Homemade 1W 445 nm laser not working right.

tornadodead · Sep 18, 2014

So I recently built my own ~1W 445nm laser in an Altoids tin following an instructable. I'm powering an M140 diode (DTR-LPF) using two 14500 3.6V batteries and a 7805 voltage regulator (no extra protection). The circuit works, but I seem to only be getting a dim spot. Anybody have any ideas as to why my system doesn't seem to be outputting anywhere near the ~1W it was intended to?

P.S. I'm new here, so if there's any other threads that address this problem, please let me know.

ARG · Sep 18, 2014

Sorry to say, but the diode is likely fried. A 7805 is a voltage regulator, and diodes should be driven by constant current.

5V was probably too much for the diode, and it died.

rhd · Sep 18, 2014

7805 is a 5V voltage regulator, and 5V is likely to correspond with a current in the diode's IV curve that is above what it can handle, meaning that you likely LED'd your diode. Sorry for your loss if so

What gave you the idea that you could drive a laser diode with a 5V voltage reg though?

3Pig · Sep 18, 2014

Either lack of heat sinking or too much current, probably too much current

sorry for your loss

alessnilsen · Sep 19, 2014

Sorry for your loss! :cryyy:

If you think is worth you can test your LD with a real driver (LM317 that is cheap) to see if your LD is really fried!
You have a very very small chance of it being below the lasing threshold.

Is possible to use a 7805 as a driver, but I think it will not be a efficient choice.

ARG · Sep 19, 2014

alessnilsen said:
Is possible to use a 7805 as a driver, but I think it will not be a efficient choice.

It is possible, but it's an extremely bad idea. Especially when an LM317 is only pennies more than a 7805.

alessnilsen · Sep 19, 2014

ARG said:
It is possible, but it's an extremely bad idea. Especially when an LM317 is only pennies more than a 7805.

If you know how to use I think isn't too bad only for a simple test, you will use it as a current regulator:

The formula will be I=5/R.

ARG · Sep 19, 2014

alessnilsen said:
If you know how to use I think isn't too bad only for a simple test, you will use it as a current regulator:

The formula will be I=5/R.

I never thought of it that way. Didn't know it was possible to use it as a current regulator with the reference voltage as 5V.

+rep

tornadodead · Sep 19, 2014

I looked at the documentation for the 7805 voltage regulator and deduced that it could be used to keep the current relatively constant at 1A.....Apparently I was mistaken if my diode is indeed fried.

rhd · Sep 19, 2014

ARG said:
I never thought of it that way. Didn't know it was possible to use it as a current regulator with the reference voltage as 5V.

+rep

Ya, as inefficient as that may be, it's an interesting insight. I'd rep you too if LPF worked properly on mobile and let me.

That configuration reminds me of the time benmw taught us all that 317s could be configured as continuous positive current regulators.

bdgreenb · Sep 19, 2014

alessnilsen said:
The formula will be I=5/R.

Sorry for being a dummy, but with a 4ohm resistor you would get 1.25A with that setup?

alessnilsen · Sep 19, 2014

bdgreenb said:
Sorry for being a dummy, but with a 4ohm resistor you would get 1.25A with that setup?

It should work! :yh:

Remember that the power dissipated by the resistor will be W=IxV. At 1.25A it will dissipate 6.25W.

Because of this I said that isn't too efficient.

:beer:

tornadodead · Sep 23, 2014

3Pig said:
Either lack of heat sinking or too much current, probably too much current sorry for your loss

I have dedicated heat sinks for the regulator and the diode. Nothing was hot when tested. How easy do you think it is it to fry regulators/drivers when soldering?

tornadodead · Sep 23, 2014

ARG said:
It is possible, but it's an extremely bad idea. Especially when an LM317 is only pennies more than a 7805.

How do I get ~1A from an LM317 driver setup?

alessnilsen · Sep 23, 2014

tornadodead said:
How do I get ~1A from an LM317 driver setup?

You will need a 1.25R 2W resistor.

Because 1.25V/1.25R=1A.

I have dedicated heat sinks for the regulator and the diode. Nothing was hot when tested. How easy do you think it is it to fry regulators/drivers when soldering?

If you use a too powerful soldering iron you can fry the regulator, here I use a 40W soldering iron and never fried a IC. There`s a chance of your IC already came fried, happened to me, A LM317 with 5A of output, all the resistors burned, but since I ever test any driver with a test load before use, It don`t fried any LD.

Eudaimonium · Sep 24, 2014

tornadodead said:
How do I get ~1A from an LM317 driver setup?

Works with LM317 and LM1117 chips.

1117 is bit more efficient but lower current limit, I think they top out at 800mA or so.

Also, don't forget the dummy load test:

Homemade 1W 445 nm laser not working right.

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