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)

current in n-type > current in p-type

Answer 2

$I_{N}=$ Current in N-type semiconductor
$=\left(\mu_{ e } N _{ e }+\mu_{ n } n _{ n }\right) eAE$
$I_{P}=$ Current in P-type semiconductor
$=\left(\mu_{n} N_{n}+\mu_{e} n_{e}\right) e A E$
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 $I_{N}>I_{P}$