Concentration of minority carriers in a pn junction

  • Thread starter jeffy
  • Start date
  • #1
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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
 

Answers and Replies

  • #3
17
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ya but is the set up correct, cuz 139.24cm^-3 is low comaring to 1.18E18cm^-3, but then again they are minority carriers
 

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