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 solve this question, we need to compare the currents in an n-type semiconductor and a p-type semiconductor under an applied electric field. Both semiconductors have the same charge carrier concentration.

1. Understanding Charge Carrier Concentrations:
   • In an n-type semiconductor, the majority charge carriers are electrons.
   • In a p-type semiconductor, the majority charge carriers are holes.
2. Current in a Semiconductor:
   • The drift current (\( I \)) in a semiconductor is given by the formula:
     • \(n\): Charge carrier concentration (number of charge carriers per unit volume)
     • \(q\): Charge of an electron (\( 1.6 \times 10^{-19} \) C)
     • \(v_d\): Drift velocity of charge carriers
     • \(A\): Cross-sectional area of the material
3. Comparison of Electron and Hole Mobilities:
   • The mobility (\( \mu \)) of electrons is generally greater than that of holes. This implies:
   • \(\mu_n > \mu_p\)
   • The drift velocity (\( v_d \)) is related to mobility as \(v_d = \mu E\), where \(E\) is the electric field.
   • Thus, the drift velocity of electrons (\( v_{d,n} \)) is greater than that of holes (\( v_{d,p} \)) under the same electric field.
4. Conclusion:
   • Both semiconductors have equal charge carrier concentrations, but due to higher electron mobility, the electrons in the n-type semiconductor will produce a larger current compared to the holes in the p-type semiconductor.
   • Therefore, the current in the n-type semiconductor is greater than the current in the p-type semiconductor.