Chapter 11, Problem 11.90P

Consider the multistage bipolar circuit in Figure P 11.90, in which de base currents are negligible. Assume the transistor parameters are $\beta =120$ $V_{BE}(\text{on})=0.7\text{V},$ and $V_A=\infty .$ The output resistance of the constant current source is $R_o=200\text{k}\Omega .$ (a) For $v_1=v_2=-1.5\text{V},$ design the circuit such that $v_{O2}=v_O=0,I_{CQ3}=0.25\text{mA},$ and $I_{C{Q}^4}=2\text{mA}$ (b) Assuming $C_E$ acts as a short circuit, determine the differential-mode voltage gains $A_{d1}=v_{o2}/v_d$ and $A_d=v_o/v_d.$ (c) Determine the common mode gains $A_{cm1}=v_{o2}/v_d$ and $A_{cm}=v_o/v_d,$ and the overall ${\text{CMRR}}_{\text{dB}}$ .

To determine

The design parameters for the circuit.

Answer to Problem 11.90P

The value of the resistances to design the circuit is $R_{E1}=17.2\text{k}\Omega$ , $R_C=17.2\text{k}\Omega$ , $R_{E2}=2.5\text{k}\Omega$ and $R=20\text{k}\Omega$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

The given values are $v_{O2}=v_O=0$ , $I_{CQ3}=0.25\text{mA}$ and $I_{CQ4}=2\text{mA}$ .

Calculation:

The collector for both the transistors are equal and is given by,

  $\begin{array}{c}{I}_C=\frac{I_Q}{2}\\ =\frac{0.5\text{mA}}{2}\\ =0.25\text{mA}\end{array}$

The expression for the voltage $v_{O2}$ is given by,

  $v_{O2}=5\text{V}-I_{C2}R$

Substitute $0\text{V}$ for $v_{O2}$ , $0.25\text{mA}$ for $I_{C2}$ in the above equation.

  $\begin{array}{l}0=5\text{V}-\left(0.25\text{mA}\right)R\\ R=20\text{k}\Omega \end{array}$

The value of the voltage $v_{O3}$ is calculated as,

  $\begin{array}{l}{V}_{BE4}=V_{B4}-V_{E4}\\ 0.7\text{V}=v_{O3}-v_O\\ 0.7\text{V}=v_{O3}-0\\ v_{O3}=0.7\text{V}\end{array}$

The value of the resistance $R_{E2}$ is given by,

  $\begin{array}{l}{v}_O=-5\text{V}-\left(-I_{CQ4}\right)R_{E2}\\ 0=-5\text{V}+2\text{mA}\left(R_{E2}\right)\\ R_{E2}=2.5\text{k}\Omega \end{array}$

The value of the resistance $R_C$ is calculated as,

  $\begin{array}{c}{R}_C=\frac{5\text{V}-\text{0}\text{.7}\text{V}}{I_{CQ3}}\\ =\frac{5\text{V}-\text{0}\text{.7}\text{V}}{0.25\text{mA}}\\ =17.2\text{k}\Omega \end{array}$

The value of the resistance $R_{E1}$ is calculated as,

  $\begin{array}{l}{V}_{BE3}=V_{B3}-V_{E3}\\ 0.7\text{V}=v_{O2}-v_{E3}\\ 0.7\text{V}=0-\left(-5\text{V}+\left(0.25\text{mA}\right)R_{E1}\right)\\ R_{E1}=17.2\text{k}\Omega \end{array}$

Conclusion:

Therefore, the value of the resistances to design the circuit is $R_{E1}=17.2\text{k}\Omega$ , $R_C=17.2\text{k}\Omega$ , $R_{E2}=2.5\text{k}\Omega$ and $R=20\text{k}\Omega$ .

To determine

The value of the differential mode voltage gain $A_{d1}=\frac{v_{O2}}{v_d}$ and $A_d=\frac{v_O}{v_d}$ .

Answer to Problem 11.90P

The value of the differential voltage gain is $-5752$ .

Explanation of Solution

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

The expression for the gain $A_{d1}$ is evaluated as,

  $\begin{array}{c}{A}_{d1}=\frac{v_{O2}}{\left|v_1-v_2\right|}\\ =\frac{-g_{m1}\left(R\left|\right|r_{\pi 3}\right)v_{BE2}}{v_{id}}\\ =\frac{-\left(\frac{I_{C2}}{0.026\text{V}}\right)\left(R\left|\right|\left(\frac{\beta V_T}{I_{CQ3}}\right)\right)\left(\frac{-v_{id}}{}\right)}{v_{id}}\\ =\frac{-\left(\frac{I_{C2}}{0.026\text{V}}\right)\left(\frac{R\left(\frac{\beta V_T}{I_{CQ3}}\right)}{R+\left(\frac{\beta V_T}{I_{CQ3}}\right)}\right)v_{BE2}}{v_{id}}\\ =\frac{I_Q}{4\left(0.026\text{V}\right)}\frac{R\left(\frac{\beta V_T}{I_{CQ3}}\right)}{R+\left(\frac{\beta V_T}{I_{CQ3}}\right)}\end{array}$

The value of the gain $A_{d1}$ is calculated as,

  $\begin{array}{c}{A}_{d1}=\frac{0.5\text{mA}}{4\left(0.026\text{V}\right)}\frac{\left(20\text{k}\Omega \right)\left(\frac{\left(120\right)\left(0.026\text{V}\right)}{0.25\text{mA}}\right)}{20\text{k}\Omega +\left(\frac{\left(120\right)\left(0.026\text{V}\right)}{0.25\text{mA}}\right)}\\ =\frac{0.5\text{mA}}{4\left(0.026\text{V}\right)}\frac{\left(20\text{k}\Omega \right)\left(12.48\text{k}\Omega \right)}{20\text{k}\Omega +\left(\frac{\left(120\right)\left(0.026\text{V}\right)}{0.25\text{mA}}\right)}\\ =36.94\end{array}$

The value of the resistance $R_{i4}$ is calculated as,

  $\begin{array}{c}{R}_{i4}=r_{\pi 4}+\left(1+\beta \right)R_{E2}\\ =\frac{\beta \left(0.026\text{V}\right)}{I_{CQ4}}+\left(1+\beta \right)R_{E2}\\ =\frac{\left(120\right)\left(0.026\text{V}\right)}{2\text{mA}}+\left(1+120\right)\left(2.5\text{k}\Omega \right)\\ =304\text{k}\Omega \end{array}$

The value of the gain $A_3$ is calculated as,

  $\begin{array}{c}{A}_3=-g_{m3}\left(R_C\left|\right|R_{i4}\right)\\ =-\frac{I_{CQ3}}{0.026\text{V}}\left(\frac{\left(17.2\text{k}\Omega \right)\left(304\text{k}\Omega \right)}{17.2\text{k}\Omega +\left(304\text{k}\Omega \right)}\right)\\ =-\frac{0.25\text{mA}}{0.026\text{V}}\left(\frac{\left(17.2\text{k}\Omega \right)\left(304\text{k}\Omega \right)}{17.2\text{k}\Omega +\left(304\text{k}\Omega \right)}\right)\\ =-156.5\end{array}$

The value of the gain $A_4$ is calculated as,

  $\begin{array}{c}{A}_3=\left(\frac{\left(1+\beta \right)R_{E2}}{r_{\pi 4}+\left(1+\beta \right)R_{E2}}\right)\\ =\frac{\left(1+120\right)2.5\text{k}\Omega }{\frac{\beta \left(0.026\text{V}\right)}{I_{CQ4}}+\left(1+120\right)2.5\text{k}\Omega }\\ =\frac{\left(1+120\right)2.5\text{k}\Omega }{\frac{\left(120\right)\left(0.026\text{V}\right)}{2\text{mA}}+\left(1+120\right)2.5\text{k}\Omega }\\ =0.995\end{array}$

The value of the voltage gain $A_d$ is given by,

  $A_d=A_{d1}{A}_3{A}_4$

Substitute $36.94$ for $A_{d1}$ , $-156.5$ for $A_{d3}$ and $0.995$ for $A_4$ in the above equation.

  $\begin{array}{c}{A}_d=\left(36.94\right)\left(-156.5\right)\left(0.995\right)\\ =-5752\end{array}$

Conclusion:

Therefore, the value of the differential voltage gain is $-5752$

To determine

The value of $A_{cm1}$ , $A_{cm}$ and ${\text{CMRR}}_{\text{dB}}$ .

Answer to Problem 11.90P

The values of the gain are $A_{cm1}=-0.01905$ and $A_{cm}=2.966$ . The value of ${\text{CMRR}}_{\text{dB}}$ is $65.8\text{dB}$ .

Explanation of Solution

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

The value of the voltage $A_{cm1}$ is calculated as,

  $\begin{array}{c}{A}_{cm1}=\frac{-g_m\left(\frac{R{r}_{\pi 3}}{R+r_{\pi 3}}\right)}{1+\frac{2\left(1+\beta \right)R_O}{r_{\pi 1}}}\\ =\frac{-\left(9.615\text{mA}/\text{V}\right)\left(\frac{\left(20\text{k}\Omega \right)\left(12.48\text{k}\Omega \right)}{\left(20\text{k}\Omega \right)+\left(12.48\text{k}\Omega \right)}\right)}{1+\frac{2\left(1+120\right)200\text{k}\Omega }{12.48\text{k}\Omega }}\\ =-0.01905\end{array}$

The value of the common mode voltage $A_{cm}$ is calculated as,

  $\begin{array}{c}{A}_{cm}=\frac{v_O}{V_{cm}}\\ =\frac{v_{O2}}{V_{cm}}\frac{v_{O3}}{V_{O2}}\frac{v_O}{V_{O3}}\\ =A_{cm1}{A}_3{A}_4\\ =\left(-0.01905\right)\left(-156.5\right)\left(0.995\right)\end{array}$

Solve further as,

  $A_{cm}=2.966$

The value of the common mode rejection ration is calculated as,

  $\begin{array}{c}{\text{CMRR}}_{\text{dB}}=20{\log }_{10}\left|\frac{A_d}{A_{cm}}\right|\\ =20{\log }_{10}\left|\frac{-5752}{2.96}\right|\\ =65.8\text{dB}\end{array}$

Conclusion:

Therefore, the values of the gain are $A_{cm1}=-0.01905$ and $A_{cm}=2.966$ . The value of ${\text{CMRR}}_{\text{dB}}$ is $65.8\text{dB}$ .