Chapter 13, Problem 13.20P

To determine

The overall small-signal voltage gain of the amplifier.

Answer to Problem 13.20P

The overall small-signal voltage gain of the amplifier is

  $A_v=341715$

Explanation of Solution

Given:

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

  

Calculation:

The reference current is.

  $\begin{array}{l}{I}_{REF}=\frac{V^{+}-V_{EB12}-V_{BE11}-V^{-}}{R_5}\\ =\frac{5-0.6-0.6-(-5)}{40}\\ =\frac{10-1.2}{40}\end{array}$

  $\begin{array}{l}=\frac{8.8}{40}\hfill \\ =0.22\text{mA}\hfill \end{array}$

Current $I_{C10}$ is.

  $\begin{array}{l}{I}_{C10}{R}_4=V_T\ln \left(\frac{I_{RFF}}{I_{C10}}\right)\\ I_{C10}(5)=(0.026)\ln \left(\frac{0.22}{I_{C10}}\right)\\ I_{C10}=\frac{(0.026)}{5}\ln \left(\frac{0.22}{I_{C10}}\right)\\ =\left(5.2\times {10}^{-3}\right)\ln \left(\frac{0.22}{I_{C10}}\right)\\ I_{C10}=14.20\mu \text{A}\end{array}$

The bias currents in the input stage are,

  $\begin{array}{l}{I}_{C1}=I_{C2}=I_{C3}=I_{C4}=\frac{I_{C10}}{2}\\ =\frac{14.20\times {10}^{-6}}{2}\\ =7.10\mu \text{A}\end{array}$

The voltage at the collector of $Q_6$ is,

  $\begin{array}{l}{V}_{C6}=V_{BE}+V_{BE6}+I_{C6}{R}_2+V^{-}\\ =0.6+0.6+\left(0.0071\times {10}^{-3}\right)\left(1\times {10}^3\right)+(-5)\\ =1.2+0.0071-5\\ =-3.79\\ V_{C6}\cong -3.8\text{V}\end{array}$

The collector current in $Q_{17}$ is

  $\begin{array}{l}{I}_{C17}=I_{C13B}=0.75I_{REF}\\ =0.75\times 0.22\times {10}^{-3}\\ =0.165\text{mA}\end{array}$ .

Now, current $I_{C13A}$ is,

  $\begin{array}{l}{I}_{C13A}=(0.25)I_{REF}\\ =(0.25)\left(0.22\times {10}^{-3}\right)\\ =0.055\text{mA}\end{array}$

Assuming ${\beta }_n=200$ for the npn device, the base current in $Q_{17}$ is

  $\begin{array}{l}{I}_{B17}=\frac{I_{C17}}{{\beta }_n}\\ =\frac{0.165}{200}\\ =0.000825\text{mA}\end{array}$

The current in $Q_{16}$ is

  $I_{C16}=I_{B17}+\frac{I_{E17}{R}_8+V_{BF17}}{R_9}$

  $\begin{array}{l}=0.000825\times {10}^{-3}+\frac{\left(0.165\times {10}^{-3}\right)\left(0.1\times {10}^3\right)+0.6}{50\times {10}^3}\hfill \\ =0.000825\times {10}^{-3}+\frac{0.6165}{50\times {10}^3}\hfill \\ =0.000825\times {10}^{-3}+0.01233\times {10}^3\hfill \\ I_{C16}=0.0132\text{mA}\hfill \end{array}$

The input resistance to the gain stage is found as

  $\begin{array}{l}{r}_{\pi 17}=\frac{{\beta }_n{V}_T}{I_{C17}}\\ =\frac{200\times 0.026}{0.165\times {10}^{-3}}=31.51\text{kΩ}\end{array}$

Therefore, the emitter resistor $R_E^{\prime }$ is

  $\begin{array}{l}{R}_E^{\prime }=R_9\parallel \left[r_{\pi 17}+\left(1+{\beta }_n\right)R_8\right]\\ =50\times {10}^3\parallel \left[31.5\times {10}^3+(1+200)\left(0.1\times {10}^3\right)\right]\\ =50\times {10}^3\parallel \left[31.5\times {10}^3+(201)\left(0.1\times {10}^3\right)\right]\end{array}$

  $\begin{array}{l}=50\times {10}^3\parallel \left[31.5\times {10}^3+20.1\times {10}^3\right]\hfill \\ =50\times {10}^3\parallel \left[51.6\times {10}^3\right]\hfill \\ =\frac{\left(50\times {10}^3\right)\left(51.6\times {10}^3\right)}{50\times {10}^3+51.6\times {10}^3}\hfill \\ =25.39\times {10}^3\hfill \\ R_E^{\prime }=25.4\text{kΩ}\hfill \end{array}$

Also

  $\begin{array}{l}{r}_{\pi 16}=\frac{{\beta }_n{V}_T}{I_{C16}}\\ =\frac{(200)(0.026)}{0.0132\times {10}^{-3}}\\ =394\text{kΩ}\end{array}$

Consequently, the resistor value $R_{i2}$ is

  $\begin{array}{l}{R}_{i2}=r_{\pi 16}+\left(1+{\beta }_n\right)R_E^{\prime }\\ =394\times {10}^3+(1+200)25.4\times {10}^3\\ =394\times {10}^3+(201)25.4\times {10}^3\\ =394\times {10}^3+5105.4\times {10}^3\\ =5499.4\times {10}^3\\ R_{i2}=5.50\text{MΩ}\end{array}$

Now the resistance of the active load is determined as

  $\begin{array}{l}{r}_{\pi 6}=\frac{{\beta }_n{V}_T}{I_{C6}}\\ =\frac{200\times 0.026}{0.0071\times {10}^{-3}}=732\text{kΩ}\\ g_{\pi 6}=\frac{I_{C6}}{V_r}\\ =\frac{0.0071\times {10}^{-3}}{0.026}=0.273\text{mA}/\text{V}\end{array}$

And

  $\begin{array}{l}{r}_{o6}=\frac{V_A}{I_{C6}}\\ =\frac{50}{0.0071\times {10}^{-3}}=7.04\text{MΩ}\end{array}$

Then

  $\begin{array}{l}{R}_{oct}=r_{o6}\left[1+g_{m6}\left(R_2\parallel r_{\pi 6}\right)\right]\\ =\left(7.04\times {10}^6\right)\left[1+0.273\times {10}^{-3}\left(1\times {10}^3\parallel 732\times {10}^3\right)\right]\end{array}$

  $=\left(7.04\times {10}^6\right)\left[1+\left(0.273\times {10}^{-3}\right)\left(\frac{1\times {10}^3\times 732\times {10}^3}{1\times {10}^3+732\times {10}^3}\right)\right]$

  $\begin{array}{l}=\left(7.04\times {10}^6\right)\left[1+\left(0.273\times {10}^{-3}\right)\left(\frac{732}{733}\times {10}^3\right)\right]\hfill \\ =\left(7.04\times {10}^6\right)[1+(0.273)(0.998)]\hfill \end{array}$

  $\begin{array}{l}=\left(7.04\times {10}^6\right)(1+0.2726)\hfill \\ =\left(7.04\times {10}^6\right)(1.2726)\hfill \end{array}$

  $R_{oct}=8.96\text{MΩ}$

Now the resistance $r_{o4}$ is

  $\begin{array}{l}{r}_{o4}=\frac{V_A}{I_{C4}}\\ =\frac{50}{0.0071\times {10}^{-3}}=7.04\text{MΩ}\end{array}$

The small signal differential voltage is

  $\begin{array}{l}{A}_d=-g_{m1}\left(r_{o4}\parallel R_{oct1}\parallel R_{i2}\right)\\ =-\left(\frac{7.10\times {10}^{-6}}{0.026}\right)\left(7.04\times {10}^6\parallel 8.96\times {10}^6\parallel 5.5\times {10}^6\right)\end{array}$

  $=-\left(273.07\times {10}^{-3}\right)\left(\frac{\left(7.04\times {10}^6\right)\left(8.96\times {10}^6\right)}{7.04\times {10}^6+8.96\times {10}^6}\parallel 5.5\times {10}^6\right)$

  $\begin{array}{l}=-\left(273.07\times {10}^{-3}\right)\left(\frac{63.0784\times {10}^6}{16}\parallel 5.5\times {10}^6\right)\hfill \\ =-\left(273.07\times {10}^{-3}\right)\left(3.9424\times {10}^6\parallel 5.5\times {10}^6\right)\hfill \end{array}$

  $\begin{array}{l}=-\left(273.07\times {10}^{-3}\right)\left(\frac{\left(3.9424\times {10}^6\right)\left(5.5\times {10}^6\right)}{3.9424\times {10}^6+5.5\times {10}^6}\right)\hfill \\ =-\left(273.07\times {10}^{-3}\right)\left(\frac{21.6832}{9.4424}\times {10}^6\right)\hfill \\ =-\left(273.07\times {10}^{-3}\right)\left(2.2963\times {10}^6\right)\hfill \\ =-627\hfill \end{array}$

The effective resistance of the active load is the resistance looking into the collector of $Q_{13B}$ is

  $\begin{array}{l}{R}_{oct2}=r_{o13B}=\frac{V_A}{I_{C13B}}\\ =\frac{100}{0.165\times {10}^{-3}}=303\text{kΩ}\end{array}$

The small signal voltage gain is

  $A_{v2}=\frac{-{\beta }_n\left(1+{\beta }_n\right)R_9\left(R_{oct2}\parallel R_{o17}\right)}{R_{i2}\left\{R_9+\left[r_{\pi 17}+\left(1+{\beta }_n\right)R_8\right]\right\}}$

  $=\frac{-200(1+200)50\times {10}^3\left(303\times {10}^3\parallel 303\times {10}^3\right)}{5500\times {10}^3\left\{50\times {10}^3+\left[31.5\times {10}^3+(1+200)\left(0.1\times {10}^3\right)\right]\right\}}$

  $=\frac{-200(201)\left(50\times {10}^3\right)\left(\frac{303\times 303\times {10}^3}{303+303}\right)}{5500\times {10}^3\left\{50\times {10}^3+\left[31.5\times {10}^3+(201)\left(0.1\times {10}^3\right)\right]\right\}}$

  $=\frac{-40200\times 50\times {10}^3\left(\frac{91809}{606}\times {10}^3\right)}{5500\times {10}^3\left\{50\times {10}^3+\left[31.5\times {10}^3+20.1\times {10}^3\right]\right\}}$

  $=\frac{-\left(402\right)\left(50\right)\left(151.5\times {10}^3\right)}{55\left\{50\times {10}^3+51.6\times {10}^3\right\}}$

  $\begin{array}{l}=\frac{-\left(402\right)\left(7575\times {10}^3\right)}{55\left\{101.6\times {10}^3\right\}}\hfill \\ =\frac{-3045150\times {10}^3}{5588\times {10}^3}\hfill \end{array}$

  $=-544.9$

  $A_{v2}=-545$

The overall gain is

  $\begin{array}{l}{A}_v=A_d\times A_n\\ =(-627)(-545)\\ =341715\end{array}$

  $A_v=341715$