Chapter 1, Problem 1.47P

Find I and $V_O$ in each circuit shown in Figure P1.47 if (i) $V_{\gamma }=0.7\text{V}$ and (ii) $V_{\gamma }=0.6\text{V}$ .

1. 

Figure P1.47

To determine

The values of $I$ and $V_o$ in the given circuit.

Answer to Problem 1.47P

For $V_{\gamma }=0.7\text{V}$ , $I=0.215\text{mA,}{V}_o=0.7\text{V}$ .

For $V_{\gamma }=0.6\text{V}$ , $I=0.22\text{mA},V_o=0.6\text{V}$ .

Explanation of Solution

Given Information:

The given circuit is shown below.

  

Two values of $V_{\gamma }$ are

  $\begin{array}{l}{V}_{\gamma }=0.7\text{V}\\ V_{\gamma }=0.6\text{V}\end{array}$

Calculation:

For $V_{\gamma }=0.7\text{V}$ ,

The given circuit diagram is:

  

Assuming the diode is in ON state and the modified circuit is:

  

Applying Kirchhoff’s voltage law:

  $\begin{array}{l}-5+I\left(20\text{k}\right)+0.7=0\\ I=\frac{5-0.7}{20\text{k}}\\ I=0.215\text{mA}\end{array}$

The direction of current is from p-region to n-region of diode. Hence, the assumption is correct.

Applying Kirchhoff’s voltage law:

  

  $\begin{array}{l}-V_o+0.7\text{V}=0\\ V_o=0.7\text{V}\end{array}$

For $V_{\gamma }=0.6\text{V}$ ,

The given circuit diagram is:

  

Assuming the diode is in ON state and the modified circuit is:

  

Applying Kirchhoff’s voltage law:

  $\begin{array}{l}-5+I\left(20\text{k}\right)+0.6=0\\ I=\frac{5-0.6}{20\text{k}}\\ I=0.22\text{mA}\end{array}$

The direction of current is from p-region to n-region of diode. Hence, the assumption is correct.

Applying Kirchhoff’s voltage law:

  

  $\begin{array}{l}-V_o+0.6\text{V}=0\\ V_o=0.6\text{V}\end{array}$

To determine

The values of $I$ and $V_o$ in the given circuit.

Answer to Problem 1.47P

For $V_{\gamma }=0.7\text{V}$ , $I=0.2325\text{mA},V_o=-0.35\text{V}$ .

For $V_{\gamma }=0.6\text{V}$ , $I=0.235\text{mA},V_o=-0.3\text{V}$ .

Explanation of Solution

Given Information:

The given circuit is shown below.

  

Two values of $V_{\gamma }$ are

  $\begin{array}{l}{V}_{\gamma }=0.7\text{V}\\ V_{\gamma }=0.6\text{V}\end{array}$

Calculation:

For $V_{\gamma }=0.7\text{V}$ ,

The given circuit diagram is:

  

Assuming the diode is in ON state and the modified circuit is:

  

Applying Kirchhoff’s voltage law:

  $\begin{array}{l}-\text{5+}I\left(\text{20}\text{k}\right)+0.7+I\left(\text{20}\text{k}\right)+\left(-5\right)=0\\ \left(\text{40}\text{k}\right)I=10-0.7\\ I=\frac{9.3}{\text{40}\text{k}}\\ I=0.2325\text{mA}\end{array}$

The direction of current is from p-region to n-region of diode. Hence, the assumption is correct.

Applying Kirchhoff’s voltage law in the following circuit:

  

  $\begin{array}{l}-V_o+I\left(\text{20}\text{k}\right)+\left(-5\right)=0\\ V_o=\left(0.2325\text{mA}\right)\left(\text{20}\text{k}\right)-5\\ V_o=4.65-5\\ V_o=-0.35\text{V}\end{array}$

For $V_{\gamma }=0.6\text{V}$ ,

The given circuit diagram is:

  

Assuming the diode is in ON state and the modified circuit is:

Applying Kirchhoff’s voltage law:

  $\begin{array}{l}-\text{5+}I\left(\text{20}\text{k}\right)+0.6+I\left(\text{20}\text{k}\right)+\left(-5\right)=0\\ \left(\text{40}\text{k}\right)I=10-0.6\\ I=\frac{9.4}{\text{40}\text{k}}\\ I=0.235\text{mA}\end{array}$

The direction of current is from p-region to n-region of diode. Hence, the assumption is correct.

Applying Kirchhoff’s voltage law in the following circuit:

  

To determine

The values of $I$ and $V_o$ in the given circuit.

Answer to Problem 1.47P

For $V_{\gamma }=0.7\text{V}$ , $I=0.372\text{mA},V_o=0.14\text{V}$

For $V_{\gamma }=0.6\text{V}$ , $I=0.376\text{mA},V_o=0.12\text{V}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

Two values of $V_{\gamma }$ are

  $\begin{array}{l}{V}_{\gamma }=0.7\text{V}\\ V_{\gamma }=0.6\text{V}\end{array}$

Calculation:

For $V_{\gamma }=0.7\text{V}$ ,

The given circuit diagram is:

  

Assuming the diode is in ON state and the modified circuit is :

Applying Kirchhoff’s voltage law:

  $\begin{array}{l}-\text{2+}I\left(\text{5}\text{k}\right)+0.7+I\left(20\text{k}\right)+\left(-8\right)=0\\ \left(25\text{k}\right)I=10-0.7\\ I=\frac{9.3}{25\text{k}}\\ I=0.372\text{mA}\end{array}$

The direction of current is from p-region to n-region of diode. Hence, the assumption is correct.

Applying Kirchhoff’s voltage law in the following circuit:

  

  $\begin{array}{l}-V_o+0.7+I\left(\text{20}\text{k}\right)+\left(-8\right)=0\\ V_o=0.7+\left(0.372\text{mA}\right)\left(\text{20}\text{k}\right)-8\\ V_o=-7.3+7.44\\ V_o=0.14\text{V}\end{array}$

For $V_{\gamma }=0.6\text{V}$

The given circuit diagram is:

  

Assuming the diode is in ON state and the modified circuit is:

  

Applying Kirchhoff’s voltage law:

  $\begin{array}{l}-\text{2+}I\left(\text{5}\text{k}\right)+0.6+I\left(20\text{k}\right)+\left(-8\right)=0\\ \left(25\text{k}\right)I=10-0.6\\ I=\frac{9.4}{25\text{k}}\\ I=0.376\text{mA}\end{array}$

The direction of current is from p-region to n-region of diode. Hence, the assumption is correct.

Applying Kirchhoff’s voltage law in the following circuit:

  

  $\begin{array}{l}-V_o+0.6+I\left(\text{20}\text{k}\right)+\left(-8\right)=0\\ V_o=0.6+\left(0.376\text{mA}\right)\left(\text{20}\text{k}\right)-8\\ V_o=-7.4+7.52\\ V_o=0.12\text{V}\end{array}$

To determine

The values of $I$ and $V_o$ in the given circuit.

Answer to Problem 1.47P

For $V_{\gamma }=0.7\text{V}$ , $I=0,V_o=-5\text{V}$

For $V_{\gamma }=0.6\text{V}$ , $I=0,V_o=-5\text{V}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

Two values of $V_{\gamma }$ are

  $\begin{array}{l}{V}_{\gamma }=0.7\text{V}\\ V_{\gamma }=0.6\text{V}\end{array}$

Calculation:

For $V_{\gamma }=0.7\text{V}$ ,

The given circuit diagram is:

  

Assuming the diode is in OFF state. Hence, the modified circuit is:

  

From the circuit:

  $\begin{array}{l}{V}_D<V_{\gamma }\\ -10\text{V}<0.7\text{V}\end{array}$

In this circuit, the diode is reverse biased, so it is in OFF mode. Hence, the assumption is correct and the value of current $I$ is zero.

Applying Kirchhoff’s voltage law in the following circuit:

  

  $\begin{array}{l}-V_o+I\left(\text{20}\text{k}\right)+\left(-5\right)=0\\ -V_o+0\times \left(\text{20}\text{k}\right)+\left(-5\right)=0\\ V_o=-5\text{V}\end{array}$

For $V_{\gamma }=0.6\text{V}$

The given circuit diagram is:

  

Assuming the diode is in OFF state. Hence, the modified circuit is:

  

From the circuit:

  $\begin{array}{l}{V}_D<V_{\gamma }\\ -10\text{V}<0.6\text{V}\end{array}$

In this circuit, the diode is reverse biased, so it is in OFF mode. Hence, the assumption is correct and the value of current $I$ is zero.

Applying Kirchhoff’s voltage law in the following circuit:

  

  $\begin{array}{l}-V_o+I\left(\text{20}\text{k}\right)+\left(-5\right)=0\\ -V_o+0\times \left(\text{20}\text{k}\right)+\left(-5\right)=0\\ V_o=-5\text{V}\end{array}$