Chapter 2, Problem 2.57P

Let $V_{\gamma }=0.7\text{V}$ for the diode in the circuit in Figure P2.57. Determine $I_D,V_D,V_A$ , and $V_B$ for (a) $V_1=V_1\text{=6V}$ ; (b) $V_1=2\text{V,}{V}_2\text{=5V}$ ; (c) $V_1=5\text{V,}{V}_2\text{=4V}$ ; and (d) $V_1=2\text{V,}{V}_2\text{=8V}$ .

Figure P2.57

To determine

The values of $I_D,V_D,V_A,V_B$ in the circuit diagram.

Answer to Problem 2.57P

  $\begin{array}{l}{I}_D=0\text{A,}{V}_D=0\text{V}\\ V_A=3\text{V,}{V}_B=3\text{V}\end{array}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $V_1=V_2=6\text{V}$

Calculation:

Assuming the diode is in cutoff mode.

Applying voltage division rule:

  $\begin{array}{l}{V}_A=\left(\frac{10}{10+10}\right)V_1\\ V_A=\frac{1}{2}\times 6\\ V_A=3\text{V}\\ V_B=\left(\frac{10}{10+10}\right)V_2\\ V_B=\frac{1}{2}\times 6\\ V_B=3\text{V}\end{array}$

The voltage across diode:

  $\begin{array}{l}{V}_D=V_B-V_A\\ V_D=3-3\\ V_D=0\text{V}\\ V_D<V_{\gamma }\\ 0<0.7\end{array}$

Hence, the diode is biased in reverse region. The assumption is correct and the value of diode current will be zero.

  $I_D=0$

To determine

The values of $I_D,V_D,V_A,V_B$ in the circuit diagram.

Answer to Problem 2.57P

  $\begin{array}{l}{I}_D=0.08\text{mA,}{V}_D=0.7\text{V}\\ V_A=1.4\text{V,}{V}_B=2.1\text{V}\end{array}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $V_1=2\text{V,}{V}_2=5\text{V}$

Calculation:

Assuming the diode is in cutoff region.

Applying voltage division rule:

  $\begin{array}{l}{V}_A=\left(\frac{10}{10+10}\right)V_1\\ V_A=\frac{1}{2}\times 2\\ V_A=1\text{V}\\ V_B=\left(\frac{10}{10+10}\right)5\\ V_B=\frac{1}{2}\times 5\\ V_B=2.5\text{V}\end{array}$

The voltage across diode:

  $\begin{array}{l}{V}_D=V_B-V_A\\ V_D=2.5-1\\ V_D=1.5\text{V}\\ V_D>V_{\gamma }\\ 1.5>0.7\end{array}$

Hence, the diode is biased in forward region. The assumption is incorrect.

Replacing the diode with its cut-in voltage:

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

The modified figure is:

  

  $V_B-V_A=0.7\text{V}\mathrm{...}\left(1\right)$

The node AB is a super node. Applying Kirchhoff’s current law at super node:

  $\begin{array}{l}\frac{V_A-V_1}{10}+\frac{V_A}{10}+\frac{V_B}{10}+\frac{V_B-V_2}{10}=0\\ V_A-2+V_A+V_B+V_B-5=0\\ 2V_B+2V_A=7\\ V_B+V_A=3.5\mathrm{...}\left(2\right)\end{array}$

Adding equation 1 and 2:

  $\begin{array}{l}2V_B=4.2\\ V_B=2.1\text{V}\end{array}$

From equation (1):

  $\begin{array}{l}2.1-V_A=0.7\\ V_A=2.1-0.7\\ V_A=1.4\text{V}\end{array}$

Applying Kirchhoff’s current law at node B:

  $\begin{array}{l}{I}_D+\frac{V_B}{\text{10}\text{k}}+\frac{V_B-V_2}{\text{10}\text{k}}=0\\ I_D+\frac{2.1}{\text{10}\text{k}}+\frac{2.1-5}{\text{10}\text{k}}=0\\ I_D=\left(0.29-0.21\right)\text{mA}\\ I_D=0.08\text{mA}\end{array}$

To determine

The values of $I_D,V_D,V_A,V_B$ in the circuit diagram.

Answer to Problem 2.57P

  $\begin{array}{l}{I}_D=0\text{A,}{V}_D=-0.5\text{V}\\ V_A=2.5\text{V,}{V}_B=2\text{V}\end{array}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{V}_1=5\text{V}\\ V_2=4\text{V}\end{array}$

Calculation:

Assuming the diode is in cutoff mode.

Applying voltage division rule:

  $\begin{array}{l}{V}_A=\left(\frac{10}{10+10}\right)V_1\\ V_A=\frac{1}{2}\times 5\\ V_A=2.5\text{V}\\ V_B=\left(\frac{10}{10+10}\right)V_2\\ V_B=\frac{1}{2}\times 4\\ V_B=2\text{V}\end{array}$

The voltage across diode:

  $\begin{array}{l}{V}_D=V_B-V_A\\ V_D=2-2.5\\ V_D=-0.5\text{V}\\ V_D<V_{\gamma }\\ -0.5\text{V}<0.7\text{V}\end{array}$

Hence, the diode is biased in reverse region. The assumption is correct and the value of diode current will be zero.

  $I_D=0$

To determine

The values of $I_D,V_D,V_A,V_B$ in the circuit diagram.

Answer to Problem 2.57P

  $\begin{array}{l}{I}_D=0.23\text{mA,}{V}_D=0.7\text{V}\\ V_A=2.15\text{V,}{V}_B=2.85\text{V}\end{array}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $V_1=2\text{V,}{V}_2=8\text{V}$

Calculation:

Assuming the diode is in cutoff region.

Applying voltage division rule:

  $\begin{array}{l}{V}_A=\left(\frac{10}{10+10}\right)V_1\\ V_A=\frac{1}{2}\times 2\\ V_A=1\text{V}\\ V_B=\left(\frac{10}{10+10}\right)8\\ V_B=\frac{1}{2}\times 8\\ V_B=4\text{V}\end{array}$

The voltage across diode:

  $\begin{array}{l}{V}_D=V_B-V_A\\ V_D=4-1\\ V_D=3\text{V}\\ V_D>V_{\gamma }\\ 3>0.7\end{array}$

Hence, the diode is biased in forward region. The assumption is incorrect.

Replacing the diode with its cut-in voltage:

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

The modified figure is:

  

  $V_B-V_A=0.7\text{V}\mathrm{...}\left(1\right)$

The node AB is a super node. Applying Kirchhoff’s current law at super node:

  $\begin{array}{l}\frac{V_A-V_1}{10}+\frac{V_A}{10}+\frac{V_B}{10}+\frac{V_B-V_2}{10}=0\\ V_A-2+V_A+V_B+V_B-8=0\\ 2V_B+2V_A=10\\ V_B+V_A=5\mathrm{...}\left(2\right)\end{array}$

Adding equation 1 and 2:

  $\begin{array}{l}2V_B=5.7\\ V_B=2.85\text{V}\end{array}$

From equation (1):

  $\begin{array}{l}2.85-V_A=0.7\\ V_A=2.85-0.7\\ V_A=2.15\text{V}\end{array}$

Applying Kirchhoff’s current law at node B:

  $\begin{array}{l}{I}_D+\frac{V_B}{\text{10}\text{k}}+\frac{V_B-V_2}{\text{10}\text{k}}=0\\ I_D+\frac{2.85}{\text{10}\text{k}}+\frac{2.85-8}{\text{10}\text{k}}=0\\ I_D=\left(0.515-0.285\right)\text{mA}\\ I_D=0.23\text{mA}\end{array}$