Appendix; Appendix 1 How To Make Accurate Measurements - YOKOGAWA WT332E User Manual

Appendix

Appendix 1 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 this instrument is approximately
2 MΩ. If the input voltage is 600 V, i
or more (the load resistance is 200 Ω or less), the effect of i
or less. If the input voltage is 100 V and the current is 5 A, i
i on the measurement accuracy is 0.001% (0.05 mA/5 A).
V
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.
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
For example, the input resistance of the current measurement circuit of the WT332E/WT333E is
approximately 6 mΩ. If the load resistance is 600 Ω, the effect on the measurement accuracy is
approximately 0.001% (6 mΩ/600 Ω). For the input resistances of the WT310E and WT310EH, see
chapter 7.
SOURCE
IM WT310E-02EN
is the current flowing through the voltage measurement circuit.
V
is approximately 0.3 mA (600 V/2 MΩ). If the load current i
V
V
LOAD
U
±
C
±
I
0.01% effect
Smaller effect
3 5 10 15 20
Measured current (A)
V
U LOAD
±
± C
I
and i . i is the current flowing through the load of
L V L
, only i
L
on the measurement accuracy is 0.01%
V
= 0.05 mA (100 V/2 MΩ), so the effect of
V
SOURCE
i
V
±
C
This instrument
0.001% effect
0.0001% effect
25 30 35 40
and e . e
L I
reduces measurement accuracy.
I
SOURCE
V
±
±
This instrument
reduces measurement
V
is 3 A
L
V
LOAD
±
i
L
is the load voltage, and
L
LOAD
e
L
C
e
I
App-1
1
2
3
4
5
6
7
App
Index