A silicon device maintained at 300 K is characterized by the foll
A silicon device maintained at 300 K is characterized by the following energy band diagram.
Assume that the intrinsic concentration of silicon is 1010 cm-3, EG = 1.12 eV, and kT = 25 mV. At x = x2, what is the value of hole concentration p? (Assume the Fermi Level at the center of the band gap)
A. 7.63 × 10<sup>6</sup>/cm<sup>3</sup>
B. 1.74 × 10<sup>13</sup>/cm<sup>3</sup>
C. 10<sup>10</sup>/cm<sup>3</sup>
D. 1.72 × 10<sup>16</sup>/cm<sup>3</sup>
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
Concept:
The hole concentration for a p-type semiconductor for the given intrinsic carrier concentration is given by:
\({p_0} = {n_i}{e^{\frac{{\left( {{E_{{Fi}}} - {E_F}} \right)}}{{kT}}}}\)
EFi = Intrinsic Fermi Level
ni = Intrinsic carrier concentration
Application:
Given: ni = 1010 cm-3
At x = x2 , we have:
\({E_i} - {E_F} = \frac{{{E_G}}}{2} - \frac{{{E_G}}}{3}\)
\({E_i} - {E_F} = \frac{{{E_G}}}{6}\)
Now, the hole concentration at x = x2 will be:
\(p = {n_i}{e^{\left( {\frac{{{E_i} - {E_F}}}{{KT}}} \right)}}\)
\(p = {10^{10}}{e^{\left( {\frac{{{E_G}}}{{6KT}}} \right)}}\)
\(p = {10^{10}}{e^{\left( {\frac{{1.12}}{{6 \times 0.025}}} \right)}}\)
p = 1.74 × 1013 cm-3