Chapter 17, Problem 17.12EP

(a)

To determine

The base and the collector current in each of the transistor. Also, the power dissipation in the gate.

Answer to Problem 17.12EP

The currents in the transistors are, $i_{C2}$ is $0$ , $i_{B2}$ is $0$ , $i_{BO}$ is $0$ , $i_{C5}$ is $0$ , $i_{B5}$ is $0$ , $i_{CO}$ is $0$ , $i_{C4}$ is $0$ , $i_{B4}$ is $0$ , $i_{B3}$ is $0$ and $i_{C3}$ is $0$ . The power dissipated in the circuit is $495\text{μW}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

The expression to determine the value of the voltage $v_1$ is given by,

  $v_1=v_X+V_{\gamma }$

Substitute $0.4\text{V}$ for $v_X$ and $0.3\text{V}$ for $V_{\gamma }$ in the above equation.

  $\begin{array}{c}{v}_1=0.4\text{V}+0.3\text{V}\\ =0.7\text{V}\end{array}$

The expression to determine the value of the current $i_1$ is given by,

  $i_1=\frac{5\text{V}-v_1}{40\text{k}\Omega }$

Substitute $0.7\text{V}$ for $v_1$ in the above equation.

  $\begin{array}{c}{i}_1=\frac{5\text{V}-0.7\text{V}}{40\text{k}\Omega }\\ =107.5\times {10}^{-3}\text{mA}\end{array}$

The conversion from $1\text{mA}$ into $\text{μA}$ is given by,

  $1\text{mA}={10}^3\text{μA}$

The conversion from $107.5\times {10}^{-3}\text{mA}$ into $\text{μA}$ is given by,

  $107.5\times {10}^{-3}\text{mA}=107.5\text{μA}$

The second transistor and the output transistor are in cut off region and the current in the circuit are given by,

  $\begin{array}{l}{i}_{B5}=0\\ i_{C5}=0\\ i_{B2}=0\\ i_{C2}=0\end{array}$

The expression for the value of the collector current and the base current of the output transistor is given by,

  $\begin{array}{l}{i}_{CO}=0\\ i_{BO}=0\end{array}$

The expression for the base and the collector current of the third and the fourth transistor are given by,

  $\begin{array}{l}{i}_{B3}=0\\ i_{C3}=0\\ i_{B4}=0\\ i_{C4}=0\end{array}$

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

  $P=i_1\left(5\text{V}-v_X\right)$

Substitute $107.5\text{μA}$ for $i_1$ and $0.4\text{V}$ for $v_X$ in the above equation.

  $\begin{array}{c}P=\left(107.5\text{μA}\right)\left(5\text{V}-0.4\text{V}\right)\\ =495\text{μW}\end{array}$

Conclusion:

Therefore, the currents in the transistors are, $i_{C2}$ is $0$ , $i_{B2}$ is $0$ , $i_{BO}$ is $0$ , $i_{C5}$ is $0$ , $i_{B5}$ is $0$ , $i_{CO}$ is $0$ , $i_{C4}$ is $0$ , $i_{B4}$ is $0$ , $i_{B3}$ is $0$ and $i_{C3}$ is $0$ . The power dissipated in the circuit is $495\text{μW}$ .

(b)

To determine

The base and the collector current in each of the transistor. Also, the power dissipation in the gate.

Answer to Problem 17.12EP

The currents in the transistors are, $i_{C2}$ is $325\text{μA}$ , $i_{B2}$ is $90\text{μA}$ , $i_{BO}$ is $415\text{μA}$ , $i_{C5}$ is $0$ , $i_{B5}$ is $0$ , $i_{CO}$ is $0$ , $i_{C4}$ is $0$ , $i_{B4}$ is $0$ , $i_{B3}$ is $0$ and $i_{C3}$ is $0$ . The power dissipated in the circuit is $2.08\text{mW}$ .

Explanation of Solution

Calculation:

The expression to determine the value of the voltage $v_1$ is given by,

  $v_1=V_{BE}{\left(\text{on}\right)}_{Q_O}+V_{BE}{\left(\text{on}\right)}_{Q_2}$

Substitute $0.7\text{V}$ for $V_{BE}{\left(\text{on}\right)}_{Q_O}$ and $0.7\text{V}$ for $V_{BE}{\left(\text{on}\right)}_{Q_2}$ in the above equation.

  $\begin{array}{c}{v}_1=0.7\text{V}+0.7\text{V}\\ =1.4\text{V}\end{array}$

The expression for the voltage $v_{C2}$ is given by,

  $v_{C2}=V_{BE}{\left(\text{on}\right)}_{Q_O}+V_{CE}{\left(\text{on}\right)}_{Q_2}$

Substitute $0.7\text{V}$ for $V_{BE}{\left(\text{on}\right)}_{Q_O}$ and $0.4\text{V}$ for $V_{CE}{\left(\text{on}\right)}_{Q_2}$ in the above equation.

  $\begin{array}{c}{v}_{C2}=0.7\text{V}+0.4\text{V}\\ =1.1\text{V}\end{array}$

The expression to determine the value of the current $i_1$ is given by,

  $i_1=\frac{5\text{V}-v_1}{40\text{k}\Omega }$

Substitute $1.4\text{V}$ for $v_1$ in the above equation.

  $\begin{array}{c}{i}_1=\frac{5\text{V}-1.4\text{V}}{40\text{k}\Omega }\\ =90\times {10}^{-3}\text{mA}\end{array}$

The expression to determine the value of the current $i_2$ is given by,

  $i_2=\frac{5\text{V}-v_{C2}}{12\text{k}\Omega }$

Substitute $1.1\text{V}$ for $v_{C2}$ in the above equation.

  $\begin{array}{c}{i}_2=\frac{5\text{V}-1.1\text{V}}{12\text{k}\Omega }\\ =325\times {10}^{-3}\text{mA}\end{array}$

The conversion from $90\times {10}^{-3}\text{mA}$ into $\text{μA}$ is given by,

  $90\times {10}^{-3}\text{mA}=90\text{μA}$

The conversion from $325\times {10}^{-3}\text{mA}$ into $\text{μA}$ is given by,

  $325\times {10}^{-3}\text{mA}=325\text{μA}$

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

  $i_{B2}=i_1$

Substitute $90\text{μA}$ for $i_1$ in the above equation.

  $i_{B2}=90\text{μA}$

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

  $i_{C2}=i_2$

Substitute $325\text{μA}$ for $i_2$ in the above equation.

  $i_{C2}=325\text{μA}$

The fifth transistor is in the cut off region and the current in the circuit are given by,

  $\begin{array}{l}{i}_{B5}=0\\ i_{C5}=0\end{array}$

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

  $i_{BO}=i_{B2}+i_{C2}$

Substitute $325\text{μA}$ for $i_{C2}$ and $90\text{μA}$ for $i_{B2}$ in the above equation.

  $\begin{array}{c}{i}_{BO}=90\text{μA}+325\text{μA}\\ =415\text{μA}\end{array}$

The expression for the base and the collector current of the third and the fourth transistor are given by,

  $\begin{array}{l}{i}_{B3}=0\\ i_{C3}=0\\ i_{B4}=0\\ i_{C4}=0\end{array}$

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

  $P=\left(i_1+i_2\right)\left(5\text{V}\right)$

Substitute $90\text{μA}$ for $i_1$ and $325\text{μA}$ for $i_2$ in the above equation.

  $\begin{array}{c}P=\left(325\text{μA}+90\text{μA}\right)\left(5\text{V}\right)\\ =2.08\times {10}^3\text{μW}\end{array}$

The conversion from $\text{μW}$ into $\text{mW}$ is given by,

  $1\text{μW}={10}^{-3}\text{mW}$

The conversion from $2.08\times {10}^3\text{μW}$ into $\text{mW}$ is given by,

  $2.08\times {10}^3\text{μW}=2.08\text{mW}$

Conclusion:

Therefore, the currents in the transistors are, $i_{C2}$ is $325\text{μA}$ , $i_{B2}$ is $90\text{μA}$ , $i_{BO}$ is $415\text{μA}$ , $i_{C5}$ is $0$ , $i_{B5}$ is $0$ , $i_{CO}$ is $0$ , $i_{C4}$ is $0$ , $i_{B4}$ is $0$ , $i_{B3}$ is $0$ and $i_{C3}$ is $0$ . The power dissipated in the circuit is $2.08\text{mW}$ .