Appendix 3 How To Make Accurate Measurements - YOKOGAWA WT1800 User Manual

Appendix 3 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 DC power supply (SOURCE) and a load resistance (LOAD)
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
the circuit under measurement, and i
Because the current flowing through the circuit under measurement is i
accuracy. The input resistance of the voltage measurement circuit of the WT1800 is approximately 2
MΩ. If the input voltage is 1000 V, i
A or more (load resistance is 200 Ω or less), the effect of i
less. If the input voltage is 100 V and the current is 5 A, i
on the measurement accuracy is 0.001% (0.05 mA/5 A).
SOURCE
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
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
e is the voltage drop across the current measurement circuit. Only e
I
The input resistance of the current measurement circuit of the WT1800 is approximately 100 mΩ for
the 5 A input terminals and approximately 2 mΩ for the 50 A input terminals. If the load resistance is 1
kΩ, the effect of e
input terminals and approximately 0.0002% (2 mΩ/1 kΩ) for the 50 A input terminals.
SOURCE
App-20
V
is approximately 0.5 mA (1000 V/2 MΩ). If the load current i
V
U
U
±
± I
I
0.01% effect
5 10 15 20
on the measurement accuracy is approximately 0.01% (100 mΩ/1 kΩ) for the 5 A
I
U
U
±
± I
I
and i . i
L V L
is the current flowing through the voltage measurement circuit.
on the measurement accuracy is 0.01% or
V
= 0.05 mA (100 V/2 MΩ), so the effect of i
V
SOURCE
LOAD
Smaller effect
25 30 35 40
Measured current (A)
SOURCE
LOAD
is the current flowing through the load of
, only i reduces measurement
L V
U
LOAD
i
V
±
± I
WT1800
0.001% effect
0.0001% effect
45 50
and e . e is the load voltage, and
L I L
reduces measurement accuracy.
I
LOAD
U
±
I
±
e
I
WT1800
is 5
L
V
i
L
e
L
IM WT1801-03EN