Question: Q4 Consider a pn junction diode at 300 K in which the p doping density is 1024 m 3 while the n doping density is

 Q4 Consider a pn junction diode at 300 K in which

Q4 Consider a pn junction diode at 300 K in which the p doping density is 1024 m 3 while the n doping density is 1022 m". The thickness of the p-doped region is 10 um while the thickness of the n-doped region is 20 um and the cross-sectional area of the diode is 2 um . Take kT=0.025 V, kT/q=0.025 eV, the intrinsic carrier concentration n; to be 1.5x1015 m" at 300K, the diffusion constant for the electrons De to be 0.0038 m /s and the diffusion constant for the holes Da to be 0.0012 m-/s. c) Apply a forward voltage of 0.6 V to the pn junction. i ) Calculate the contact potential with this forward bias and the total depletion layer thickness. How much does the depletion layer thickness extend into each region? ii) Explain in words how current flow arises in the pn junction with forward bias applied as opposed to the equilibrium case when no current flows. (Hint: Discuss in terms of drift and diffusion) iii) Using a short diode approximation calculate the current for the diode with this forward voltage applied

Step by Step Solution

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related Physics Questions!