CALIFORNIA STATE UNIVERSITY, NORTHRIDGE
DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING
ECE340 – ELECTRONICS I
FALL 2009
HOMEWORK ASSIGNMENT #2
SOLUTIONS
Problem One
Problem 2.10.
N D  5  1017 cm 3
a 
N A  4  1016 cm 3
 nn  5  1017 cm  3
nn  N D
1.08  10

cm  3
5  1017 cm  3
n2
pn  i
ND

pn
pp  N A

p p  4  1016 cm  3

1.08  10

n2
np  i
NA
b 


ni  1.08  1010 cm 3
10
cm  3
4  1016 cm  3
10
pn
 

2


pn  233.28 cm - 3
2
 n p  2916 cm - 3

19
3


ni 250 K  5.2  1015 250  2 exp  1.792  10 J
 23

1.38  10 J / T 250 K 


ni 250 K  1.08  108 cm - 3



ni 300 K  1.08  1010 cm - 3




ni 350 K  2.99  1011 cm -3
b 


ni T   5.2  1015 T

3
2
exp 



ni 250  K  1.08  10 8 cm - 3
VO 
kT  N A N D
ln 
q  ni2
VO
VO
1.38  10
300 K  
VO
1.38  10
350 K  
T
J
 23
J
1.6  10

 VO
 23
1.6  10

 23
J
1.6  10


kT 


ni 300  K  1.08  1010 cm -3

ni 350  K  2.99  1011 cm - 3




1.38  10
250 K  

 EG
250 K  ln  5  10
19
J




17
350 K  ln  5  10
17
19
J




K
17
300 K  ln  5  10

K
19
J










cm 3 4  1016 cm 3 

  VO 250 K  0.9053V
8
-3 2
1.08  10 cm


K



cm 3 4  1016 cm 3 

  VO 300 K  0.848V
10
-3 2
1.08  10 cm





cm 3 4  1016 cm 3 

  VO 350 K  0.789 V
11
-3 2
2.99  10 cm




Problem Two
Problem 2.12
N D  3  1016 cm 3
a 
VO 
C j0 
C j0 
N A  2  1015 cm 3
kT  N D N A 
  VO  0.698V
ln 
q  ni2 
q Si  N A N D 


2VO  N A  N D 
1.6  10
19
C j0
V
1 R
VO





C 11.7  8.85  1014 F cm 3  1016 cm  3 2  1015 cm  3
20.698V  3  1016 cm  3  2  1015 cm  3
C j0  14.92  10-9 F
Cj 
VR  1.6V


cm 2
 Cj 
14.92  10-9 F
cm 2
1.6V
1
0.698V
 C j  8.22  10-9 F
cm 2
b 
Cj 
C j0
1
C 2j 
VR
VO
 C 2j 
C 2j 0
 VR 
1  
 VO 
q Si N A N D

kT  N D N A 

2 N A  N D VR 
ln 
2
q
n
i



 C 2j 
 Cj 
q Si N A N D
2VO  VR N A  N D 
q Si N A N D

kT  N D N A 

2N A  N D VR 
ln 
2
q
n
i



Plotting Cj vs. NA:
NA (cm-3)
2x1015
4x1015
6x1015
8x1015
1x1016
Cj (F/cm2)
8.22x10-9
11.24x10-9
13.34x10-9
14.97x10-9
16.3x10-9
Thus, NA needs to increase by a factor of 5 in order to double the junction capacitance.
Problem Three
Problem 2.17
D1
D2
+ VD1 -
+ VD2 -
IB
VB
I B  I D1  I D 2
V
I B  I S 1 exp  D1
 VT
VB  VD1  VD 2



I 
VD1  VT ln  B 
 I S1 
V
I B  I S 2 exp  D 2
 VT



 I 
I 
 I 
VD 2  VT ln  B   VB  VT ln  B   ln  B 
 IS 2 
 I S 2 
  I S1 
 I B  I B 
 I B2 
V
VB
 
  exp  B
VB  VT ln  
 ln 
VT
 VT
 I S1I S 2 
 I S 1  I S 2 
V
I B2  I S 1I S 2 exp  B
 VT

 V
  I B  I S 1I S 2 exp  B

 2VT




I B2
 
 I S1I S 2
 I I
I 
VD1  VT ln  B   VD1  VT ln  S 1 S 2
 I S 1
 I S1 
  I
VD1  VT ln  S 2
  I S 1
VD1 


  ln exp  VB
 2V

T



 IS 2 
1
  VB 
VT ln 
2
 I S1 


 exp  VB
 2V

T





1  I  V 

  VD1  VT  ln  S 2   B 

 2  I S 1  2VT 
VD 2 

 I S1 
1
  VB 
VT ln 
2
 IS 2 

Problem Four
Problem 2.21
IX
R1
2kΩ
VX
D1
VX  2V
I D1 
VD1  0.85V
VX  VD1
R1
 I D1 
I X  I D1
2V  0.85V
2kΩ
 I D1  0.575mA
 V 
  0.85V 
-18
I S  I D1 exp  D1   I S  575  10- 6 Aexp 
  I S  3.64  10 A
 0.026V 
 ηVT 
Problem Five
Problem 3.4
R1
VX
D1
VB
VX  VO cos t
VB  1V
VO
(VO+1)/R1
VX
IX
VX  VO cos t
VB  1V
VO
(VO-1)/R1
VX
IX
Problem Six
Problem 3.47
vout
2V
0.5
-2V
vin
2V
0.5
-2V
R1
vout
vin
D1
D2
R2
R3
2V
R1  R2  R3  1k
2V
Problem Seven
D1
IO
D2
+
10mA
R1
D3
R3
VO
-
R2
VO  9V
R3 
I O  1mA@ T  25 C
VD1  0.7V
9V
 R3  9k
1mA
 0.7 V
I D1  5  1015 A exp 
 0.026V



  I D1  2.5 mA

I R1  10 mA  2.5 mA  I R1  7.5 mA
R1 
I S  5  1015 A
VR1  0.7V  9V
 VR1  9.7V
9.7V
 R1  1.29kΩ
7.5 mA
I D 2  2.5 mA  1mA  I D 2  1.5 mA
 1.5  10 3 A 
  VD 2  0.6865 V
VD 2  0.026V ln 
15
 5  10 A 
R2 
9V  20.6865V 
 R2  5.2123kΩ
1.5 mA
VD 3  0.6865 V
Problem Eight
25kΩ
D1
vA
+
10kΩ
2kΩ
-
5kΩ
-
vo
+
vs(t)
3kΩ
10kΩ
3kΩ
For the circuit shown above, the diode and operational amplifiers are ideal. The input voltage
is a sinusoid given by the following equation:
vS  15 sin( 2 1000t ) V
a) Sketch the input and output voltage waveforms as a function of time.
Positive cycle representative circuit:
25kΩ
10kΩ
va
+
2kΩ
vs(t)
va
5kΩ
-
-
vo
3kΩ
+
10kΩ
3kΩ
3k


va  
 vs
 2k  3k 
 va  0.615 sin( 2 1000t ) V   va  9 sin( 2 1000t ) V
 25k 
vo  
9 sin( 2 1000t ) V   vo  45 sin( 2 1000t ) V
 5k 
Negative cycle representative circuit:
25kΩ
va
+
2kΩ
vs(t)
10kΩ
va
-
vo
3kΩ
+
10kΩ
va  0  vo  0
5kΩ
3kΩ
vs(V)
15
t(s)
vo(V)
t(s)
-45
15
10
vs
0
-10V
-15V
10.8
vo
0
-25.0V
-45 V
-48.2V
0s
0.5m s
1.0m s
1.5m s
2.0m s
2.5m s
Tim e
3.0m s
3.5m s
4.0m s
4.5m s
b) Redo part a using a piece-wise linear model for the diode with rd  10 and Vbi  0.7V .
Positive cycle representative circuit:
25k
0.7V
10k
5k
+
2k
0.01k
vs(t)
-
vo
-
3k
+
10k
3k
 v  0.7 
 v  0.7 
i t    s
  va   s
3k   va  0.598815 sin 2 1000t V   0.7V 
 5.01k 
 5.01k 
vO   50.598815 sin 2 1000t V   0.7V   vO  2.994 15 sin 2 1000t V   0.7V 
vO  44.91sin 2 1000t V   2.1V 
vs(V)
15
t(s)
vo(V)
t(s)
-42.8
20
0
-20V
50
V(R1:1)
0
-50V
0s
V(R5:2)
0.5m s
1.0m s
1.5m s
2.0m s
s
2.5m s
Tim e
3.0m s
3.5m s
4.0m s
4.5m s
s
5.0m s
Problem Nine
300Ω
D1
D3
vs
vo
D4
D2
2.7kΩ
4.7uF
For the circuit shown above, let vS  10 sin 377t V . Each diode is to be modeled as a piece-wise
linear device with the following parameters:
V  0.7V
rd  26
a) Sketch v S and vO as a function of time.
Circuit Model:
300Ω
0.7V
0.7V
26Ω
26Ω
vs
vo
2.7kΩ
0.7V
0.7V
26Ω
26Ω
4.7μF
Positive half cycle:
300Ω
0.7V
26Ω
vs
vo
2.7kΩ
0.7V
26Ω
10  300i  0.7  26i  2700i  0.7  26i  i 
vO , MAX  2.82mA2.7k   vO , MAX  7.6V
Negative half cycle produces the same results.
4.7μF
26Ω
10V  1.4V
3052
 i  2.82mA
vs(V)
10
t(s)
vo(V)
7.6
t(s)
b) Calculate the ripple voltage of vO and determine what percentage it is of the peak output
voltage.
VP  10V
VD , on  0.7V
  377 rad sec 
VR 
V
P
VR  5V
V300  0.846V
f  60 Hz
 2VD , on  V300 
10V  1.4V  0.846V 
 VR 
2700  4.7  10 6 F 260 Hz 
2 RLCL f


c) Repeat part a with a PSPICE simulation and compare your results.
10
7.02 V
4.88 V
5
0
-5V
-10V
0s
5m s
10m s
15m s
20m s
25m s
Tim e
30m s
35m s
40m s
45m s
50m s
Problem Ten
1kΩ
+
D1
vs(t)
D3
vo(t)
D4
D2
-
VC
vS t   15 sin 377t V
In this problem, diodes D2 and D3 are non-ideal diodes that can be characterized by using the
piece-wise linear model. These devices have the following model parameters:
Vbi  0.7V
rd  26
D1 and D4 are zener diodes that have zener voltages of 4.7V and 5V respectively. The
incremental resistance of the zener diodes are 5Ω and 10Ω respectively.
a) Draw an equivalent circuit showing the incremental models for each diode.
1k
VCC
5
26
+
4.7V
0.7V
vo
vs
0.7V
5V
26
10
VCC
b) Sketch the output waveforms for vs(t) and vo(t) on the same graph as a function of time.
15  1000i  0.7  26i  5  10i  i 
15V  5.7V
1036
 i  8.98mA
vO , MAX  8.98mA0.036k   5V  0.7V
 vO , MAX  6.023V
15  1000i  0.7  26i  4.7V  5i  i 
15V  5.4V
1031
vO , MIN  9.31mA0.031k   4.7V  0.7V
 i  9.31mA
 vO , MIN  5.69V
vs(V), vO(V)
6.02
t(s)
-5.69
c) Verify your results using PSPICE.
15V
10V
(4.1620m,5.5680)
5V
0V
(12.503m,-5.5680)
-5V
-10V
-15V
0s
V(V1:+)
5ms
V(D2:1)
10ms
15ms
20ms
25ms
Time
30ms
35ms
40ms
45ms
50ms