# 9.7 Semiconductor devices  (Page 2/12)

 Page 2 / 12

We can estimate the mathematical relationship between the current and voltage for a diode using the electric potential concept. Consider N negatively charged majority carriers (electrons donated by impurity atoms) in the n -type material and a potential barrier V across the p-n junction. According to the Maxwell-Boltzmann distribution, the fraction of electrons that have enough energy to diffuse across the potential barrier is $N{e}^{\text{−}eV\text{/}{k}_{\text{B}}T}$ . However, if a battery of voltage ${V}_{b}$ is applied in the forward-bias configuration, this fraction improves to $N{e}^{\text{−}e\left(V-{V}_{b}\right)\text{/}{k}_{\text{B}}T}$ . The electric current due to the majority carriers from the n -side to the p -side is therefore

$I=N{e}^{\text{−}eV\text{/}{k}_{\text{B}}T}\phantom{\rule{0.2em}{0ex}}{e}^{e{V}_{b}\text{/}{k}_{\text{B}}T}={I}_{0}{e}^{e{V}_{b}\text{/}{k}_{\text{B}}T},$

where ${I}_{0}$ is the current with no applied voltage and T is the temperature. Current due to the minority carriers (thermal excitation of electrons from the valence band to the conduction band on the p- side and subsequent attraction to the n -side) is $\text{−}{I}_{0}$ , independent of the bias voltage. The net current is therefore

${I}_{\text{net}}={I}_{0}\left({e}^{e{V}_{b}\text{/}{k}_{\text{B}}T}-1\right).$

A sample graph of the current versus bias voltage is given in [link] . In the forward bias configuration, small changes in the bias voltage lead to large changes in the current. In the reverse bias configuration, the current is ${I}_{\text{net}}\approx \text{−}{I}_{0}$ . For extreme values of reverse bias, the atoms in the material are ionized which triggers an avalanche of current. This case occurs at the breakdown voltage    . Current versus voltage across a p-n junction (diode). In the forward bias configuration, electric current flows easily. However, in the reverse bias configuration, electric current flow very little.

## Diode current

Attaching the positive end of a battery to the p -side and the negative end to the n -side of a semiconductor diode produces a current of $4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-1}\phantom{\rule{0.2em}{0ex}}\text{A}\text{.}$ The reverse saturation current is $2.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}\phantom{\rule{0.2em}{0ex}}\text{A}\text{.}$ (The reverse saturation current is the current of a diode in a reverse bias configuration such as this.) The battery voltage is 0.12 V. What is the diode temperature?

## Strategy

The first arrangement is a forward bias configuration, and the second is the reverse bias configuration. In either case, [link] gives the current.

## Solution

The current in the forward and reverse bias configurations is given by

${I}_{\text{net}}={I}_{0}\left({e}^{e{V}_{b}\text{/}{k}_{\text{B}}T}-1\right).$

The current with no bias is related to the reverse saturation current by

${I}_{0}\approx -{I}_{\text{sat}}=2.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}.$

Therefore

$\frac{{I}_{\text{net}}}{{I}_{0}}=\frac{4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-1}\phantom{\rule{0.2em}{0ex}}\text{A}}{2.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}\phantom{\rule{0.2em}{0ex}}\text{A}}=2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8}.$

$\frac{{I}_{\text{net}}}{{I}_{0}}+1={e}^{e{V}_{b}\text{/}{k}_{\text{B}}T}.$

This ratio is much greater than one, so the second term on the left-hand side of the equation vanishes. Taking the natural log of both sides gives

$\frac{e{V}_{b}}{{k}_{\text{B}}T}=19.$

The temperature is therefore

$T=\frac{e{V}_{b}}{{k}_{\text{B}}}\left(\frac{1}{19}\right)=\frac{e\left(0.12\phantom{\rule{0.2em}{0ex}}\text{V}\right)}{8.617\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}\text{eV}\text{/}\text{K}}\left(\frac{1}{19}\right)=73\phantom{\rule{0.2em}{0ex}}\text{K}\text{.}$

## Significance

The current moving through a diode in the forward and reverse bias configuration is sensitive to the temperature of the diode. If the potential energy supplied by the battery is large compared to the thermal energy of the diode’s surroundings, ${k}_{\text{B}}T,$ then the forward bias current is very large compared to the reverse saturation current.

Check Your Understanding How does the magnitude of the forward bias current compare with the reverse bias current?

The forward bias current is much larger. To a good approximation, diodes permit current flow in only one direction.

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