voltage reduction

[Woodsman]

I have a laser diode module that normally operates from two AAA batteries that produce a measured 1.27 mA at 3 V and I want to run it off of a 7-cell NiMH battery pack that produces a measured 5.3 mA at 9.56 V at full charge. (It's for an integral aiming device for an airsoft gun, in case you're wondering.) Given that I = V/R then 7k ohms of resistance should give me the proper amperage, but how do I reduce the voltage to the necessary level?

Dan

 

[ericgibbs]

hi woodsman,
To run your laser from 9v battery rather than a 3v battery, providing the
current the laser draws from the battery is 1.27mA.[this seems low ???]

Then you have to drop 6v across a series resistor, so 6/0.00127 = 4724 ohms
the nearest preferred value is 4700R.

Place a 4K7 resistor in series with battery positive to the laser positive input.

Regards
EricG

Note: dont assume that AA batteries give 1.5v , alkaline will only give 1.2v
also check your 9v battery, [ PP3 type] for voltage when you have got your resistor connected.

 

[audioguru]

A small laser pointer operates at about 20mA to 60mA. The internal resistance of its button battery cells limits the current. Your current-measuring meter also limited the current so the laser probably didn't work.

Maybe your "laser" is realy just a focussed LED.

AAA cells have double the capacity of the AAAA cells inside a 9V battery. You will be throwing away 6V of battery power.

A 9V alkaline battery starts at nearly 10V when its load current is low then quickly drops to 8V. Then its voltage continues dropping when it is loaded.