Hi all,

I purchased two laser diodes (link below) as part of a triggering circuit. When I applied 4V to them, the laser diode stayed very dim (barely visible when placed 3inches away from a surface). Am I doing something wrong or is the diode faulty?

http://www.ebay.co.uk/itm/161027652865?ssPageName=STRK:MEWNX:IT&_trksid=p3984.m1439.l2649

 

datasheet, where?

 

The seller did not provide any datasheet. Most probably these are cheap China made products so I will not find any datasheet for them.

 

Based upon the eBay listing, these diodes require 20mA of current. Have you set up proper current limiting?

 

Use a 9 volt battery (excess of 5 volts) so to drop 5 volts at 20 mA, use a 220 ohm resistor in series with the unit (from the battery).

It will adjust to the voltage needed and current will be limited to 20 mA. It will be good enough for a test.

 

Hi,

Some tutorials suggest the use of a driver like the one in the link below. Is there the need to do one or will a single resistor do?

http://laserpointerforums.com/attac...laser-diode-driver-lm317_components01_001.jpg

 

Hi,

Some tutorials suggest the use of a driver like the one in the link below. Is there the need to do one or will a single resistor do?

http://laserpointerforums.com/attac...laser-diode-driver-lm317_components01_001.jpg

A single resistor essentially does the same thing..

 

The driver you show will allow control of the current without depending upon the exact forward voltage drop of the diode as well as compensating for variation in input voltage. A resistor, although effective, will not compensate for those variables resulting in minor variations in current. As an example, a resistor selected to give 20mA with 9V applied and a diode voltage drop of 1.2v will give less than 20mA as the battery ages and the voltage begins to drop.

 

The driver you show will allow control of the current without depending upon the exact forward voltage drop of the diode as well as compensating for variation in input voltage. A resistor, although effective, will not compensate for those variables resulting in minor variations in current. As an example, a resistor selected to give 20mA with 9V applied and a diode voltage drop of 1.2v will give less than 20mA as the battery ages and the voltage begins to drop.

An IR laser would appear to be only producing a very dim glow on the surface of the chip - but that's only because the light is a longer wavelength than the human eye can see.

That's my impression from the laser out of a scrap laser printer - these ones could be hefty enough to illuminate the "dim glow" on a nearby surface.

 

Seems its a module rather than just a diode. I am experimenting with similar modules but 1mW power. Module refers to a diode and driver board in one housing. Sometimes the tiny driver board (about 10 x 10 mm in size) is visible but often not. So no need to limit current, just experiment by varying voltage only.

A diode alone would look like the first pic below while a module looks like the second. If the module is cut up there you will find a lens, laser diode, driver board...all encased in that tiny cylindrical case.

Your problem may be that the power source is not giving enough current, measure current to make sure its at least 20mA. Yours is a red (650nm) diode so the light should be visible, don't look directly at it.

 

I have a few of the same lasers. They are indeed modules, they have a basic driver on board. All you need to do is apply the required voltage.

 

An IR laser would appear to be only producing a very dim glow on the surface of the chip - but that's only because the light is a longer wavelength than the human eye can see.

That's my impression from the laser out of a scrap laser printer - these ones could be hefty enough to illuminate the "dim glow" on a nearby surface.

The OP's laser is bright ReD.

 

The OP's laser is bright ReD.

Apparently not so much bright - or the OP wouldn't have posted.