Chapter 17, Problem 17.5P

(a)

To determine

The value of $R_{C2}$ .

Answer to Problem 17.5P

The value of the resistance is $0.758\text{k}\Omega$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

Mark the currents and redraw the circuit.

The required diagram is shown in Figure 2

  

The transistor number two is off and the value of the current $i_{C2}$ is equal to $i_E$ and is given by,

  $i_{C2}=i_E$   ........ (1)

The expression for the value of the emitter voltage $v_{E2}$ is given by,

  $v_{E2}=v_{B2}-V_{BE2}$

Substitute $-1.2\text{V}$ for $v_{B2}$ and $0.7\text{V}$ for $V_{BE2}$ in the above equation.

  $\begin{array}{c}{v}_{E2}=-1.2\text{V}-0.7\text{V}\\ =-1.9\text{V}\end{array}$

The expression for the value of the emitter current $i_E$ is given by,

  $i_E=\frac{v_{E2}-\left(-5.2\text{V}\right)}{2.5\text{k}\Omega }$

Substitute $-1.9\text{V}$ for $v_{E2}$ in the above equation.

  $\begin{array}{c}{i}_E=\frac{-1.9\text{V}-\left(-5.2\text{V}\right)}{2.5\text{k}\Omega }\\ =1.32\text{mA}\end{array}$

Substitute $1.32\text{mA}$ for $i_E$ in equation (1).

  $i_{C2}=1.32\text{mA}$

The expression for the value of the capacitance of resistance $R_{C2}$ is given by,

  $R_{C2}=\frac{-v_2}{i_{C2}}$

Substitute $1.32\text{mA}$ for $i_{C2}$ and $-1\text{V}$ for $v_2$ in the above equation.

  $\begin{array}{c}{R}_{C2}=\frac{-\left(-1\text{V}\right)}{1.32\text{mA}}\\ =0.758\text{k}\Omega \end{array}$

Conclusion:

Therefore, the value of the resistance is $0.758\text{k}\Omega$ .

(b)

To determine

The value of the resistance $R_{C1}$ .

Answer to Problem 17.5P

The value of the resistance is $0.658\text{k}\Omega$ .

Explanation of Solution

Calculation:

The first transistor is on and the second is off.

The expression for the voltage $v_{E1}$ is given by,

  $v_{E1}=2V_{BE}$

Substitute $-0.7\text{V}$ for $V_{BE}$ in the above equation.

  $\begin{array}{c}{v}_{E1}=2\left(-0.7\text{V}\right)\\ =-1.4\text{V}\end{array}$

The expression for the value of the emitter current $i_E$ is given by,

  $i_E=\frac{v_{E1}-\left(-5.2\text{V}\right)}{2.5\text{k}\Omega }$

Substitute $-1.4\text{V}$ for $v_{E1}$ in the above equation.

  $\begin{array}{c}{i}_E=\frac{-1.4\text{V}-\left(-5.2\text{V}\right)}{2.5\text{k}\Omega }\\ =1.52\text{mA}\end{array}$

The expression for the value of the current $i_{C1}$ is given by,

  $i_{C1}=i_E$

Substitute $1.52\text{mA}$ for $i_E$ in the above equation.

  $i_{C1}=1.52\text{mA}$

The expression for the value of the resistance $R_{C1}$ is given by,

  $R_{C1}=\frac{-\left(v_1\right)}{i_{C1}}$

Substitute $1.52\text{mA}$ for $i_{C1}$ and $-1\text{V}$ for $v_1$ in the above equation.

  $\begin{array}{c}{R}_{C1}=\frac{-\left(-1\text{V}\right)}{1.52\text{mA}}\\ =0.658\text{k}\Omega \end{array}$

Conclusion:

Therefore, the value of the resistance is $0.658\text{k}\Omega$ .

(c)

To determine

The value of the output voltage $v_{O1}$ and $v_{O2}$ .

Answer to Problem 17.5P

The case when the input voltage is $-0.7\text{V}$ voltage $v_{O1}$ is $-1.7\text{V}$ and $v_{O2}$ is $-0.7\text{V}$ . The case when the input voltage is $-1.7\text{V}$ voltage $v_{O1}$ is $-1.7\text{V}$ and $v_{O2}$ is $-0.7\text{V}$

Explanation of Solution

Calculation:

Consider the case when the input voltage is $-0.7\text{V}$ .

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

  $v_{O1}=v_2-V_{BE2}$

Substitute $0$ for $v_2$ and $0.7\text{V}$ for $V_{BE2}$ in the above equation.

  $\begin{array}{c}{v}_{O1}=0-0.7\text{V}\\ =-0.7\text{V}\end{array}$

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

  $v_{O2}=v_1-V_{BE1}$

Substitute $0$ for $v_1$ and $0.7\text{V}$ for $V_{BE1}$ in the above equation.

  $\begin{array}{c}{v}_{O2}=0\text{V}-0.7\text{V}\\ =-0.7\text{V}\end{array}$

Consider the case when the input voltage is $-1.7\text{V}$ .

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

  $v_{O1}=v_1-V_{BE1}$

Substitute $-1\text{V}$ for $v_2$ and $0.7\text{V}$ for $V_{BE2}$ in the above equation.

  $\begin{array}{c}{v}_{O1}=-1\text{V}-0.7\text{V}\\ =-1.7\text{V}\end{array}$

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

  $v_{O2}=v_1-V_{BE1}$

Substitute $0$ for $v_2$ and $0.7\text{V}$ for $V_{BE2}$ in the above equation.

  $\begin{array}{c}{v}_{O2}=0\text{V}-0.7\text{V}\\ =-0.7\text{V}\end{array}$

Conclusion:

Therefore, the case when the input voltage is $-0.7\text{V}$ voltage $v_{O1}$ is $-1.7\text{V}$ and $v_{O2}$ is $-0.7\text{V}$ . The case when the input voltage is $-1.7\text{V}$ voltage $v_{O1}$ is $-1.7\text{V}$ and $v_{O2}$ is $-0.7\text{V}$

(d)

To determine

The expression for the power dissipated in the circuit.

Answer to Problem 17.5P

The value of the power dissipated in the circuit for the input voltage $-0.7\text{V}$ is $21.7784\text{mW}$ and for the input voltage of $-1.7\text{V}$ is $20.7\text{mW}$ .

Explanation of Solution

Calculation:

Consider the case when the input voltage is $-0.7V$ .

The expression for the value of the current through the resistance $R_3$ is given by,

  $i_{R_3}=\frac{v_{O2}-\left(-5.2\text{V}\right)}{3\text{k}\Omega }$

Substitute $-1.7\text{V}$ for $v_{O2}$ is given by,

  $\begin{array}{c}{i}_{R3}=\frac{-1.7\text{V}-\left(-5.2\text{V}\right)}{3\text{k}\Omega }\\ =1.167\text{mA}\end{array}$

The expression for the value of the current through the resistance $R_2$ is given by,

  $i_{R_2}=\frac{v_{O1}-\left(-5.2\text{V}\right)}{3\text{k}\Omega }$

Substitute $-0.7\text{V}$ for $v_{O1}$ is given by,

  $\begin{array}{c}{i}_{R_2}=\frac{-0.7\text{V}-\left(-5.2\text{V}\right)}{3\text{k}\Omega }\\ =1.5\text{mA}\end{array}$

The expression for the value of the power dissipated in the circuit is given by,

  $P_D=\left(i_E+i_{R_2}+i_{R_3}\right)\left(-V^{-}\right)$

Substitute $5.2\text{V}$ for $V^{-}$ , $1.5\text{mA}$ for $i_{R_2}$ , $1.167\text{mA}$ for $i_{R3}$ and $1.52\text{mA}$ for $i_E$ in the above equation.

  $\begin{array}{c}{P}_D=\left(1.52\text{mA}+1.5\text{mA}+1.167\text{mA}\right)\left(-5.2\text{V}\right)\\ =21.7784\text{mW}\end{array}$

Consider the case when the input voltage is $-1.7V$ .

The expression for the value of the current through the resistance $R_3$ is given by,

  $i_{R_3}=\frac{v_{O2}-\left(-5.2\text{V}\right)}{3\text{k}\Omega }$

Substitute $-0.7\text{V}$ for $v_{O2}$ is given by,

  $\begin{array}{c}{i}_{R3}=\frac{-0.7\text{V}-\left(-5.2\text{V}\right)}{3\text{k}\Omega }\\ =1.5\text{mA}\end{array}$

The expression for the value of the current through the resistance $R_2$ is given by,

  $i_{R_2}=\frac{v_{O1}-\left(-5.2\text{V}\right)}{3\text{k}\Omega }$

Substitute $-1.7\text{V}$ for $v_{O1}$ is given by,

  $\begin{array}{c}{i}_{R_2}=\frac{-1.7\text{V}-\left(-5.2\text{V}\right)}{3\text{k}\Omega }\\ =1.167\text{mA}\end{array}$

The expression for the value of the power dissipated in the circuit is given by,

  $P_D=\left(i_E+i_{R_2}+i_{R_3}\right)\left(-V^{-}\right)$

Substitute $5.2\text{V}$ for $V^{-}$ , $1.167\text{mA}$ for $i_{R_2}$ , $1.5\text{mA}$ for $i_{R3}$ and $1.32\text{mA}$ for $i_E$ in the above equation.

  $\begin{array}{c}{P}_D=\left(1.32\text{mA}+1.5\text{mA}+1.167\text{mA}\right)\left(-5.2\text{V}\right)\\ =20.7\text{mW}\end{array}$

Conclusion:

Therefore, the value of the power dissipated in the circuit for the input voltage $-0.7\text{V}$ is $21.7784\text{mW}$ and for the input voltage of $-1.7\text{V}$ is $20.7\text{mW}$ .