Transcript 投影片 1 - Advanced Silicon Device and Process Laboratory

EXAMPLE 5.1
OBJECTIVE
Calculate the built-in potential barrier of a pn junction.
Consider a silicon pn junction at T = 300 K with doping
concentrations of Na = 2  1016 cm-3 and Nd = 5  1015 cm-3.
 Solution
The built-in potential barrier is determined from
Vbi  Vt
 2 1016 5 1015 

Na Nd 
ln
 0.0259 ln


2
i
n
1.5 10 
10 2


or
Vbi = 0.695 V
 Comment
The built-in potential barrier changes only slightly as the doping
concentrations change by orders of magnitude because of the logarithmic
dependence.
EXAMPLE 5.2
OBJECTIVE
Calculate the space charge widths and peak electric field in a pn junction.
Consider a silicon pn junction at T = 300 K with uniform doping concentrations of
Na = 2  1016 cm-3 and Nd = 5  1015 cm-3. Determine xn , xp , W , and max .
 Solution
In Example 5.1, we determined the built-in potential barrier, for these same doping
concentrations, to be Vbi = 0.695 V.
 2 V
xn   s bi
 e
 N a 

1



 N  N  N 

d 
 d  a

or

 211.7  8.85 1014 0.695  2  1016


19
 5  1015
1
.
6

10



1


 2  1016  5  1015




1/ 2
xn = 0.379  10-4 cm = 0.379 m
The space charge width extending into the p region is found to be
 2 V
x p   s bi
 e
 N d 

1



 N a  N a  N d 

or
1/ 2

1/ 2
 211.7  8.85 1014 0.695  5 1015


19
16
1
.
6

10
 2 10


1

16
15
 2 10  5 10
xp = 0.0948  10-4 cm = 0.0948 m



1/ 2
EXAMPLE 5.2
The total space charge width, using Equation (5.31), is
 2 V
W   s bi
 e



N a  N d 

N a N d 


1/ 2
 211.7  8.851014 0.695  2  1016  5  1015


16
15
1.6  1019
 2 10 5 10

or


1/ 2




W = 0.474  10-4 cm = 0.474 m
We can note that the total space charge width can also be found from
W = xn + xp = 0.379 + 0.0948 = 0.474 m
The maximum or peak electric field can be determined from, for example,
 max 
or
eNd xn
s
1.6 10 5 10 0.37910 

11.78.8510 
19
4
15
14
max = 2.93  104 V/cm
 Comment
We can note from the space charge width calculations that the depletion region extends farther
into the lower-doped region. Also, a space charge with on the order of a micrometer is very
typical of depletion region widths. The peak electric field in the space charge region is fairly
large. However, to a good first approximation, there are no mobile carriers in this region so there
is no drift current. (We will modify this statement slightly in Chapter 9.)
EXAMPLE 5.3
OBJECTIVE
Calculate width of the space charge region in a pn junction when a reverse-bias voltage
is applied.
Again, consider the silicon pn junction at T = 300 K with uniform doping
concentrations of Na = 2  1016 cm-3 and Nd = 5  1015 cm-3. Assume a reverse-bias
voltage of VR = 5 V is applied.
 Solution
From Example 5.1, the built-in potential was found to be Vbi = 0.695 V. The total space
charge width is determined to be
 2 s Vbi  VR   N a  N d

W 
e
 Na Nd


or




1/ 2
 211.7  8.8510 0.695 5  2 10  5 10


19
16
15
1
.
6

10
2

10
5

10


14
16

15

1/ 2




W = 1.36  10-4 cm = 1.36 m
 Comment
The space charge width has increased from 0.474 m to 1.36 m at a reverse bias
voltage of 5 V.
EXAMPLE 5.4
OBJECTIVE
Design a pn junction to meet a maximum electric field specification at particular reverse-bias voltage.
Consider a silicon pn junction at T = 300 K with a p-type doping concentration of Na = 1018 cm-3. Determine the n-type
doping concentration such that the maximum electric field in the space charge region is max = 105 V/cm at a reverse bias
voltage of VR = 10 V .
The maximum electric field is given by
 max
 2eVbi  VR   N a N d


N N
s
a
d






1/ 2
Since Vbi is also a function of Na through the log term, this equation is transcendental in nature and cannot be solved
analytically. However, as an approximation, we will assume that Vbi  0.75 V.
We can then write
which yields




 2 1.6 1019 0.75  10  1018  N d   
5
10  
 18

14


11
.
7
8
.
85

10
10

N
d 




1/ 2
Nd = 3.02  1015 cm-3
We can note that the built-in potential for this value of Nd is
Vbi


 1018 3.021015
 0.0259 ln 
10 2
1.5 10




  0.783V


Which is very close to the assumed value used in the calculation. So the calculated value of Nd is a very good approximation.
 Comment
A smaller value of Nd than calculated results in a smaller value of max for a given reverse-bias voltage. The value of Nd
determined in this example, then, is the maximum value that will meet the specifications.
EXAMPLE 5.5
OBJECTIVE
Calculate the junction capacitance of a pn junction.
Consider the same pn junction as described in Example 5.3. Calculate the junction
capacitance at VR = 5 V assuming the cross-sectional area of the pn junction is A = 10-4
cm2.
 Solution
The built-in potential was found to be Vbi = 0.695 V. The junction capacitance per unit
1/ 2
area is found to be


e s N a N d
C  




2
V

V
N

N
bi
R
a
d 


or






 1.6  1019 11.7  8.85 1014 2  1016 5  1015 


20.695 5 2  1016  5  1015



1/ 2
C = 7.63  10-9 F/cm2
The total junction is found as
C = AC = (10-4) (7.63  10-9)
or
C = 7.63  10-2 F = 0.763 pF
 Comment
The value of the junction capacitance for a pn junction is usually in the pF range, or
even smaller.
EXAMPLE 5.6
OBJECTIVE
Determine the impurity concentrations in a p+n junction given the parameters from
Figure 5.12.
Consider a silicon p+n junction at T = 300 K. Assume the intercept of the curve on
the voltage axis in Figure 5.12 gives Vbi = 0.742 V and that the slope is 3.92  1015
(F/cm2)-2/V.
 Solution
The slope of the curve in Figure 5.12 is given by 2/esNd , so we canwrite
or
Nd 
2
2

e s slope
1.6 1019 11.7  8.851014 3.921015

Solving for Na , we find
 Comment



Nd = 3.08  1015 cm-3
The built-in potential is given by
or

 Na Nd
Vbi  Vt ln
 n2
i







 Vbi 
ni2
1.5 1010
 0.742 

Na 
exp

exp


 V  3.081015
Nd
0
.
0259


 t 
Na = 2.02  1017 cm-3
The results of this example show that Na >> Nd ; therefore the assumption of a onesided junction was valid.
EXAMPLE 5.7
OBJECTIVE
Determine the diode current in a silicon pn junction dilde.
Consider a silicon pn junction diode at T = 300 K. The reverse-saturation current is
IS = 10-14 A. Determine the forward-bias diode current at VD = 0.5 V, 0.6 V, and 0.7 V.
 Solution
The diode current is found from
 VD 
 VD 
14


I D  I S exp   10 exp
 A 
 0.0259
 Vt 

so for VD = 0.5 V,
and for VD = 0.6 V,
and for VD = 0.7 V,
 Comment

ID = 2.42 m
ID = 0.115 m
ID = 5.47 m
Because of the exponential function, reasonable diode currents can be achieved even
though the reverse-saturation current is a small value.
EXAMPLE 5.8
OBJECTIVE
Calculate the forward-bias voltage required to generate a forward-bias current density
of 10A/cm2 in a Schottky diode and a pn junction diode.
Consider diodes with parameters JsT = 6  10-5 A/cm2 and JS = 3.5  10-11 A/cm2.
 Solution
For the Schottky diode, we hare
  VD  
J  J sT exp   1
  Vt  
Neglecting the (1) term, we can solve for the forward-bias voltage. We find
 J 
 10 



VD  Vt ln

0
.
0259
ln
 0.311 V

5 
J 
6

10


 sT 
For the pn junction diode, we have
 J 
10





VD  Vt ln

0
.
0259
ln
 0.683 V

11 
J 
 3.5 10 
 S 
 Comment
A comparison of the two forward-bias voltages shows that the schottky diode has an
effective turn-on voltage that, in this case, is approximately 0.37 V smaller than the
turn-on voltage of the pn junction diode.
EXAMPLE 5.9
OBJECTIVE
Calculate the space charge width for a Schottky barrier on a heavily doped
semiconductor.
Consider silicon at T = 300 K doped at Nd = 7  1018 cm-3. Assume a
Schottky barrier with B0 = 0.67 V. For this case, we can assume that Vbi  B0.
 Solution
For a one-sided junction, we have for zero applied bias
 2 sVbi
xn  
 eNd
or
1/ 2





 211.7  8.8510 0.67 


19
18
1.6 10
7 10



14


xn = 1.1  10-6 cm = 110 Ǻ
 Comment
In a heavily doped semiconductor, the depletion width is on the order of
angstroms, so that tunneling is now a distinct possibility. For these types of
barrier widths, tunneling may become the dominant current mechanism.
1/ 2