Chapter 11, Problem 11.1EP

The circuit parameters for the differential amplifier shown in Figure 11.2 are $V^{+}=5\text{V},V^{-}=-5\text{V},I_Q=0.3\text{mA},$ and $R_C=20\text{k}\Omega .$ The transistor Parameters are $\beta =180,V_{BE}(\text{on})=0.7\text{V},$ and $V_A=\infty .$ Determine the voltages $v_E,v_{C1},v_{C2},v_{CE1},$ and $v_{CE2}$ for $(\text{a})v_1=v_2=0,(\text{b})v_1=v_2=-1\text{V}$ and (c) $v_1=v_2=+1$ V. (Ans. (a) $v_E=-0.7\text{V},v_{C1}=v_{C2}=2\text{V},v_{CE1}=$ $v_{CE2}=2.7\text{V};$ (b) $v_E=-1.7\text{V},v_{C1}=v_{C2}=2\text{V},v_{CE1}=v_{CE2}=3.7\text{V}$ (c) $v_E=+0.3\text{V},v_{C1}=v_{C2}=2\text{V},v_{CE1}=v_{CE2}=1.7\text{V})$ .

To determine

The collector voltages, collector to emitter voltages and the emitter voltage for the given input values of a differential amplifier.

Answer to Problem 11.1EP

  $v_E=-0.7\text{ V}$ , $v_{C1}=2\text{ V }$ , $v_{C2}=2\text{ V}$ , $v_{CE1}=2.7\text{ V}$ , $v_{CE2}=2.7\text{ V}$

Explanation of Solution

Given Information:

The given values are:

  $\begin{array}{l}{V}^{+}=5\text{ V, }{V}^{-}=-5\text{ V,}{I}_Q=0.3\text{ mA,}{R}_C=20\text{ k}\Omega ,\beta \text{=180,}{V}_{BE(on)}=0.7\text{ V,}{V}_A=\infty \\ v_1=v_2=0\end{array}$

The given circuit is shown below.

  

Calculation:

Voltage at emitter of each transistor

  $v_E=0-V_{BE(on)}$

  $v_E=-0.7\text{ V}$

Assume transistors are matched. Then,

  $i_{E1}=i_{E2}=\frac{I_Q}{2}$

Also assume that the bias currents are negligible. Then,

  $i_{C1}\cong i_{E1}=\frac{I_Q}{2},i_{C2}\cong =i_{E2}=\frac{I_Q}{2}$

Now

  $v_{C1}=v_{C2}=V^{+}-\frac{I_Q}{2}{R}_C=5-0.15\times 20\text{ V}$

  $v_{C1}=2\text{ V, }{v}_{C2}=2\text{ V}$

Then,

  $v_{CE1}=v_{C1}-v_E=2-(-0.7)$

  $v_{CE1}=2.7\text{ V}$

Similarly,

  $v_{CE2}=v_{C2}-v_E=2-(-0.7)$

  $v_{CE2}=2.7\text{ V}$

To determine

The collector voltages, collector to emitter voltages and the emitter voltage for the given values of a differential amplifier.

Answer to Problem 11.1EP

  $v_E=-1.7\text{ V}$ , $v_{C1}=2\text{ V }$ , $v_{C2}=2\text{ V}$ , $v_{CE1}=3.7\text{ V}$ , $v_{CE2}=3.7\text{ V}$

Explanation of Solution

Given Information:

The given values are:

  $\begin{array}{c}{V}^{+}=5\text{ V, }{V}^{-}=-5\text{ V,}{I}_Q=0.3\text{ mA,}{R}_C=20\text{ k}\Omega ,\beta \text{=180,}{V}_{BE(on)}=0.7\text{ V,}{V}_A=\infty \\ v_1=v_2=-1\text{ V}\end{array}$

The given circuit is shown below.

  

Calculation:

Voltage at emitter of each transistor

  $\begin{array}{r}{v}_E=v_1-V_{BE(on)}\\ =-1-0.7\text{ V}\end{array}$

  $v_E=-1.7\text{ V}$

Assume transistors are matched. Then,

  $i_{E1}=i_{E2}=\frac{I_Q}{2}$

Also assume the bias currents are negligible. Then,

  $i_{C1}\cong i_{E1}=\frac{I_Q}{2},i_{C2}\cong i_{E2}=\frac{I_Q}{2}$

Now

  $v_{C1}=v_{C2}=V^{+}-\frac{I_Q}{2}{R}_C=5-0.15\times 20\text{ V}$

  $v_{C1}=2\text{ V, }{v}_{C2}=2\text{ V}$

Then,

  $v_{CE1}=v_{C1}-v_E=2-(-1.7)$

  $v_{CE1}=3.7\text{ V}$

Similarly,

  $v_{CE2}=v_{C2}-v_E=2-(-0.7)$

  $v_{CE2}=3.7\text{ V}$

To determine

The collector voltages, collector to emitter voltages and the emitter voltage for the given values of differential amplifier.

Answer to Problem 11.1EP

  $v_E=0.3\text{ V}$ , $v_{C1}=2\text{ V }$ , $v_{C2}=2\text{ V}$ , $v_{CE1}=1.7\text{ V}$ , $v_{CE2}=1.7\text{ V}$

Explanation of Solution

Given Information:

The given values are:

  $\begin{array}{c}{V}^{+}=5\text{ V, }{V}^{-}=-5\text{ V,}{I}_Q=0.3\text{ mA,}{R}_C=20\text{ k}\Omega ,\beta \text{=180,}{V}_{BE(on)}=0.7\text{ V,}{V}_A=\infty \\ v_1=v_2=+1\text{ V}\end{array}$

The given circuit is shown below.

  

Calculation:

Voltage at emitter of each transistor

  $v_E=v_1-V_{BE(on)}=1-0.7\text{ V}$

  $v_E=0.3\text{ V}$

Assume transistors are matched. Then,

  $i_{E1}=i_{E2}=\frac{I_Q}{2}$

Also assume the bias currents are negligible. Then,

  $i_{C1}\cong i_{E1}=\frac{I_Q}{2},i_{C2}\cong i_{E2}=\frac{I_Q}{2}$

Now

  $v_{C1}=v_{C2}=V^{+}-\frac{I_Q}{2}{R}_C=5-0.15\times 20\text{ V}$

  $v_{C1}=2\text{ V, }{v}_{C2}=2\text{ V}$

Then,

  $v_{CE1}=v_{C1}-v_E=2-(0.3)$

  $v_{CE1}=1.7\text{ V}$

Similarly,

  $v_{CE2}=v_{C2}-v_E=2-(0.3)$

  $v_{CE2}=1.7\text{ V}$