Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

EXPERIMENT 8: Measuring the band gap of a semiconductor

Objectives
 To determine the gap energy for a semiconductor material by measuring the
resistance of a thermistor as a function of temperature
...

 The thermistor was placed in an oil bath (silicon oil)
...

 The oil was heated gradually, stirring gently
...

 The temperature was converted from 0C to K and a graph of lnR against 1 was
𝑇



plotted
...


Theory
 “Band gap” or “energy gap” refers to an energy range in a solid where no electron
states can exist
...

Equivalently it is the energy required to free an outer shell electron from its orbit
about the nucleus to become a mobile charge carrier, able to move freely within the
solid material
...

 The band gap energy of insulators is large (> 4eV), but lower for semiconductors (<
3eV)
...
Conductors
either have very small band gaps or none, because the valence and conduction
bands overlap
...
The probability of the occupation of an
energy level is based on the Fermi function
...
When an electron in an intrinsic
semiconductor gets enough energy, it can go to the conduction band and leave
behind a hole
...

 Extrinsic semiconductors are made by introducing different atoms, called dopant
atoms, into the crystal
...
It is weakly dependent on temperature
...
381×10−23J/K = 8
...

1
Equation 3 represents a linear relationship between lnR and from which Eg is found



T

from the gradient
...

 Point to note: Slight impurities in a semiconductor hugely impacts on the conductivity
...

 For intrinsic semiconductors, the resistance is very sensitive to slight changes in
temperature
...

Expectations
 From equation (4), a linear graph is expected after plotting the points
...

 Therefore a larger gradient slope of the “line of best fit” is expected
...


Results
T
/C
0

16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
Page | 2

T
/K
289
...
15
291
...
15
293
...
15
295
...
15
297
...
15
299
...
15
301
...
15
303
...
15

R
/kΩ
124
...
7
113
...
4
104
...
5
86
83
...
2
78
...
4
70
...
3
64
...
1
55
...
46
3
...
43
3
...
41
3
...
39
3
...
37
3
...
34
3
...
32
3
...
3
3
...
82
4
...
73
4
...
65
4
...
45
4
...
4
4
...
31
4
...
21
4
...
13
4
...
15
333
...
15
335
...
15
337
...
15
339
...
15
341
...
15
343
...
15
345
...
15
347
...
7
15
...
5
13
...
4
12
...
3
11
...
2
10
...
1
9
...
4
9
8
...
3

X=

𝟏

𝐓

*10-3
3
...
99
2
...
97
2
...
96
2
...
94
2
...
92
2
...
91
2
...
89
2
...
75
2
...
67
2
...
6
2
...
51
2
...
42
2
...
31
2
...
24
2
...
16
2
...
15
306
...
15
308
...
15
310
...
15
312
...
15
314
...
15
316
...
15
318
...
15
320
...
15
322
...
15
324
...
15
326
...
15
328
...
15
330
...
15

51
...
1
43
...
2
42
...
2
39
...
4
35
33
31
...
3
28
...
5
26
...
2
24
...
2
22
...
1
20
...
3
18
...
6
16
...
3

3
...
27
3
...
25
3
...
22
3
...
2
3
...
18
3
...
16
3
...
14
3
...
12
3
...
1
3
...
08
3
...
07
3
...
05
3
...
03
3
...
94
3
...
83
3
...
77
3
...
72
3
...
62
3
...
5
3
...
41
3
...
31
3
...
23
3
...
14
3
...
05
3
...
96
2
...
87
2
...
79

75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100

348
...
15
350
...
15
352
...
15
354
...
15
356
...
15
358
...
15
360
...
15
362
...
15
364
...
15
366
...
15
368
...
15
370
...
15
372
...
15

8
...
7
7
...
2
6
...
6
6
...
1
5
...
7
5
...
3
5
...
9
4
...
6
4
...
2
4
...
9
3
...
6
3
...
4
3
...
2

2
...
86
2
...
85
2
...
83
2
...
82
2
...
8
2
...
78
2
...
77
2
...
75
2
...
74
2
...
72
2
...
71
2
...
69
2
...
68

Analysis
Graph of lnR against


𝟏

analysis:

𝑻



number of variables, N = 85; ∑ 𝑥 = 0
...
75; ∑ 𝑥2 = 0
...
94; ∑ 𝑥𝑦 = 0
...
258 = 0
...
75 = 2
...
000788 −

0
...
36*10 ;

𝑁
85
(∑ 𝑦)2
242
...
94 − 85 = 96
...
258∗242
...
0205;
= 0
...
86 – 4708*0
...
44;

Page | 3

∑ 𝑦2

−

𝑥𝑦

𝑠𝑥𝑥

=

0
...
36∗10−6

= 4 708;

2
...
04
2
...
97
1
...
89
1
...
81
1
...
74
1
...
67
1
...
59
1
...
53
1
...
44
1
...
36
1
...
28
1
...
22
1
...
16





𝑠𝑦𝑦− 𝑚 𝑠2𝑥 𝑥

standard deviation about regression line, s = √
r

0
...
69− 47082∗4
...
02162

= 10
...
36∗10−6
∑ 𝑥2

=

2

𝑁 ∑ 𝑥2−(∑ 𝑥)

0
...
0216 √
* 85∗0
...
2582 =

0
...
44 and m= 4708
...
44 + 4708 
Eg
,
2𝐾𝐵

2𝐾𝐵 T

T

m=


therefore Eg = m*2𝐾𝐵 = 4 708*2*8
...
811 eV
...

The gradient was found to be large as well
...
811 eV
...
11 eV and that for
Germanium (Ge) is 0
...

Therefore the unknown semiconductor is Germanium
...

 Sources of errors in the experiment were possibly due to:
o Random errors in the reading of the temperature from the analogue
thermometer
...

o There could have been impurities on the semiconductor to affect the
conductivity
...

 Even though there were some sources of errors, the standard deviation about
regression line, which represents the error, is so small that the results can be
considered valid
...
811 eV
...

References
1) CJ Sheppard, S Jacobs, & B
...
Doyle
...
PHYSICS 2A
PRACTICAL GUIDE
...
RSA
...
chembio
...
ca/educmat/chm729/band/fl
...
ece
...
edu/~dilli/courses/enee313_spr09/files/supplement1_carrierconc

---