Why Does N-Type Doping Result in Higher Hole than Electron Concentration?

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SUMMARY

The discussion centers on the behavior of an n-type doped semiconductor with a bandgap energy of 0.1 eV and a density of states of 1018 cm-3 for both conduction and valence bands. When doped with shallow donors at a concentration of Nd = 1017 cm-3, the calculations revealed a higher hole concentration than electron concentration, which contradicts the expectations for n-type materials. The confusion arose from a mix-up between the acceptor concentration (Na) and donor concentration (Nd), leading to the conclusion that under these conditions, the electron concentration (n) should indeed be greater than the hole concentration (p), specifically n ~ 2p.

PREREQUISITES
  • Understanding of semiconductor physics, particularly n-type doping.
  • Familiarity with the equations for intrinsic carrier concentration (ni), electron concentration (n), and hole concentration (p).
  • Knowledge of bandgap energy and its impact on semiconductor behavior.
  • Basic proficiency in solving quadratic equations in the context of semiconductor calculations.
NEXT STEPS
  • Study the effects of varying donor and acceptor concentrations on semiconductor behavior.
  • Learn about the temperature dependence of intrinsic carrier concentration in semiconductors.
  • Explore the role of bandgap energy in determining electron and hole concentrations.
  • Investigate the mathematical derivation of the equations used in semiconductor doping analysis.
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Students and professionals in semiconductor physics, electrical engineering, and materials science who are analyzing doping effects in semiconductor materials.

heehar
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Homework Statement


Suppose an unknown semiconductor material has bandgap energy of .1 eV. Density of states for conduction and valence bands are both = 10^18cm^-3.
If the material is doped with shallow donors (n-type) Nd=10^17cm^-3, what is electron and hole concentrations?


Homework Equations


ni=sqrt(Nc*Nv)*exp(-Eg/2kT)
n+Na=p+Nd
p=ni^2/n

The Attempt at a Solution


so I set Na = 0. and solve the quadratic equation but I end up with a hole concentration greater than an electron concentration...why is that the case? isn't the fact that its doped n-type mean the n concentration is supposed to be greater?
 
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You did the math wrong. With those conditions, I get n>p, with n~2*p. Did you mix up Na and Nd?
 
you are right haha thanks
 

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