Sarthaks Test
0 votes
794 views
in Physics by (38.1k points)

Suppose a ‘n’-type wafer is created by doping Si crystal having 5 × 1028 atoms/m3 with 1ppm concentration of As. On the surface 200 ppm Boron is added to create ‘P’ region in this wafer. Considering ni = 1.5 × 1016 m–3, (i) Calculate the densities of the charge carriers in the n & p regions. (ii) Comment which charge carriers would contribute largely for the reverse saturation current when diode is reverse biased.

1 Answer

0 votes
by (63.6k points)
selected by
 
Best answer

(i) In ‘n’ region; number of e– is due to As:

Similarly, when Boron is implanted a ‘p’ type is created with holes

nh = NA = 200 × 10–6 × 5 × 1028

= 1 × 1025/m 

This is far greater than e– that existed in ‘n’ type wafer on which Boron was diffused.

Therefore, minority carriers in created ‘p’ region

(ii) Thus, when reverse biased 0.45 × 1010/m3, holes of ‘n’ region would contribute more to the reverse saturation current than 2.25× 107/m3 minority e– of p type region.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...