Answer:
a) 6.8 Volt
b) 1.21Цm
Explanation:
We are given from the question that
The zero -bias depletion layer width([tex]W_{do}[/tex]) is 0.4Цm
The built in voltage φj is 0.85V
Now to calculate the reverse voltage( [tex]V_{R}[/tex]) that would be required to triple the depletion - layer width.
The depletion - layer width ([tex]W_{d}[/tex]) of the diode has the formula
[tex]W_{d} = W_{do} \sqrt{1 + \frac{V_{R} }{Qj} }[/tex]
For three times of [tex]W_{d}[/tex] we have
[tex]3W_{d} = W_{do} \sqrt{1 +\frac{V_{R} }{Qj} }[/tex]
=> [tex]\frac{V_{R} }{Qj} = 3^{2} -1[/tex]
=> [tex]V_{R} = 8Qj[/tex]
Substituting value of φj
We have
[tex]V_{R} = 8(0.85V)[/tex]
= 6.8 V
The required bias voltage [tex]V_{R}[/tex] is 6.8 V
The solution for the b part of the question is uploaded on first image
