Product specifications

Page 66 of 146 198-1600-00 Rev F
Loads
The load should be a resistor or set of resistors of approximately twice the required wattage to
prevent injury to personnel or equipment and to prevent significant temperature drift. The resistor
size is calculated so both the current and voltage can be calibrated with the same load. This can
be done because the calibration for the high end values takes place at 95% of rating.
EXAMPLE LOAD CALCULATION:
Power supply module rating is 24-750 or 24 Volts at 750 Amps.
The voltage developed across load should be approximately 24 volts at 750 amps current. To
determine the resistor value use Ohms law:
24Volts / 750Amps = 0.032 ohms
To determine the wattage use:
24 volts x 750 amps = 18kW
To calibrate the unit it will be necessary to connect a load with a value of 0.032 ohms and 36000
watts (twice calculated wattage).
Optional Load Method (lower power resistors, 18kW vs. 36kW)
This method uses 2 different load values. When chosen properly the same set of resistors
can be rewired to work for both current and voltage calibration.
When calibrating the current settings, a value should be chosen which allows
approximately 50% of the voltage at the rated current of the power supply.
When calibrating the voltage settings, the value should allow for 50% of the current at the
rated voltage of the power supply.
Example Load Calculations:
Power supply module rating is 24-750 or 24 Volts at 750 Amps.
Current Calibration Load: The voltage developed across the load should be
approximately 12 volts at 750 amps average current. To determine the load resistance
value use Ohms law:
12 Volts / 750 Amps = 0.016 ohms
To determine the wattage use:
12 Volts x 750 Amps = 9000 watts
To calibrate the current settings it will be necessary to connect a load with a resistance
value of 0.016 ohms and 18000 watts (twice calculated wattage).