Chapter 2, Problem 2.47P

Consider the circuit shown in Figure P2.47. Assume each diode cut−in voltage is $V_{\gamma }=0.6\text{V}$ . (a) Determine $R_1,R_2$ , and $R_3$ such that $I_{D1}=0.2\text{mA}$ , $I_{D2}=0.3\text{mA}$ , and $I_{D3}=0.5\text{mA}$ . (b) Find $V_1,V_2$ , and each diode current for $R_1=10\text{k}\Omega$ , $R_2=4\text{k}\Omega$ , and $R_3=2.2\text{k}\Omega$ . (c) Repeat part (b) for $R_1=3\text{k}\Omega$ , $R_2=6\text{k}\Omega$ , and $R_3=2.5\text{k}\Omega$ . (d) Repeat part (b) for $R_1=6\text{k}\Omega$ , $R_2=3\text{k}\Omega$ , and $R_3=6\text{k}\Omega$ .

Figure P2.47

To determine

The values of $R_1$ , $R_2$ , and $R_3$

Answer to Problem 2.47P

The required values are,

  $R_1$ is $25k\Omega$ .

  $R_2$ is $10k\Omega$ .

  $R_3$ is $4.4k\Omega$

Explanation of Solution

Given:

Diode’s cut-in voltage = $V_{\gamma }=0.6\text{V}$

  $I_{D_1}=0.2\text{mA}$ , $I_{D_2}=0.3\text{mA}$ , and $I_{D_3}=0.5\text{mA}$

  

Calculation:

Assume all diode are conducting.

Draw the circuit diagram with node voltages and cut-in voltages.

  

Figure 1

From Figure 1, the voltage at node $V_1$ is,

  $\begin{array}{c}{V}_1=5-0.6\\ =4.4\text{V}\end{array}$

The voltage at node $V_2$ is,

  $V_2=-0.6\text{V}$

Write the expression for current following through diode $D_1$ .

  $I_{D_1}=\frac{10-0-V_1}{R_1}$

Substitute $0.2\text{mA}$ for $I_{D_1}$ and $4.4V$ for $V_1$ .

  $\begin{array}{l}0.2\times {10}^{-3}=\frac{10-0.6-4.4}{R_1}\\ 0.2\times {10}^{-3}=\frac{5}{R_1}\\ R_1=\frac{5}{0.2\times {10}^{-3}}\\ =25k\Omega \end{array}$

Therefore, the value of the resistor, $R_1$ is $25k\Omega$ .

Apply Kirchhoff’s current law at node $V_1$ .

  $\begin{array}{l}{I}_{R_2}-I_{D_1}-I_{D_2}=0\\ I_{R_2}=I_{D_1}+I_{D_2}\end{array}$

Substitute $0.2\text{mA}$ for $I_{D_1}$ and $0.3\text{mA}$ for $I_{D_2}$ .

  $\begin{array}{c}{I}_{R_2}=0.2\text{mA}+0.3\text{mA}\\ =0.5\text{mA}\end{array}$

Calculate the value of resistor $R_2$ using ohm’s law.

  $I_{R_2}=\frac{V_1-V_2}{R_2}$

Substitute $0.5\text{mA}$ for $I_{R_2}$ , $4.4\text{V}$ For $V_1$ and $-0.6\text{V}$ for $V_2$ .

  $\begin{array}{l}0.5\times {10}^{-3}=\frac{4.4-(-0.6)}{R_2}\\ R_2=\frac{5}{0.5\times {10}^{-3}}\\ R_2=10k\Omega \end{array}$

Therefore, the value of the resistor $R_2$ is $10k\Omega$ .

Apply Kirchhoff’s current law at node $V_2$ .

  $\begin{array}{l}-I_{R_2}-I_{D_3}+I_{R_3}=0\\ I_{R_3}=I_{R_2}+I_{D_3}\end{array}$

Substitute $0.5\text{mA}$ for $I_{R_2}$ and $0.5\text{mA}$ for $I_{D_3}$ .

  $\begin{array}{c}{I}_{R_3}=0.5\text{mA}+0.5\text{mA}\\ =1\text{mA}\end{array}$

Calculate the value of resistor $R_3$ using ohm’s law.

  $R_3=\frac{V_2-(-5)}{I{R}_3}$

Substitute $1\text{mA}$ for $I_{R_3}$ and $-0.6\text{V}$ for $V_2$ .

  $\begin{array}{l}{R}_3=\frac{-0.6-(-5)}{1\times {10}^{-3}}\\ =\frac{4.4}{1\times {10}^{-3}}\\ =4.4k\Omega \end{array}$

Therefore, the value of the resistor, $R_3$ is $4.4k\Omega$

Conclusion:

Therefore, the required values are $R_1$ is $25k\Omega$ , $R_2$ is $10k\Omega$ and $R_3$ is $4.4k\Omega$

To determine

To find: The values of $V_1$ , $V_2$ and diode current for each.

Answer to Problem 2.47P

The required values are,

  $V_1$ is $4.4\text{V}$ .

  $V_2$ is $-0.6\text{V}$

  $I_{D_3}$ is $0.75\text{mA}$

  $I_{D_2}$ is $0.75\text{mA}$

  $I_{D_1}$ is $0.5\text{mA}$

Explanation of Solution

Given:

Diode’s cut-in volt- age is $V_{\gamma }=0.6\text{V}$

  $R_1=10k\Omega$ , $R_2=4k\Omega$ , and $R_3=2.2k\Omega$ .

  

Calculation:

Assume at diodes are conducting.

From Figure 1, the voltage at node $V_1$ is,

  $\begin{array}{l}{V}_1=5-0.6\\ =4.4\text{V}\end{array}$

The voltage at node $V_2$ is,

  $V_2=-0.6\text{V}$

Calculate the current, $I_{D_1}$ .

  $I_{D_1}=\frac{10-0.6-V_1}{R_1}$

Substitute $10k\Omega$ for $R_1$ and $4.4V$ for $V_1$ .

  $\begin{array}{c}{I}_{D_1}=\frac{10-0.6-4.4}{10\times {10}^{-3}}\\ =\frac{5}{10\times {10}^{-3}}\\ =0.5\text{mA}\end{array}$

Therefore, the current following through diode $D_1$ , $I_{D_1}$ is $0.5\text{mA}$ .

Calculate the current following through resistor $R_2$ .

  $I_{R_2}=\frac{V_1-V_2}{R_2}$

Substitute $4k\Omega$ for $R_2$ , $4.4\text{V}$ for $V_1$ and $-0.6\text{V}$ for $V_2$ .

  $\begin{array}{c}{I}_{R_2}=\frac{4.4-(-0.6)}{4\times {10}^{-3}}\\ =\frac{5}{4\times {10}^{-3}}\\ =1.25\text{mA}\end{array}$

Apply Kirchhoff’s current law at node $V_1$ .

  $\begin{array}{l}{I}_{R_2}-I_{D_1}-I_{D_2}=0\\ I_{D_2}=I_{R_2}-I_{D_1}\end{array}$

Substitute $0.5\text{mA}$ for $I_{D_1}$ and $1.25\text{mA}$ for $I_{R_2}$ .

  $\begin{array}{c}{I}_{D_2}=1.25\text{mA}-0.5\text{mA}\\ =0.75\text{mA}\end{array}$

Therefore, the current following diode $D_2$ , $I_{D_2}$ is $0.75\text{mA}$ .

Calculate the current following through resistor, $R_3$ .

  $I_{R_3}=\frac{V_2-(-5)}{R_3}$

Substitute $2.2k\Omega$ for $R_3$ and $-0.6\text{V}$ for $V_2$ .

  $\begin{array}{l}{I}_{R_3}=\frac{-0.6+5}{2.2\times {10}^3}\\ =\frac{4.4}{2.2\times {10}^3}\\ =2\text{mA}\end{array}$

Apply Kirchhoff’s current law at node $V_2$ .

  $\begin{array}{l}-I_{R_2}-I_{D_3}+I_{R_3}=0\\ I_{D_3}=I_{R_3}-I_{R_2}\end{array}$

Substitute $1.25\text{mA}$ for $I_{R_2}$ and $2\text{mA}$ for $I_{R_3}$ .

  $\begin{array}{c}{I}_{D_3}=2\text{mA}-1.25\text{mA}\\ =0.75\text{mA}\end{array}$

Therefore, the current following diode $D_3$ , $I_{D_3}$ is $0.75\text{mA}$ .

Conclusion:

Therefore, the required values are $V_1$ is $4.4\text{V}$ , $V_2$ is $-0.6\text{V}$ , $I_{D_3}$ is $0.75\text{mA}$ , $I_{D_2}$ is $0.75\text{mA}$ and $I_{D_1}$ is $0.5\text{mA}$

To determine

To find: The values of $V_1$ , $V_2$ and diode current for each.

Answer to Problem 2.47P

The require values are

  $I_{D_2}$ is $0\text{A}$ .

  $V_1$ is $6.07\text{V}$

  $V_2$ is $-0.6\text{V}$

  $I_{D_1}$ is $1.11\text{mA}$

  $I_{D_3}$ is $0.65\text{mA}$

Explanation of Solution

Given:

Diode’s cut-in volt- age is $V_{\gamma }=0.6\text{V}$

  $R_1=3k\Omega$ , $R_2=6k\Omega$ , and $R_3=2.5k\Omega$

  

Calculation:

Assume diode $D_2$ is OFF and remaining diodes are ON.

Draw the current diagram with node voltage and cut-in voltages.

  

Figure 2

In Figure 2, the Diode $D_2$ is reversed biased. Therefore, the current following through diode $D_2$ is zero.

That is, $I_{D_2}=0\text{A}$

Therefore, the current following through diode $D_2$ , $I_D{}_{{}_2}$ is $0\text{A}$ .

From Figure 2, the voltage at node $V_2$ is,

  $V_2=-0.6\text{V}$

Apply Kirchhoff’s current law at node $V_1$ .

  $\frac{V_1-(-0.6)-10}{R_1}+\frac{V_1-V_2}{R_2}=0$

Substitute $-0.6V$ for $V_2$ , $3k\Omega$ for $R_1$ and $6k\Omega$ for $R_2$ .

  $\begin{array}{l}\frac{V_1-(-0.6)-10}{3k}+\frac{V_1-(-0.6)}{6k}=0\\ \frac{V_1-9.4}{3k}+\frac{V_1+0.6}{6k}=0\\ \frac{2V_1-18.8+V_1+0.6}{6}=0\\ 3V_1-18.2=0\\ 3V_1=18.2\\ V_1=6.07\text{V}\end{array}$

Calculate the current $I_{D_1}$ .

  $I_{D_1}=\frac{10-0.6-V_1}{R_1}$

Substitute $3k\Omega$ for $R_1$ and $6.07\text{V}$ for $V_1$ .

  $\begin{array}{l}{I}_{D_1}=\frac{10-0.6-6.07}{3\times {10}^3}\\ =\frac{3.33}{3\times {10}^3}\\ =1.11\text{mA}\end{array}$

Therefore, the current following diode $D_1$ , $I_{D_1}$ is $1.11\text{mA}$ .

From Figure 2,

  $\begin{array}{l}{I}_{R_2}=I_{D_1}\\ =1.11\text{mA}\end{array}$

Calculate the current following through resistor, $R_3$ .

  $I_{R_3}=\frac{V_2-(-5)}{R_3}$

Substitute $2.5k\Omega$ for $R_3$ and $-0.6\text{V}$ for $V_2$ .

  $\begin{array}{l}{I}_{R_3}=\frac{0.6+5}{2.5\times {10}^3}\\ =\frac{4.4}{2.5\times {10}^3}\\ =1.76\text{mA}\end{array}$

Apply Kirchhoff’s current law at node $V_2$ .

  $\begin{array}{l}-I_{R_2}-I_{D_3}+I_{R_3}=0\\ I_{D_3}=I_{R_3}-I_{R_2}\end{array}$

Substitute $1.11\text{mA}$ for $I_{R_2}$ and $1.76\text{mA}$ for $I_{R_3}$

  $\begin{array}{l}{I}_{D_3}=1.76\text{mA}-1.11\text{mA}\\ =0.65\text{mA}\end{array}$

Therefore, the current following diode $D_3$ , $I_{D_3}$ is $0.6\text{5mA}$ .

To determine

To find: The values of $V_1$ , $V_2$ and diode current for each.

Answer to Problem 2.47P

The required values are

  $I_{D_3}$ is $0\text{A}$

  $V_1$ is $4.4\text{V}$

  $V_2$ is $1.27\text{V}$

  $I_{D_1}$ is $0.833\text{mA}$

  $I_{D_2}$ is $0.211\text{mA}$

Explanation of Solution

Given:

Diode’s cut-in volt- age is $V_{\gamma }=0.6\text{V}$

  $R_1=6k\Omega$ , $R_2=3k\Omega$ , and $R_3=6k\Omega$

  

Calculation:

Assume diode $D_3$ is OFF and remaining diodes are ON.

Draw the circuit diagram with node voltages and cut-in voltages.

  

Figure 3

In Figure 3, the Diode $D_3$ is reversed biased. Therefore, the current following through diode $D_3$ is zero.

That is, $I_{D_3}=0\text{A}$

Therefore, the current following through diode $D_3$ , $I_{D_3}$ is $0A$ .

From Figure 1, the voltage at node $V_1$ is,

  $\begin{array}{c}{V}_1=5-0.6\\ =4.4\text{V}\end{array}$

Therefore, the node voltage, $V_1$ is $4.4\text{V}$

Apply Kirchhoff’s current law at node $V_2$ .

  $\frac{V_2-(-5)}{R_3}+\frac{V_2-V_1}{R_2}=0$

Substitute $4.4\text{V}$ for $V_1$ , $3k\Omega$ for $R_2$ and $6k\Omega$ for $R_3$ .

  $\begin{array}{l}\frac{V_2-(-5)}{6k}+\frac{V_2-4.4}{3k}=0\\ \frac{V_2+5+2V_2-8.8}{6k}=0\\ 3V_2-3.8=0\\ V_2=1.27\text{V}\end{array}$

Calculate the current $I_{D_1}$ .

  $I_{D_1}=\frac{10-0.6-V_1}{R_1}$

Substitute $6k\Omega$ for $R_1$ and $4.4\text{V}$ for $V_1$ .

  $\begin{array}{l}{I}_{D_1}=\frac{10-0.6-4.4}{6\times {10}^3}\\ =\frac{5}{6\times {10}^3}\\ =0.833\text{mA}\end{array}$

Therefore, the current following diode $D_1$ , $I_{D_1}$ is $0.833\text{mA}$ .

Calculate the current following through resistor, $R_2$ .

  $I_{R_2}=\frac{V_1-V_2}{R_2}$

Substitute $3k\Omega$ for $R_2$ , $4.4\text{V}$ for $V_1$ and $1.27\text{V}$ for $V_2$ .

  $\begin{array}{l}{I}_{R_2}=\frac{4.4-1.27}{3\times {10}^3}\\ =\frac{3.13}{3\times {10}^3}\\ =1.044\text{mA}\end{array}$

Apply Kirchhoff’s current law at node $V_1$ .

  $\begin{array}{l}-I_{R_2}-I_{D_1}+I_{D_2}=0\\ I_{D_2}=I_{R_2}-I_{D_1}\end{array}$

Substitute $0.833\text{mA}$ for $I_{D_1}$ and $1.044\text{mA}$ for $I_{R_2}$ .

  $\begin{array}{l}{I}_{D_2}=1.044\text{mA}-0.833\text{mA}\\ =0.211\text{mA}\end{array}$