Chapter 12, Problem 12.11TYU

Consider the BJT feedback circuit in Figure $12.37(\text{a}).$ The transistor parameters are $h_{FE}=180,V_{BE}(\text{ on })=0.7\text{V},$ and $V_A=\infty .$ (a) Determine the voltage $\mathrm{gain}{A}_v=v_o/v_i$ for (i) $R_F=\infty$ and (ii) $R_F=60\text{k}\Omega .$ (b) Repeat part (a) if $h_{FE}$ decreases to $h_{FE}=120.$ (c) By what percent do the magnitudes of the voltage gains change from part (a) to part (b)? (Ans. (a) (i) $A_v=-48.19,$ (ii) $A_{vf}=-5.212$ ; (b) (i) $A_v=-41.72,$ (ii) $A_{vf}=-5.111;$ (c) (i) $-13.4\%$ , (ii) $-1.93\%$ )

To determine

The value of the voltage gain for the given values of $R_F=\infty$ and $R_F=60\text{k}\Omega$ .

Answer to Problem 12.11TYU

The value of the voltage gain for $R_F=\infty$ is $-48.19$ and for $R_F=60\text{k}\Omega$ is $-5.12$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

  $R_F=\infty$ and $R_F=60\text{k}\Omega$ .

Calculation:

Apply KVL to the base emitter terminal of the circuit.

  $V_{TH}-R_{TH}{I}_B-V_{BE}=I_E{R}_E$

The value of the Thevenin equivalent voltage is calculated as,

  $\begin{array}{c}{V}_{TH}=\frac{10\text{V}\left(5.5\text{k}\Omega \right)}{5.5\text{k}\Omega +51\text{k}\Omega }\\ =0.973\text{V}\end{array}$

The value of the Thevenin equivalent resistance is calculated as,

  $\begin{array}{c}{R}_{TH}=\frac{R_1{R}_2}{R_1+R_2}\\ =\frac{\left(5.5\text{k}\Omega \right)\left(51\text{k}\Omega \right)}{\left(5.5\text{k}\Omega \right)+\left(51\text{k}\Omega \right)}\\ =4.96\text{k}\Omega \end{array}$

The emitter current is calculated as,

  $I_E=\left(1+\beta \right)I_B$

The value of the base current is calculated as,

  $\begin{array}{c}{I}_B=\frac{V_{TH}-V_{BE}}{R_{TH}+\left(1+\beta \right)R_E}\\ =\frac{0.973\text{V}-0.7\text{V}}{4.96\text{k}\Omega +\left(180+1\right)\left(0.5\text{k}\Omega \right)}\\ =2.859\text{μA}\end{array}$

The value of the collector current is calculated as,

  $\begin{array}{c}{I}_C=\beta I_B\\ =\left(180\right)\left(2.859\text{μA}\right)\\ =0.515\text{mA}\end{array}$

The value of the small signal resistance is given by,

  $\begin{array}{c}{r}_{\pi }=\frac{V_T}{I_B}\\ =\frac{26\text{mV}}{2.859\text{μA}}\\ =9.09\text{k}\Omega \end{array}$

The expression to determine the voltage gain of the amplifier is calculated as,

  $\begin{array}{c}{A}_V=-g_m{R}_C\\ =\left(\frac{-I_C}{0.026\text{V}}\right)\left(R_C\right)\\ =\left(\frac{-0.515\text{mA}}{0.026\text{V}}\right)\left(10\text{k}\Omega \right)\\ =-198.07\end{array}$

The value of the input resistance is calculated as,

  $\begin{array}{c}{R}_i=r_{\pi }\left|\right|R_{TH}\\ =\frac{\left(9.09\text{k}\Omega \right)\left(4.96\text{k}\Omega \right)}{\left(9.09\text{k}\Omega \right)+\left(4.96\text{k}\Omega \right)}\\ =3.03\text{k}\Omega \end{array}$

The value of the small signal voltage gain is calculated as,

  $\begin{array}{c}{A}_{vs}=\frac{A_V{R}_i}{R_i+R_S}\\ =-198.07\left(\frac{3.03\text{k}\Omega }{3.03\text{k}\Omega +10\text{k}\Omega }\right)\\ =-48.19\end{array}$

By miller theorem the value of the resistance $R_m$ is given by,

  $R_m=\frac{R_f}{1-\frac{1}{A_V}}$

For $A_V\gg 1$ the value $R_m=R_F$ .

From DC analysis the value of the base current is calculated as,

  $\begin{array}{c}{I^{\prime }}_B=I_B+I_F\\ \cong I_B\\ \cong 2.43\text{mA}\end{array}$

The value of the transconductance is calculated as,

  $\begin{array}{c}{g}_m=\frac{I_C}{0.026\text{V}}\\ =\frac{0.515\text{mA}}{0.026\text{V}}\\ =0.0198\text{A}/\text{V}\end{array}$

The value of voltage gain is calculated as,

  $\begin{array}{c}{A}_V=-g_m\left(R_C\left|\right|R_F\right)\\ =-0.0198\text{A}/\text{V}\left(\frac{\left(10\text{k}\Omega \right)\left(82\text{k}\Omega \right)}{\left(10\text{k}\Omega \right)+\left(82\text{k}\Omega \right)}\right)\\ =-\left(0.0198\text{A}/\text{V}\right)8.913\text{k}\Omega \\ =-176.47\end{array}$

The value of the input resistance is calculated as,

  $\begin{array}{c}{R}_i=r_{\pi }\left|\right|R_{TH}\left|\right|R_m\\ =\left(\frac{r_{\pi }{R}_{TH}}{r_{\pi }+R_{TH}}\right)\left|\right|\left(\frac{R_F}{1-A_V}\right)\\ =\left(3.36\text{k}\Omega \right)\left|\right|\left(\frac{60\text{k}\Omega }{1+176.47}\right)\\ =\frac{\left(3.36\text{k}\Omega \right)\left(0.338\text{k}\Omega \right)}{\left(3.36\text{k}\Omega \right)+\left(0.338\text{k}\Omega \right)}\\ =0.307\text{k}\Omega \end{array}$

The voltage gain is calculated as,

  $\begin{array}{c}{A}_{VS}=A_v\left(\frac{R_i}{R_i+R_S}\right)\\ =-176.47\left(\frac{0.307\text{k}\Omega }{0.307\text{k}\Omega +10\text{k}\Omega }\right)\\ =-5.2\end{array}$

Conclusion:

Therefore, the value of the voltage gain for $R_F=\infty$ is $-48.19$ and for $R_F=60\text{k}\Omega$ is $-5.2$ .

To determine

The value of the voltage gain for the given values of $R_F$ .

Answer to Problem 12.11TYU

The value of the voltage gain for $R_F=\infty$ is $-41.69$ and for $R_F=60\text{k}\Omega$ is $-5.11$ .

Explanation of Solution

Given:

  $R_F=\infty$ and $R_F=60\text{k}\Omega$ .

  $h_{FE}=120$

Calculation:

Consider $R_F=\infty$ .

The value of the base current is calculated as,

  $\begin{array}{c}{I}_B=\frac{V_{TH}-V_{BE}}{R_{TH}+\left(\beta +\right)R_E}\\ =\frac{0.973-0.7}{4.96+121\times 0.5}\\ =4.14\text{μA}\end{array}$

The value of the collector current is calculated as,

  $\begin{array}{c}{I}_C=\left(1+\beta \right)I_B\\ =\left(121\right)4.17\text{μA}\\ =0.5\text{mA}\end{array}$

The value of the transconductance is calculated as,

  $\begin{array}{c}{g}_m=\frac{I_C}{V_T}\\ =\frac{0.5\text{mA}}{0.026\text{V}}\\ =0.0192\text{A}/\text{V}\end{array}$

The value of the voltage gain is calculated as,

  $\begin{array}{c}{A}_V=-g_m{R}_C\\ =-0.0192\text{A}/\text{V}\left(10\text{k}\Omega \right)\\ =-192\end{array}$

The value of the voltage gain is calculated as,

  $\begin{array}{c}{A}_V=A_V\left(\frac{R_i}{R_i+R_S}\right)\\ =-192\left(\frac{\left(\frac{r_{\pi }{R}_{TH}}{r_{\pi }+R_{TH}}\right)}{\left(\frac{\left(\frac{\beta }{g_m}\right)R_{TH}}{\frac{\beta }{g_m}+R_{TH}}\right)+R_S}\right)\\ =-192\frac{\left(\frac{\left(\frac{\beta }{g_m}\right)R_{TH}}{\frac{\beta }{g_m}+R_{TH}}\right)}{\left(\frac{\left(\frac{\beta }{g_m}\right)R_{TH}}{\frac{\beta }{g_m}+R_{TH}}\right)+R_S}\\ =\frac{\left(\frac{\left(\frac{120}{0.0192}\right)R_{TH}}{\frac{120}{0.0192}+R_{TH}}\right)}{\left(\frac{\left(\frac{120}{0.0192}\right)R_{TH}}{\frac{120}{0.0192}+R_{TH}}\right)+10\text{k}\Omega }\end{array}$

Solve further as,

  $\begin{array}{c}{A}_{Vs}=\frac{\left(\frac{\left(\frac{120}{0.0192}\right)R_{TH}}{\frac{120}{0.0192}+R_{TH}}\right)}{\left(\frac{\left(6.25\text{k}\Omega \right)\left(4.96\text{k}\Omega \right)}{\left(6.25\text{k}\Omega \right)+\left(4.96\text{k}\Omega \right)}\right)+10\text{k}\Omega }\\ =-41.69\end{array}$

For $R_F=60\text{k}\Omega$

By miller theorem the value of the resistance $R_m$ is given by,

  $R_m=\frac{R_f}{1-\frac{1}{A_V}}$

For $A_V\gg 1$ the value $R_m=R_F$ .

The value of voltage gain is calculated as,

  $\begin{array}{c}{A}_V=-g_m\left(R_C\left|\right|R_F\right)\\ =-0.0192\text{A}/\text{V}\left(\frac{\left(10\text{k}\Omega \right)\left(60\text{k}\Omega \right)}{\left(10\text{k}\Omega \right)+\left(60\text{k}\Omega \right)}\right)\\ =-164.57\end{array}$

The value of the input resistance is calculated as,

  $\begin{array}{c}{R}_i=r_{\pi }\left|\right|R_{TH}\left|\right|R_m\\ =\left(\frac{r_{\pi }{R}_{TH}}{r_{\pi }+R_{TH}}\right)\left|\right|\left(\frac{R_F}{1-A_V}\right)\\ =\left(0.36\text{k}\Omega \right)\left|\right|\frac{60\text{k}\Omega }{1-\left(-164.57\right)}\\ =\frac{\left(3.36\text{k}\Omega \right)\left(0.36\text{k}\Omega \right)}{\left(3.36\text{k}\Omega \right)+\left(0.36\text{k}\Omega \right)}\\ =0.32\text{k}\Omega \end{array}$

The voltage gain is calculated as,

  $\begin{array}{c}{A}_{VS}=A_v\left(\frac{R_i}{R_i+R_S}\right)\\ =-164.57\left(\frac{0.32\text{k}\Omega }{0.32\text{k}\Omega +10\text{k}\Omega }\right)\\ =-5.11\end{array}$

Conclusion:

Therefore, the value of the voltage gain for $R_F=\infty$ is $-41.69$ and for $R_F=60\text{k}\Omega$ is $-5.11$ .

To determine

The percent change in the magnitude of the voltage gain from part (a) to (b).

Answer to Problem 12.11TYU

The percent change in the magnitude for $R_F=\infty$ is $-13.4\%$ and for $R_F=60\text{k}\Omega$ is $-1.93\%$

Explanation of Solution

Given:

  $R_F=\infty$ and $R_F=60\text{k}\Omega$ .

Calculation:

The change in the magnitude of the voltage gain for $R_F=\infty$ is calculated as,

  $\begin{array}{c}\%A=\left(\frac{\left|-41.72\right|-48.19}{48.19}\right)\times 100\\ =-13.4\%\end{array}$

  $R_F=60\text{k}\Omega$

The change in the magnitude of the voltage gain for $R_F=60\text{k}\Omega$ is calculated as,

  $\begin{array}{c}\%A=\left(\frac{\left|-5.111\right|-\left(5.212\right)}{5.212}\right)\times 100\\ =-1.93\%\end{array}$

  $-1.93\%$

Conclusion:

Therefore, the percent change in the magnitude for $R_F=\infty$ is $-13.4\%$ and for $R_F=60\text{k}\Omega$ is $-1.93\%$