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# How to calculate resistor size for lasers in parallel - for dummies Answered

I have four laser modules taken from laser pointers. Normally each module runs off of three hearing aid batteries in series, 1.5 volts per battery 4.5 volts total. I want to run all four laser modules off of two D size batteries which are 1.5 volts each 3 volts total in series.The problem is they will not light up when connected to the two D size batteries in series ?  When I use a power supply they will light up even at only 1.5 volts  ( although dim ) up to 5.5 volts,after that they become permanently dim, with the brightest at 3 volts. It seems like I have to limit the amps not necessarily the volts. Each module has its own resistor  - the # on it is 680. When I measure the resistance it shows 68 ohms. What do I need to do to get this to work ? There must be some way to simply figure this out.... By the way I have about 40 more laser modules and I might like to use them all at once to make say..... an unquestionably visible turn signal bulb for my truck or maybe a very illuminating flood light for my backyard .... If I can figure this out the possibilities are endless !

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## Discussions

You need to get a minimum voltage on the diode to get it to strike, 3V, even with an appropriate resistor, isn't enough. Does the same laser still light on the 4.4 V ?

Yes, you've destroyed them. Laser diodes are JUST like LEDs, in that they have a very low resistance and need current limiting to protect them.

Put the units on their original, 4.5 volt supply and measure the voltage drop across the 68 ohm resistor. I= V/R, so you can deduce the running current, and the forward volt drop, since V supply - Vdropped = Vlaser.

Now, take your new supply voltage (3V) and R= (Vsupply-Vdiode)/ I

Using lasers as "light bulbs" is likely to get you thrown in prison

Steve

Thanks for making it simple.... I did the calculations with 3 hearing aid batteries slightly decharged ( 3.96 V ) when I measured across the resister I got a voltage drop of 1 , when I used a set of slightly newer batteries ( 4.40 V ) I got a voltage drop of 1.26 ? If I have 3 Volts - 1Vdiode = 2 ohm resister? I was curious so I put 2 D batteries together, connected them to the laser ( with its original 68 ohm resistor ) and the voltage drop was 0 the laser did not work. I even bypassed the resistor and the laser DIDN'T light up or burn out. I tried the same with 4 AA batteries No light and did not burn out ? So how come with 3 hearing aid batteries the laser will light but nothing with the D or AA batteries. Shouldn't I at least be able to make the laser light or burn it out with the larger batteries? As for Prison -Your right while my neighbors are used to my projects I'm not sure the rest of the world would be ........