Chapter 1, Problem 1.7TYU

A silicon pn junction diode at $T=300\text{K}$ has a reverse−saturation current of $I_S={10}^{-16}\text{A}$ . (a) Determine the forward−bias diode current for (i) $V_D=0.55\text{V}$ , (ii) $V_D=0.65\text{V}$ , and (iii) $V_D=0.75\text{V}$ , (b) Find the reverse−bias diode current for (i) $V_D=-0.55\text{V}$ and (ii) $V_D=-2.5\text{V}$ . (Ans. (a) (i) $0.154\mu \text{A}$ , (ii) $7.20\mu \text{A}$ , (iii) 0.337 mA; (b) (i) $-{10}^{-16}\text{A}$ , (ii) $-{10}^{-16}\text{A}$ ).

To determine

The forward-bias diode current for different values of $V_D$ .

Answer to Problem 1.7TYU

For $V_D=0.55\text{V}$ , the value of forward-bias diode current is $I_D=0.154\text{μA}$ .

For $V_D=0.65\text{V}$ , the value of forward-bias diode current is, $I_D=7.2\text{μA}$

For $V_D=0.75\text{V}$ , the value of forward-bias diode current is, $I_D=0.337\text{mA}$

Explanation of Solution

Given Information:

  $\begin{array}{l}T=300\text{K}\\ I_s={10}^{-16}\text{A}\end{array}$

  $\begin{array}{l}\left(i\right).V_D=0.55\text{V}\\ \left(ii\right).V_D=0.65\text{V}\\ \left(iii\right).V_D=0.75\text{V}\end{array}$

Calculation:

For $V_D=0.55\text{V}$:

The diode current equation is:

  $I_D=I_S\left(e^{\frac{V_D}{\eta V_T}}-1\right)$

  $I_D$ = forward-bias diode current

  $I_S$ = reverse saturation current

  $V_D$ = forward biased voltage

  $V_T$ = thermal voltage

  $\eta$ = recombination factor (material constant)

  $\eta =1$

The thermal voltage is determined as:

  $V_T=\frac{kT}{q}$

k = Boltzmann’s constant $\left(1.38\times {10}^{-23}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{K}$}\right.\right)$

  $\begin{array}{l}q=1.6\times {10}^{-19}\text{C}\\ V_T=\frac{T\left(\text{K}\right)}{11,600}\\ V_T=\frac{300}{11,600}\\ V_T\approx 26\text{mV}\end{array}$

The value of diode current is:

  $\begin{array}{l}{I}_D=I_S\left(e^{\frac{V_D}{\eta V_T}}-1\right)\\ =\left({10}^{-16}\right)\left(e^{\frac{550}{26}}-1\right)\\ =1.538\times {10}^{-7}\\ =0.154\times {10}^{-6}\\ =0.154\text{μA}\end{array}$

Hence, the value of forward bias diode current is, $I_D=0.154\text{μA}$ .

For $V_D=0.65\text{V}$:

The diode current equation is:

  $I_D=I_S\left(e^{\frac{V_D}{\eta V_T}}-1\right)$

  $I_D$ = forward-bias diode current

  $I_S$ = reverse saturation current

  $V_D$ = forward biased voltage

  $V_T$ = thermal voltage

  $\eta$ = recombination factor (material constant)

  $\eta =1$

The thermal voltage is determined as:

  $V_T=\frac{kT}{q}$

k = Boltzmann’s constant $\left(1.38\times {10}^{-23}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{K}$}\right.\right)$

  $\begin{array}{l}q=1.6\times {10}^{-19}\text{C}\\ V_T=\frac{T\left(\text{K}\right)}{11,600}\\ V_T=\frac{300}{11,600}\\ V_T\approx 26\text{mV}\end{array}$

The value of diode current is:

  $\begin{array}{l}{I}_D=I_S\left(e^{\frac{V_D}{\eta V_T}}-1\right)\\ =\left({10}^{-16}\right)\left(e^{\frac{650}{26}}-1\right)\\ =7.2\times {10}^{-6}\\ =7.2\text{μA}\end{array}$

Hence, the value of forward bias diode current is, $I_D=7.2\text{μA}$ .

For $V_D=0.75\text{V}$:

The diode current equation is:

  $I_D=I_S\left(e^{\frac{V_D}{\eta V_T}}-1\right)$

  $I_D$ = forward-bias diode current

  $I_S$ = reverse saturation current

  $V_D$ = forward biased voltage

  $V_T$ = thermal voltage

  $\eta$ = recombination factor (material constant)

  $\eta =1$

The thermal voltage is determined as:

  $V_T=\frac{kT}{q}$

k = Boltzmann’s constant $\left(1.38\times {10}^{-23}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{K}$}\right.\right)$

  $\begin{array}{l}q=1.6\times {10}^{-19}\text{C}\\ V_T=\frac{T\left(\text{K}\right)}{11,600}\\ V_T=\frac{300}{11,600}\\ V_T\approx 26\text{mV}\end{array}$

The value of diode current is:

  $\begin{array}{l}{I}_D=I_S\left(e^{\frac{V_D}{\eta V_T}}-1\right)\\ =\left({10}^{-16}\right)\left(e^{\frac{750}{26}}-1\right)\\ =3.37\times {10}^{-4}\\ =0.337\times {10}^{-3}\\ =0.337\text{mA}\end{array}$

Hence, the value of forward bias diode current is, $I_D=0.337\text{mA}$ .

To determine

The reverse-bias diode current for different values of $V_D$ .

Answer to Problem 1.7TYU

For $V_D=-0.55\text{V}$ , the value of reverse-bias diode current is, $I_D=-{10}^{-16}\text{A}$ .

For $V_D=-2.5\text{V}$ , the value of reverse-bias diode current is, $I_D=-{10}^{-16}\text{A}$ .

Explanation of Solution

Given Information:

  $\begin{array}{l}T=300\text{K}\\ I_s={10}^{-16}\text{A}\end{array}$

  $\begin{array}{l}\left(i\right).V_D=-0.55\text{V}\\ \left(ii\right).V_D=-2.5\text{V}\end{array}$

Calculation:

For $V_D=-0.55\text{V}$ :

The diode current equation is:

  $I_D=I_S\left(e^{\frac{V_D}{\eta V_T}}-1\right)$

  $V_D$ = reverse-biased voltage

  $I_D$ = Reverse-bias diode current

  $I_S$ = reverse saturation current

  $V_T$ = thermal voltage

  $\eta$ = recombination factor (material constant)

  $\eta =1$

The thermal voltage is determined as:

  $V_T=\frac{kT}{q}$

k = Boltzmann’s constant $\left(1.38\times {10}^{-23}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{K}$}\right.\right)$

  $\begin{array}{l}q=1.6\times {10}^{-19}\text{C}\\ V_T=\frac{T\left(\text{K}\right)}{11,600}\\ V_T=\frac{300}{11,600}\\ V_T\approx 26\text{mV}\end{array}$

The value of diode current is:

  $\begin{array}{l}{I}_D=I_S\left(e^{\frac{V_D}{\eta V_T}}-1\right)\\ =\left({10}^{-16}\right)\left(e^{\frac{-550}{26}}-1\right)\\ =\left(-0.99\right)\left({10}^{-16}\right)\\ =-{10}^{-16}\text{A}\end{array}$

Hence, the value of reverse-bias diode current is, $I_D=-{10}^{-16}\text{A}$ .

For $V_D=-2.5\text{V}$:

The diode current equation is:

  $I_D=I_S\left(e^{\frac{V_D}{\eta V_T}}-1\right)$

  $V_D$ = reverse-biased voltage

  $I_D$ = Reverse-bias diode current

  $I_S$ = reverse saturation current

  $V_T$ = thermal voltage

  $\eta$ = recombination factor (material constant)

  $\eta =1$

The thermal voltage is determined as:

  $V_T=\frac{kT}{q}$

k = Boltzmann’s constant $\left(1.38\times {10}^{-23}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{K}$}\right.\right)$

  $\begin{array}{l}q=1.6\times {10}^{-19}\text{C}\\ V_T=\frac{T\left(\text{K}\right)}{11,600}\\ V_T=\frac{300}{11,600}\\ V_T\approx 26\text{mV}\end{array}$

The value of diode current is:

  $\begin{array}{l}{I}_D=I_S\left(e^{\frac{V_D}{\eta V_T}}-1\right)\\ =\left({10}^{-16}\right)\left(e^{\frac{-2500}{26}}-1\right)\\ =\left(-1\right)\left({10}^{-16}\right)\\ =-{10}^{-16}\text{A}\end{array}$

Hence, the value of reverse-bias diode current is, $I_D=-{10}^{-16}\text{A}$ .