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Consider a silicon pn junction at T = 300 K with doping densities Na = 1 × 10¹⁶ cm⁻³ and Nd = 1 × 10¹⁵ cm⁻³. Assume that ni = 1.5 x 10¹⁰ cm⁻³.

a) Find out the built-in potential barrier in pn junction.

User Deroccha
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Final answer:

To calculate the built-in potential barrier of a pn junction with specific doping densities at room temperature, a formula involving the Boltzmann constant, temperature, charge of electron, and doping densities is used, and the resulting electric field at the junction is essential for the diode function.

Step-by-step explanation:

To find out the built-in potential barrier in a pn junction for a silicon semiconductor at T = 300 K with doping densities Na = 1 × 1016 cm−3 and Nd = 1 × 1015 cm−3, assuming that ni = 1.5 x 1010 cm−3, we use the contact potential formula:

Vbi = (kT/q)ln((Na · Nd)/(ni2))

Where:

  • Vbi is the built-in potential
  • k is the Boltzmann constant (8.617 x 10−5 eV/K)
  • T is the temperature in Kelvin (300 K)
  • q is the charge of an electron (1.602 x 10−19 coulombs)
  • Na is the acceptor doping density
  • Nd is the donor doping density
  • ni is the intrinsic carrier concentration

Plugging in the values, we have:

Vbi = (8.617 x 10−5 eV/K · 300 K / 1.602 x 10−19 coulombs) · ln((1 × 1016 cm−3 · 1 × 1015 cm−3) / (1.5 x 1010 cm−3)2)

Calculating the built-in potential gives us the voltage that separates charges at the pn junction, forming the depletion region and establishing an electric field that prevents further migration of carriers thereby creating a balance.

User Ryandawkins
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