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Old 04-05-2008, 01:32 AM #1
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Default Questions about this laser diode

Subject - G&P 467A chinese airsoft red lasersight



Red wire is +3.27v from the main circuit board.

I pried up the little diode board to look under it. The diode has 3 pins, all sticking up through solder points on the edge of the board.

The red wire connects to Pin 2, which is also common with the case. That pin is labeled "A-LASER" on the board. I assume that means "Anode-Laser"??

All tests below were performed with the laser on.

Current through the red supply wire is 59mA. Current does not change when the pot is turned.

In the picture, the pin to the "right" of the pot is labeled "PD". With the red DMM lead on Pin 2 and the black lead on the PD pin, I get a variable voltage of +2.94 to +3.27v, depending on pot setting.

In the picture, the pin to the "left" of the pot is labeled "LD". With the red lead on Pin 2 and the black lead on the LD pin, I get a fixed voltage of +2.35, regardless of pot setting.

So it appears that the pot is causing variable voltage on the leg that is labeled PD.

Is that how it is supposed to work? Is this thing labeled correctly?

I guess I had assumed that the variable voltage would be on the LD leg, not the PD leg.

I'm assuming that this must be an N-type diode because the power supply is into Pin 2, making Pin 2 the LD Anode, and N-type is the only type where Pin 2 is the LD anode??

So. Am I on the right track, or totally derailed? ;D


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Old 04-05-2008, 09:41 PM #2
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Default Re: Questions about this laser diode

Let me try again. *

I've got a red laser i want to convert to IR. *

I measured some voltages and current with the laser on. *Then i took it apart. *

Here's the top of the board. *It has a pot. *



I arbitrarily numbered the pin-holes to help with my questions.

Here's the bottom of the board. *Same numbers (arbitrary).



The two black bits are both labeled " 1Mt ".

A red wire carried +3.27v from the big board to Pin 2 on this driver board, at 59mA. *A black wire soldered to the spring was the -leg back to the big board. *

With the laser on:
1 - I got +2.35 volt from Pin 2 to Pin 1, regardless of pot position.

2 - I got variable voltage +2.94 to +3.27 from Pin 2 to Pin 3, depending on pot position. *

My questions:

1 - based on the info i have provided, is it possible to determine what type of laser diode this is? *(N,P, or M, or some other type that I've never heard of)?

2 - If so, what type is it, and how did you determine that? *

3 - According to the labels on the top of the board, the fixed +2.35v from Pin 2 is to the LD leg. *Am I correct to assume that this is the laser diode leg? *

4 - According to the labels, the variable pot-adjusted voltage (2.94 to 3.27) is from Pin 2 to the PD leg. *Am I correct to assume that this is the photodiode leg? *

5 - Newbie question: *why is the variable pot-adjusted voltage on the PD leg? *Is this how lasers generally work? *I expected the variable voltage to be on the LD leg.

6 - Is this laser operating according to your general expectations? *59mA 3.3v supply, with the pot controlling voltage on the PD leg? *Or, does it seem "mislabeled"?

7 - OR, did I miss a measurement somehwere? *I measured current on the supply line, and I measured volage on the supply line and across both sides of the diode. *That's all i measured. *Should I have measured something else? *

If you need it, here's a pic of the diode. *I have been calling the middle pin = Pin 2. *It's common to case. The diode is 5.6mm diameter.



That's it. *I have searched and read a lot, but I haven't found any concise summary of how to spec out an unknown diode, or determine what "Type" it is.

Thanks in advance for any help you can offer. *

Bonus question for anyone who is interested: *identify each bit on the board, and tell me how the board works. *

THANKS!
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Old 04-05-2008, 11:03 PM #3
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Default Re: Questions about this laser diode

PD must be the photo diode for optical feed back too control the current.
The black things are transistors, I think.

Do you want to use the board for the IR? Maybe you need to make a new driver? I'm trying to help, but I'm so confused. What diode is on the picture, the IR?. Do you have the specs for the IR? On what diode did you mesure? What do you mean with diode types? "N, P, M" I've never heard of such a thing. :-/
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Old 04-06-2008, 07:51 AM #4
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Default Re: Questions about this laser diode

Hi thanks for the reply. *

The diode in the picture is the red diode. *All the measurements are for the red diode. *

Yes i want to use the board for the new IR laser. *

The specs for the new IR diode are 904nm 10mw, operating current *= 50 to 70mA max, forward voltage = *2.0 to 2.5 max. *

I'll have to find a way to reduce voltage. *Other than that, it looks pretty straightforward. *

Sorry for the confusion. *I thought pin specs ("type&quot were a common issue when modding lasers. *

An N-type laser diode uses Pin 2 for the laser anode, Pin 1 for the laser cathode, and Pin 3 for the photodiode anode.

P-type and M-type: *Pin 1 is the laser diode anode, pin 2 is the cathode. *

P-type - Pin 3 is the PD cathode.
M-type - Pin 3 is the PD anode. *

I'm pretty sure that my red diode is an N-type. *

But I didn't completely understand the voltage specs i got, and I'm not completely sure about the type. *

I'm asking how to spec an unknown laser to determine its type. *

I'm also asking if the pot should be behaving like mine did: *it alters voltage on the PD leg, but not on the LD leg. *

I'm pretty sure that Pin 2 on a 3-pin laser diode is always the "middle" pin. *But I don't know how to determine which diode pin is Pin 1 and which is Pin 3. *

Lots of people have bought chinese lasers and changed the diode in them. Unless I'm mistaken, you gotta match the diode type with the old board, or use a whole new board. If you stick an N-type diode on a board that was driving a different type, my understanding is that you will kill the diode.

Thanks!
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Old 04-06-2008, 08:53 AM #5
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Default Re: Questions about this laser diode

I can't say this enough. Sort out current and let voltage deal with itself. It will. The nature of a diode is such that it will always have a similar voltage drop.
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