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Adjusting LED brightness on Side-A of N8102 (20ma current sources)
For certain applications it may be desirable to have an LED or LEDs connected to this board operate at less than full brightness (less than 20 ma current). For this reason, solder pads are included (see note 7 on board layout) for placement of a resistor to compensate current source output down from 20 ma. Fig. 1 below shows a resistor placed on these solder pads).
Figure 1
To calculate the resistor value, divide the LED device voltage by the milliamps (ma) you wish to reduce. This will give the value in ohms of the resistor.
Example: Reduce the current output from 20 ma to 10 ma with a yellow LED having a device voltage of 2.0. Divide 2.0 by 0.010 (10 ma). The result equals 200. Use a 200-ohm resistor. To reduce current through a series of LEDs, add the device voltages in the series together then divide by the amount of ma current reduction.
© 2008 Ngineering