HP 4284A - Page 344

Example of C-D
Accuracy Calculation
Measurement Conditions
Frequency : 1 kHz
C measured : 100 nF
Test Signal Voltage : 1V
rms
Integration Time : MEDIUM
Cable Length : 0m
Then ;
A=0 : 05
j Z
m
j =
1
2  2 1 2 10
3
2 100 2 10
0 9
= 1590 []
K
a
=
1 2 10
0 3
1590

1+
200
1000

=7 : 5 2 10
0 7
K
b
= 1590 2 1 2 10
0 9

1+
70
1000

=1 : 70 2 10
0 6
K
c
=0
Therefore ;
C
accuracy
= 6
2
0 : 05 +
0
7 : 5 2 10
0 7
+1 : 70 2 10
0 6
1
2 100
3
6 0 : 05 [%]
D
accuracy
= 6
0 : 05
100
= 6 0 : 0005
General Information 9-11