3-6
IM 253421-01E
3.4
Improving the Measurement Accuracy
By wiring the circuit to match the load, you can minimize the effect of the power loss on
the measurement accuracy. We will consider a circuit consisting of a current source
(SOURCE) and load resistance (LOAD) below.
• When the measurement current is relatively large
In this case, the voltage measurement circuit is connected to the load side. The
current measurement circuit measures the sum of the current that flows through the
load of the circuit under measurement (i
L
) and the current that flows through the
voltage measurement circuit (i
V
). Since the current flowing through the circuit under
measurement is i
L
, i
V
is the error. The input resistance of the voltage measurement
circuit is approximately 2 M
Ω
. For a 600-V input signal, i
V
is approximately 0.3 mA
(600 V/2 M
Ω
). If the load current i
L
is greater than or equal to 3 A (load resistance is
200
Ω
or less), the effect of the voltage measurement circuit on the measurement
accuracy is less than or equal to 0.01%. As another example, if the input signal is
200 V and 10 A, i
V
= 0.1 mA (200 V/ 2 M
Ω
). The effect on the measurement accuracy
is 0.001% (0.1 mA/10 A) in this case.
SOURCE
LOAD
V
A
i
V
i
L
SOURCE
LOAD
A
V
Input terminal
WT200
±
±
±
±
The following figure shows the relationship between the voltage and current that leads
to 0.1%, 0.01%, and 0.001% errors.
600
500
400
300
200
100
0
0
2
4
6
8
10
12
14
16
18
20
300 mA 3 A
Effect or 0.1%
Effect of 0.01%
Effect of 0.001%
Measured voltage (V)
Less effect
Measured current (A)
• When the measurement current is relatively small
Connect the current measurement circuit to the load side. In this case, the voltage
measurement circuit measures the sum of the load voltage (e
L
) and the voltage drop
of the current measurement circuit (e
A
). The voltage drop e
A
is the error. The input
resistance of the current measurement circuit is approximately 6 m
Ω
.
If the load
resistance is 600
Ω
, for example, the effect on the measurement accuracy is
approximately 0.001% (6 m
Ω
/600
Ω
).
SOURCE
LOAD
V
A
e
L
e
A
WT200
±
±
