Chapter 13, Problem 13.24P

To determine

The output resistance of the 741 op-amp if $Q_{14}$ is conducting and $Q_{20}$ is cut off.

Answer to Problem 13.24P

The output resistance is

  $R_0=69.2\text{Ω}$

Explanation of Solution

Given:

Consider the 741 op-amp having bias voltage ±5 V

  

Calculation:

The resistance at $Q_{14}$ can be define as

  $R_{e14}=\frac{r_{\pi 14}+R_{01}}{1+{\beta }_P}\dots \dots (1)$

And the output resistance is defined as

  $R_0=R_{e14}+R_6\mathrm{...........}(2)$

Now for solving equation (1) we have

Assuming series resistance of $Q_{18}$ and $Q_{19}$ is small then

  $R_{01}=r_{o13A}\parallel R_{e22}\dots \dots (3)$

Where,

  $R_{e22}=\frac{r_{\pi 22}+R_{o17}\parallel r_{o13B}}{1+{\beta }_P}\mathrm{.........}(4)$

And

  $R_{o17}=r_{o17}\left[1+g_{m17}\left(R_8\parallel r_{m17}\right)\right]\dots \dots (5)$

Where,

  $r_{\pi 17}=\frac{{\beta }_n{V}_T}{I_{C17}}\dots \dots (6)$

  $r_{017}=\frac{V_A}{I_{C17}}\dots \dots (7)$

The collector current in $Q_{17}$ is

  $I_{C17}=0.75I_{REF}$

The reference current is.

  $\begin{array}{l}{I}_{REF}=\frac{V^{+}-V_{EB12}-V_{BE11}-V^{-}}{R_5}\\ =\frac{15-0.6-0.6-(-15)}{40}\\ =0.72\text{mA}\end{array}$

The collector current in $Q_{17}$ is

  $\begin{array}{l}{I}_{C17}=0.75I_{REF}\\ =0.75\times 0.72\times {10}^{-3}\\ =0.54\text{mA}\end{array}$ .

Assuming ${\beta }_n=200$

  $\begin{array}{l}{r}_{\pi 17}=\frac{{\beta }_n{V}_T}{I_{C17}}\\ =\frac{200\times 0.026}{0.54\times {10}^{-3}}=9.63\text{kΩ}\end{array}$

And from equation (7),

  $\begin{array}{l}{r}_{o17}=\frac{V_A}{I_{C17}}\\ =\frac{50}{0.54\times {10}^{-3}}=\frac{50}{0.54\times {10}^{-3}}\end{array}$

Now, the current $I_{C13A}$ is

  $\begin{array}{l}{I}_{C13A}=0.25I_{REF}\\ =0.25\times 0.72\times {10}^{-3}=0.18\text{mA}\end{array}$

And the current $I_{C13B}$ is

  $\begin{array}{l}{I}_{C13B}=0.75\times I_{REF}\\ =0.75\times 0.72\times {10}^{-3}=0.54\text{mA}\end{array}$

Now the resistance $r_{\pi 22}$ is

  $\begin{array}{l}{r}_{\pi 22}=\frac{{\beta }_p{V}_T}{I_{C13A}}\\ \text{Where}{\beta }_p=50\\ r_{\pi 2}=\frac{50\times 0.026}{0.18\times {10}^{-3}}=7.22\text{kΩ}\end{array}$

The resistance $r_{o13B}$ is

  $\begin{array}{l}{r}_{o13B}=\frac{V_A}{I_{C13B}}\\ =\frac{50}{0.54\times {10}^{-3}}=92.6\text{kΩ}\end{array}$

  $\begin{array}{l}\text{And the resistance}{r}_{o13A}\text{is}\\ r_{o13A}=\frac{V_A}{I_{C13A}}=\frac{50}{0.18\times {10}^{-3}}=278\text{kΩ}\end{array}$

Now from the equation (5),

  $g_{m17}=20.8\text{mA}/\text{V}$ then

  $\begin{array}{l}{R}_{017}=r_{017}\left[1+g_{m17}\left(R_8\parallel r_{\pi 17}\right)\right]\\ =92.6\times {10}^3\left[1+\left(20.8\times {10}^{-3}\right)\left(0.1\times {10}^3\parallel 9.63\times {10}^3\right)\right]\\ =92.6\times {10}^3\left[1+\left(20.8\times {10}^{-3}\right)\left(\frac{\left(0.1\times {10}^3\right)\left(9.63\times {10}^3\right)}{0.1\times {10}^3+9.63\times {10}^3}\right)\right]\end{array}$

  $\begin{array}{l}=\left(92.6\times {10}^3\right)\left[1+\left(20.8\times {10}^{-3}\right)\left(\frac{0.963\times {10}^3}{9.73}\right)\right]\hfill \\ =\left(92.6\times {10}^3\right)\left[1+\left(20.8\times {10}^{-3}\right)\left(0.098\times {10}^3\right)\right]\hfill \\ =\left(92.6\times {10}^3\right)[1+2.058]\hfill \\ =\left(92.6\times {10}^3\right)(3.058)\hfill \\ R_{017}=283\text{kΩ}\hfill \end{array}$

Now from equation (4),

  $\begin{array}{l}{R}_{e22}=\frac{r_{\pi 22}+R_{017}\parallel r_{013B}}{1+{\beta }_P}\\ =\frac{7.22\times {10}^3+\left(283\times {10}^3\parallel 92.6\times {10}^3\right)}{1+50}\\ =\frac{7.22\times {10}^3+\frac{\left(283\times {10}^3\right)\left(92.6\times {10}^3\right)}{283\times {10}^3+92.6\times {10}^3}}{51}\end{array}$

  $\begin{array}{l}=\frac{7.22\times {10}^3+\frac{26205.8\times {10}^3}{375.6}}{51}\hfill \\ =\frac{7.22\times {10}^3+69.77\times {10}^3}{51}\hfill \\ =\frac{76.99\times {10}^3}{51}\hfill \\ R_{e22.}=1.51\text{kΩ}\hfill \end{array}$

From the equation (3),

  $\begin{array}{l}{R}_{01}=r_{o13A}\parallel R_{e22}\\ =278\times {10}^3\parallel 1.51\times {10}^3\\ =278\times {10}^{3}01.51\times {10}^3\\ =\frac{\left(278\times {10}^3\right)\left(1.51\times {10}^3\right)}{278\times {10}^3+1.51\times {10}^3}\\ =\frac{419.78\times {10}^3}{279.51}\\ R_{01}=1.50\text{kΩ}\end{array}$

Now from the equation (1),

  $R_{e14}=\frac{r_{\pi 14}+R_{01}}{1+{\beta }_P}$

  $=\frac{0.65\times {10}^3+1.50\times {10}^3}{1+50}$

  $=\frac{2.15\times {10}^3}{51}$

  $R_{e14}=42.2\text{Ω}$

Now from the equation (2),

  $\begin{array}{l}{R}_0=42.2+27\hfill \\ R_0=69.2\text{Ω}\hfill \end{array}$