# ch02 ```2 The electrons in a piece of n‐type semiconductor take 10 ps to cross from one end to
another end when a potential of 1 V is applied across it. Find the length of the
semiconductor bar.
Solution
L
 10 ps
V
V
L
2
L  10 ps  nV  10 ps  1350  1
But
V  n E and
E
L  1.16  104 cm  1.16  m
current of 0.05 μA flows through an n‐type silicon bar of length 0.2 μm and cross
section area of 0.01 μm  0.01 μm when a voltage of 1 V is applied across it. Find the doping
level at room temperature.
3 A
Solution
I  0.05 A
L  0.2 m
A  0.01 m  0.01 m  1  1012 cm 2
I tot  I drift  q  nn  p  p  AE
n 2 1.08  10 
p i 
n
n
2
n  1350cm / (V.s)
----(1)
10 2
 p  480 cm 2 /(V.s)
V
1V

 5  104 V / cm
L 0.2 m
Substituting these values in Eq. (1) we get
E

(1.08  1010 ) 2  480  12
4
0.05  106 A  1.6  1019  n  1350 
 10  5  10
n


n  4.63  1015 cm 3 .
4. Repeat Problem 2.3 for Ge. Assume μn=3900cm2/(V&middot;s) and
μp =1900cm2/(V&middot;s).
In order to obtain the doping level proceed as follows. We use the formula

1
q ( n n  p  p )
For n-type semiconductor n&gt;&gt;p, and hence
 1
nqn
Given:I  0.05 A
l  0.2 m
A  (0.01 m)2
V  1V
n  3900cm 2 / (V s)
 p  1900cm 2 / (V s)
Resistance of the Ge bar;
R
V
I
1V
0.05 A
 20MΩ

Therefore resistivity,
R
A
l
 20MΩ 
(0.01 m)2
0.2 m
 1cm
Again,  
or, n 
1
 e n
1
nen
Substituting the values, we get
1
1cm 1.6 10  3900cm 2 / (V s)
1

6.24 1016 cm3
 1.6 1015 cm3
ND  n 
19
Since at room temperature all the impurities are ionized and hence carrier
concentration n is equal to the doping density.
A Si semiconductor cube with side equal to 1 μm is doped with 4  1017 cm–3
phosphorous impurities. Calculate the drift current when a voltage of 5 V is applied across it.
9
Solution
n  4  1017 cm 3
ni2 (1.08  1010 ) 2

 291.6
n
4  1017
n  1350 cm 2 /(V.s) and  p  480 cm 2 /(V.s)
p
E
V
5V

 5  104 V /cm
L 1m
A  1  m  1  m  108 cm 2
Since nn   p , we can write
I tot  qnn AE  0.0432 A
One side of pn junction is doped with pentavalent impurities which gives a net
doping of 5  1018 cm–3. Find the doping concentration to be added to the other side to get a
built‐in potential of 0.6 V at room temperature.
10
Solution
V0  VT ln
N A ND
ni2
N D  5  1018 cm 3 , VT  26 mV, ni  1.08  1010 cm 3 , V0  0.6 V
 N  5  1018 
0.6  0.026 ln  A
 N A  2.45  1011 cm 3
10 2 
 (1.08  10 ) 
The built‐in potential of an equally doped pn junction is 0.65 V at room temperature.
When the doping level in n‐side is doubled, keeping the doping of the p‐side unchanged,
calculate the new built‐in potential and the doping level in P and N region.
11
Solution
V0  VT ln
N A ND
ni2
N A  N D , VT  26 mV, ni  1.08  1010 cm 3 , V0  0.65 V
 NA  NA 
0.65  0.026 ln 
 N A  2.89  1015 cm 3
10 2 
(1.08

10
)


 2  2.89  1015  2.89  1015 
V0  0.026 ln 
  0.668 V
(1.08  1010 ) 2


A silicon pn junction diode having ND = 1015 cm–3 and NA= 1017 cm–3 gives a total
depletion capacitance of 0.41 pF/m2. Determine the voltage applied across the diode.
12
Solution
V0  VT ln
C j0 
Cj 
NA ND
1015  1017

0.026
ln
 0.7144 V
(1.08  1010 ) 2
ni2
 si q N A N D
2 N A  ND

1
11.7  8.854  1014  1015  1017

 1.072  108 F/cm 2
2  (1015  1017 )  0.7144
V0
C j0
1
VR
V0
 C  2 
 1.072  108 
j0
 1  0.7144 
 4.167 V
VR  V0 

8 
 C j 

 0.41  10 


Two identical pn junction diodes are connected in series. Calculate the current
flowing through each diode when a forward bias of 1 V is applied across the series
combination. Assume the reverse saturation current is 1.44  10–17 A for each diode.
14
Solution
.5
 VVF

 0.0026

17
T



I D  I S e  1  1.44  10  e
 1  3.23  10 9 A






Consider a pn junction in forward bias. Initially a current of 1 mA flows through it,
and the current increases by 10 times when the forward voltage is increased by 1.5 times.
Determine the initial bias applied and reverse saturation current.
16
Solution
VF
I D  I S e VT
1 mA  I S e
VF
VT
10 mA  I S e
1.5VF
VT
VT
ln(100)  0.239 V
0.5
From Eq. (1), we get I S  107 A
VF 
(1)
(2)
```