I have a device that is usually powered by soldered on 1.2v 80maH NiMH coin cell. The device is difficult to take apart and replace the battery, so I want to modify it to receive power from an external source. The source is 6v which I've been told the device will operate on, but nobody has done long term testing to see if the device will burn out eventually, so I want to drop the incoming voltage to about 1.5v to be safe. On an 80maH battery the device can stay running for over 12hrs so its current draw is quite small. I've done some testing with an LM317 with an R1 of 220ohms, and an R2 of 47ohms (which should get me 1.5v roughly) and with my multimeter with or without the device connected I'm seeing 4.5v. My understanding is that the LM317 requires a 10maH load to work correctly, so apparently my device isn't creating enough of a load.
My question is, how do I create a large enough load to get the correct voltage?
I read somewhere that if I divide my resistor values by 10 and use those new values it will drop the voltage to what I need without a load. Tried this and the voltage does come close at 1.8v, but it doesn't power my device.
This device is less than 3/4" square and quite lite and I'd like to keep the package small because it will be held in place by velcro. The LM317 and the 2 1/4 watt resistors weight much less than the battery itself and current housing (that I wont be using) gives me some weight to work with (1 or 2 ounces maybe) so I can add some weight I just don't want to add anything to bulky.
Not great at this kind of stuff so any help is appreciated.
My question is, how do I create a large enough load to get the correct voltage?
I read somewhere that if I divide my resistor values by 10 and use those new values it will drop the voltage to what I need without a load. Tried this and the voltage does come close at 1.8v, but it doesn't power my device.
This device is less than 3/4" square and quite lite and I'd like to keep the package small because it will be held in place by velcro. The LM317 and the 2 1/4 watt resistors weight much less than the battery itself and current housing (that I wont be using) gives me some weight to work with (1 or 2 ounces maybe) so I can add some weight I just don't want to add anything to bulky.
Not great at this kind of stuff so any help is appreciated.