Chapter 11, Problem 11.62P

Consider the diff-amp shown in Figure P 11.62 . The circuit parameters are $V^{+}=3\text{V},V^{-}=-3\text{V},$ and $I_Q=0.4\text{mA}.$ The npn transistor parameters are ${\beta }_{\text{npn}}=180,V_{BE}(\text{on})=0.7\text{V},$ and $V_{AN}=120\text{V},$ and the pnp transistor parameters are ${\beta }_{\text{pnp}}=120,V_{EB}(\text{on})=0.7\text{V},$ and $V_{AP}=80\text{V}$ Sketch the small-signal equivalent circuit for the diff-amp assuming an ideal differential-mode input signal. (b) Determine the one-sided differential-mode gain $A_{d1}=\Delta v_{O1}/v_d.$ (c) Determine the one-sided differential mode gain $A_{d2}=\Delta v_{O2}/v_d.$ (d) Find the two-sided differential-mode gain $A_{d3}=\Delta \left(v_{O2}-v_{O1}\right)/v_d$ .

To determine

To sketch: The small signal equivalent circuit for the differential amplifier.

Answer to Problem 11.62P

The small signal circuit is shown in Figure 2.

Explanation of Solution

Given:

The given circuit is shown below.

  

Calculation:

The small signal equivalent circuit for the above circuit is shown in Figure 1.

  

Figure 1

Because of symmetry the value of the currents $I_{EQ1}$ , $I_{EQ2}$ are given by,

  $\begin{array}{c}{I}_E=\frac{I_Q}{2}\\ =\frac{0.4\text{mA}}{2}\\ =0.2\text{mA}\end{array}$

The value of the currents $I_{CQ1}=I_{CQ2}$ which are equal to $I_{C1}$ and is given by,

  $\begin{array}{c}{I}_{C1}=\frac{{\beta }_m}{1+{\beta }_m}{I}_{E1}\\ =\frac{180}{1+180}\left(0.2\text{mA}\right)\\ \approx 0.2\text{mA}\end{array}$

The value of the currents $I_{EQ4}=I_{EQ4}$ which are equal to $I_{C1}$ and is given by,

  $I_{EQ4},I_{EQ4}=0.2\text{mA}$

The value of the currents $I_{CQ3}=I_{CQ4}$ which are equal to $I_{C1}$ and is given by,

  $\begin{array}{c}{I}_{CQ3}=I_{CQ4}\\ =\frac{I_{E2}{\beta }_{mp}}{{\beta }_{mp}+1}\\ =\frac{120}{121}\left(0.2\text{mA}\right)\\ \approx 0.2\text{mA}\end{array}$

For differential up circuit emitter current $Q_1$ or $Q_2$ is grounded in incremental circuit because of the symmetry of the circuit.

Modify the circuit, the required diagram is shown in Figure 2

  

Figure 2

Conclusion:

Therefore, the small signal circuit is shown in Figure 3

To determine

The value of one sided differential voltage gain.

Answer to Problem 11.62P

The value of the differential mode voltage gain is $-0.4996$ .

Explanation of Solution

Given:

The given circuit is shown below.

  Calculation:

Consider the left portion of the circuit as,

  

Figure 3

Modify the circuit as shown in Figure 4

  

Figure 4

The expression for the output voltage of the above circuit is given by,

  $\begin{array}{c}{v}_O=-\left(g_{m1}{V}_{\pi 1}+g_{m3}{V}_{\pi 2}\right)\left(r_{O1}\left|\right|r_{O3}\left|\right|r_{\pi 3}\right)\\ =-\left[g_{m1}\left(\frac{v_d}{2}\right)+g_{m3}{v}_{O1}\right]\left[\frac{1}{\frac{1}{r_{O1}}+\frac{1}{r_{O3}}+\frac{1}{r_{\pi 3}}}\right]\end{array}$

The emitter current of the transistor is calculated as,

  $\begin{array}{c}{I}_{E1}=\frac{I_Q}{2}\\ =\frac{0.4\text{mA}}{2}\\ =0.2\text{mA}\end{array}$

The expression for the collector current of the transistor $I_{C1}$ is evaluated as,

  $\begin{array}{l}{I}_{E1}=I_{C1}+I_{B1}\\ I_{E1}=I_{C1}+\frac{I_{C1}}{{\beta }_{\text{npn}}}\\ I_{C1}=\left(\frac{{\beta }_{\text{npn}}}{1+{\beta }_{\text{npn}}}\right)I_{E1}\end{array}$

Substitute $180$ for ${\beta }_{\text{npn}}$ and $0.2\text{mA}$ for $I_{E1}$ in the above equation.

  $\begin{array}{c}{I}_{C1}=\left(\frac{180}{1+180}\right)0.2\text{mA}\\ \text{=0}\text{.1989}\text{mA}\end{array}$

Apply KCL at the output node $v_{O1}$ .

  $\begin{array}{c}{I}_{C1}=I_{C3}+I_{B3}\\ I_{C1}=I_{C3}+\frac{I_{C1}}{{\beta }_{\text{pnp}}}\\ I_{C3}=\left(\frac{{\beta }_{\text{pnp}}}{{\beta }_{\text{pnp}}+1}\right)I_{C1}\end{array}$

Substitute $120$ for ${\beta }_{\text{pnp}}$ and $0.1989\text{mA}$ for $I_{C1}$ in the above equation. ${\beta }_{\text{pnp}}$

  $\begin{array}{c}{I}_{C3}=\left(\frac{120}{120+1}\right)0.1989\text{mA}\\ I_{C3}=0.1973\text{mA}\end{array}$

The transconductance of transistor $Q_1$ is calculated as,

  $\begin{array}{c}{g}_{m1}=-\frac{I_{C1}}{0.026\text{V}}\\ =-\frac{0.1989\text{mA}}{0.026\text{V}}\\ =7.956\text{mA}/\text{V}\end{array}$

The transconductance of transistor $Q_3$ is calculated as,

  $\begin{array}{c}{g}_{m3}=-\frac{I_{C3}}{0.026\text{V}}\\ =-\frac{0.1973\text{mA}}{0.026\text{V}}\\ =7.892\text{mA}/\text{V}\end{array}$

The small signal output resistance of $Q_1$ is calculated as,

  $\begin{array}{c}{r}_{O1}=\frac{V_{AB}}{I_{C1}}\\ =\frac{120\text{V}}{0.1989\text{mA}}\\ =603.3182\text{k}\Omega \end{array}$

The small signal output resistance of $Q_3$ is calculated as,

  $\begin{array}{c}{r}_{O3}=\frac{V_{AP}}{I_{C3}}\\ =\frac{80\text{V}}{0.1973\text{mA}}\\ =405.4739\text{k}\Omega \end{array}$

The small signal resistance $r_{\pi 3}$ is calculated as,

  $\begin{array}{c}{r}_{\pi 3}=\frac{V_T{\beta }_{\text{pnp}}}{I_{C3}}\\ =\frac{\left(120\right)\left(0.026\text{V}\right)}{0.1973\text{mA}}\\ =15.2052\text{k}\Omega \end{array}$

The value of the voltage $v_{O1}$ is calculated as,

  $\begin{array}{l}{v}_{O1}=-\left[\left(7.965\text{mA}/\text{V}\right)\left(\frac{v_d}{2}\right)+\left(7.892\text{mA}/\text{V}\right)v_{O1}\right]\left[\frac{1}{\begin{array}{l}\frac{1}{603.3182\text{k}\Omega }+\frac{1}{\mathrm{4.5.4739}}\\ +\frac{1}{15.2052\text{k}\Omega }\end{array}}\right]\\ 113.9219v_{O1}=-56.9188\\ \frac{v_{O1}}{v_d}=-0.4996\end{array}$

Conclusion:

Therefore, the value of the differential mode voltage gain is $-0.4996$ .

To determine

The differential voltage gain of both the sides.

Answer to Problem 11.62P

The value of differential voltage gain is $0.4996$ .

Explanation of Solution

Given:

The given circuit is shown below.

  

Calculation:

The expression for the output voltage of the above circuit is given by,

  $\begin{array}{c}{v}_{O2}=-\left(g_{m1}{V}_{\pi 1}+g_{m3}{V}_{\pi 2}\right)\left(r_{O1}\left|\right|r_{O3}\left|\right|r_{\pi 3}\right)\\ =-\left[g_{m2}\left(\frac{-v_d}{2}\right)+g_{m34}{v}_{O1}\right]\left[\frac{1}{\frac{1}{r_{O2}}+\frac{1}{r_{O4}}+\frac{1}{r_{\pi 4}}}\right]\end{array}$

The value of the voltage $v_{O1}$ is calculated as,

  $\begin{array}{l}{v}_{O1}=-\left[\left(7.965\text{mA}/\text{V}\right)\left(\frac{-v_d}{2}\right)+\left(7.892\text{mA}/\text{V}\right)v_{O1}\right]\left[\frac{1}{\begin{array}{l}\frac{1}{603.3182\text{k}\Omega }+\frac{1}{\mathrm{4.5.4739}}\\ +\frac{1}{15.2052\text{k}\Omega }\end{array}}\right]\\ 113.9219v_{O1}=-56.9188\\ \frac{v_{O1}}{v_d}=0.4996\end{array}$

Conclusion:

Therefore, the value of differential voltage gain is $0.4996$ .

To determine

The value of the differential voltage gain of both the sides.

Answer to Problem 11.62P

The value of differential voltage gain is $0.9992$ .

Explanation of Solution

Given:

The given circuit is shown below.

  

Calculation:

The expression to determine the value of the two sided differential voltage gain is given by,

  $\begin{array}{c}{A}_{d3}=\frac{\Delta \left(v_{O2}-v_{O1}\right)}{v_d}\\ =\frac{\Delta v_{O2}}{v_d}0-\frac{\Delta v_{O1}}{v_d}\\ =0.4996-\left(-0.4996\right)\\ =0.9992\end{array}$

Conclusion:

Therefore, the value of differential voltage gain is $0.9992$ .