Improving The Measurement Accuracy - YOKOGAWA WT200 User Manual

3.4

Improving the Measurement Accuracy

3-6
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
voltage measurement circuit (i
measurement is i , i
L V
circuit is approximately 2 MΩ. For a 600-V input signal, i
(600 V/2 MΩ). If the load current i
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 = 0.1 mA (200 V/ 2 MΩ). The effect on the measurement accuracy
V
is 0.001% (0.1 mA/10 A) in this case.
SOURCE
±
V
A
±
Input terminal
The following figure shows the relationship between the voltage and current that leads
to 0.1%, 0.01%, and 0.001% errors.
Effect or 0.1%
Effect of 0.01%
600
500
400
Less effect
300
200
100
0
0 2 4
300 mA 3 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
of the current measurement circuit (e
resistance of the current measurement circuit is approximately 6 mΩ. If the load
resistance is 600 Ω, for example, the effect on the measurement accuracy is
approximately 0.001% (6 mΩ/600 Ω).
SOURCE
V
±
±
WT200
) and the current that flows through the
L
). Since the current flowing through the circuit under
V
is the error. The input resistance of the voltage measurement
is greater than or equal to 3 A (load resistance is
L
LOAD
SOURCE
6 8 10 12
Measured current (A)
). The voltage drop e
A
LOAD
e
L
A
e
A
is approximately 0.3 mA
V
V
i
V
±
±
A
WT200
Effect of 0.001%
14 16 18 20
) and the voltage drop
L
is the error. The input
A
LOAD
i
L
IM 253421-01E