App-21
IM 760201-01E
Appendix
3
2
1
4
5
6
7
8
9
10
11
12
13
14
App
Index
Appendix 4
How to Make Accurate Measurements
Effects of Power Loss
By wiring a circuit to match the load, you can minimize the effects of power loss on
measurement accuracy. We will discuss the wiring of the current source and load
resistance shown in the figure below.
When the Measured Current Is Relatively Large
Connect the voltage measurement circuit between the current measurement circuit and
the load. The current measurement circuit measures the sum of i
L
and i
V
. i
L
is the current
flowing through the load of the circuit under measurement, and i
V
is the current flowing
through the voltage measurement circuit. Because the current flowing through the circuit
under measurement is i
L
, only i
V
reduces measurement accuracy. The input resistance of
the voltage measurement circuit of the WT500 is approximately 2 M
Ω
. If the input voltage
is 1000 V, i
V
is approximately 0.5 mA (1000 V/2 M
Ω
). If the load current i
L
, is 5 A or more
(load resistance is 200
Ω
or less), the effect of i
V
on the measurement accuracy is 0.01%
or less. If the input voltage is 100 V and the current is 5 A, iV = 0.05 mA (100 V/2 MΩ), so
the effect of i
V
on the measurement accuracy is 0.001% (0.05 mA/5 A).
SOURCE
LOAD
WT500
U
I
i
V
i
L
LOAD
SOURCE
U
I
I
U
±
±
±
±
As a reference, the relationships between the voltages and currents that produce effects
of 0.01%, 0.001%, and 0.0001% are shown in the figure below.
1000
800
600
400
200
0
0
5
10
15
20
25
30
35
40
0.001% effect
0.0001% effect
Measured current (A)
Measured voltage (V)
0.01% effect
Smaller effect
When the Measured Current Is Relatively Small
Connect the current measurement circuit between the voltage measurement circuit and
the load. In this case, the voltage measurement circuit measures the sum of e
L
and
e
I
. e
L
is the load voltage, and e
I
is the voltage drop across the current measurement
circuit. Only e
I
reduces measurement accuracy. The input resistance of the current
measurement circuit of the WT500 is approximately 5 m
Ω
. If the load resistance is 1 k
Ω
,
the effect of e
I
on the measurement accuracy is approximately 0.0005% (5 m
Ω
/1 k
Ω
).
SOURCE
LOAD
U
I
WT500
e
L
e
I
SOURCE
LOAD
U
I
I
U
±
±
±
±
