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Parallelogram method - revision
• Drawing the best fit line
• Drawing the parallelogram
• Modified parallelogram method
• What if the data is “too good” ?
• The modified parallelogram method
• Calculating and using the “final” results
Drawing the best fit line
Fig 1: Current vs Voltage
Voltage [V]
-2 0 2 4 6 8 10 12 14 16 18 20 22
Cu
rren
t [m
A]
64.8
65.0
65.2
65.4
65.6
65.8
66.0
66.2
Fig 1: Current vs Voltage
Voltage [V]
-2 0 2 4 6 8 10 12 14 16 18 20 22
Cu
rren
t [m
A]
64.8
65.0
65.2
65.4
65.6
65.8
66.0
66.2
Drawing the parallelogram
The best fit line is in the centre of the two parallel lines above and below it
2/3 of the data points should lie within the parallelogramm
Fig 1: Current vs Voltage
Voltage [V]
-2 0 2 4 6 8 10 12 14 16 18 20 22
Cu
rren
t [m
A]
64.8
65.0
65.2
65.4
65.6
65.8
66.0
66.2
Determining the diagonals
Draw the diagonals in the parallelogram and determine their gradients
What if your data is not “bad enough” ?
Fig 1: Current vs Voltage
Voltage [V]
0 2 4 6 8 10 12 14 16 18 20
Cu
rren
t [m
A]
-1
0
1
2
3
4
5
6
7
8
V
mA 0.345
V20
mA 6.9bestm
mA 0bestc
bestbestbest cVmI
Fig. 2: Current vs Voltage
What if your data is not “bad enough” ?
bestbestbest cVmI
Fig 1: Current vs Voltage
Voltage [V]
14.5 15.0 15.5 16.0 16.5
I [m
A]
5.0
5.2
5.4
5.6
Look at the difference between
fit line and measured data
Fig. 3: Current vs Voltage, detail of the data shown in Fig. 2
Modified parallelogram method
V [V] Imeas[mA] Imeas-Ibest
0 0.010 0.0101 0.343 -0.00172 0.697 0.00673 1.020 -0.01504 1.373 -0.00675 1.697 -0.02836 2.050 -0.02007 2.413 ...8 2.747 …9 3.100 …. . …. . …. . …
)( bestbestmeasbestmeas cVmIII
Modified parallelogram method
Fig 1: Current vs Voltage
Voltage [V]
0 2 4 6 8 10 12 14 16 18 20
I mea
s -
I fit [
mA
]
-0.06
-0.04
-0.02
0.00
0.02
0.04Fig. 4: Modified Current vs Voltage
Modified parallelogram method
Fig 1: Current vs Voltage
Voltage [V]
0 2 4 6 8 10 12 14 16 18 20
I mea
s -
I fit [
mA
]
-0.06
-0.04
-0.02
0.00
0.02
0.04Fig 1: Current vs Voltage
Voltage [V]
0 2 4 6 8 10 12 14 16 18 20
I mea
s -
I fit [
mA
]
-0.06
-0.04
-0.02
0.00
0.02
0.04Fig. 4: Modified Current vs Voltage
V
mA 0.001|
bestm
V
mA 0.0004|
maxm
V-0.001|bestc
V-0.011|minc
V-0.010|maxc
V
mA 0.0025|
minm
Using those results…
Using the modified fit result to correct the initial best fit line:
mS 0.000121
mS) 0.0025 mS 0.0004-
n
(|min
|max mm
m
Calculating the uncertainties, where n is the #data points:
mA 0.00121
mA) (-0.011-mA 0.010
n
minmax ccc
mS 0.0001)0.3440 (m
mA 0.001)-0.001(c
mA -0.001mA 0.001 mA 0 |bestbestfit ccc
Make sure these values haven’t
been rounded in a way that
compromises the accuracy
mS 0.344mS 0.001 mS 0.345 |bestbestfit mmm
Questions ?
