317\end{array}$

Apply nodal at middle part of the circuit,

  $\begin{array}{l}\frac{-V_{gs}-\left(-V_{\pi 2}\right)}{r_{02}}{g}_{m2}{V}_{\pi 2}=0\\ V_{gs}\left(\frac{1}{r_{02}}\right)+V_{\pi 2}\left(\frac{1}{r_{02}}+g_{m2}\right)=0\\ V_{\pi 2}=\left(\frac{1}{1+g_{m2}{r}_{02}}\right)V_{gs}\\ \text{put}\text{the}\text{values}\text{,}\\ V_{\pi 2}=\left(\frac{1}{1+0.9615\times {10}^{-3}\left(2\times {10}^6\right)}\right)\left(-\frac{V_o}{317}\right)\\ V_{\pi 2}=-V_o/725348\end{array}$

  $\begin{array}{l}{V}_o=-117.28\times {10}^7{V}_i\\ V_{\pi }=1923V_i\\ V_{gs}=3699852V_i\end{array}$

  $\begin{array}{l}{A}_v=V_o/V_i\\ A_v=-5.80\times {10}^7\end{array}$

Find $V_{\pi }$ ,

  $\begin{array}{l}{V}_{\pi }=-I_x\left(r_{01}\left|\right|r_{\pi }\right)\\ V_{\pi }=-I_x\left(\frac{r_{01}{r}_{\pi }}{r_{01}+r_{\pi }}\right)\\ V_{\pi }=-I_x\left(\frac{\left(2\times {10}^6\right)\left(104\times {10}^3\right)}{\left(2\times {10}^6\right)+\left(104\times {10}^3\right)}\right)\\ V_{\pi }=-9868\times {10}^3{I}_x\end{array}$

Apply nodal analysis at node $V_x$ ,

  $\begin{array}{l}{I}_x=g_{m2}{V}_{\pi }+\frac{V_x-\left(-V_{\pi }\right)}{r_{02}}\\ I_x=g_{m2}{V}_{\pi }+\frac{V_x+V_{\pi }}{r_{02}}\\ I_x=-\left(0.9615\times {10}^{-3}\right)\left(98.86\times {10}^3\right)I_x+\frac{V_x-\left(98.86\times {10}^3\right)I_x}{\left(2\times {10}^6\right)}\\ \frac{V_x}{I_x}=R_o=96.1\left(2\times {10}^6\right)\\ R_o=192.2M\Omega \end{array}$

Hence,

  $\begin{array}{l}{A}_v=-5.80\times {10}^7\\ R_o=192.2M\Omega \end{array}$