ELE533 Bipolar Devices

Mid Term Exam
July 19, 2002

Name_____________________________________

1. Consider a silicon pn junction. The p side is uniformly doped with 10¹⁶ acceptors/cm³. The n side is doped with 5x10¹⁸ donors/cm³.
   The contact on the p-side is 2 microns from the junction. The contact on the n-side is 10 microns from the junction. The junction area is 10 microns by 10 microns.
   What is the built in potential?

   V_bi = V_t ln[N_AN_D/n_i²] = 0.86V

   What is the position of the Fermi level relative to the intrinsic level on the p side of the junction?

   E_i - E_f = V_t ln[N_A/n_i] = 0.347V

   What is the saturation current, Is, in the equation,

   I = Is*e^V/Vt

   Is = Aqn_i² u_n V_t / N_AW_p = 3.315x10^-16A

   Draw the equilibrium energy band diagram for this junction, including numerical values for the Fermi level position relative to the intrinsic level on each side. Show the built-in potential on the diagram.

   
   V_n = 0.51V, V_p = 0.347V, V_bi = 0.8556V

   For an applied forward voltage of 0.6V, find the,

   1. Diffusion capacitance.

      C_diff = w_p² I / 2 D_n V_t

      where

      I = Is e^V/V_t = 3.81 uA

      C_diff = 3.81x10^-6(2x10^-4)²/ (2 *800*0.259) = 0.142 pF

   2. Junction capacitance.

      C_J = C_J0/[1 - V/V_bi]^½

      where

      C_J0 = A [qeN_A / 2V_bi ]^½ = 10^-6[1.6x10^-19 10^-12 10¹⁶ / (2*0.8586) = 3x10^-14 = 30 f F

      C_J = C_J0/[1 - V/V_bi]^½ = 30 f F/[1-0.6/0.8586]^½ = 55.6 f F = 0.0556 pF