Question

23/5/2019 Final 2019 - 5 shown 2 Q21Consider the silicon Sample with the impurity doping profile that widiths of neutral regions are much greater than di Husion lengthe A corriers. al calculate the built in voltage between region 1 and 2, Zand 3 and Land 3 for Zero bios condition Il region 13 ad region 2nd region bil calowate the excess carrier concentration for a forward bois douvat the edge of depletin region sl calculate the excess carrier concentration and de dron and hole concentration for a reverse bias & 1.5 Vat He alge of depletion region.

you can say me just correct answer I don't need any step by step solution

Answer 1

Given: concentration in region 1$\small =10^{17}$ (p+ doped region)

concentration of region 2 $\small =10^{11}$ (p type doped region)

concentration of region 3 $\small =10^{18}$ (n type doped region)

a) No depletion region forms between region 1 and 2. Hence built in voltage, Vo= 0.

For zero bias condition,

Built in voltage between region 2 and 3 is

NAND = 0.026 x In 2 ny

0.026 x In 1018 x 1011 (1.5 x 1010)

=0.5177V

Therefore coltage between 1 and 3 is 0.5177V

b)When a forward voltage of 0.4V is applied, junction voltage, Vj= Vo+V

where, V is the applied voltage

Therefore, Vj= 0.5177-0.4 =0.1177V

excess carrier concentration on p side, $\small n_{p}= n_{n}e^{-V_{j}/V_{T}}$

$\small n_{p}= 10^{18}e^{-0.1177/0.026}$  

$\small =1.08\times10^{16}$

excess carrier concentration on n side, $\small p_{n}= p_{p}e^{-V_{j}/V_{T}}$

$\small p_{n}= 10^{11}e^{-0.1177/0.026}$

$\small = 1.08\times10^{9}$

c) When reverse biased by 1.5V, junction voltage, Vj= 2.0177

during reverse bias, no majority charge carriers flows between p and n regions. Hence excess carriers are negligible and majority carrier concentration remains the same