Chapter 17, Problem 17.1EP

Consider the differential amplifier circuit in Figure 17.1 biased at $V^{+}=1.8\text{V}$ , $V^{-}=-1.8\text{V}$ , and ${\upsilon }_2=0$ . Assume $V_{BE}\left(\text{on}\right)=0.7\text{V}$ and neglect base currents. (a) Design the circuit such that $i_E=0.11\text{mA}$ and ${\upsilon }_{O1}={\upsilon }_{O2}=1.45\text{V}$ when ${\upsilon }_1=0$ . (b) Using the results of part (a), determine $i_E$ , ${\upsilon }_{O1}$ , and ${\upsilon }_{O2}$ for (i) ${\upsilon }_1=+0.5\text{V}$ and (ii) ${\upsilon }_1=-0.5\text{V}$ . (c) Using the results of parts (a) and (b), calculate the power dissipated in the circuit for (i) ${\upsilon }_1=+0.5\text{V}$ and (ii) ${\upsilon }_1=-0.5\text{V}$ .(Ans. (a) $R_E=10\text{k}\Omega$ , $R_C=6.364\text{k}\Omega$ ; (b) (i) $i_E=0.16\text{mA}$ , ${\upsilon }_{O1}=0.782\text{V}$ , ${\upsilon }_{O2}=1.8\text{V}$ ; (ii) $i_E=0.11\text{mA}$ , ${\upsilon }_{O1}=1.8\text{V}$ , ${\upsilon }_{O2}=1.10\text{V}$ ; (c) (i) $P=0.576\text{mW}$ , (ii) $P=0.396\text{mW}$ )

To determine

The design parameters of the circuit for the given values.

Answer to Problem 17.1EP

The value of the resistances to design the circuit are for $R_E$ is $10\text{k}\Omega$ and for $R_C$ is $6.364\text{k}\Omega$ .

Explanation of Solution

Calculation:

The given circuit is shown in Figure 1.

  

The expression for the voltage at the emitter terminals is given by,

$v_E=v_1-V_{BE1\left(\text{on}\right)}$

Substitute $0$ for $v_1$ and $0.7\text{V}$ for $V_{BE1\left(\text{on}\right)}$ in the above equation.

  $\begin{array}{c}{v}_E=0-0.7\text{V}\\ \text{=}-0.7\text{V}\end{array}$

The expression to determine the value of the current $i_E$ is given by,

$i_E=\frac{v_E-V^{-}}{R_E}$

Substitute $0.11\text{mA}$ for $i_E$

$-0.7\text{V}$ for $v_E$ and $-1.8\text{V}$ for $V^{-}$ in the above equation.

  $\begin{array}{l}0.11\text{mA}=\frac{0.7\text{V}-\left(-1.8\text{V}\right)}{R_E}\\ R_E=10\text{k}\Omega \end{array}$

The expression to determine the value of the collector current of the first transistor is given by,

$i_{C1}=\frac{i_E}{2}$

Substitute $0.11\text{mA}$ for $i_E$ in the above equation.

$\begin{array}{c}{i}_{C1}=\frac{0.11\text{mA}}{2}\\ =0.055\text{mA}\end{array}$

The expression for the value of the collector current of the second transistor is given by,

$i_{C2}=i_{C1}$

Substitute $0.055\text{mA}$ for $i_{C1}$ in the above equation.

$i_{C2}=0.055\text{mA}$

The expression for the value of the voltage $v_{O1}$ is given by,

$v_{O1}=V^{+}-i_C{R}_C$

Substitute $1.45\text{V}$ for $v_{O1}$ , $0.055\text{mA}$ for $i_C$ and $1.8\text{V}$ for $V^{+}$ in the above equation.

$\begin{array}{l}1.45\text{V}=1.8\text{V}-\left(0.055\text{mA}\right)R_C\\ R_C=6.364\text{k}\Omega \end{array}$

Conclusion:

Therefore, the value of the resistances to design the circuit are for $R_E$ is $10\text{k}\Omega$ and for $R_C$ is $6.364\text{k}\Omega$ .

To determine

The value of the current $i_E$ , voltage $v_{O1}$ and $v_{O2}$ for the given input voltage values.

Answer to Problem 17.1EP

The value of the current and the voltage for the value of the input voltage of $+0.5\text{V}$ are $i_E$ is $0.16\text{mA}$ , $v_{O1}$ is $0.782\text{V}$ and $v_{O2}$ is $1.8\text{V}$ . The value of the current and the voltage for the value of the input voltage of $+0.5\text{V}$ are $i_E$ is $0.11\text{mA}$ , $v_{O1}$ is $1.8\text{V}$ and $v_{O2}$ is $1.10\text{V}$ .

Explanation of Solution

Calculation:

The expression for the voltage at the emitter terminals is given by,

$v_E=v_1-V_{BE1\left(\text{on}\right)}$

Substitute $0.5\text{V}$ for $v_1$ and $0.7\text{V}$ for $V_{BE1\left(\text{on}\right)}$ in the above equation.

  $\begin{array}{c}{v}_E=0.5\text{V}-0.7\text{V}\\ \text{=}-0.2\text{V}\end{array}$

The expression to determine the value of the current $i_E$ is given by,

$i_E=\frac{v_E-V^{-}}{R_E}$

Substitute $10\text{k}\Omega$ for $R_E$

$-0.2\text{V}$ for $v_E$ and $-1.8\text{V}$ for $V^{-}$ in the above equation.

  $\begin{array}{c}{i}_E=\frac{-0.2\text{V}-\left(-1.8\text{V}\right)}{10\text{k}\Omega }\\ =0.16\text{mA}\end{array}$

The expression to determine the value of the collector current of the first transistor is given by,

$i_{C1}=i_E$

Substitute $0.16\text{mA}$ for $i_E$ in the above equation.

$i_{C1}=0.16\text{mA}$

The expression for the value of the collector current of the second transistor is given by,

$i_{C2}=0\text{A}$

The expression for the value of the voltage $v_{O1}$ is given by,

  $v_{O1}=V^{+}-i_{C1}{R}_C$

Substitute $6.364\text{k}\Omega$ for $R_C$ , $0.16\text{mA}$ for $i_{C1}$ and $1.8\text{V}$ for $V^{+}$ in the above equation.

  $\begin{array}{c}{v}_{O1}=1.8\text{V}-\left(0.055\text{mA}\right)\left(6.364\text{k}\Omega \right)\\ =0.782\text{V}\end{array}$

The expression for the value of the voltage $v_{O2}$ is given by,

$v_{O2}=V^{+}-i_{C2}{R}_C$

Substitute $6.364\text{k}\Omega$ for $R_C$ , $0$ for $i_{C2}$ and $1.8\text{V}$ for $V^{+}$ in the above equation.

$\begin{array}{c}{v}_{O2}=1.8\text{V}-\left(0\right)\left(6.364\text{k}\Omega \right)\\ =1.8\text{V}\end{array}$

The expression for the voltage at the emitter terminals is given by,

$v_E=v_2-V_{BE1\left(\text{on}\right)}$

Substitute $0\text{V}$ for $v_2$ and $0.7\text{V}$ for $V_{BE1\left(\text{on}\right)}$ in the above equation.

  $\begin{array}{c}{v}_E=0\text{V}-0.7\text{V}\\ \text{=}-0.7\text{V}\end{array}$

The expression to determine the value of the current $i_E$ is given by,

$i_E=\frac{v_E-V^{-}}{R_E}$

Substitute $10\text{k}\Omega$ for $R_E$

$-0.7\text{V}$ for $v_E$ and $-1.8\text{V}$ for $V^{-}$ in the above equation.

  $\begin{array}{c}{i}_E=\frac{-0.7\text{V}-\left(-1.8\text{V}\right)}{10\text{k}\Omega }\\ =0.11\text{mA}\end{array}$

The expression to determine the value of the collector current of the first transistor is given by,

$i_{C1}=0\text{A}$

The expression for the value of the collector current of the second transistor is given by,

$i_{C2}=i_E$

Substitute $0.11\text{mA}$ for $i_E$ in the above equation.

$i_{C2}=0.11\text{mA}$

The expression for the value of the voltage $v_{O1}$ is given by,

  $v_{O1}=V^{+}-i_{C1}{R}_C$

Substitute $6.364\text{k}\Omega$ for $R_C$ , $0$ for $i_{C1}$ and $1.8\text{V}$ for $V^{+}$ in the above equation.

  $\begin{array}{c}{v}_{O1}=1.8\text{V}-\left(0\right)\left(6.364\text{k}\Omega \right)\\ =1.8\text{V}\end{array}$

The expression for the value of the voltage $v_{O2}$ is given by,

$v_{O2}=V^{+}-i_{C2}{R}_C$

Substitute $6.364\text{k}\Omega$ for $R_C$ , $0.11\text{mA}$ for $i_{C2}$ and $1.8\text{V}$ for $V^{+}$ in the above equation.

$\begin{array}{c}{v}_{O2}=1.8\text{V}-\left(0.11\text{mA}\right)\left(6.364\text{k}\Omega \right)\\ =1.10\text{V}\end{array}$

Conclusion:

Therefore, the value of the current and the voltage for the value of the input voltage of $+0.5\text{V}$ are $i_E$ is $0.16\text{mA}$ , $v_{O1}$ is $0.782\text{V}$ and $v_{O2}$ is $1.8\text{V}$ . The value of the current and the voltage for the value of the input voltage of $+0.5\text{V}$ are $i_E$ is $0.11\text{mA}$ , $v_{O1}$ is $1.8\text{V}$ and $v_{O2}$ is $1.10\text{V}$ .

To determine

The value of the power dissipated in the circuit for the different value of the input voltage.

Answer to Problem 17.1EP

The value of the power consumed in the circuit for the input voltage of $+5\text{V}$ is $0.576\text{mW}$ and for the value of $-0.5\text{V}$ is $0.396\text{mW}$ .

Explanation of Solution

Calculation:

The expression for the value of the power dissipated in the circuit.

$P=\left(v_{O1}-v_E\right)i_{C1}+i_{C1}^2{R}_{C1}+\left(v_{O2}-v_E\right)i_{C2}+i_{C2}^2{R}_{C2}+i_E^2{R}_E$ ......... (1)

Substitute $0$ for $i_{C2}$ , $0.782\text{V}$ for $v_{O1}$ , $-0.2\text{V}$ for $v_E$ , $0.16\text{mA}$ for $i_{C1}$ , $6.364\text{k}\Omega$ for $R_{C1}$ , $1.8\text{V}$ for $v_{O2}$ , $10\text{k}\Omega$ for $R_E$ and $0.16\text{mA}$ for $i_E$ in the above equation.

$\begin{array}{c}P=\left[\begin{array}{l}\left(0.782\text{V}-\left(-0.2\text{V}\right)\right)\left(0.16\text{mA}\right)+\left(0.16\text{mA}\right)\left(6.364\text{k}\Omega \right)\\ +\left(1.8\text{V}-\left(-0.2\text{V}\right)\right)\left(0\right)+\left(0\right)\left(6.364\text{k}\Omega \right)+{\left(0.16\text{mA}\right)}^2\left(10\text{k}\Omega \right)\end{array}\right]\\ =0.576\text{mW}\end{array}$

Substitute $0.11\text{mA}$ for $i_{C2}$ , $1.8\text{V}$ for $v_{O1}$ , $-0.7\text{V}$ for $v_E$ , $0$ for $i_{C1}$ , $6.364\text{k}\Omega$ for $R_{C1}$ , $1.10\text{V}$ for $v_{O2}$ , $10\text{k}\Omega$ for $R_E$ and $0.11\text{mA}$ for $i_E$ in equation (1).

  $\begin{array}{c}P=\left[\begin{array}{l}\left(1.8\text{V}-\left(-0.7\text{V}\right)\right)\left(0\right)+\left(0\right)\left(6.364\text{k}\Omega \right)+\left(1.10\text{V}-\left(-0.7\text{V}\right)\right)\left(0.11\text{mA}\right)\\ +{\left(0.11\text{mA}\right)}^2\left(6.364\text{k}\Omega \right)+{\left(0.11\text{mA}\right)}^2\left(10\text{k}\Omega \right)\end{array}\right]\\ =0.396\text{mW}\end{array}$

Conclusion:

Therefore, the value of the power consumed in the circuit for the input voltage of $+5\text{V}$ is $0.576\text{mW}$ and for the value of $-0.5\text{V}$ is $0.396\text{mW}$ .