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Macro shots of ebay greenie modules

ryansoh3

ryansoh3

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Feb 23, 2012
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I just thought it might be interesting to share some macro photos of the "5mW" ebay green laser pointer pens, and I hope you enjoy them!


Note these don't have pots. :mad:

cxc7.jpg


i4j.jpg


26.JPG


pct9.jpg


9z2a.jpg


jvwa.jpg


Thanks for looking!
 





I'm surprised it uses a TL431 for the voltage reference which is a little more expensive (and better performing) than zener diodes which I've seen in other pens.

R2+R6 and R3+R5 are the voltage dividers. varying those will change the set current.
 
R2 is 20k, R5 is 1.2K. R3 and R6 are in a weird code so I don't know their values but to change the current,

You decrease the value of R2 or R6 or increase the value of R3 or R5 to raise the current.

You might be able to get away by shorting R6 but I don't know how much it will raise the current as I have no idea of its (and R3's) value.
 
Thanks. After some research, I found out that R3 = 383 Ohms and R6 = 8.25 kOhms. (EIA-96 notation)

I'll do some messing around, thanks for your help! +1

Edit: Nope, won't let me. haha
 
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According to some quick calculation, the current is set fixed to 402mA. It can go up/down a few % due to parts tolerance.

If you replace R6 with a short and leave everything else alone, current will be at 556mA.

If you replace R3 with a short and leave everything else alone, current will be at 309mA.

If you short both R3 and R6, current will be 422mA.
 
djQUAN said:
According to some quick calculation, the current is set fixed to 402mA. It can go up/down a few % due to parts tolerance.

If you replace R6 with a short and leave everything else alone, current will be at 556mA.

If you replace R3 with a short and leave everything else alone, current will be at 309mA.

If you short both R3 and R6, current will be 422mA.
Click to expand...

This is way to much for a 5mW greenie and for the small R4 shunt resistor as well. My 100mW greenie is driven at 450mA and my shunt resistor is huge compared to this one but is getting damn hot at this current.

However this shunt resistor has 33 Ohm. In my greenies they are way lower rated. So this shunt is also limiting the current.

The 808nm laserdiodes used in 5mW greenies are rated at 300mW max and being driven with less than 300mA.
 
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Keep in mind most greenies are overspec. :)

The shunt resistor is "R330" which is 0.33 ohm. "R" represents the decimal point. P=IR so P = 0.402 * 0.33 so power dissipation in the shunt resistor is about 0.13W which I think it can handle.

Current can also drop if the op amp cannot drive the pass transistor (2SD882) hard enough to carry the current. The calculations are based solely on the reference voltage and shunt resistor values. ;)
 
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Ahh ok. Didn´t care about the "R" :) But well. The wattage of the shunt will be still to small for about 400mA and this current would probably also kill a 300mW 808nm diode as well.
 
A 1206 size resistor (if I remember correctly) is rated at 1/4W or 0.25W. 0.13W dissipation is within that. It will get warm but won't burn up. I'm not sure about the IR diode though. But we know that laser diodes are rated conservatively as a lot of folks here overdrive their laser diodes. ;)
 
djQUAN said:
A 1206 size resistor (if I remember correctly) is rated at 1/4W or 0.25W. 0.13W dissipation is within that. It will get warm but won't burn up. I'm not sure about the IR diode though. But we know that laser diodes are rated conservatively as a lot of folks here overdrive their laser diodes. ;)
Click to expand...

About the resistor itself you are right but must also take into account there is no heat transfer inside a host and the air surrounding the driver and the resistor acts like a isulator. Therefore they are mostly using 0.5W shunt resistors, which still are getting pretty hot. So driving them without a host would be ok at such currents, but inside a host they will get hot very quickly resulting in very short on/off cycles.

Well at least the small 808nm laserdiodes are very anal about the max. current to my experience. They are not the same like a 1W 808nm or the famous 445nm diodes you can push way above the spec.
 
Hey man P=IV and V=IR, therefore P = (I^2)R

But anyways, it should handle it just fine. Its not like it will be on for very long. I would be worried about the 808nm diode not the shunt resistor haha.
 
benmwv said:
Hey man P=IV and V=IR, therefore P = (I^2)R

But anyways, it should handle it just fine. Its not like it will be on for very long. I would be worried about the 808nm diode not the shunt resistor haha.
Click to expand...

Lol missed that one. Been doing a bunch of stuff I incorrectly remembered ohms law. :wtf: +1 for the refresher course. Either way, the power dissipation is now 0.05W with the correct formula. Well within the ratings.
 
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