Chapter 17, Problem 17.15P

(a)

To determine

The value of the current and voltage $i_1,i_2,i_3,i_4,i_D$ and $v_O$ for the given value of $v_X$ and $v_Y$ .

Answer to Problem 17.15P

The value of the current $i_D$ is $0.7\text{4}\text{mA}$ , $i_1$ is $1.4\text{mA}$ , $i_2$ is $0.8\text{mA}$ , $i_3$ is $0.14\text{mA}$ , $i_4$ is $0.14\text{mA}$ and the value of voltage $v_O$ is $-4\text{V}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1.

  

The expression for the current $i_1$ is given by,

  $i_1=\frac{\left(-0.9\text{V}-V_{BE}\right)-\left(-3\text{V}\right)}{1\text{k}\Omega }$

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

  $\begin{array}{c}{i}_1=\frac{\left(-0.9\text{V}-0.7\text{V}\right)-\left(-3\text{V}\right)}{1\text{k}\Omega }\\ =1.4\text{mA}\end{array}$

The expression to determine the value of the current $i_3$ is given by,

  $i_3=\frac{\left(-0.2\text{V}-V_{BE}\right)-\left(-3\text{V}\right)}{15\text{k}\Omega }$

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

  $\begin{array}{c}{i}_3=\frac{\left(-0.2\text{V}-0.7\text{V}\right)-\left(-3\text{V}\right)}{15\text{k}\Omega }\\ =0.14\text{mA}\end{array}$

The expression to determine the value of the current $i_4$ is given by,

  $i_4=\frac{\left(-0.2\text{V}-V_{BE}\right)-\left(-3\text{V}\right)}{15\text{k}\Omega }$

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

  $\begin{array}{c}{i}_4=\frac{\left(-0.2\text{V}-0.7\text{V}\right)-\left(-3\text{V}\right)}{15\text{k}\Omega }\\ =0.14\text{mA}\end{array}$

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

  $i_2=\frac{V_{\gamma }}{0.5\text{k}\Omega }$

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

  $\begin{array}{c}{i}_2=\frac{0.4\text{V}}{0.5\text{k}\Omega }\\ =0.8\text{mA}\end{array}$

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

  $i_D=i_1+i_3-i_2$

Substitute $1.4\text{mA}$ for $i_1$ , $0.14\text{mA}$ for $i_3$ and $0.8\text{mA}$ for $i_2$ in the above equation.

  $\begin{array}{c}{i}_D=1.4\text{mA}+0.14\text{mA}-0.8\text{mA}\\ =0.7\text{4}\text{mA}\end{array}$

The output voltage is the voltage of the diode with opposite polarity and is given by,

  $v_O=-4\text{V}$

Conclusion:

Therefore, the value of the current $i_D$ is $0.7\text{4}\text{mA}$ , $i_1$ is $1.4\text{mA}$ , $i_2$ is $0.8\text{mA}$ , $i_3$ is $0.14\text{mA}$ , $i_4$ is $0.14\text{mA}$ and the value of voltage $v_O$ is $-4\text{V}$ .

(b)

To determine

The value of the current and voltage $i_1,i_2,i_3,i_4,i_D$ and $v_O$

Answer to Problem 17.15P

The value of the current $i_D$ is $0$ , $i_1$ is $1.4\text{mA}$ , $i_2$ is $0.153\text{mA}$ , $i_3$ is $0.153\text{mA}$ , $i_4$ is $0.153\text{mA}$ and voltage $v_O$ is $-0.0765\text{V}$ .

Explanation of Solution

Calculation:

The diode $D$ is cut off and transistors $Q_1$ is cutoff, therefore the diode current is given by,

  $i_D=0$

The expression for the current $i_1$ is given by,

  $i_1=\frac{\left(-0.9\text{V}-V_{BE}\right)-\left(-3\text{V}\right)}{1\text{k}\Omega }$

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

  $\begin{array}{c}{i}_1=\frac{\left(-0.9\text{V}-0.7\text{V}\right)-\left(-3\text{V}\right)}{1\text{k}\Omega }\\ =1.4\text{mA}\end{array}$

The expression to determine the value of the current $i_3$ is given by,

  $i_3=\frac{\left(v_X-V_{BE}\right)-\left(-3\text{V}\right)}{15\text{k}\Omega }$

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

  $\begin{array}{c}{i}_3=\frac{\left(0\text{V}-0.7\text{V}\right)-\left(-3\text{V}\right)}{15\text{k}\Omega }\\ =0.153\text{mA}\end{array}$

The expression to determine the value of the current $i_4$ is given by,

  $i_4=\frac{\left(v_X-V_{BE}\right)-\left(-3\text{V}\right)}{15\text{k}\Omega }$

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

  $\begin{array}{c}{i}_4=\frac{\left(0-0.7\text{V}\right)-\left(-3\text{V}\right)}{15\text{k}\Omega }\\ =0.153\text{mA}\end{array}$

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

  $i_2=i_4-i_D$

Substitute $0$ for $i_D$ and $0.153\text{mA}$ for $i_4$ in the above equation.

  $\begin{array}{c}{i}_2=0.153\text{mA}-0\\ =0.153\text{mA}\end{array}$

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

  $v_O=-i_2\left(0.5\text{k}\Omega \right)$

Substitute $0.153\text{mA}$ for $i_2$ in the above equation.

  $\begin{array}{c}{v}_O=-\left(0.153\text{mA}\right)\left(0.5\text{k}\Omega \right)\\ =-0.0765\text{V}\end{array}$

Conclusion:

Therefore, the value of the current $i_D$ is $0$ , $i_1$ is $1.4\text{mA}$ , $i_2$ is $0.153\text{mA}$ , $i_3$ is $0.153\text{mA}$ , $i_4$ is $0.153\text{mA}$ and $v_O$ is $-0.0765\text{V}$ .

(c)

To determine

The value of the current and voltage $i_1,i_2,i_3,i_4,i_D$ and $v_O$

Answer to Problem 17.15P

The value of the current $i_D$ is $0$ , $i_1$ is $1.6\text{mA}$ , $i_2$ is $0.14\text{mA}$ , $i_3$ is $0.14\text{mA}$ , $i_4$ is $0.14\text{mA}$ and $v_O$ is $-0.07\text{V}$ .

Explanation of Solution

Calculation:

The transistor $Q_3$ and $Q_6$ are cut off and the first transistor is in the active region.

The expression for the current $i_1$ is given by,

  $i_1=\frac{\left(v_Y-2V_{BE}\right)-\left(-3\text{V}\right)}{1\text{k}\Omega }$

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

  $\begin{array}{c}{i}_1=\frac{\left(0\text{V}-0.7\text{V}-\text{0}\text{.7V}\right)-\left(-3\text{V}\right)}{1\text{k}\Omega }\\ =1.6\text{mA}\end{array}$

The expression to determine the value of the current $i_3$ is given by,

  $i_3=\frac{\left(-0.2\text{V}-V_{BE}\right)-\left(-3\text{V}\right)}{15\text{k}\Omega }$

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

  $\begin{array}{c}{i}_3=\frac{\left(-0.2\text{V}-0.7\text{V}\right)-\left(-3\text{V}\right)}{15\text{k}\Omega }\\ =0.14\text{mA}\end{array}$

The expression to determine the value of the current $i_4$ is given by,

  $i_4=\frac{\left(-0.2\text{V}-V_{BE}\right)-\left(-3\text{V}\right)}{15\text{k}\Omega }$

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

  $\begin{array}{c}{i}_4=\frac{\left(-0.2\text{V}-0.7\text{V}\right)-\left(-3\text{V}\right)}{15\text{k}\Omega }\\ =0.14\text{mA}\end{array}$

The diode is in the cut off and the diode current is given by,

  $i_D=0$

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

  $i_2+i_D=i_3$

Substitute $0$ for $i_D$ and $0.14\text{mA}$ for $i_3$ in the above equation.

  $i_2=0.14\text{mA}$

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

  $v_O=-i_2\left(0.5\text{k}\Omega \right)$

Substitute $0.14\text{mA}$ for $i_2$ in the above equation.

  $\begin{array}{c}{v}_O=-\left(0.14\text{mA}\right)\left(0.5\text{k}\Omega \right)\\ =-0.07\text{V}\end{array}$

Conclusion:

Therefore, the value of the current $i_D$ is $0$ , $i_1$ is $1.6\text{mA}$ , $i_2$ is $0.14\text{mA}$ , $i_3$ is $0.14\text{mA}$ , $i_4$ is $0.14\text{mA}$ and $v_O$ is $-0.07\text{V}$ .

(d)

To determine

The value of the current and voltage $i_1,i_2,i_3,i_4,i_D$ and $v_O$

Answer to Problem 17.15P

The value of the current $i_D$ is $0.953\text{mA}$ , $i_1$ is $1.6\text{mA}$ , $i_2$ is $0.8\text{mA}$ , $i_3$ is $0.153\text{mA}$ , $i_4$ is $0.153\text{mA}$ and $v_O$ is $-0.4\text{V}$ .

Explanation of Solution

Calculation:

The transistor $Q_1$ , $Q_3$ and $Q_6$ are on.

The expression for the current $i_1$ is given by,a

  $i_1=\frac{\left(v_Y-2V_{BE}\right)-\left(-3\text{V}\right)}{1\text{k}\Omega }$

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

  $\begin{array}{c}{i}_1=\frac{\left(0\text{V}-0.7\text{V}-\text{0}\text{.7V}\right)-\left(-3\text{V}\right)}{1\text{k}\Omega }\\ =1.6\text{mA}\end{array}$

The expression to determine the value of the current $i_3$ is given by,

  $i_3=\frac{\left(v_X-V_{BE}\right)-\left(-3\text{V}\right)}{15\text{k}\Omega }$

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

  $\begin{array}{c}{i}_3=\frac{\left(0\text{V}-0.7\text{V}\right)-\left(-3\text{V}\right)}{15\text{k}\Omega }\\ =0.153\text{mA}\end{array}$

The expression to determine the value of the current $i_4$ is given by,

  $i_4=\frac{\left(v_X-V_{BE}\right)-\left(-3\text{V}\right)}{15\text{k}\Omega }$

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

  $\begin{array}{c}{i}_4=\frac{\left(0-0.7\text{V}\right)-\left(-3\text{V}\right)}{15\text{k}\Omega }\\ =0.153\text{mA}\end{array}$

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

  $i_2=\frac{V_{\gamma }}{0.5\text{k}\Omega }$

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

  $\begin{array}{c}{i}_2=\frac{0.4\text{V}}{0.5\text{k}\Omega }\\ =0.8\text{mA}\end{array}$

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

  $i_D=i_1+i_3-i_2$

Substitute $1.6\text{mA}$ for $i_1$ , $0.153\text{mA}$ for $i_3$ and $0.8\text{mA}$ for $i_2$ in the above equation.

  $\begin{array}{c}{i}_D=1.6\text{mA}+0.153\text{mAA}-0.8\text{mA}\\ =0.953\text{mA}\end{array}$

The output voltage is the voltage of the opposite polarity and is given by,

  $v_O=-0.4\text{V}$ .

Conclusion:

Therefore, the value of the current $i_D$ is $0.953\text{mA}$ , $i_1$ is $1.6\text{mA}$ , $i_2$ is $0.8\text{mA}$ , $i_3$ is $0.153\text{mA}$ , $i_4$ is $0.153\text{mA}$ and $v_O$ is $-0.4\text{V}$ .