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In a $p$ - $n$ junction diode, the current $I$ can be expressed as $\mathrm{I}=\mathrm{I}_0 \exp \left(\frac{\mathrm{eV}}{2 \mathrm{k}_{\mathrm{B}} \mathrm{T}}-1\right)$, where $I_0$ is called the reverse saturation current, $V$ is the voltage across the diode and is positive for forward bias and negative for reverse bias, and $I$ is the current through the diode, $k_{\mathrm{B}}$ is the Boltzmann constant $\left(8.6 \times 10^{-5} \mathrm{e}^{\mathrm{V} / \mathrm{K}}\right)$ and $T$ is the absolute temperature. If for a given diode $I_0=5 \times 10^{-12} \mathrm{~A}$ and $T=300 \mathrm{~K}$, then
(a) What will be the forward current at a forward voltage of $0.6 \mathrm{~V}$ ?
(b) What will be the increase in the current if the voltage across the diode is increased to $0.7 \mathrm{~V}$ ?
(c) What is the dynamic resistance?
(d) What will be the current if reverse bias voltage changes from $1 \mathrm{~V}$ to $2 \mathrm{~V}$ ?
(a) What will be the forward current at a forward voltage of $0.6 \mathrm{~V}$ ?
(b) What will be the increase in the current if the voltage across the diode is increased to $0.7 \mathrm{~V}$ ?
(c) What is the dynamic resistance?
(d) What will be the current if reverse bias voltage changes from $1 \mathrm{~V}$ to $2 \mathrm{~V}$ ?
Solution:
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Verified Answer
The current I through a junction diode is given as
$$
I=I_0\left[\exp \left(\frac{e V}{k_B T}\right)-1\right]
$$
where $I_0=5 \times 10^{-12} A, T=300 \mathrm{~K}$,
$$
\begin{aligned}
k_B &=8.6 \times 10^{-5} \mathrm{eVK}^{-1} \\
&=8.6 \times 10^{-5} \times 1.6 \times 10^{-19} \mathrm{JK}^{-1}
\end{aligned}
$$
(a) When $V=0.6 \mathrm{~V}$
$$
\begin{aligned}
&\frac{e V}{k_B T}=\frac{1.6 \times 10^{-19} \times 0.6}{8.6 \times 1.6 \times 10^{-24} \times 300} \\
&=\frac{600}{8.6 \times 3}=23.26 \\
&\therefore \quad I=I_0=\left[\exp \left(\frac{\mathrm{eV}}{k_B T}\right)-1\right] \\
&\quad=5 \times 10^{-12}[\operatorname{ex}(23.26)-1] \mathrm{A} \\
&=5 \times 10^{-12} \times\left[1.2586 \times 10^{10}-1\right] \mathrm{A} \\
&=5 \times 10^{-12} \times 1.2586 \times 10^{-10} \mathrm{~A} \\
&=0.06293 \mathrm{~A} .
\end{aligned}
$$
(b) When $\mathrm{V}=0.7 \mathrm{~V}$,
$$
\begin{aligned}
& \frac{e V}{k_B T}=\frac{1.6 \times 10^{-19} \times 0.7}{8.6 \times 1.6 \times 10^{-24}}=27.13 \\
\therefore I &=I_0\left[\exp \left(\frac{\mathrm{eV}}{k_B T}\right)-1\right]=5 \times 10^{-12}[\exp (27.13)-1] \mathrm{A} \\
&=5 \times 10^{-12} \times\left[6.07 \times 10^{11}-1\right] \mathrm{A} \\
&=5 \times 10^{-12} \times 6.07 \times 10^{11} \mathrm{~A}=3.035 \mathrm{~A}
\end{aligned}
$$
Increase in current, $\Delta I=3.035-0.06293=2.972 \mathrm{~A}$.
(c) For $\Delta V=0.7-0.6=0.1 \mathrm{~V}, \Delta I=2.972 \mathrm{~A}$
Dynamic resistance,
$$
r_d=\frac{\Delta V}{\Delta I}=\frac{0.1}{2.972}=0.0336 \Omega
$$
(d) For both the voltages, the current $I$ will be almost equal to $\mathrm{I}_0$, showing almost infinite dynamic resistance in the reverse bias.
$$
I \simeq-I_0=-5 \times 10^{-12} \mathrm{~A} .
$$
$$
I=I_0\left[\exp \left(\frac{e V}{k_B T}\right)-1\right]
$$
where $I_0=5 \times 10^{-12} A, T=300 \mathrm{~K}$,
$$
\begin{aligned}
k_B &=8.6 \times 10^{-5} \mathrm{eVK}^{-1} \\
&=8.6 \times 10^{-5} \times 1.6 \times 10^{-19} \mathrm{JK}^{-1}
\end{aligned}
$$
(a) When $V=0.6 \mathrm{~V}$
$$
\begin{aligned}
&\frac{e V}{k_B T}=\frac{1.6 \times 10^{-19} \times 0.6}{8.6 \times 1.6 \times 10^{-24} \times 300} \\
&=\frac{600}{8.6 \times 3}=23.26 \\
&\therefore \quad I=I_0=\left[\exp \left(\frac{\mathrm{eV}}{k_B T}\right)-1\right] \\
&\quad=5 \times 10^{-12}[\operatorname{ex}(23.26)-1] \mathrm{A} \\
&=5 \times 10^{-12} \times\left[1.2586 \times 10^{10}-1\right] \mathrm{A} \\
&=5 \times 10^{-12} \times 1.2586 \times 10^{-10} \mathrm{~A} \\
&=0.06293 \mathrm{~A} .
\end{aligned}
$$
(b) When $\mathrm{V}=0.7 \mathrm{~V}$,
$$
\begin{aligned}
& \frac{e V}{k_B T}=\frac{1.6 \times 10^{-19} \times 0.7}{8.6 \times 1.6 \times 10^{-24}}=27.13 \\
\therefore I &=I_0\left[\exp \left(\frac{\mathrm{eV}}{k_B T}\right)-1\right]=5 \times 10^{-12}[\exp (27.13)-1] \mathrm{A} \\
&=5 \times 10^{-12} \times\left[6.07 \times 10^{11}-1\right] \mathrm{A} \\
&=5 \times 10^{-12} \times 6.07 \times 10^{11} \mathrm{~A}=3.035 \mathrm{~A}
\end{aligned}
$$
Increase in current, $\Delta I=3.035-0.06293=2.972 \mathrm{~A}$.
(c) For $\Delta V=0.7-0.6=0.1 \mathrm{~V}, \Delta I=2.972 \mathrm{~A}$
Dynamic resistance,
$$
r_d=\frac{\Delta V}{\Delta I}=\frac{0.1}{2.972}=0.0336 \Omega
$$
(d) For both the voltages, the current $I$ will be almost equal to $\mathrm{I}_0$, showing almost infinite dynamic resistance in the reverse bias.
$$
I \simeq-I_0=-5 \times 10^{-12} \mathrm{~A} .
$$
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