I recently bought a bag of 100 5mw 4.5V laser diodes (the spec said they have "driver: out"). My idea was to connect two and two of these in series to get something that needs 9V, and then connect about 70 of these diodes in parallel and run it off a 9V battery.
This didn't work as expected.
Just using 6-8 diodes, the light was already getting visibly dimmer. I figured there could be a short somewhere, but it seemed everything was in order. The 9V battery was used, but only for driving a few LEDs for a few hours.
So I decided to try something else. By using a battery pack of 6 1.5V batteries, the laser diodes were shining at full force again. I tried using about 20 of the diodes in the same setup as above, and still everything seemed fine.
Thus it seems I'm stuck with AA batteries for my setup.
Anyway, my questions are:
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Why is this happening?
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Are there not enough amperes to drive the lasers?
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Why is the battery pack able to drive the lasers when the 9V battery isn't?
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Is there any way I can calculate the approximate number of lasers the batteries can drive?
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Will an amplifier or extra driver be able to fix this so I can still use the 9V battery?
Bonus question: If the answer to 5 is "yes", can I find such an item on eBay? I just need an example of what to look for since I'm just a hobby electrician.
EDIT:
Since you good folks asked for it, here's all the data I have regarding the laser diodes:
http://www.ebay.com/itm/110887381932
EDIT 2:
A brand new 9V battery was able to drive the diode arrangement after all. There appears to be a small resistor on the diode board, so I took my chances that the diodes would not blow just using them as described above (and I had spent way too much time on this Halloween cannon already to bother soldering 40 or so more things into the setup). Last night I used the cannon for a full entire night, and the lasers were still jolly good by the end of the evening.
Plus the cannon looked damn good in a dark nightclub with some artificial fog 🙂
Best Answer
First thing to say is that whenever using laser diodes (or LEDs for that matter) you need to limit the current into the device. You can't rely on the battery voltage for producing exactly 4.5V x 2 because if it produces 4.6V x 2 then you might find that the current into the laser diodes (2 in series) would go from 50mA to over 100mA or much higher leading to destruction of the laser diode.
Like LEDs, laser diodes have a typical forward volt drop (4.5V in your case) at the rated current (say 50mA) and some may be only 4.4V and if you had a bunch of these they may well have died and you'd be regretting spending the money so use a series resistor or an active current control mechanism.
Without seeing the data sheet I can imagine that full 5mW laser output is achieved with round about 50mA of current and that your battery arrangement has just enough internal series resistance to limit the current to safe levels.
If a pair of lasers in series took 50mA then 4 pairs would be taking 200mA and already a PP3 battery would start to reduce its output voltage and dim the lasers. Look at the discharge curve for an energizer 522 9V battery: -
With 200mA load after 2 hours the voltage will be at 4.8V. You could estimate that after 40 minutes it would be 7.6V and well below the 2 x 4.5V needed for the laser diode. You could estimate that each minute that passes the battery terminal voltage drops 35mV and after 2 hours this drop is 4.2V hence only 4.8V left at the terminals!!
In fact, if the laser diodes were exactly 4.5V then after a few minutes there would be not enough voltage to activate them. And this was a battery that had been used to drive a few LEDs for a few hours!
The battery pack made from several 1.5V batteries can generate much more current for a longer time period. All the same, if I was using a battery pack I'd make it 6V and have individual current limit resistors for lasers i.e. don't wire in series.
To get more out of the 9V battery (and you won't get a whole lot more) you could drive each one individually from the 9V with a current limit resistor. You need to find out what the forward current of the laser is. I can guess at 50mA and this would mean a dropper resistor would be: -
\$\dfrac{9V - 4.5V}{50mA} = 90 ohms\$