Chapter 11, Problem 11.92P

a.

To determine

The differential-mode voltage gain $A_d=v_{o3}/v_d$ and the common-mode voltage gain $A_{cm}=v_{o3}/v_d$

Answer to Problem 11.92P

  $\begin{array}{l}{A}_d=-176\\ A_{CM}=0.114\end{array}$

Explanation of Solution

Given:

The given circuit is,

  

  $\beta =200,V_{BE\left(on\right)}=0.7V,V_A=80V$

Calculation:

Consider the given figure,

  $\begin{array}{l}{I}_1=\frac{V^{+}-V_{BE}-V^{-}}{R_1}\\ I_1=\frac{12-0.7-\left(-12\right)}{12\times {10}^3}\\ I_1=1.94mA\\ I_{c4}=I_1=1.94mA\\ I_{cs}=I_{c4}=1.94mA\end{array}$

  $\begin{array}{l}{I}_{c1}=I_{c2}=\frac{I_{c5}}{2}\\ I_{c1}=\frac{1.94mA}{2}=0.97mA\\ g_{m2}=\frac{I_{c2}}{V_T}\\ g_{m2}=\frac{0.97\times {10}^{-3}}{0.026}=37.3mA/V\end{array}$

The voltage $v_{02}$ is,

  $\begin{array}{l}{v}_{02}=V^{+}-I_{c2}{R}_C\\ =12-\left(0.97\times {10}^{-3}\right)\left(8\times {10}^3\right)\\ =4.24V\end{array}$

  $\begin{array}{l}{I}_{c3}=I_{E3}\\ =\frac{v_{o2}-V_{BE}}{R_E}\\ =\frac{4.24-0.7}{3.3\times {10}^3}\\ =1.07mA\end{array}$

  $\begin{array}{l}{r}_{\pi 3}=\frac{\beta V_T}{I_{c3}}\\ =\frac{200\left(0.026\right)}{1.07\times {10}^{-3}}\\ =4.86k\Omega \end{array}$

The input resistance of the CE amplifier is,

  $\begin{array}{l}{R}_{in3}=r_{\pi 3}+\left(1+\beta \right)R_E\\ =4.86\times {10}^3+\left(1+200\right)\left(3.3\times {10}^3\right)\\ =668k\Omega \\ \text{The}\text{gain}\text{is}\\ A_{CE}=\frac{-\beta R_{c2}}{r_{\pi 3}+\left(1+\beta \right)R_E}\\ =\frac{-200\left(8\times {10}^3\right)}{4.86\times {10}^3+\left(1+200\right)\left(3.3\times {10}^3\right)}\\ =-1.197\end{array}$

  $\begin{array}{l}{A}_{d1}=\frac{1}{2}{g}_{m2}\left(R_c\left|\right|R_{in3}\right)\\ A_{d1}=\frac{1}{2}{g}_{m2}\left(\frac{R_c{R}_{in3}}{R_c+R_{in3}}\right)\\ A_{d1}=\frac{1}{2}\left(37.3\times {10}^{-3}\right)\left(\frac{8\times {10}^3\left(668\times {10}^3\right)}{8\times {10}^3+\left(668\times {10}^3\right)}\right)\\ A_{d1}=147.43\\ A_d=\frac{v_{03}}{v_d}\\ =A_{d1}{A}_{CE}\\ =\left(147.43\right)\left(-1.197\right)\\ A_d=-176\end{array}$

Now calculate the output resistance,

  $\begin{array}{l}{R}_o=r_{05}\\ R_o=\frac{V_A}{I_{c5}}=\frac{80}{1.94\times {10}^{-3}}\\ R_o=41.2k\Omega \end{array}$

Now calculate $r_{\pi 2}$ ,

  $\begin{array}{l}{r}_{\pi 2}=\frac{\beta V_T}{I_{c2}}\\ =\frac{200\left(0.026\right)}{0.97\times {10}^{-3}}\\ =5.36k\Omega \\ \text{now,}{A}_{cm1}\\ A_{cm1}=\frac{-g_{m2}\left(R_c\left|\right|R_{in3}\right)}{1+\frac{2\left(1+\beta \right)R_o}{r_{\pi 2}}}\\ A_{cm1}=\frac{-\left(37.3\times {10}^{-3}\right)\frac{8\times {10}^3\left(668\times {10}^3\right)}{8\times {10}^3+\left(668\times {10}^3\right)}}{1+\frac{2\left(1+200\right)\left(41.2\times {10}^3\right)}{5.36\times {10}^3}}\\ A_{cm1}=-0.09539\\ \text{now}{A}_{cm},\\ A_{cm}=A_{cm1}{A}_{CE}\\ A_{cm}=\left(-0.09539\right)\left(-1.197\right)\\ A_{cm}=0.114\end{array}$

Hence,

  $\begin{array}{l}{A}_d=-176\\ A_{CM}=0.114\end{array}$

b.

To determine

The output voltage $v_{03}$ .

To compare: The obtained result with the ideal output.

Answer to Problem 11.92P

  $v_{03}=-5.28\sin \omega tV$

The output voltage obtained for ideal case is greater than the actual result obtained.

Explanation of Solution

Given:

The given circuit is,

  

  $\beta =200,V_{BE\left(on\right)}=0.7V,V_A=80V$

  $v_1=2.015\sin \omega tV\text{and}{v}_2=1.985\sin \omega tV$

Calculation:

Consider the given voltage,

  $v_1=2.015\sin \omega tV\text{and}{v}_2=1.985\sin \omega tV$

The differential input voltage will be,

  $\begin{array}{l}{v}_d=v_1-v_2\\ =2.015\sin \omega t-1.985\sin \omega t\\ =0.03\sin \omega tV\\ \text{now}{v}_{cm}\text{is,}\\ v_{cm}=\frac{v_1+v_2}{2}=\frac{2.015\sin \omega t+1.985\sin \omega t}{2}\\ v_{cm}=2\sin \omega tV\end{array}$

Now calculate output voltage,

  $\begin{array}{l}{v}_{03}=A_d{v}_d+A_{cm}{v}_{cm}\\ =-176\left(0.03\sin \omega t\right)+0.114\left(2\sin \omega t\right)\\ =-5.052\sin \omega tV\end{array}$

Now calculate output voltage for ideal case,

Put $A_{cm}=0$ for ideal case,

  $\begin{array}{l}{v}_{03}=A_d{v}_d+A_{cm}{v}_{cm}\\ =-176\left(0.03\sin \omega t\right)+0\left(2\sin \omega t\right)\\ =-5.28\sin \omega tV\end{array}$

Hence,

  $v_{03}=-5.28\sin \omega tV$

It is observed that both the values are slightly different. The output voltage obtained for ideal case is greater than the actual result obtained.

c.

To determine

The differential-mode and common-mode input resistances.

Answer to Problem 11.92P

  $\begin{array}{l}{R}_{id}=10.72k\Omega \\ R_{icm}=4.15M\Omega \end{array}$

Explanation of Solution

Given:

The given circuit is,

  

  $\beta =200,V_{BE\left(on\right)}=0.7V,V_A=80V$

Calculation:

Let calculatethe differential-mode input resistances,

  $\begin{array}{l}{R}_{id}=r_{\pi 1}+r_{\pi 2}\\ =2r_{\pi 2}\left[\because r_{\pi 1}=r_{\pi 2}\right]\\ =2\left(5.36\times {10}^3\right)\\ =10.72k\Omega \end{array}$

  $\begin{array}{l}{r}_{02}=\frac{V_A}{I_{c2}}\\ =\frac{80}{0.97\times {10}^{-3}}\\ =82.5k\Omega \end{array}$

The common-mode input resistances will be,

  $\begin{array}{l}2R_{icm}=\left[\left(1+\beta \right)2R_0\right]\left|\right|\left[\left(1+\beta \right)r_{02}\right]\\ 2R_{icm}=\left[\left(1+200\right)2\left(41.2\times {10}^3\right)\right]\left|\right|\left[\left(1+200\right)\left(82.5\times {10}^3\right)\right]\\ 2R_{icm}=16.6\times {10}^6\left|\right|16.6\times {10}^6\\ 2R_{icm}=\frac{16.6\times {10}^6\times 16.6\times {10}^6}{16.6\times {10}^6+16.6\times {10}^6}\\ 2R_{icm}=8.3\\ R_{icm}=8.3/2\\ R_{icm}=4.15M\Omega \end{array}$

Hence,

  $\begin{array}{l}{R}_{id}=10.72k\Omega \\ R_{icm}=4.15M\Omega \end{array}$