Concentration of minority carriers in a pn junction

AI Thread Summary
The discussion focuses on calculating the majority and minority carrier concentrations in a p-n junction with doping concentrations of Na=1E18 cm^-3 and Nd=1E16 cm^-3. The user applied the equations ni^2=(npo)(Na) and ni^2=(Pno)Nd to find npo and minority carrier concentrations, resulting in values of 139.24 cm^-3 for electrons in the p region and 13924 cm^-3 for holes in the n region. There is concern about the low values of these concentrations compared to the majority carrier concentrations. However, it is noted that these low values are expected for minority carriers. The calculations are confirmed as correct, emphasizing the nature of minority carrier concentrations in semiconductors.
jeffy
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Homework Statement




A semiconductor p-n junction is fabricated where the doping concentration on the p and n sides of the junction is Na=1E18cm^-3 and Nd=1E16cm^-3 respectively. Given that ni=1.18E10cm^-3, T=300k calculate:

a) The majority and minority carrier concentrations in the neutral bulk regions on either side of the junction.

Homework Equations


ni^2=(npo)(Na) and ni^2=(Pno)Nd
where npo is the concentration of electrons in the p doped region and pno is the concentration of holes in the n region

The Attempt at a Solution


using the above equations i got npo=139.24 cm^-3 and 13924cm^-3 for the minority concentrations. I want to ask if i did this right because i think these concentrations are kinda low


thanks for reading
 
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ya but is the set up correct, because 139.24cm^-3 is low comaring to 1.18E18cm^-3, but then again they are minority carriers
 
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