I am trying to external power a keychain laser.

Using a 4.8 volt power supply, once I power up the laser it does not shine near as bright even if I switch back to the 3 1.5v watch batteries. It's like the laser is now faulty. Tried this on 2 lasers so far.

Does the laser voltage have to be exactly 4.5V?

Thanks.

 

They are a current device. The current generation use the internal resistance of the batteries as far as I know (not sure). You are probably frying the device. Try measuring the current with a DVM, then match it.

 

Diode lasers are extremely sensitive to overcurrent. If they get too much current for even a very small fraction of a second, they will be permanently damaged or destroyed.

Keychain laser pointers are by necessity (competition) designed with minimal protection. The tiny batteries used might be able to put out around 100mA if there were a dead short across them; much less when the laser is operating.

Your power supply may have "ripple" voltage on it. Try connecting an electrolytic capacitor (470uF @ 16v or larger) across the output terminals and measure it again using a DMM (digital multimeter).

If you wish to try it again with another laser, you could use forward-biased diodes in series with your power supply to drop the voltage down. A common 1N4002 rectifier diode will drop about 0.6v across itself with a current of 2mA through it, 0.7v with 20mA, and around 0.8v with 100mA. Therefore, a single 1N4002 diode would reduce the supply voltage to 4.0v to 4.2v, depending upon how much current the laser diode consumes.

A better choice would be a Schottky diode like a 1N5817; these have a much lower Vf (forward voltage) start at around 0.32v. You could start off with three in series to see how well that works.

Your 4.8v supply has so little "headroom" (voltage above the batterys' voltage) that you really couldn't use a normal regulator circuit without going to something comparatively expensive, like a buck/boost DC-DC converter.

 

The laser diode operates at about 3.2V. The internal resistance of the tiny battery cells reduces the current to the operating amount of about 40mA at 3.2V.
A 4.8V power supply would fry the laser diode so quickly that it will not even flash.

 

Awesome thanks for the replies!