Chapter 10, Problem 10.81P

(a)

To determine

The small-signal voltage gain for given ${\mathrm{R}}_{\mathrm{L}}$ .

Answer to Problem 10.81P

  ${\mathrm{A}}_{\mathrm{v}}=-1978$

Explanation of Solution

Given:

  $\begin{array}{l}{\mathrm{R}}_1=47\mathrm{k}\mathrm{\Omega }\\ {\mathrm{V}}_{\mathrm{A}\mathrm{N}}=120\mathrm{V}\\ {\mathrm{V}}_{\mathrm{A}\mathrm{P}}=90\mathrm{V}\\ {\mathrm{V}}^{+}=3\mathrm{V}\\ {\mathrm{V}}_{\mathrm{E}\mathrm{B}(\mathrm{o}\mathrm{n})}=0.6\mathrm{V}\end{array}$

Calculation:

The given circuit is,

  

The transistor ${\mathrm{Q}}_1$ and ${\mathrm{Q}}_2$ are matched so expression for reference current will be,

  ${\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=\frac{{\mathrm{V}}^{+}-{\mathrm{V}}_{\mathrm{E}\mathrm{B}(\mathrm{o}\mathrm{n})}}{{\mathrm{R}}_1}$

  $\begin{array}{l}{\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=\frac{3-0.6}{47\times {10}^3}\\ {\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=\frac{2.4}{47\times {10}^3}\\ {\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=51.06\mathrm{\mu }\mathrm{A}\end{array}$

Now calculate the small-signal voltage gain,

  ${\mathrm{A}}_{\mathrm{v}}=\frac{-\left(\frac{{\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}}{{\mathrm{V}}_{\mathrm{T}}}\right)}{\left(\frac{{\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}}{{\mathrm{V}}_{\mathrm{A}\mathrm{N}}}+\frac{1}{{\mathrm{R}}_{\mathrm{L}}}+\frac{{\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}}{{\mathrm{V}}_{\mathrm{A}\mathrm{P}}}\right)}$

Substitute the given values,

  $\begin{array}{l}{\mathrm{A}}_{\mathrm{v}}=\frac{-\left(\frac{51.06\times {10}^{-6}}{0.026}\right)}{\left(\frac{51.06\times {10}^{-6}}{120}+\frac{1}{{\mathrm{R}}_{\mathrm{L}}}+\frac{51.06\times {10}^{-6}}{90}\right)}\\ {\mathrm{A}}_{\mathrm{v}}=\frac{-(1.964\times {10}^{-3})}{\left(4.255\times {10}^{-7}+\frac{1}{{\mathrm{R}}_{\mathrm{L}}}+5.673\times {10}^{-7}\right)}\\ {\mathrm{A}}_{\mathrm{v}}=\frac{-(1.964\times {10}^{-3})}{\left(9.928\times {10}^{-7}+\frac{1}{{\mathrm{R}}_{\mathrm{L}}}\right)}\left(1\right)\end{array}$

Substitute ${\mathrm{R}}_{\mathrm{L}}=\mathrm{\infty }$ in equation (1)

  $\begin{array}{l}{\mathrm{A}}_{\mathrm{v}}=\frac{-(1.964\times {10}^{-3})}{\left(9.928\times {10}^{-7}+\frac{1}{\mathrm{\infty }}\right)}\\ {\mathrm{A}}_{\mathrm{v}}=\frac{-(1.964\times {10}^{-3})}{\left(9.928\times {10}^{-7}+0\right)}\\ {\mathrm{A}}_{\mathrm{v}}=\frac{-(1.964\times {10}^{-3})}{\left(9.928\times {10}^{-7}\right)}\end{array}$

  ${\mathrm{A}}_{\mathrm{v}}=-1978$

Conclusion:

  ${\mathrm{A}}_{\mathrm{v}}=-1978$

(b)

To determine

The small-signal voltage gain for given ${\mathrm{R}}_{\mathrm{L}}$ .

Answer to Problem 10.81P

  ${\mathrm{A}}_{\mathrm{v}}=-454$

Explanation of Solution

Given:

  $\begin{array}{l}{\mathrm{R}}_1=47\mathrm{k}\mathrm{\Omega }\\ {\mathrm{V}}_{\mathrm{A}\mathrm{N}}=120\mathrm{V}\\ {\mathrm{V}}_{\mathrm{A}\mathrm{P}}=90\mathrm{V}\\ {\mathrm{V}}^{+}=3\mathrm{V}\\ {\mathrm{V}}_{\mathrm{E}\mathrm{B}(\mathrm{o}\mathrm{n})}=0.6\mathrm{V}\end{array}$

Calculation:

The given circuit is,

  

The transistor ${\mathrm{Q}}_1$ and ${\mathrm{Q}}_2$ are matched so expression for reference current will be,

  ${\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=\frac{{\mathrm{V}}^{+}-{\mathrm{V}}_{\mathrm{E}\mathrm{B}(\mathrm{o}\mathrm{n})}}{{\mathrm{R}}_1}$

  $\begin{array}{l}{\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=\frac{3-0.6}{47\times {10}^3}\\ {\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=\frac{2.4}{47\times {10}^3}\\ {\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=51.06\mathrm{\mu }\mathrm{A}\end{array}$

Now calculate the small-signal voltage gain,

  ${\mathrm{A}}_{\mathrm{v}}=\frac{-\left(\frac{{\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}}{{\mathrm{V}}_{\mathrm{T}}}\right)}{\left(\frac{{\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}}{{\mathrm{V}}_{\mathrm{A}\mathrm{N}}}+\frac{1}{{\mathrm{R}}_{\mathrm{L}}}+\frac{{\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}}{{\mathrm{V}}_{\mathrm{A}\mathrm{P}}}\right)}$

Substitute the given values,

  $\begin{array}{l}{\mathrm{A}}_{\mathrm{v}}=\frac{-\left(\frac{51.06\times {10}^{-6}}{0.026}\right)}{\left(\frac{51.06\times {10}^{-6}}{120}+\frac{1}{{\mathrm{R}}_{\mathrm{L}}}+\frac{51.06\times {10}^{-6}}{90}\right)}\\ {\mathrm{A}}_{\mathrm{v}}=\frac{-(1.964\times {10}^{-3})}{\left(4.255\times {10}^{-7}+\frac{1}{{\mathrm{R}}_{\mathrm{L}}}+5.673\times {10}^{-7}\right)}\\ {\mathrm{A}}_{\mathrm{v}}=\frac{-(1.964\times {10}^{-3})}{\left(9.928\times {10}^{-7}+\frac{1}{{\mathrm{R}}_{\mathrm{L}}}\right)}\left(1\right)\end{array}$

Substitute ${\mathrm{R}}_{\mathrm{L}}=300\mathrm{k}\mathrm{\Omega }$ in equation (1)

  $\begin{array}{l}{\mathrm{A}}_{\mathrm{v}}=\frac{-(1.964\times {10}^{-3})}{\left(9.928\times {10}^{-7}+\frac{1}{300\times {10}^3}\right)}\\ {\mathrm{A}}_{\mathrm{v}}=\frac{-(1.964\times {10}^{-3})}{\left(9.928\times {10}^{-7}+3.33\times {10}^{-6}\right)}\\ {\mathrm{A}}_{\mathrm{v}}=\frac{-(1.964\times {10}^{-3})}{\left(4.326\times {10}^{-6}\right)}\end{array}$

  ${\mathrm{A}}_{\mathrm{v}}=-454$

Conclusion:

  ${\mathrm{A}}_{\mathrm{v}}=-454$

(c)

To determine

The small-signal voltage gain for given ${\mathrm{R}}_{\mathrm{L}}$ .

Answer to Problem 10.81P

  ${\mathrm{A}}_{\mathrm{v}}=-256$

Explanation of Solution

Given:

  $\begin{array}{l}{\mathrm{R}}_1=47\mathrm{k}\mathrm{\Omega }\\ {\mathrm{V}}_{\mathrm{A}\mathrm{N}}=120\mathrm{V}\\ {\mathrm{V}}_{\mathrm{A}\mathrm{P}}=90\mathrm{V}\\ {\mathrm{V}}^{+}=3\mathrm{V}\\ {\mathrm{V}}_{\mathrm{E}\mathrm{B}(\mathrm{o}\mathrm{n})}=0.6\mathrm{V}\end{array}$

Calculation:

The given circuit is,

  

The transistor ${\mathrm{Q}}_1$ and ${\mathrm{Q}}_2$ are matched so expression for reference current will be,

  ${\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=\frac{{\mathrm{V}}^{+}-{\mathrm{V}}_{\mathrm{E}\mathrm{B}(\mathrm{o}\mathrm{n})}}{{\mathrm{R}}_1}$

  $\begin{array}{l}{\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=\frac{3-0.6}{47\times {10}^3}\\ {\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=\frac{2.4}{47\times {10}^3}\\ {\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}=51.06\mathrm{\mu }\mathrm{A}\end{array}$

Now calculate the small-signal voltage gain,

  ${\mathrm{A}}_{\mathrm{v}}=\frac{-\left(\frac{{\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}}{{\mathrm{V}}_{\mathrm{T}}}\right)}{\left(\frac{{\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}}{{\mathrm{V}}_{\mathrm{A}\mathrm{N}}}+\frac{1}{{\mathrm{R}}_{\mathrm{L}}}+\frac{{\mathrm{I}}_{\mathrm{R}\mathrm{E}\mathrm{F}}}{{\mathrm{V}}_{\mathrm{A}\mathrm{P}}}\right)}$

Substitute the given values,

  $\begin{array}{l}{\mathrm{A}}_{\mathrm{v}}=\frac{-\left(\frac{51.06\times {10}^{-6}}{0.026}\right)}{\left(\frac{51.06\times {10}^{-6}}{120}+\frac{1}{{\mathrm{R}}_{\mathrm{L}}}+\frac{51.06\times {10}^{-6}}{90}\right)}\\ {\mathrm{A}}_{\mathrm{v}}=\frac{-(1.964\times {10}^{-3})}{\left(4.255\times {10}^{-7}+\frac{1}{{\mathrm{R}}_{\mathrm{L}}}+5.673\times {10}^{-7}\right)}\\ {\mathrm{A}}_{\mathrm{v}}=\frac{-(1.964\times {10}^{-3})}{\left(9.928\times {10}^{-7}+\frac{1}{{\mathrm{R}}_{\mathrm{L}}}\right)}\left(1\right)\end{array}$

Substitute ${\mathrm{R}}_{\mathrm{L}}=150\mathrm{k}\mathrm{\Omega }$ in equation (1)

  $\begin{array}{l}{\mathrm{A}}_{\mathrm{v}}=\frac{-(1.964\times {10}^{-3})}{\left(9.928\times {10}^{-7}+\frac{1}{150\times {10}^3}\right)}\\ {\mathrm{A}}_{\mathrm{v}}=\frac{-(1.964\times {10}^{-3})}{\left(9.928\times {10}^{-7}+6.66\times {10}^{-6}\right)}\\ {\mathrm{A}}_{\mathrm{v}}=\frac{-(1.964\times {10}^{-3})}{\left(7.659\times {10}^{-6}\right)}\end{array}$

  ${\mathrm{A}}_{\mathrm{v}}=-256$

Conclusion:

  ${\mathrm{A}}_{\mathrm{v}}=-256$