a Calculate the majority and minority carriers for each side

(a) Calculate the majority and minority carriers for each side of an N+P junction if ND = 2 x 1017/cm3 for the n-side, and NA = 1014/cm3 for the p-side. Assume the semiconductor is Si and the temperature is 300K. (b) In which direction are the minority carriers moving for each side? (c) Which minority carriers will result in the greatest current due to the electric field (electrons or holes)?

Solution

here,

Part a:

Nd = 2*10^17 cm^-3
Na = 10^14 cm^-3

Using the Law of Mass Action
for n type
No = Nd , Po = ni^2 / Nd

where
Ni = Intrinsic Carrier Concentration
Ni for Silicon at 300 K = 9.65*10^9 cm^-3

Therefore

Po = 9.65*10^9 / (2*10^17)

Po for n type is 4.825*10^-8

Now
Using the Law of Mass Action
for Ptype: Po = Na , No = ni^2 / Na

No = 9.65*10^9 / 10^14

No = 9.6*10^-5

Part B:

Minority charge carrier will try to move to n side as P side is always at a higher level ( It has to do with the Fermi level)

Part C:

Electrons are majority carriers