In semiconductor physics, the concept of depletion width is central to understanding the behavior of p-n junctions, which are fundamental components of various electronic devices such as diodes, transistors, and solar cells. The depletion region width describes the region around the p-n junction where mobile charge carriers (free electrons and holes) have been swept away, leaving behind a zone of fixed charged ions. This article delves into the definition of depletion width, provides its formula, derives this formula, and presents a solved example to illustrate the concept.

### Definition of Depletion Width

In semiconductor physics, particularly in the study of p-n junctions, the depletion width refers to the region where mobile charge carriers (free electrons and holes) have been swept away due to the diffusion across the junction. This leaves behind a zone with only fixed charged ions, creating an electric field. The PN junction width is crucial in defining the electrical behavior of diodes, transistors, and other semiconductor devices.

### Depletion Width Formula

The depletion width (π) in a p-n junction can be expressed by the formula:

Where:

*Ο΅*is the permittivity of the semiconductor material,- π
_{0β}is the built-in potential of the junction, - π is the externally applied voltage (positive for forward bias and negative for reverse bias),
- π is the elementary charge (approximately1.6Γ10
^{β19}Coulombs), - π
_{π΄}and π_{π·}β are the acceptor and donor doping concentrations, respectively.

**Depletion Width Formula Derivation**

The derivation of the depletion width formula in a semiconductor junction (like a PN junction) involves understanding the behavior of charge carriers and electric fields at the interface between P-type and N-type semiconductors. The depletion region is where mobile charge carriers (electrons and holes) are absent, creating an area dominated by the immobile ionized donor and acceptor impurities. Here’s a step-by-step derivation:

**Step 1: Basic Assumptions**

**Uniform Doping**: Assume the P-type and N-type regions are uniformly doped with acceptor density ππ΄β and donor density ππ·β, respectively.**One-dimensional Analysis**: Consider only the one-dimensional case along the x-axis across the PN junction.

**Step 2: Charge Neutrality**

The total charge in the depletion region must be zero for overall electrical neutrality: ππ+ππ=0 where:

- π
_{π}=βππ_{π΄}π_{π}ββ is the charge in the P-type depletion region. - π
_{π}=ππ_{π·}π_{π}β is the charge in the N-type depletion region. - π
_{π}β and π_{π}β are the widths of the depletion regions in the P-type and N-type materials, respectively. - π is the elementary charge.

**Step 3: Relationship Between Wpβ and Wn**β

Using charge neutrality:

π_{π΄}π_{π}=π_{π·}π_{π}

This can be rearranged to relate the widths of the depletion regions:

**Step 4: Electric Field and Potential Drop**

The electric field πΈ reaches its maximum at the metallurgical junction with:

Where π is the permittivity of the semiconductor.

By integrating the electric field across the depletion region, the built-in voltage *V*_{0}β (also called the contact potential) is found:

**ββStep 5: Total Depletion Width π**

The total width π

Using the charge neutrality and the relationship between ππ and ππβ:

**Step 6: Solving for π**

Substitute the expression for ππ or ππβ from the potential drop:

Then, the total depletion width *W* is given by:

If external voltage (V) is applied to the PN junction and the diode is in reverse bias, the depletion region is,

If external voltage (V) is applied to the PN junction and the diode is in forward bias, the depletion region is,

This formula quantifies the width of the depletion region as a function of the doping concentrations, the permittivity of the material, and the built-in voltage across the junction. It’s fundamental in understanding how PN junctions behave under different electrical and environmental conditions.

## Solved Problem

A silicon PN junction at room temperature and its width are used as an example to solidify understanding of the theoretical derivation.

### Problem-1

A silicon PN junction at room temperature (300 K) has a uniform acceptor concentration π_{π΄}=10^{16}βcm^{β3} in the P-type region and a donor concentration π_{π·}=5Γ10^{16}βcm^{β3} in the N-type region. Assume the built-in potential π_{0}β is 0.7 V. Calculate the depletion width of this junction.

**Constants**

- Elementary charge, π=1.6Γ10
^{β19}βC - Relative permittivity of silicon, π
_{π}=11.7 - Permittivity of free space, π
_{0}=8.854Γ10^{β14}βF/cm - Permittivity of silicon, π=π
_{π}π_{0}

**Solution**

#### Step 1: Calculate the permittivity of silicon

#### Step 2: Apply the PN Junction width formula

Using the formula:

We substitute the given values:

### Problem-2

Consider a silicon PN junction with an acceptor concentration π_{π΄}=10^{17}βcm^{β3} in the P-type region and a donor concentration π_{π·}=10^{16}βcm^{β3} in the N-type region. The built-in potential π_{0} is 0.8 V. Calculate the depletion width when a reverse bias of 5 V is applied to the junction.

### Constants

- Elementary charge, π=1.6Γ10
^{β19}βC - Relative permittivity of silicon, π
_{π}=11.7 - The permittivity of free space, π
_{0}=8.854Γ10^{β14}βF/cm

**Solution**

#### Step 1: Calculate the permittivity of silicon

Step 2: Calculate W

### Conclusion

In conclusion, the study of depletion width in PN junctions provides a fundamental insight into the operational principles of semiconductor devices, from defining what depletion width is and why it forms to deriving its quantitative formula and applying this knowledge through solved examples.