Need help figuring out driver :/

[Zorael]

Long story short, I have a 20mW red 658nm diode that I'm trying to use with a small driver. And I don't know where to solder what!

There are several solder surfaces, and the voltages and polarities between them vary (upon connecting a battery to the two that logically must be for DC in... I think). Power comes from a normal CR2023 3V button battery. The diode is rated for 3V and <38mA, while the driver can handle 3V and 20-40mA.

There is a potentiometer, but turning it seems to do jack squat except at a very precise angle where voltage between certain surfaces drops considerably. It can be turned around and around and around to no end, so there's no way of knowing where "zero" and "max" is.

The driver seller is Chinese, and while he does know some English the language barrier is considerable. There were no specs as to what surfaces on the driver are for what, and he isn't very good at explaining. From the item description;

Pin definition: designed pad especially for 650nm with a PD function of the driver board, you just need to solder pins on the triangle to the natural order, not the same as the other board, to cut off one foot, or staggered welding!

After a few emails I managed to get him to produce this for me;

Please see these pictures where I've labeled the surfaces. I would have shown them here using tags, but they're a bit big for that. Do feel free to ridicule my soldering! ;3 [I](note to self: buy new soldering iron)[/I]
[list][*][url=http://i.imgur.com/463Kv.jpg]Front (empty)[/url]
[*][url=http://i.imgur.com/1seaa.jpg]Back (empty)[/url]
[*][url=http://i.imgur.com/rahM7.jpg]Front (with cabling)[/url]
[*][url=http://i.imgur.com/lrbST.jpg]Back (with cabling)[/url][/list]
I [i]*assume*[/i] that surfaces [b](a)[/b] and [b](b)[/b] are the same as surfaces [b](e)[/b] [b](f)[/b]. Stop me if I'm wrong.

[list][*]The 3V battery is connected at [b](a)+[/b], [b](b)-[/b].
[*]Between [b](g)[/b] and [b](c)[/b], there is a potential of about +2.7V. If I reverse the polarity of the batteries, this potential reads at <0.3V, so I think I have the batteries right.
[*]Between [b](c)[/b] and [b](d)[/b], there is a potential of about -2V (iirc).
[*]Between [b](g)[/b] and [b](d)[/b], there is a potential of about +0.6V (iirc).
[*]There is negligible resistance (~0.3 Ohm) between [b](c)[/b] and the pot.
[*]Turning the pot, the resistance between it and [b](d)[/b] varies, from which follows that the resistance between [b](c)[/b] and [b](d)[/b] varies.[/list]

What am I missing? Where is the "triangle" to which I can solder "to the natural order"? [b](c)[/b] + [b](d)[/b] + pot? To begin with why doesn't the pot seem to be doing much at all? Broken? (I bought an extra driver so if necessary I can bin this one.)


Thanks in advance for any help whatsoever, I'm really stuck here.

 

[MadEye]

I just had the same issue with them, I also mailed them but the only thing they told me was this "blabla natural order"-stuff.

I used a case positive diode. I just had to solder it to the side with the 3 pads... just like its natural order...

.O.....O
---....---
....---
.....O

The O are your laser diodes pins and the --- are the solderpads (Ignore dots, forum doesnt let me do space on this...). The triangle is just the 3 Pins on your laser diode.

 

[Zorael]

Thanks for replying! It's actually very reassuring to know I'm not alone with this thing. :3


If we go by my camera pictures, when you say "solder it to the side with the 3 pads", do you mean the back side and its three pads (e), (f) and (g)?

Or do you mean that you tilted the driver itself sideways, and soldered two pins on the front side ((c) and (d)) and one on the back ((g))?

 

[MadEye]

Acutally, there is only one way to solder your diode to the driver... And there is only one side with 3 pads...

Diode to c, d and g on your pictures...

The pads for the battery I just tried out... Cant remember which one was + and -, sry, but the driver took no damage from trying it out^^

My problem was I wanted to know if the driver is for case negative or positive diodes, but I never got a matching reply...

One more tip: If you want to know if 2 pads are the same (a&b and e&f), just meassure the resistance between them , but you can also see they are connected through the PCB (the small holes on them)

 

[Zorael]

Thank you! Now it works. In my head I had pictured mounting the diode perpendicular to the driver board, but this makes more sense.

However, the light is very *very* weak. Confusingly so. :<

Recap:

[u]       item    voltage   current[/U]
     CR2032       3V     ?mA max
 diode spec       3V     <38mA --.___
driver spec       3V     20-40mA ----`-- checks out
  lit diode     2.6V     ?mA drain

Actual light output is easily <1mW. My weakest cat toy with its tired batteries puts this one to utter shame. I could keep it right infront of my eye for an hour without any remote worries.

Turning the potentiometer does raise resistance between the three diode pins, but there's no noticeable difference in diode brightness except for that particular angle at which the diode basically goes dark.

I'm very green to this area, but it seems to me that if I tweak the potentiometer to let the driver push out its maximum 40mA, that should max out the <38mA diode too, right?

Where did I plan wrong?

 

[AnthoT]

Thank you! Now it works. In my head I had pictured mounting the diode perpendicular to the driver board, but this makes more sense.

However, the light is very *very* weak. Confusingly so. :<

Recap:

[u]       item    voltage   current[/U]
     CR2032       3V     ?mA max
 diode spec       3V     <38mA --.___
driver spec       3V     20-40mA ----`-- checks out
  lit diode     2.6V     ?mA drain

Actual light output is easily <1mW. My weakest cat toy with its tired batteries puts this one to utter shame. I could keep it right infront of my eye for an hour without any remote worries.

Turning the potentiometer does raise resistance between the three diode pins, but there's no noticeable difference in diode brightness except for that particular angle at which the diode basically goes dark.

I'm very green to this area, but it seems to me that if I tweak the potentiometer to let the driver push out its maximum 40mA, that should max out the <38mA diode too, right?

Where did I plan wrong?

You should test the actual output of the driver if it is where it should be then you probably got a bad diode

-Anthony

 

[DTR]

Actual light output is easily <1mW. My weakest cat toy with its tired batteries puts this one to utter shame. I could keep it right infront of my eye for an hour without any remote worries.

Turning the potentiometer does raise resistance between the three diode pins, but there's no noticeable difference in diode brightness except for that particular angle at which the diode basically goes dark.

This sounds like your diode has gone LED.:cryyy: Did you test the driver on a test load first to set the current?

 

[Zorael]

:<

Did you test the driver on a test load first to set the current?

I had naïvely assumed I didn't need to worry! Seeing as the max current output of the driver was close enough to the max tolerable current of the diode, I figured the chance of it frying the diode was negligible...

To my growing dismay I don't have an ammeter; my ancient multimeter can only measure voltage and resistance, and test diodes. Supposedly it can test capacitors too but I've done that.

I have one more diode I can try, and I can use this borked setup to make sure the potentiometer is set to throttle as much as it can before soldering it all together.

Is there any way to conclusively tell if the diode is blown? Anything else I'm missing? Is the setup sound otherwise, or is the driver underpowered for this diode?

 

[DTR]

To my growing dismay I don't have an ammeter; my ancient multimeter can only measure voltage and resistance, and test diodes.

If it can measure voltage with a resolution of 1mV or even 10mV then you can test the driver with a test load. You measure mV across the resistor of the test load. 1mV=1mA.