Question

A new semiconductor has Nc = 1019 /cm3 and Nv=5 x 1018, and Eg = 2eV. If it is doped with 1017 donors (fully ionized), calculate the electron, hole, and intrinsic carrier concentrations at 627K.​

Answer 1

The intrinsic concentration at 627K =  1.82 * $10^{13}$ $cm^{-3}$

Step-by-step Explanation:

Given data's are:

$N_{c}$ = $10^{13}$$cm^{-3}$

$N_{v}$= 5 x $10^{18}$

$E_{g}$ = 2eV

Converting 627°C into kelvin

$T_{k}$ = $T_{c}$ + 273K

= 627 + 273

= 900K

valu of kT at 300K = 0.0259eV

Valu of kT at 900K = 0.0259 * 900/ 300

= 0.078eV

Intrinsic concentration (n1) = $\sqrt{NcNve} \ ^{Eg/2Kt}$  

= $10^{19}$ (5* $10^{18}$) $e^{2/0.078 * 2}$

Intrinsic concentration = 1.82 * $10^{13}$ $cm^{-3}$