I'm working on a small project that needs 1.5v DC but only have in my possession a wall wart that pumps out 8.25v DC. Is there a way to (simply) reduce the voltage of the wart so that I don't fry the electronics that I'm working with? I tried linking up resistors and that changed nothing.
The circuit (a small radio) needs 1.5v / 20ma. The wart outputs
It's probably simpler to get a different wallwart. You need a regulator to bring down the voltage and keep it steady. However it will dissipate about 140mW. Or you can use one of the many switching regulator modules.
What do you mean you tried "linking up resistors?" A 330 ohm 1/2 watt resistor in series with the wall wart will probably work. If the current is as you say, it will drop the voltage to 1.5 volts. Place three diodes in series across the 1.5 volt point, forward biased, to common. This will insure that the voltage never goes above about 2 volts no matter what the radio draws. It should protect it.
This scheme has very poor regulation but I doubt the radio will care. It should work and is about as simple as you can do. Of course, regulators with more advanced electronics would be better but probably not necessary.
BTW, measure the wall wart voltage. If it is rated at 8.25 volts chances are that it will put out 12 or more volts unloaded assuming it is un-regulated. Most are. That would require a different resistor.
Use a ~300 ohm R in series from the (+) lead and 2 1n4001 or similar, diodes in series to clamp the output of the R going to the radio. The cathode side will go to the (-) side of the wallwart!
The diode's will break down to ~ 1.5 volts and release as the drain from the radio increases there by, regulating it to some point.
It's a good idea to check the voltage with a meter before connecting the radio to it.
easiest way is to just use a dry cell. ZN414/MK484?
use 40 ohms in series with 10 ohms take the 1.5 in parallel with the 10 ohms. 8.25V ---[40]---+-- +1.5V @ 20ma | [10] | 0V -----------+-- 0V (40 could be 22 in series with 18)
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.