LPC 826 driver with LM317 circuit

Hi all

i am old member but i am little speak english. i was read many many topics at forum. i research how i build burning laser... And last....

i was buy a diode LPC 826 and the host to this two link..

This is LED link,

660nm 300 400mW CW Red Laser Diode LPC 826 | eBay

This is modul host

http://www.robotmalzemeleri.com/index.php?route=product/product&product_id=109

Now i am waiting LED from sales guy.. i take a host. But i have not driver LED. i was read how i build LM317 driver. But i have a question.

i am simulate with proteus LM317. Supply 12 volt. i build a constant current circuit for 400 ma ( 0.40 amp ). it is ok but i'm confused. if i run the circuit in simulate the LED diode voltage be high 2.2 volt. is it problem? 5.13 volt to high for LPC 826. This current is ok and this circuit is constant current circuit.

The sales guy write the ebuy page the supply voltage must be 2.2 volt max.

Please explane to me how i drive the diode? Whitch circuit use for LPC826 ?

if i am supply diode with this circuit do i have a problems? Please help me..

This is simulated circuit for diode driver

And this is last question..

How i build a two stage LED driver ?

For example one stage 150 mw for pointer second stage is 300 mw for burning pointer? Do you have a driver circuit this function?

What is it a supply voltage LM317 ? if i use only 1 lithium battery ok (3.7 volt ) or i should bu use two lithium battery? (7.2 volt )

Ignore the voltage - it is irrelevant because this is a constant current driver.

150mW is not "pointer power." You want more like 2-10mW for that.

You need two lithiums with this driver and this diode.

You can add "stages" to this driver by having a 5mW set-resistor hard-wired in the circuit, and then switch in another resistor in parallel to give a net resistance of 3.24ohm or whatever. Do not place a switch that will ever (even momentarily!) open the feedback path between ADJ and out - this will cause you problems (ie dead diodes).

Cyparagon said:

Ignore the voltage - it is irrelevant because this is a constant current driver.

150mW is not "pointer power." You want more like 2-10mW for that.

You need two lithiums with this driver and this diode.

You can add "stages" to this driver by having a 5mW set-resistor hard-wired in the circuit, and then switch in another resistor in parallel to give a net resistance of 3.24ohm or whatever. Do not place a switch that will ever (even momentarily!) open the feedback path between ADJ and out - this will cause you problems (ie dead diodes).

Click to expand...

Thank you very much ...

i give my diode today. i test run now. i use LPC826 and axiz. the driver LM317 constant current mode. i use 3.4 ohm resistor but the supply voltage for diode is high then 2.2 volt. i measure 2.8 volt and the current is 420 ma 0.42 Amp.

is it problem ?? is this voltage will kill my diode in the future or not?

i am change current resistor increase 9.5 ohm and then the diode voltage down to 2.2 volt. ok but the current down 160 ma. i have a power loss.. Please give me a idea for this operation? Help please..

i calculate the power. is it right ?

2.2 volt and 0.16 amp. Power = V x I ---> Power = 2.2 x 0.16 ---> Power = 352 miliwatt = 0.35 watt ?? is it right?

Do you have a table for LPC 826 operation voltage and amps?

You raised the current-set resistor which is supposed to lower the current. Where is the problem?

Ignore the voltage across the diode. Anything between -0 and 3.5V is normal for a red. It depends on the drive current.

Your power calculations give input power. Laser diodes are not 100% efficient. The only way to know output power is to use a laser power meter.

Use the search function at the bottom of the page so you can get quicker results. Here is a table, but please note your results may vary.

You really shouldn't think about the diode as having a voltage requirement, it should be thought of as having a voltage drop across it. As others have said, don't worry about the voltage. Also, reduce your voltage going into the driver to 7 or 8 volts, you don't need 12 volts. All the extra voltage you supply to the driver that's not dropped across the laser diode will be converted to heat. Try to think of this in terms of current with a minumum required voltage and no real max (no max TEK a degree, I wouldn't try putting 240 v across your diode). Do you have a dummy test load? If not you can build one very cheap using a 1 ohm resistor (tight tolerance of <2% preferred) in series with a string of a couple of N1004 plain diodes. Hook your driver up with the forward current going across the diodes in their pass direction and check the milivolts with a DVM across the resistor. With 1 ohm you'll get very close to 1mV = 1mA. Adjust your set resistor to the current reading you calculate from try our test load. In this build you will need x2 LiPo 3.7v batteries as your power source. Also, the LPC is case positive I believe so if your host uses the body as a current route you will need to isolate the heat sink from the body of the host if you're trying to hook it all up like most people do.

Good luck,
jM

Cyparagon said:

You raised the current-set resistor which is supposed to lower the current. Where is the problem?

Ignore the voltage across the diode. Anything between -0 and 3.5V is normal for a red. It depends on the drive current.

Your power calculations give input power. Laser diodes are not 100% efficient. The only way to know output power is to use a laser power meter.

Use the search function at the bottom of the page so you can get quicker results. Here is a table, but please note your results may vary.

Click to expand...

Thak you very much this information .. The table to much helpful. i take a idea for search function.

Jmillerdoc said:

You really shouldn't think about the diode as having a voltage requirement, it should be thought of as having a voltage drop across it. As others have said, don't worry about the voltage. Also, reduce your voltage going into the driver to 7 or 8 volts, you don't need 12 volts. All the extra voltage you supply to the driver that's not dropped across the laser diode will be converted to heat. Try to think of this in terms of current with a minumum required voltage and no real max (no max TEK a degree, I wouldn't try putting 240 v across your diode). Do you have a dummy test load? If not you can build one very cheap using a 1 ohm resistor (tight tolerance of <2% preferred) in series with a string of a couple of N1004 plain diodes. Hook your driver up with the forward current going across the diodes in their pass direction and check the milivolts with a DVM across the resistor. With 1 ohm you'll get very close to 1mV = 1mA. Adjust your set resistor to the current reading you calculate from try our test load. In this build you will need x2 LiPo 3.7v batteries as your power source. Also, the LPC is case positive I believe so if your host uses the body as a current route you will need to isolate the heat sink from the body of the host if you're trying to hook it all up like most people do.

Good luck,
jM

Click to expand...

Thank you very much for your answer. i have a 4 piece 1n4001 diode. put it serial and connect to my driver. But i don't use a 1 ohm resistor. Because i have a multimeter and the multimeter include a ampermeter. i was build circuit. i don't successful.. the 4 piece diodes very hot and the current to high. why i don't understand.. maybe i use 5 diodes.. i don't know.

But i was build a LPC circuit. i use a 9.5 ohm resistor with LM317. i measure current and voltage.. it is good. i was burn a paper and match.. Maybe i take a low power but it is safety now. i have no laser power meter.. this video is first test,, short test..

I think the 1004 series diodes are rated for 1a so any current at 400+ mA through them will likely make them very hot if left on for more than several seconds. You should consider buying a diode with a 3-5 amp rating if you are going to run current through them for more than 15-20 seconds. I don't let my dummy load run for any longer than that. It's not necessary if you have everything hooked up and ready to measure when you power up. At least that's my experience, others may do it differently.

Here is a good cheap alternative to the 1004 series diode for a dollar:
10A10 10 Amp 1000V 10A 1KV Axial Rectifier Diode 5 Pcs Lot FL USA | eBay

These are rated at 10a and should handle almost anything you will be working with.

For a typical silicon diode the voltage drop across it at its threshold (where it begins to conduct current forward) is between 0.6 and 0.7 volts (for schottey and germanium it can be 0.25 - 0.3 volts). From the threshold voltage at which the diode begins to conduct current (0.7v typ) the voltage drop across it will start to rise logarithmically (see picture below) and for most diodes when you reach the max rated current your drop will be closer to 1 volt. This phenomenon requires some careful thought to how many diodes you will want to use in your dummy load. Most red diodes have a typical voltage drop around 2.5-3 volts so if you are running the diode near its threshold you can use (n*0.7) = voltage drop. Four diodes running well below their max rated current should give you 2.8v of drop which should be a close enough approximation to your red laser diode. For higher currents you may need to reduce the number of diodes in your string (Vd=1v at max rated current). Hope that helps a little.

Oh yes, when searching this forum it is best to use the white search bar at the bottom of the page instead of the search utility labeled "search" at the top of the page. You get much more useful results with the white search bar (see attached picture). When I first started here I used the forum's search utility (tab at top of page) and would get so frustrated when members would tell me I should search before I posted. I did, and every time I used that tab to search I got everything but what I wanted. Eventually a member showed me the white bar at the bottom and everything started to make sense.