Question

The electron concentration in an n-type semiconductor is the same as the hole concentration in a p-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them

• A. No current will flow in p-type, current will only flow in n-type
• B. current in n-type=current in p-type
• C. current in p-type > current in n-type
• D. current in n-type > current in p-type

Answer 1

The Correct Option is D

To compare the currents in an n-type and a p-type semiconductor when an external electric field is applied, let's delve into the basic principles of how semiconductors conduct electricity. 

1. Understanding Carriers in Semiconductors:
   • n-type Semiconductor: In an n-type semiconductor, the majority charge carriers are electrons. The electron mobility (denoted by \(\mu_n\)) is generally higher compared to the hole mobility.
   • p-type Semiconductor: In a p-type semiconductor, the majority charge carriers are holes. The hole mobility (denoted by \(\mu_p\)) is typically less than the electron mobility.
2. Mathematically Expressing Current:
   • The current density \(J\) in a semiconductor can be expressed as: \(J = q \cdot (n \cdot \mu_n \cdot E + p \cdot \mu_p \cdot E)\) where \(q\) is the charge of an electron, \(n\) and \(p\) are electron and hole concentrations respectively, \(E\) is the applied electric field.
3. Since the electron concentration in the n-type is equal to the hole concentration in the p-type, we have \(n = p\).
4. Comparing Currents:
   • n-type Semiconductor: \(J_{\text{n-type}} = q \cdot n \cdot \mu_n \cdot E\)
   • p-type Semiconductor: \(J_{\text{p-type}} = q \cdot p \cdot \mu_p \cdot E\)
   • Given \(n = p\) and \(\mu_n > \mu_p\), it follows that \(J_{\text{n-type}} > J_{\text{p-type}}\).
5. Conclusion: The current in the n-type semiconductor is greater than the current in the p-type semiconductor due to the higher electron mobility. This justifies the answer: \(\text{current in n-type} > \text{current in p-type}\).