Chapter 1, Problem 1.30P

Plot ${\log }_{10}{I}_D$ versus $V_D$ over the range $0.1\le V_D\le 0.7\text{V}$ for (a) $I_S={10}^{-12}$ and (b) $I_S={10}^{-14}\text{A}$ .

To determine

To plot: The graph of ${\log }_{10}{I}_D$ versus $V_D$ .

Answer to Problem 1.30P

The graph of ${\log }_{10}{I}_D$ versus $V_D$ for the $I_S={10}^{-12}$ is shown in Figure 1.

Explanation of Solution

Given:

Reverse saturation current $I_S={10}^{-12}\text{A}$

Diode Voltage range, $0.1\le V_D\le 0.7$

Concept Used:

  $I_D=I_S\left(\exp \left(\frac{V_D}{V_T}\right)\right)$

Calculation:

For $I_S={10}^{-12}$

For $V_D=0.1V$

  $\begin{array}{c}{I}_D={10}^{-12}\left(\exp \left(\frac{0.1}{0.026}\right)\right)=4.68127\times {10}^{-11}\text{A}\\ {\log }_{10}\left(I_D\right)=-10.3296\end{array}$

For $V_D=0.2V$

  $\begin{array}{c}{I}_D={10}^{-12}\left(\exp \left(\frac{0.2}{0.026}\right)\right)=2.1914\times {10}^{-9}\text{A}\\ {\log }_{10}\left(I_D\right)=-8.6593\end{array}$

For $V_D=0.3\text{V}$

  $\begin{array}{c}{I}_D={10}^{-12}\left(\exp \left(\frac{0.3}{0.026}\right)\right)=1.02591\times {10}^{-7}\text{A}\\ {\log }_{10}\left(I_D\right)=-6.9889\end{array}$

For $V_D=0.4\text{V}$

  $\begin{array}{c}{I}_D={10}^{-12}\left(\exp \left(\frac{0.4}{0.026}\right)\right)=4.8023\times {10}^{-6}\text{A}\\ {\log }_{10}\left(I_D\right)=-5.3185\end{array}$

For $V_D=0.5\text{V}$

  $\begin{array}{c}{I}_D={10}^{-12}\left(\exp \left(\frac{0.5}{0.026}\right)\right)=2.248\times {10}^{-4}\text{A}\\ {\log }_{10}\left(I_D\right)=-3.6482\end{array}$

For $V_D=0.6\text{V}$

  $\begin{array}{c}{I}_D={10}^{-12}\left(\exp \left(\frac{0.6}{0.026}\right)\right)=1.05240\times {10}^{-2}\text{A}\\ {\log }_{10}\left(I_D\right)=-1.9778\end{array}$

For $V_D=0.7\text{V}$

  $\begin{array}{c}{I}_D={10}^{-12}\left(\exp \left(\frac{0.7}{0.026}\right)\right)=4.92656\times {10}^{-1}\text{A}\\ {\log }_{10}\left(I_D\right)=-0.30745\end{array}$

  

Figure 1

To determine

To plot: The graph of ${\log }_{10}{I}_D$ versus $V_D$ .

Answer to Problem 1.30P

The graph of ${\log }_{10}{I}_D$ Versus $V_D$ for the $I_S={10}^{-14}$ is shown in Figure 2.

Explanation of Solution

Given:

Reverse saturation current $I_S={10}^{-14}$

Concept Used:

Formulae;

  $I_D=I_S\left(\exp \left(\frac{V_D}{V_T}\right)\right)$

Calculation:

For $I_S={10}^{-14}$

For $V_D=0.1V$

  $\begin{array}{c}{I}_D={10}^{-14}\left(\exp \left(\frac{0.1}{0.026}\right)\right)=4.68127\times {10}^{-13}\text{A}\\ {\log }_{10}\left(I_D\right)=-12.3296\end{array}$

For $V_D=0.2V$

  $\begin{array}{c}{I}_D={10}^{-14}\left(\exp \left(\frac{0.2}{0.026}\right)\right)=2.1914\times {10}^{-11}\text{A}\\ {\log }_{10}\left(I_D\right)=-10.6593\end{array}$

For $V_D=0.3\text{V}$

  $\begin{array}{c}{I}_D={10}^{-14}\left(\exp \left(\frac{0.3}{0.026}\right)\right)=1.02591\times {10}^{-9}\text{A}\\ {\log }_{10}\left(I_D\right)=-8.9889\end{array}$

For $V_D=0.4\text{V}$

  $\begin{array}{c}{I}_D={10}^{-14}\left(\exp \left(\frac{0.4}{0.026}\right)\right)=4.8023\times {10}^{-8}\text{A}\\ {\log }_{10}\left(I_D\right)=-7.3185\end{array}$

For $V_D=0.5\text{V}$

  $\begin{array}{c}{I}_D={10}^{-14}\left(\exp \left(\frac{0.5}{0.026}\right)\right)=2.248\times {10}^{-6}\text{A}\\ {\log }_{10}\left(I_D\right)=-5.6482\end{array}$

For $V_D=0.6\text{V}$

  $\begin{array}{c}{I}_D={10}^{-14}\left(\exp \left(\frac{0.6}{0.026}\right)\right)=1.05240\times {10}^{-4}\text{A}\\ {\log }_{10}\left(I_D\right)=-3.9778\end{array}$

For $V_D=0.7\text{V}$

  $\begin{array}{c}{I}_D={10}^{-14}\left(\exp \left(\frac{0.7}{0.026}\right)\right)=4.92656\times {10}^{-3}\text{A}\\ {\log }_{10}\left(I_D\right)=-2.30745\end{array}$

  

Figure 2