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If the device requires 2.28 watts at 4.5 volts, then its current draw is 0.507 amperes. (Watts = volts times amperes) If you want to drop 0.5 volts at 0.507 amperes, you need a resistance of 0.987 ohms. (Volts = amperes times resistance)

One solution is to put a 0.987 ohm resistor in series with the device. That resistor would need to handle 0.253 watts, so a half watt resistor would be minimum.

Problem is, this depends on constant power demand by the device. If the power goes up or down, probably down, then the voltage drop across the resistor will change.

Another problem is that the closest standard resistor value to 0.987 is 1 ohm, and that would deliver slightly less than 4.5 volts to the device.

Another solution would be to place a diode in series with the device. The diode would drop about 0.7 volts, giving the device about 4.3 volts. Of course, the diode would need to be rated for 0.507 amperes and one half watt, but the output voltage would be more stable.

A regulator is probably not going to work, because you don't have enough headroom between 5.0 volts and 4.5 volts to do it without a complex design.

Best solution: Get a 4.5 volt power pack.

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Q: I have a 120v ac to 5v dc transformer and want to decrease the voltage to 4.5v dc. The device being used is rated 4.5 v 2.28 watts.?
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