Appx.12
Assembling Switching Unit for Objects to be Measured
Example of terminal for adding relay and resistor
(Eective screw depth is 3 mm)
Metal (nut)
(Used as guard shown
in gure on previous
page)
TEFLON
(Insulator)
Metal terminal
(Used as metal terminal shown
in gure above)
Thread mount TEFLON
terminal block FN-1-1
manufactured by Mac-Eight
Current-limiting resistor
Resistance value
Decide resistance values of current-limiting resistor as follows, the resistance value can be determined in the
following manner:
(1) Resistors shall be low enough to be negligible relative to the insulation resistance of objects to be
measured; however, be as high as possible.
(2) The value equivalent to 20 times the time constant with respect to the electrostatic capacity component of
objects to be measured shall be short enough relative to the measurement time.
For example, if you want to measure a insulation resistance of 10,000 M
Ω
with a electrostatic capacitance of
1,000 pF for 5 sec
Considering Item (1), the current limit resistance is calculated from the following expression:
10 000×106 ÷ 1000 = Approximately 10 M
Ω
The measurement time is calculated, as described in Item (2), from the following expression:
10 M
Ω
× 1,000 pF × 20 = 0.02 sec
Because this time is considered to be short enough compared to 5 seconds, a resistance of 10 M
Ω
is
acceptable as the current-limiting resistance.
Selecting resistor
Refer to the specications of current limiting-resistors to check if the maximum working voltages of them are
higher that a set voltage of the instrument.
In addition, resistances with power ratings that are high enough relative to a current owing even if the
terminals of an object to be measured is short-circuited with each other.
For example, for a 10 M
Ω
current-limiting resistor and a set voltage of 250 V, if the terminals of the object to
be measured are short-circuited with each other, the load power is calculated from the following expression:
250 V × 250 V ÷ 10 M
Ω
= 0.00625 W
Considering heat generated by resistance, in general, power ratings 5 times or more, ideally 10 times of the
load power is required.
(Example: KOA high voltage high resistance thick lm resistor GS1/2 10 M
Ω
)
Determine a current-limiting resistor so as to limit the current owing into the instrument to the current rating.
Test using a measurement voltage of 1000 V or less: 50 mA or less
Test using a measurement voltage exceeding 1000 V: 1.8 mA of less
The current is calculated from the following expression:
250 V ÷ 10 M
Ω
= 25 μA
The resistance value, which meets the condition above, is proved to be acceptable.
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