Physics » Condensed Matter Physics » Semiconductor Devices

# Diodes

## Semiconductor Devices

Semiconductors have many applications in modern electronics. We describe some basic semiconductor devices in this section. A great advantage of using semiconductors for circuit elements is the fact that many thousands or millions of semiconductor devices can be combined on the same tiny piece of silicon and connected by conducting paths. The resulting structure is called an integrated circuit (ic), and ic chips are the basis of many modern devices, from computers and smartphones to the internet and global communications networks.

## Diodes

Perhaps the simplest device that can be created with a semiconductor is a diode. A diode is a circuit element that allows electric current to flow in only one direction, like a one-way valve. A diode is created by joining a p-type semiconductor to an n-type semiconductor (this figure). The junction between these materials is called a p-n junction. A comparison of the energy bands of a silicon-based diode is shown in this figure (b). The positions of the valence and conduction bands are the same, but the impurity levels are quite different. When a p-n junction is formed, electrons from the conduction band of the n-type material diffuse to the p-side, where they combine with holes in the valence band. This migration of charge leaves positive ionized donor ions on the n-side and negative ionized acceptor ions on the p-side, producing a narrow double layer of charge at the pn junction called the depletion layer. The electric field associated with the depletion layer prevents further diffusion. The potential energy for electrons across the p-n junction is given by this figure.

(a) Representation of a p-n junction. (b) A comparison of the energy bands of p-type and n-type silicon prior to equilibrium.

At equilibrium, (a) excess charge resides near the interface and the net current is zero, and (b) the potential energy difference for electrons (in light blue) prevents further diffusion of electrons into the p-side.

The behavior of a semiconductor diode can now be understood. If the positive side of the battery is connected to the n-type material, the depletion layer is widened, and the potential energy difference across the p-n junction is increased. Few or none of the electrons (holes) have enough energy to climb the potential barrier, and current is significantly reduced. This is called the reverse bias configuration. On the other hand, if the positive side of a battery is connected to the p-type material, the depletion layer is narrowed, the potential energy difference across the p-n junction is reduced, and electrons (holes) flow easily. This is called the forward bias configuration of the diode. In sum, the diode allows current to flow freely in one direction but prevents current flow in the opposite direction. In this sense, the semiconductor diode is a one-way valve.

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 $$N{e}^{\text{−}e(V-{V}_{b})\text{/}{k}_{\text{B}}T}$$. The electric current due to the majority carriers from the n-side to the p-side is therefore

$$I=N{e}^{\text{−}eV\text{/}{k}_{\text{B}}T}\;{e}^{e{V}_{b}\text{/}{k}_{\text{B}}T}={I}_{0}{e}^{e{V}_{b}\text{/}{k}_{\text{B}}T},$$

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

#### Note:

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

A sample graph of the current versus bias voltage is given in this figure. 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.

## Example: 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 $$4.5\;×\;{10}^{-1}\;\text{A}\text{.}$$ The reverse saturation current is $$2.2\;×\;{10}^{-8}\;\text{A}\text{.}$$ (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, this figure gives the current.

### Solution

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

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

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

$${I}_{0}\approx -{I}_{\text{sat}}=2.2\;×\;{10}^{-8}.$$

Therefore

$$\cfrac{{I}_{\text{net}}}{{I}_{0}}=\cfrac{4.5\;×\;{10}^{-1}\;\text{A}}{2.2\;×\;{10}^{-8}\;\text{A}}=2.0\;×\;{10}^{8}.$$

This can be written as

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

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

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

The temperature is therefore

$$T=\cfrac{e{V}_{b}}{{k}_{\text{B}}}\left(\cfrac{1}{19}\right)=\cfrac{e(0.12\;\text{V})}{8.617\;×\;{10}^{-5}\;\text{eV}\text{/}\text{K}}\left(\cfrac{1}{19}\right)=73\;\text{K}\text{.}$$

### 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.

#### Resource:

Create a pn junction and observe the behavior of a simple circuit for forward and reverse bias voltages. Visit this site to learn more about semiconductor diodes.

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