The electron concentration in an \(n\)-type semiconductor is the same as hole concentration in a \(p\)-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them.
No current will flow in \(p\)-type, current will only flow in \(n\)-type
Current in \(n\)-type = current in \(p\)-type
Current in \(p\)-type > current in \(n\)-type
Current in \(n\)-type > current in \(p\)-type.
Solution:
The mobility of electrons (\(mu_e\)) is greater than the mobility of holes (\(mu_h\)). Since current is proportional to mobility for the same carrier concentration and electric field, the current in the \(n\)-type semiconductor is greater than that in the \(p\)-type.
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