Space charge width pn junction

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SUMMARY

The discussion focuses on calculating the width of the space charge region in a silicon pn junction under reverse bias conditions. Given the parameters of Na=1016 cm-3, Nd=1015 cm-3, intrinsic carrier concentration Ni=1.5*1010 cm-3, and reverse voltage VR=5V, the built-in potential was calculated to be 0.635V. The formula used for calculating the width is W=((2Ef(Vbi+Vr)/e)((Na+Nd)/((Na*Nd))))^1/2. Clarifications were made regarding the non-involvement of Fermi energy in the calculation and the identification of epsilon as the permittivity of the material.

PREREQUISITES
  • Understanding of pn junction theory
  • Familiarity with semiconductor physics at T=300K
  • Knowledge of built-in potential calculations
  • Basic understanding of permittivity in materials
NEXT STEPS
  • Study the derivation of the space charge width formula in pn junctions
  • Learn about the impact of doping concentrations on junction characteristics
  • Explore the role of permittivity in semiconductor devices
  • Investigate the effects of reverse bias on pn junction behavior
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Students and professionals in electrical engineering, particularly those focusing on semiconductor devices and pn junction analysis.

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Homework Statement



Calculate the width of the space charge region in a pn junction when a reverse biased voltage is applied. Consider a silicon pn junction at T=300K with doping concentrations of Na=10^16 cm^-3 and Nd=10^15 cm-3. Assume that Ni=1.5*10^10 cm-3 and VR=5V

Homework Equations


W=((2Ef(Vbi+Vr)/e)((Na+Nd)/((Na*Nd))))^1/2

The Attempt at a Solution



I correctly calculated the in built potential to be 0.635V however, I do not know how to calculate the fermi energy
 
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Check that formula again. It does not involve the Fermi energy.
 
phyzguy said:
Check that formula again. It does not involve the Fermi energy.

What is the value of epsilon in the formula attached?
 

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It's the permittivity of the material.
 

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