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tiny 635nm "pointer"

Bluefan

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I had a 635nm diode of 5mW laying around, so I decided to make something small out of it. Nothing fancy, just a few batteries, a switch and a resistor, but completely put into an aixiz module. Doesn't get much smaller than this, even the bullet style pointer look large compared to it. Unfortunately the batteries are nearly empty and I don't have replacements yet, so no beam shots.
 

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Kmor2004

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This is not something a guy should say but...That is one cute little pointer!
 

Kmor2004

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I'll try to explain it (just to make myself feel good)->

Batterys output a voltage and a current we all know this.

However sometimes batteries have current spikes, kind of like driving a car, not paying attention to the speed and you notice your going 30 in 45 so you pnch it to get back upto speed, in the batterys case it just guns and sometimes it goes a little bit over board.

The lasers we use can only handle so much current or else they become worthless or realy expensive LED's or inexpensive colors, we all know (or at least should).

So then if the battery guns it to say 100mA and (through testing) we know the ABSOLUTE MAX current a laser can handle is 90mA, then I think you can guees what 100mA is gonna do.

Now the other thing is lets say you have a 10440 (AAA sized battery) its MAX out put current is 350mA (roughly) and lets say you use a 445 blue LAser which can handle ~1000mA with the help of an inrush resistor (look inrush up in wikipedia) the initial current peak is limited plus the average current is limited, so then lets just say the 10440 plus a resistor gives 500mA starup spike which is well within the 445's current range, and 300mA continous the laser SHOULD be fine.

One last however if you tried this with a 18650 which outputs ~3000mA... even if you did use a resistor the laser would still die.

Hopefully that is understandable, if I forgot something I know someone will throw it out there or correct me.
 

lasersbee

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I'll try to explain it (just to make myself feel good)->

Batterys output a voltage and a current we all know this.

However sometimes batteries have current spikes, kind of like driving a car, not paying attention to the speed and you notice your going 30 in 45 so you pnch it to get back upto speed, in the batterys case it just guns and sometimes it goes a little bit over board.

The lasers we use can only handle so much current or else they become worthless or realy expensive LED's or inexpensive colors, we all know (or at least should).

So then if the battery guns it to say 100mA and (through testing) we know the ABSOLUTE MAX current a laser can handle is 90mA, then I think you can guees what 100mA is gonna do.

Now the other thing is lets say you have a 10440 (AAA sized battery) its MAX out put current is 350mA (roughly) and lets say you use a 445 blue LAser which can handle ~1000mA with the help of an inrush resistor (look inrush up in wikipedia) the initial current peak is limited plus the average current is limited, so then lets just say the 10440 plus a resistor gives 500mA starup spike which is well within the 445's current range, and 300mA continous the laser SHOULD be fine.

One last however if you tried this with a 18650 which outputs ~3000mA... even if you did use a resistor the laser would still die.

Hopefully that is understandable, if I forgot something I know someone will throw it out there or correct me.
It has nothing to do with surge... IMO

If you use 3 Hearing Aid 1.5V (4.5V @ 50mA) cells or a 18650
(4.2V @3000mA) cell... the same scenario plays out... It is the
total resistance of a circuit and the Battery Voltage that decides
the current flow in that circuit...

A Laser Diode is a Current Device... It has a certain internal
resistance (that can change with internal heat).

If a battery of enough voltage is used to drive the LD... the LD's
internal resistance and the battery's voltage will decide the current
that will be drawn by that internal resistance.
If the current is higher than the allowable specs if the LD it will
damage the LD.

To reduce the current draw you can use an external series connected
resistor to increase the total resistance in the circuit to reduce
the total current draw from the battery. This is called a Current
Limiting Resistor.


The problem starts when the LD starts to heat up and the internal
resistance starts to decrease... this in turn increases the current
through the LD and could cause it to damage or blow if the battery
voltage stays constant.

For this reason we use a Current Regulated Laser Driver that solves
this problem...

In the case of the 5mW Laser above the internal resistance will
probably not change much through heat buildup.... and a Current
Limiting Resistor would be just fine...


Jerry
 
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jcranmer

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I would love to see the inside of that one if you can get any pics. What size resistor did you end up using?
 

Kmor2004

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Like I said somebody with a little bit more Laser knowledge would correct me, At least I think I was in the ball parks parking lot with my explaination.
 




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