Question

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.

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

Answer 1

Answer: (C)

$I_N=$ Current in N-type semiconductor
$=\left({\mu }_e{N}_e+{\mu }_n{n}_n\right)eAE$
$I_P=$ Current in P-type semiconductor
$=\left({\mu }_n{N}_n+{\mu }_e{n}_e\right)eAE$
Now $N_e=N_h$ and $n_e=n_h$
Also ${\mu }_e>{\mu }_n$ and $N_e>>n_e,N_n>>n_e$
Where $N_e=$ electron concentration in $N$-type
$N_h=$ Hole concentration in P-type and $N_e=N_h$
$n_h=$ hole concentration in N-type
$n_e=$ Electron concentration in P-type
$\therefore I_N-l_p=\left[\left({\mu }_e-{\mu }_h\right)N_e+\left({\mu }_h-{\mu }_e\right)n_e\right]eAE$
$=\left({\mu }_e-{\mu }_h\right)\left(N_e-n_e\right)eAE$
Since ${\mu }_e>{\mu }_h$ and $N_e>n_e$
We conclude $I_N>I_P$