Chapter 1, Problem 1.27P

(a) The reverse−saturation current of a pn junction diode is $I_S={10}^{-11}\text{A}$ . Determine the diode current for diode voltages of 0.3, 0.5, 0.7, −0.02, −0.2, and −2 V. (b) Repeat part (a) fo $I_S={10}^{-13}\text{A}$ .

To determine

The diode current for the given diode voltages.

Answer to Problem 1.27P

Diode current for the given voltages:

0.3V: $i_D=1.025\mu \text{A}$

0.5V: $i_D=2.248\text{mA}$

0.7V: $i_D=4.926\text{A}$

  $-0.02\text{V}$ : $i_D=-5.3\times {10}^{-12}\text{A}$

  $-0.2\text{V}$ : $i_D\approx {10}^{-11}\text{A}$

  $-2\text{V}$ : $i_D={10}^{-11}\text{A}$

Explanation of Solution

Given:

The reverse saturation current of diode, $I_S={10}^{-11}\text{A}$ .

The following voltages are given as:

  $0.3,0.5,0.7,-0.02,-0.2,-2\text{V}$ .

Calculation:

The relationship between current and voltage for the p-n junction diode is written as:

  $i_D=I_s\left[e^{\left(\frac{v_D}{\eta V_T}\right)}-1\right]$

Where,

  $i_D$ is pn junction current.

  $I_s$ is the reverse bias saturation current.

  $V_T$ is thermal voltage.

  $\eta$ is the emission coefficient.

The thermal voltage at room temperature is 26 mV.

The emission coefficient in this case is 1.

Now, substituting the know values to evaluate the value of diode current.

For 0.3V:

  $\begin{array}{l}{i}_D=I_s\left[e^{\left(\frac{v_D}{\eta V_T}\right)}-1\right]\\ i_D={10}^{-11}\left[e^{\left(\frac{0.3}{1\left(0.026\right)}\right)}-1\right]\\ i_D={10}^{-11}\left(e^{11.5384}-1\right)\\ i_D={10}^{-11}\left(102585.491\right)\\ i_D=1.025\times {10}^{-6}\text{A}\\ i_D=1.025\mu \text{A}\end{array}$

For 0.5V:

  $\begin{array}{l}{i}_D=I_s\left[e^{\left(\frac{v_D}{\eta V_T}\right)}-1\right]\\ i_D={10}^{-11}\left[e^{\left(\frac{0.5}{1\left(0.026\right)}\right)}-1\right]\\ i_D={10}^{-11}\left(e^{19.2307}-1\right)\\ i_D={10}^{-11}\left(2.2481\times {10}^8\right)\\ i_D=0.002248\text{A}\\ i_D=2.248\text{mA}\end{array}$

For 0.7V:

  $\begin{array}{l}{i}_D=I_s\left[e^{\left(\frac{v_D}{\eta V_T}\right)}-1\right]\\ i_D={10}^{-11}\left[e^{\left(\frac{0.7}{1\left(0.026\right)}\right)}-1\right]\\ i_D={10}^{-11}\left(e^{26.923}-1\right)\\ i_D={10}^{-11}\left(2.2481\times {10}^8\right)\\ i_D=4.926\text{A}\end{array}$

For $-0.02\text{V}$ :

  $\begin{array}{l}{i}_D=I_s\left[e^{\left(\frac{v_D}{\eta V_T}\right)}-1\right]\\ i_D={10}^{-11}\left[e^{\left(\frac{-0.02}{1\left(0.026\right)}\right)}-1\right]\\ i_D={10}^{-11}\left(e^{-0.7692}-1\right)\\ i_D={10}^{-11}\left(-0.53\right)\\ i_D=-5.3\times {10}^{-12}\text{A}\end{array}$

For $-0.2\text{V}$ :

  $\begin{array}{l}{i}_D=I_s\left[e^{\left(\frac{v_D}{\eta V_T}\right)}-1\right]\\ i_D={10}^{-11}\left[e^{\left(\frac{-0.2}{1\left(0.026\right)}\right)}-1\right]\\ i_D={10}^{-11}\left(e^{-7.692}-1\right)\\ i_D={10}^{-11}\left(-0.999\right)\\ i_D\approx {10}^{-11}\text{A}\end{array}$

For $-2\text{V}$ :

  $\begin{array}{l}{i}_D=I_s\left[e^{\left(\frac{v_D}{\eta V_T}\right)}-1\right]\\ i_D={10}^{-11}\left[e^{\left(\frac{-2}{1\left(0.026\right)}\right)}-1\right]\\ i_D={10}^{-11}\left(e^{-76.92}-1\right)\\ i_D={10}^{-11}\left(-1\right)\\ i_D={10}^{-11}\text{A}\end{array}$

To determine

The value of the diode current for the given diode voltages.

Answer to Problem 1.27P

Diode current for the given voltages:

0.3V: $i_D=1.025\times {10}^{-8}\text{A}$

0.5V: $i_D=2.25\times {10}^{-5}\text{A}$

0.7V: $i_D=4.926\times {10}^{-5}\text{A}$

  $-0.02\text{V}$ : $i_D=-5.3\times {10}^{-14}\text{A}$

  $-0.2\text{V}$ : $i_D\approx {10}^{-13}\text{A}$

  $-2\text{V}$ : $i_D={10}^{-13}\text{A}$

Explanation of Solution

Given:

The reverse saturation current of diode, $I_S={10}^{-13}\text{A}$ .

The following voltages are given as:

  $0.3,0.5,0.7,-0.02,-0.2,-2\text{V}$ .

Calculation:

The relationship between current and voltage for the p-n junction diode is written as:

  $i_D=I_s\left[e^{\left(\frac{v_D}{\eta V_T}\right)}-1\right]$

Where,

  $i_D$ is pn junction current.

  $I_s$ is the reverse bias saturation current.

  $V_T$ is thermal voltage.

  $\eta$ is the emission coefficient.

The thermal voltage at room temperature is 26 mV.

The emission coefficient in this case is 1.

Now, substituting the know values to evaluate the value of diode current.

For 0.3V:

  $\begin{array}{l}{i}_D=I_s\left[e^{\left(\frac{v_D}{\eta V_T}\right)}-1\right]\\ i_D={10}^{-13}\left[e^{\left(\frac{0.3}{1\left(0.026\right)}\right)}-1\right]\\ i_D={10}^{-13}\left(e^{11.5384}-1\right)\\ i_D={10}^{-13}\left(102585.491\right)\\ i_D=1.025\times {10}^{-8}\text{A}\end{array}$

For 0.5V:

  $\begin{array}{l}{i}_D=I_s\left[e^{\left(\frac{v_D}{\eta V_T}\right)}-1\right]\\ i_D={10}^{-13}\left[e^{\left(\frac{0.5}{1\left(0.026\right)}\right)}-1\right]\\ i_D={10}^{-13}\left(e^{19.2307}-1\right)\\ i_D={10}^{-13}\left(2.2481\times {10}^8\right)\\ i_D=0.00002248\text{A}\\ i_D=2.25\times {10}^{-5}\text{A}\end{array}$

For 0.7V:

  $\begin{array}{l}{i}_D=I_s\left[e^{\left(\frac{v_D}{\eta V_T}\right)}-1\right]\\ i_D={10}^{-13}\left[e^{\left(\frac{0.7}{1\left(0.026\right)}\right)}-1\right]\\ i_D={10}^{-13}\left(e^{26.923}-1\right)\\ i_D={10}^{-13}\left(4.92\times {10}^8\right)\\ i_D=4.92\times {10}^{-5}\text{A}\end{array}$

For $-0.02\text{V}$ :

  $\begin{array}{l}{i}_D=I_s\left[e^{\left(\frac{v_D}{\eta V_T}\right)}-1\right]\\ i_D={10}^{-13}\left[e^{\left(\frac{-0.02}{1\left(0.026\right)}\right)}-1\right]\\ i_D={10}^{-13}\left(e^{-0.7692}-1\right)\\ i_D={10}^{-13}\left(-0.53\right)\\ i_D=-5.3\times {10}^{-14}\text{A}\end{array}$

For $-0.2\text{V}$ :

  $\begin{array}{l}{i}_D=I_s\left[e^{\left(\frac{v_D}{\eta V_T}\right)}-1\right]\\ i_D={10}^{-13}\left[e^{\left(\frac{-0.2}{1\left(0.026\right)}\right)}-1\right]\\ i_D={10}^{-13}\left(e^{-7.692}-1\right)\\ i_D={10}^{-13}\left(-0.999\right)\\ i_D\approx {10}^{-13}\text{A}\end{array}$

For $-2\text{V}$ :

  $\begin{array}{l}{i}_D=I_s\left[e^{\left(\frac{v_D}{\eta V_T}\right)}-1\right]\\ i_D={10}^{-13}\left[e^{\left(\frac{-2}{1\left(0.026\right)}\right)}-1\right]\\ i_D={10}^{-13}\left(e^{-76.92}-1\right)\\ i_D={10}^{-13}\left(-1\right)\\ i_D={10}^{-13}\text{A}\end{array}$