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Chapter 1-4答案1
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1
Chapter 1 (Odd)
1. Copper has 20 orbiting electrons with only one electron in the outermost shell. The fact that
the outermost shell with its 29
th
electron is incomplete (subshell can contain 2 electrons) and
distant from the nucleus reveals that this electron is loosely bound to its parent atom. The
application of an external electric field of the correct polarity can easily draw this loosely
bound electron from its atomic structure for conduction.
Both intrinsic silicon and germanium have complete outer shells due to the sharing (covalent
bonding) of electrons between atoms. Electrons that are part of a complete shell structure
require increased levels of applied attractive forces to be removed from their parent atom.
3.
5. 48 eV = 48(1.6 10
19
J) = 76.8 10
19
J
Q =
W
V
=
19
76.8 10 J
12 V
= 6.40 10
19
C
6.4 10
19
C is the charge associated with 4 electrons.
7. An n-type semiconductor material has an excess of electrons for conduction established by
doping an intrinsic material with donor atoms having more valence electrons than needed to
establish the covalent bonding. The majority carrier is the electron while the minority carrier
is the hole.
A p-type semiconductor material is formed by doping an intrinsic material with acceptor
atoms having an insufficient number of electrons in the valence shell to complete the covalent
bonding thereby creating a hole in the covalent structure. The majority carrier is the hole
while the minority carrier is the electron.
9. Majority carriers are those carriers of a material that far exceed the number of any other
carriers in the material.
Minority carriers are those carriers of a material that are less in number than any other carrier
of the material.
11. Same basic appearance as Fig. 1.9 since boron also has 3 valence electrons (trivalent).
13.
15. T
K
= 20 + 273 = 293
k = 11,600/n = 11,600/2 (low value of V
D
) = 5800
I
D
= I
s
1
D
K
kV
T
e
= 50 10
9
(5800)(0.6)
293
1e
= 50 10
9
(e
11.877
1) = 7.197 mA
2
17. (a) T
K
= 20 + 273 = 293
k = 11,600/n = 11,600/2 = 5800
I
D
= I
s
1
D
K
kV
T
e
= 0.1
A
(5800)( 10 V )
293
1e
= 0.1 10
6
(e
197.95
1) = 0.1 10
6
(1.07 10
86
1)
0.1 10
6
0.1
A
I
D
= I
s
= 0.1
A
(b) The result is expected since the diode current under reverse-bias conditions should equal
the saturation value.
19. T = 20C: I
s
= 0.1
A
T = 30C: I
s
= 2(0.1
A) = 0.2
A (Doubles every 10C rise in temperature)
T = 40C: I
s
= 2(0.2
A) = 0.4
A
T = 50C: I
s
= 2(0.4
A) = 0.8
A
T = 60C: I
s
= 2(0.8
A) = 1.6
A
1.6
A: 0.1
A 16:1 increase due to rise in temperature of 40C.
21. From 1.19:
75C
25C
100C
200C
V
F
@ 10 mA
I
s
1.7 V
0.1
A
1.3 V
0.5
A
1.0 V
1
A
0.65 V
2
A
V
F
decreased with increase in temperature
1.7 V: 0.65 V 2.6:1
I
s
increased with increase in temperature
2
A: 0.1
A = 20:1
23. In the forward-bias region the 0 V drop across the diode at any level of current results in a
resistance level of zero ohms – the “on” state – conduction is established. In the reverse-bias
region the zero current level at any reverse-bias voltage assures a very high resistance level
the open circuit or “off” state conduction is interrupted.
25. V
D
0.66 V, I
D
= 2 mA
R
DC
=
0.65 V
2 mA
D
D
V
I
= 325
27. V
D
= 10 V, I
D
= I
s
= 0.1
A
R
DC
=
10 V
0.1 A
D
D
V
I
= 100 M
V
D
= 30 V, I
D
= I
s
= 0.1
A
R
DC
=
30 V
0.1 A
D
D
V
I
= 300 M
As the reverse voltage increases, the reverse resistance increases directly (since the diode
leakage current remains constant).
3
29. I
D
= 10 mA, V
D
= 0.76 V
R
DC
=
0.76 V
10 mA
D
D
V
I
= 76
r
d
=
0.79 V 0.76 V 0.03 V
15 mA 5 mA 10 mA
d
d
V
I
= 3
R
DC
>> r
d
31. I
D
= 1 mA, r
d
=
26 mV
2
D
I
= 2(26 ) = 52 vs 55 (#30)
I
D
= 15 mA, r
d
=
26 mV 26 mV
15 mA
D
I
= 1.73 vs 2 (#30)
33. r
d
=
0.8 V 0.7 V 0.09 V
7 mA 3 mA 4 mA
d
d
V
I
= 22.5
(relatively close to average value of 24.4 (#32))
35. Using the best approximation to the curve beyond V
D
= 0.7 V:
r
av
=
0.8 V 0.7 V 0.1 V
25 mA 0 mA 25 mA
d
d
V
I
= 4
37. From Fig. 1.33
V
D
= 0 V, C
D
= 3.3 pF
V
D
= 0.25 V, C
D
= 9 pF
39. V
D
= 0.2 V, C
D
= 7.3 pF
X
C
=
1 1
2 2 (6 MHz)(7.3 pF)fC
= 3.64 k
V
D
= 20 V, C
T
= 0.9 pF
X
C
=
1 1
2 2 (6 MHz)(0.9 pF)fC
= 29.47 k
41.
43. At V
D
= 25 V, I
D
= 0.2 nA and at V
D
= 100 V, I
D
0.45 nA. Although the change in I
R
is
more than 100%, the level of I
R
and the resulting change is relatively small for most
applications.
4
45. I
F
= 0.1 mA: r
d
700
I
F
= 1.5 mA: r
d
70
I
F
= 20 mA: r
d
6
The results support the fact that the dynamic or ac resistance decreases rapidly with
increasing current levels.
47. Using the bottom right graph of Fig. 1.37:
I
F
= 500 mA @ T = 25C
At I
F
= 250 mA, T 104C
49. T
C
= +0.072% =
1 0
100%
( )
Z
Z
V
V T T
0.072 =
1
0.75 V
100
10 V( 25)T
0.072 =
1
7.5
25T
T
1
25 =
7.5
0.072
= 104.17
T
1
= 104.17 + 25 = 129.17
51.
(20 V 6.8 V)
(24 V 6.8 V)
100% = 77%
The 20 V Zener is therefore 77% of the distance between 6.8 V and 24 V measured from
the 6.8 V characteristic.
At I
Z
= 0.1 mA, T
C
0.06%/C
(5 V 3.6 V)
(6.8 V 3.6 V)
100% = 44%
The 5 V Zener is therefore 44% of the distance between 3.6 V and 6.8 V measured from the
3.6 V characteristic.
At I
Z
= 0.1 mA, T
C
0.025%/C
53. 24 V Zener:
0.2 mA: 400
1 mA: 95
10 mA: 13
The steeper the curve (higher dI/dV) the less the dynamic resistance.
55. Fig. 1.53 (f) I
F
13 mA
Fig. 1.53 (e) V
F
2.3 V
57. (a)
0.75
3.0
= 0.25
From Fig. 1.53 (i)
75
(b) 0.5
= 40
5
Chapter 1 (Even)
2. Intrinsic material: an intrinsic semiconductor is one that has been refined to be as pure as
physically possible. That is, one with the fewest possible number of impurities.
Negative temperature coefficient: materials with negative temperature coefficients have
decreasing resistance levels as the temperature increases.
Covalent bonding: covalent bonding is the sharing of electrons between neighboring atoms to
form complete outermost shells and a more stable lattice structure.
4. W = QV = (6 C)(3 V) = 18 J
6. GaP Gallium Phosphide E
g
= 2.24 eV
ZnS Zinc Sulfide E
g
= 3.67 eV
8. A donor atom has five electrons in its outermost valence shell while an acceptor atom has
only 3 electrons in the valence shell.
10. Same basic appearance as Fig. 1.7 since arsenic also has 5 valence electrons (pentavalent).
12.
14. For forward bias, the positive potential is applied to the p-type material and the negative
potential to the n-type material.
16. k = 11,600/n = 11,600/2 = 5800 (n = 2 for V
D
= 0.6 V)
T
K
= T
C
+ 273 = 100 + 273 = 373
(5800)(0.6 V)
/
9.33
373
K
kV T
e e e
= 11.27 10
3
I =
/
( 1)
K
kV T
s
I e
= 5
A(11.27 10
3
1) = 56.35 mA
18. (a)
x
y = e
x
0
1
1
2.7182
2
7.389
3
20.086
4
54.6
5
148.4
(b) y = e
0
= 1
(c) For V = 0 V, e
0
= 1 and I = I
s
(1 1) = 0 mA
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