Question

When electrons cross from the \(\mathrm{n}\) -type to the p-type to equalize the Fenni energy on both sides in an unbiased diode, they leave the \(n\) -type side with an excess of positive charge and give the p-type side an excess of negative. Charge layers oppose one another on either side of the depletion zone, producing, in essence, a capacitor, which harbors the so-called built-in electric field. The crossing of the electrons to equalize the Fenni energy produces the dogleg in the bands of roughly \(E_{\text {mag, }}\) and the corresponding potential difference is \(E_{\text {gap }}\) /e. The depletion zone in a typical diode is \(1 \mu \mathrm{m}\) wide, and the band gap is \(1.0 \mathrm{eV}\), How large is the buill-in electric field?

Short answer

The magnitude of the built-in electric field in the diode is \(1.0 \times 10^6 V/m\).

Step-by-step solution

Understand the concepts

In a diode, when no bias potential is applied, the charge carriers (electrons) move from high energy regions to low energy regions, attempting to achieve equilibrium. This migration generates a built-in electric field, forming a 'capacitor' across the depletion region. The potential difference across the capacitor can be assumed equal to the band gap energy divided by the charge of an electron.

Derive the formula for electric field

The formula for electric field in a capacitor is given by \( E = V/d \), where \( V \) is the potential difference and \( d \) is the distance across which the potential difference is applied. Here, \( V = E_{gap}/e \), where \( E_{gap} \) is the band gap energy and \( e \) is the electric charge. Therefore, the formula for the built-in electric field can be derived as \( E_{built-in} = (E_{gap}/e) / d \)

Substitute the given values into the formula

Now, substitute the given values into the derived formula. The depletion zone width is given as \( d = 1 \mu m = 1×10^{-6} m \), the band gap energy is \( E_{gap} = 1.0 eV = 1.0×1.6×10^{-19} J \) and the electric charge is \( e = 1.6×10^{-19} C \). So, the built-in electric field \( E_{built-in} = [1.0×1.6×10^{-19} J / 1.6×10^{-19} C] / 1×10^{-6} m = 1.0×10^6 V/m \)

Key concepts

Diode

A diode is a semiconductor device that allows current to flow in one direction while blocking it in the opposite direction.
It is composed of two regions: the p-type (positive) and n-type (negative) semiconductors, which are combined to create a p-n junction.
When a diode is unbiased (not connected to an external voltage), it does not conduct electricity due to the balance of electron and hole charges in these regions.

In an unbiased state, electrons from the n-type region move across the junction to the p-type region.
This movement aims to balance the energy difference between the two sides.
As electrons cross over, they leave behind positively charged ions on the n-type side and create negatively charged ions on the p-type side.
This forms a charged layer on each side of the junction, leading to a region called the depletion zone.
It creates an internal electric field that resists further movement of charge carriers, effectively preventing current flow.

Depletion Zone

The depletion zone is a key concept in understanding how diodes function.
It is the region around the p-n junction where mobile charge carriers are absent.
This occurs because the free electrons from the n-type region have moved into the p-type region, leaving behind fixed positive charges.

On the other side, holes from the p-type region fill these incoming electrons, creating a region with fixed negative charges.
This region lacks mobile charge carriers and hence is depleted of them, forming what is known as the depletion zone.
The width of this zone is crucial as it determines a diode's behavior in terms of switching between conductive and non-conductive states.
For a typical diode, this zone is around 1 micrometer wide.
The built-in electric field within this zone acts like a barrier, which, when overcome by an external force, allows the diode to conduct switching currents.

Band Gap Energy

Band gap energy is an important property that characterizes semiconductors.
It is the energy difference between the top of the valence band and the bottom of the conduction band.
This energy gap determines whether or not electrons can move freely to conduct electricity.
In a diode, this band gap energy plays a crucial role in determining its electrical properties.

A larger band gap generally means that more energy is required for electrons to jump from the valence band to the conduction band.
For the diode in the given scenario, the band gap energy is 1.0 eV. This energy is required per electron to push it into the conduction band where it can facilitate current flow.
This property is critical when calculating the built-in electric field, as the energy difference helps set up the potential difference across the depletion zone.

Electric Field Formula

The electric field within a diode's depletion zone is derived from the basic electric field formula for a capacitor.
This formula is given by the expression: \( E = \frac{V}{d} \), where \( E \) is the electric field, \( V \) is the potential difference, and \( d \) is the distance across which the potential difference is applied.

In the context of a diode, the potential difference \( V \) can be described by the band gap energy \( E_{gap} \) divided by the charge of an electron \( e \).
Thus, the built-in electric field \( E_{built-in} \) formula is expressed as:
\[ E_{built-in} = \frac{E_{gap} / e}{d} \]

Plugging in the provided values: the width of the depletion zone \( d = 1 \mu m = 1 \times 10^{-6} \) meters, band gap energy \( E_{gap} = 1.0 \) eV, and the charge of an electron \( e = 1.6 \times 10^{-19} \) C, we can calculate the built-in electric field as:
\[ E_{built-in} = \frac{1.0 \times 1.6 \times 10^{-19} \text{ J} / 1.6 \times 10^{-19} \text{ C}}{1 \times 10^{-6} \text{ m}} = 1.0 \times 10^{6} \text{ V/m} \]
This electric field helps prevent further flow of charge carriers across the junction, maintaining the diode’s functionality.