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Chapter 06.03
Linear Regression-More Examples
Computer Engineering
Example 1
To simplify a model for a diode, it is approximated by a forward bias model consisting of DC
voltage, Vd , and resistor Rd . Below are the current vs. voltage data that is collected for a
small signal.
Table 1 Current versus voltage for a small signal.
V
(volts)
0.6
0.7
0.8
0.9
1.0
1.1
I
(amps)
0.01
0.05
0.20
0.70
2.00
4.00
The I vs. V data is regressed to I  B1V  B0 .
Once B0 and B1 are known, Vd and Rd can be computed as
B0
1
and Rd 
B1
B1
Find the value of Vd and Rd .
Solution
Vd  
Table 2 shows the summations needed for the calculation of the constants of the regression
model.
06.03.1
06.03.2
Chapter 06.03
Table 1 Tabulation of data for calculation of needed summations.
i
−
1
2
3
4
5
6
V
Volts
0.6
0.7
0.8
0.9
1.0
1.1
I
Amperes
0.01
0.05
0.20
0.70
2.00
4.00
V2
Volts 2
0.36
0.49
0.64
0.81
1.00
1.21
V I
Volt-Amps
0.006
0.035
0.160
0.630
2.000
4.400
5.1
6.96
4.51
7.231
6

i 1
n6
6
6
6
n Vi I i  Vi  I i
B1 
i 1
i 1
i 1


n Vi   Vi 
i 1
 i 1 
6
2
6
2
67.231  5.16.96 
2
64.51  5.1
 7.5143 A/V

6
_
I
I
i 1
i
n
6.96

6
 1.16 A
6
_
V 
V
i 1
i
n
5.1

6
 0.85 V
Linear Regression-More Examples: Computer Engineering
_
_
B0  I  B1 V
 1.16  7.514 0.85
 5.2269 A
I  7.514  V  5.2269
Figure 1 Linear regression of current vs. voltage
Solving for Vd and Rd :
B
Vd   0
B1
  5.2269 
 

 7.5143 
 0.69560 Volts
Rd 
1
B1
1
7.5143
 0.13308 Ohms

06.03.3