Chapter 13, Problem 13.5P

(a)

To determine

The small signal differential-mode voltage gain.

Answer to Problem 13.5P

The overall small signal differential voltage gain $A_v=-1.59\times {10}^6$ .

Explanation of Solution

Given:

The circuit diagram of the BJT op-amp is

  

Given that

The transistor parameters are,

  $\beta (\text{npn})=120$

  $\beta (\text{pnp})=80$

  $V_A=80\text{V}$ (all transistors)

And base-emitter turn-on voltage is $V_{BE}(\text{on})=0.6\text{V}$ .

Calculation:

The differential mode voltage gain can be defined as

  $A_d=g_{m1}\left(r_{o2}\parallel r_{o4}\parallel R_{i6}\right)\dots \dots (1)$

Where

  $g_{m1}=\frac{I_{C1}}{V_T},r_{o2}=\frac{V_{A2}}{I_{C2}},r_{o_4}=\frac{V_A}{I_{C2}}$ and

  $R_{i6}=r_{\pi 6}+\left(1+{\beta }_n\right)\left[R_1\parallel r_{\pi 7}\right]$

  $\left(\text{Here},r_{\pi 6}=\frac{{\beta }_n{V}_T}{I_{C6}}\text{and }{r}_{\pi 7}=\frac{{\beta }_n{V}_T}{I_{C7}}\right)$

From the figure the quiescent collector currents in $Q_1$ through $Q_4$ are then

  $\begin{array}{l}{I}_{\text{C}}=I_{C2}=I_{C3}=I_{C4}=\frac{I_{Q1}}{2}\hfill \\ I_{\text{C}}=I_{C2}=I_{C3}=I_{C4}=\frac{40\times {10}^{-6}}{2}\hfill \\ I_{\text{C}}=I_{C2}=I_{\text{C}3}=I_{C4}=20\times {10}^{-6}\hfill \end{array}$

Hence, $I_{{\text{C}}_{\text{C}}}=I_{{\text{C}}_2}=I_{{\text{C}}_3}=I_{{\text{C}}_4}=20\mu \text{A}$

The collector current for $Q_6$ is

  $\begin{array}{cc}{I}_{c6}& =\frac{V_{AE}(\text{on})}{R_1}\\ & =\frac{0.6}{20\times {10}^3}\\ I_{C6}& =30\times {10}^{-6}\end{array}$

Therefore, the collector current for $Q_6$ is $I_{c6}=30\mu \text{A}$ .

The transconductance can be calculated as

  $\begin{array}{l}{g}_m=\frac{I_G}{V_T}\\ =\frac{20\times {10}^{-6}}{0.026}\\ g_m=0.769\times {10}^{-3}\end{array}$

  $\left(\text{Since, }{V}_T=0.026\text{V}\right)$

Therefore, the transconductance for $Q_1$ is $g_{m1}=0.769\text{mA}/\text{V}$ .

The resistance $r_{o2}$ can be calculated as

  $\begin{array}{l}{r}_{o2}=\frac{V_{A2}}{I_{C2}}\\ =\frac{80}{20\times {10}^{-6}}\\ r_{02}=4\times {10}^6\end{array}$

Therefore, the resistance $r_{o2}$ is $r_{o2}=4\text{MΩ}$

The resistance $r_{o4}$ can be calculated as

  $\begin{array}{l}{r}_{o4}=\frac{V_{A4}}{I_{C4}}\\ =\frac{80}{20\times {10}^{-6}}\\ r_{04}=4\times {10}^6\end{array}$

Therefore, the resistance $r_{o4}$ is $r_{o4}=4\text{MΩ}$

The resistance $r_{\pi 6}$ can be calculated as

  $\begin{array}{l}{r}_{\pi 6}=\frac{{\beta }_m{V}_T}{I_{C6}}\\ =\frac{(120)(0.026)}{30\times {10}^{-6}}=\frac{3.12}{30\times {10}^{-6}}\\ r_{\pi 6}=104\times {10}^3\end{array}$

Therefore, the resistance $r_{\pi 6}$ is $r_{\pi 6}=104\text{kΩ}$

The resistance $r_{\pi 7}$ can be calculated as

  $\begin{array}{l}{r}_{\pi 7}=\frac{{\beta }_m{V}_T}{I_{C7}}\\ =\frac{(120)(0.026)}{200\times {10}^{-6}}=\frac{3.12}{200\times {10}^{-6}}\\ r_{\pi 7}=15.6\times {10}^3\end{array}$

Therefore, the resistance $r_{\pi 7}$ is $r_{\pi 7}=15.6\text{kΩ}$

Substitute $r_{\pi 6},r_{\pi 7},R_1$ and ${\beta }_n$ in $R_{i6}$ equation

  $\begin{array}{l}{R}_{i6}=r_{\pi 6}+\left(1+{\beta }_n\right)\left[R_1\parallel r_{\pi 7}\right]\\ =104\times {10}^3+(1+120)\left[20\times {10}^3\parallel 15.6\times {10}^3\right]\end{array}$.

  $\begin{array}{l}=104\times {10}^3+(121)\left[\frac{\left(20\times {10}^3\right)\left(15.6\times {10}^3\right)}{20\times {10}^3+15.6\times {10}^3}\right]\hfill \\ =104\times {10}^3+(121)\left[8.764\times {10}^3\right]\hfill \\ =104\times {10}^3+1060.444\times {10}^3\hfill \end{array}$

  $R_{i6}=1.16\times {10}^6$

Hence, $R_{i6}=1.16\text{MΩ}$

Substitute $g_{m1},r_{o2},r_{o4}$ and $R_{i6}$ in equation-(1)

  $\begin{array}{l}{A}_d=g_{m1}\left(r_{o2}\parallel r_{ot}\parallel R_{r6}\right)\\ =0.769\times {10}^{-3}\left(4\times {10}^6\parallel 4\times {10}^6\parallel 1.16\times {10}^6\right)\end{array}$

  $\begin{array}{l}=0.769\times {10}^{-3}\left(4\times {10}^6\parallel \left(\frac{\left(4\times {10}^6\right)\left(1.16\times {10}^6\right)}{4\times {10}^6+1.16\times {10}^6}\right)\right)\hfill \\ =0.769\times {10}^{-3}\left(4\times {10}^6\parallel \left(0.899\times {10}^6\right)\right)\hfill \end{array}$

  $\begin{array}{l}=0.769\times {10}^{-3}\left(\frac{\left(4\times {10}^6\right)\left(0.899\times {10}^6\right)}{4\times {10}^6+0.899\times {10}^6}\right)\hfill \\ =0.769\times {10}^{-3}\left(0.734\times {10}^6\right)\hfill \end{array}$

  $A_d=564.45$

Therefore, the differential mode voltage gain $A_d=564.45$ .

The small signal voltage gain is

  

Now,

  $\begin{array}{l}{V}_0=-I_{C7}{r}_{07}\hfill \\ V_0=-\left({\beta }_n{I}_{b7}\right)r_{07}\hfill \\ =-\left({\beta }_n\left(\frac{R_1}{R_1+r_{\pi 7}}\right)I_{C6}\right)r_{07}\hfill \\ =-{\beta }_n{r}_{07}\left(\frac{R_1}{R_1+r_{\pi 7}}\right)I_{C6}\hfill \end{array}$

  $\begin{array}{l}=-{\beta }_n{r}_{07}\left(\frac{R_1}{R_1+r_{\pi 7}}\right)\left(1+{\beta }_n\right)I_{b6}\hfill \\ =-{\beta }_n{r}_{07}\left(1+{\beta }_n\right)\left(\frac{R_1}{R_1+r_{\pi 7}}\right)I_{n6}\hfill \\ V_0=-{\beta }_n{r}_{07}\left(1+{\beta }_n\right)\left(\frac{R_1}{R_1+r_{\pi 7}}\right)\frac{V_{o1}}{R_{i6}}\hfill \end{array}$

  $\Rightarrow \frac{V_o}{V_{\text{o1}}}=-\frac{{\beta }_s{r}_{07}\left(1+{\beta }_n\right)}{R_{i6}}\left(\frac{R_1}{R_1+r_{\pi 7}}\right)$

Therefore,

  $A_n=\frac{V_e}{V_{o1}}=-\frac{{\beta }_n{r}_{o7}\left(1+{\beta }_n\right)}{R_{i6}}\left(\frac{R_1}{R_1+r_{\pi 7}}\right)\mathrm{.........}(2)$

Where the resistance $r_{o7}$ is

  $\begin{array}{l}{r}_{o7}=\frac{V_A}{I_{C7}}=\frac{80}{200\times {10}^{-6}}\\ r_{a7}=400\times {10}^3\end{array}$

Hence, $r_{o7}=400\text{kΩ}$

Equation(2) becomes

  $\begin{array}{l}{A}_2=-\frac{{\beta }_n{r}_{07}\left(1+{\beta }_n\right)}{R_{66}}\left(\frac{R_1}{R_1+r_{n7}}\right)\\ =-\frac{(120)\left(400\times {10}^3\right)(1+120)}{1.16\times {10}^6}\left(\frac{20\times {10}^3}{20\times {10}^3+15.6\times {10}^3}\right)\end{array}$

  $\begin{array}{l}=-\frac{(120)\left(400\times {10}^3\right)(121)}{1.16\times {10}^6}\left(\frac{20\times {10}^3}{35.6\times {10}^3}\right)\hfill \\ =-\frac{5.808\times {10}^9}{1.16\times {10}^6}\left(\frac{20}{35.6}\right)\hfill \end{array}$

  $A_{v2}=-2812.86$

Therefore, the small signal voltage gain is $A_{v2}=-2812.86$

Now the overall small signal differential voltage gain is

  $\begin{array}{l}{A}_v=A_d{A}_{v2}\\ =(565)(-2813)\\ A_v=-1.59\times {10}^6\end{array}$

Therefore, the overall small signal differential voltage gain $A_v=-1.59\times {10}^6$ .

(b)

To determine

The differential-mode input resistance.

Answer to Problem 13.5P

The differential-mode input resistance is $R_{id}=208\times {10}^3$ .

Explanation of Solution

Given:

The circuit diagram of the BJT op-amp is

Given that

The transistor parameters are,

  $\beta (\text{npn})=120$

  $\beta (\text{pnp})=80$

  $V_A=80\text{V}$ (all transistors)

And base-emitter turn-on voltage is $V_{BE}(\text{on})=0.6\text{V}$

Calculation:

The differential-mode input resistance is given as

  $R_{id}=2r_{\pi 1}$

Where

  $\begin{array}{l}{r}_{\pi 1}=\frac{{\beta }_p{V}_T}{I_{C1}}=\frac{(80)(0.026)}{20\times {10}^{-6}}\\ =\frac{2.08}{20\times {10}^{-6}}\\ r_{\pi 1}=104\times {10}^3\end{array}$

Hence, $r_{\pi 1}=104\text{kΩ}$

Now, the differential-mode input resistance is

  $\begin{array}{l}{R}_{id}=2r_{\pi 1}\\ =2\left(104\times {10}^3\right)\\ R_{id}=208\times {10}^3\end{array}$

Therefore, the differential-mode input resistance is $R_{id}=208\times {10}^3$ .

(c)

To determine

Theunity-gain bandwidth.

Answer to Problem 13.5P

The gain bandwidth product is $\text{GBW}=12.23\text{MHz}$ .

Explanation of Solution

Given:

The circuit diagram of the BJT op-amp is

  

Given that

The transistor parameters are,

  $\beta (\text{npn})=120$

  $\beta (\text{pnp})=80$

  $V_A=80\text{V}$ (all transistors)

And base-emitter turn-on voltage is $V_{BE}(\text{on})=0.6\text{V}$

Calculation:

The unity-gain bandwidth product is $\text{GBW}=\left|A_{v2}\right|f_{PD}$ .

Here, the dominant pole frequency is given as

  $\begin{array}{l}{f}_{PD}=\frac{1}{2\pi R_{eq}{C}_M}\hfill \\ \text{Where}{C}_M=C_F\left(1+\left|A_{v2}\right|\right)\hfill \\ C_M=10\times {10}^{-12}(1+|-2813|)\hfill \\ =10\times {10}^{-12}(1+2813)\hfill \\ =10\times {10}^{-12}(2814)\hfill \\ C_M=28140\times {10}^{-12}\hfill \end{array}$

Hence, $C_M=28140\times {10}^{-12}$

And

  $\begin{array}{l}{R}_{eq}=r_{a2}\parallel r_{o4}\mid R_{r6}\\ =4\times {10}^6\parallel 4\times {10}^6\parallel 1.16\times {10}^6\end{array}$

  $\begin{array}{l}=4\times {10}^6\parallel \left(\frac{\left(4\times {10}^6\right)\left(1.16\times {10}^6\right)}{4\times {10}^6+1.16\times {10}^6}\right)\hfill \\ =4\times {10}^6\parallel \left(0.899\times {10}^6\right)\hfill \\ =\frac{\left(4\times {10}^6\right)\left(0.899\times {10}^6\right)}{4\times {10}^6+0.899\times {10}^6}\hfill \end{array}$

  $R_{eq}=0.734\times {10}^6$

Hence, $R_{eq}=0.734\text{M}\Omega$

Now the dominant pole frequency we obtain as

  $\begin{array}{l}{f}_{PD}=\frac{1}{2\pi R_{eq}{C}_M}\\ =\frac{1}{2\pi \left(0.734\times {10}^6\right)\left(28140\times {10}^{-12}\right)}\end{array}$

  $f_{PD}=7.71$

Therefore, the dominant pole frequency $f_{PD}=7.71\text{Hz}$

The unity-gain bandwidth product is

  $\begin{array}{l}\text{GBW}=\left|A_{v2}\right|f_{PD}=\left(1.59\times {10}^6\right)(7.71)\\ \text{GBW}=12.26\times {10}^6\end{array}$

Therefore, the gain bandwidth product is $\text{GBW}=12.23\text{MHz}$ .