Active Load

Implementation

Testing a power supply requires loading and overloading each output (on a multiple output unit). This tests the operation of the regulators and protection circuitry. Using fixed resistors as a load is adequate for general operation but requires additional loading to test over current protection.

Datel recently manufactured a programmable load unit (DTL2A-LC) that is specifically design as a current load for testing batteries and power supplies. It is a 12 bit programmable current sink capable of dissipating 100Watts. The units are isolated and serially programmed making them ideal for testing our multi-output power supplies

To handle each type of power supply 16 devices will be available to connect as needed to provide the current draw for testing. The control unit for these devices will be handled by a PC over a parallel port. Software behind an Excel spreadsheet will provide a simple user interface.

There are two methods that can be employed to use the Datel active load.

1. One method is to supplement a fixed load with the active load, a parallel combination, such that the active portion simply adds to the fixed load current. One drawback of this arrangement is that there is a minimum current that cannot go to zero. Additionally when the voltage being loaded is above 5 volts the active load device must be de-rated to prevent over powering the unit. The maximum power the active device can dissipate is 100 watts. The maximum current is 20Amps.
2. The second method used to load a power supply is the series connection. Attaching the active device in series with an additional resistive load will allow full control of the load current from zero to maximum. The disadvantage to this method relates to the maximum current desired for the load. Since the device can only handle 20 amps maximum whenever the current exceeds that value an additional device must the used. Devices must be added in 20-ampere intervals.

When the load power is less than 100 Watts the active load can be used directly without any additional resistors. It is, however, suggested that a resistor is added to limit the maximum current flow for loads that are small in value, such as currents less than 5 amps.

The method used, series or parallel, will determine the configuration of the resistors required for the load according to the power supply being tested.

Select a fixed resistor for each power supply output using the following criteria.
• Subtract 2 volts from the output. This will provide the device with some working voltage.
• Divide that voltage by the maximum current the supply can provide. Use the value for the resistance.
• Calculate the total power dissipated by the resistor. v*i=watts

For example:	5V @ 20A 

		Subtract 1 volt and divide by the current 4/20=0.2Ohms

		4*20=80watts

or	

		13V @ 20A

		12/20=0.6Ohms

		12*20=240watts

• If the output is greater than 20 amps calculate R the same way but use two or more devices to handle the current.

A spreadsheet was created to aid the resistor selection process. If four 0.15 Ohm resistors are added in series with the device and the proper resistor is tapped supply outputs from (2-15.5)volts can be loaded over the full 20Amp range. The addition of these resistors increases the power handling capability of the assembly raising the power from 1600 to 4800 watts.