Help me understand what I did wrong

So I finally got my LPC-815 from Modwerx - awesome!

I spent the evening creating a driver for the laser according to the instructions at Laser driver - It can be done.

I was successful in setting up the circuit almost exactly as displayed and hooking up a LED with a 9 volt. Voila, let there be light! I did omit the 100 ohm variable resistor because Radioshack no longer carries them (people were using them to rig cable boxes?). However, I did the math and with 2x 10 ohm resistors, I would have 5 ohms of resistance, leaving me with 250 mA. Perfect for the red laser from the LPC-815, right?

I was not successful in harvesting the LPC-815 diode from the heatsink using the small amount of tools I have available. I decided to set the diode at the end of the driver circuit anyway using the 9V battery power source. I received a large red flash of light from the diode (it is not installed in the housing yet because it is still stuck in the heatsink) and then it went off. I have not been able to get the diode to produce light again although I did see a faint very dim red light. It seems as if the diode is burned out.

My question is why? According to that website cited earlier (and referenced in many places on this forum), "You could put 12v or more in for either laser diode all that will happen is that the regulator turns any volts it cannot use into heat.". I assumed I would be fine with the rest of the power being turned into heat but I'm new to all this.

In short, did my 9v battery kill the laser or did I wire the circuit wrong?

Doh, just remembered that I didn't short out the capacitor before I connected it to the diode after playing with the LED light - could this have shorted the diode out?

Either the charged capacitor killed it or you wired the diode + pin to the OUT pin of the lm317 instead of the ADJ pin (which is a common mistake).

lm317 is a voltage regulator when you draw power out of it's OUT pin, but when you draw power out of the ADJUST pin with a fixed resistance between OUT and ADJ it becomes a current regulator. So, if you accidentally used the OUT pin for powering the diode instead you sent 6V at unregulated current to the diode and fried it.

Hey sigthur, thanks for the quick reply! Nope, I definitely have my two resistors (gold end/tolerance end) wired to the middle/OUT pin of the LM317 and the load wired to the ADJ pin. It's a bummer that my capacitor killed the diode. Oh well, live and learn! But was I correct about the 9V being safe to use still? Will the regulator shed the heat from the LM317 unit or will it send the extra voltage along the connection?

A 9V should be fine. A good way to think of it is a lm317 uses 3V to do it's job and then sends the rest through to the thing it is powering. The Diode will see 6V, but this should not be enough to hurt it. The Vf of a diode is the minimum voltage to turn it on, not the maximum it can handle, and most diodes can handle 2x Vf in the forward direction (that means not hooking it up backwards).

Caps kill diodes. I NEVER put my caps on the diode side of a lm317 regulator. It will do it's job just as well if you put the cap across the battery side of the lm317 and doesn't have any chance of blowing up the diode.

Sigurthr said:

A 9V should be fine. A good way to think of it is a lm317 uses 3V to do it's job and then sends the rest through to the thing it is powering. The Diode will see 6V, but this should not be enough to hurt it. The Vf of a diode is the minimum voltage to turn it on, not the maximum it can handle, and most diodes can handle 2x Vf in the forward direction (that means not hooking it up backwards).

Caps kill diodes. I NEVER put my caps on the diode side of a lm317 regulator. It will do it's job just as well if you put the cap across the battery side of the lm317 and doesn't have any chance of blowing up the diode.

Click to expand...

Thanks a lot for the info. I wish I had known that I could put on the IN pin (w/ the battery/switch) two hours ago :beer:

Aye, it's just there to stop self-oscillation of the regulator chip by shunting high frequencies which may ring up. It certainly is good practice to have a cap across the output side in most applications, but with Laser diodes you need to be wary of what is normally good practice. "finiky fings dem laser bits are" Live and learn!

You could also place a high value resistor across the output side of the lm317 along with the cap that is there now if you don't want to remake the entire driver circuit. Anything over 6kOhms will do, normal values for this type of application are typically in the hundreds of thousands of ohms though. This is called a bleeder resistor and serves to discharge the capacitor automatically when power is removed. The higher the value the less power it steals from the laser diodewhile power is on. At 6k it will only take 1mA away, which is nothing serious.

Did you build the test load as the page suggests and confirm that the math gave the current you were desiring in reality?





If it lit up when you powered it after connecting the diode I don't see how the cap could have killed it. I would not think it would have lit up if it were already dead/LED from a charged capacitor.

Why are we suggesting going without the cap as there should only be one point where it can kill the diode and that is when connecting the diode to the driver unless we are saying it does nothing anyway.

DTR said:

Did you build the test load as the page suggests and confirm that the math gave the current you were desiring in reality?

If it lit up when you powered it after connecting the diode I don't see how the cap could have killed it. I would not think it would have lit up if it were already dead/LED from a charged capacitor.

Why are we suggesting going without the cap as there should only be one point where it can kill the diode and that is when connecting the diode to the driver unless we are saying it does nothing anyway.

Click to expand...

To answer that:

jordan said:

I decided to set the diode at the end of the driver circuit anyway using the 9V battery power source. I received a large red flash of light from the diode (it is not installed in the housing yet because it is still stuck in the heatsink) and then it went off. I have not been able to get the diode to produce light again although I did see a faint very dim red light. It seems as if the diode is burned out.

Click to expand...

The way I read it is that he assembled the driver, powered it up at some point, perhaps inadvertently momentarily, and THEN connected the diode. Thus the cap was already charged (even if the circuit wasn't powered when he actually connected the diode). It is possible the flash was due to the diode lasing at full output until normal current finished off the diode in its weakened state after the cap charge damaged it.

I know it is bad practice but you really don't need a test load for a lm317 circuit if you are experienced enough with circuitry to build a lm317 without errors. I'm not saying the OP is experienced as such, just that it is an unnecessary step if you are certain you built the thing correctly and used solid math.

Anyway...

If the cap didn't kill it, there is still the possibility of ESD. Aside from that only circuit flaw remains, possibly resistor value error, but the OP already said this is not the case.

Hi, I'm confused by what DTR is trying to say but just to clarify: I assembled the driver and then ran a LED light at the end of the circuit several times to test the functionality of the circuit. I'm assuming that during the time that the LED was powered the capacitor became charged and burned out the diode after it was powered, but I do not know if that is the case.

I did not use a test load because I am positive my math is accurate on the ohms but in the future I plan to. Impatience bit me in the butt. Perhaps a resistor was not functioning properly or the diode was not functioning properly, who knows; all possibilities without having the load tested. I will test it tomorrow.

It sounded to me like he said he connected the diode to the driver. At this point no flash of light. Then he applied 9V to the circuit and it lit up full tilt for a brief second then died. That sounds more like the circuit may have been giving too much current.

If the cap had killed it I would think it would go like this. He connected the diode to the driver and it flashed(or not as I have seen charged caps kill a diode and you don't see squat). Then he went to power the circuit and the diode just gave LED level light.


Even when I buy premade linear driver's I have always put on a test load before connecting a diode to make sure they are putting out the current I know they should. I have had a linear driver I bought that was supposed to be a 1.8A and when I tested it was 2.3A(most likely completely unregulated). It had all the right resistors and looked fine but I think somewhere there was a short in the driver.

Aye as I said it isn't good practice, but on paper a test load isn't needed for a circuit as simple as he is using.

I completely agree that the flash was likely the diode lasing before complete destruction. I simply pose that perhaps the cap did not completely destroy the diode, merely only weaken it's current capapbility enough that full current killed it.

There is a simple test if you have a multimeter; disconnect the LD and connect the meter in series with the output of the driver circuit, then put the LED you used earlier between the meter's negative lead and the ground/return/negative of the battery. You'll read whatever current the driver is putting out.

Alright. The 9V battery was measuring as having 7.5 V charge in the volt meter. After setting the LED up on the driver circuit with the 9V and measuring the voltage, the LED was receiving 4.3-4.4 V. This makes sense to me because the regulator is using 3V (4.4V+3V = 7.4V, close enough). The ammeter measured 91 mA at the end of the driver circuit (the LED).

However, I'm confused - aren't the resistors/regulator supposed to stop current past 3V? The guide i followed said that the regulator will turn any voltage past 3V into heat - but it seems that everything the regulator isn't using is being sent through the circuit instead. If I sent 4.4 V to the diode, would that have fried it?

You must stop thinking in Volts. If you set the LM317 with a 5 Ohm resistor, that regulator will only provide 250mA to the laser diode. Not matter the voltage. That why it is called current regulator.
So with a 9V battery the LM317 will output 250mA with the voltage needed by the laser diode. All the rest of the voltage will be dump as heat in the LM317.
Of course the battery voltage must be higher than the Vf diode + Vf LM317.

jordan said:

However, I'm confused - aren't the resistors/regulator supposed to stop current past 3V? The guide i followed said that the regulator will turn any voltage past 3V into heat - but it seems that everything the regulator isn't using is being sent through the circuit instead. If I sent 4.4 V to the diode, would that have fried it?

Click to expand...

The resistors and regulator stop current past the set amount determined by the resistance between ADJ and OUT, they have no effect on the voltage that gets through. The regulator doesn't remove all the left over voltage and turn it in to heat, but it does generate more heat the more excess voltage there is left over. It doesn't get removed but it does cause an effect. Output voltage is always Volts In minus 3.

If the diode doesn't use all of that so be it, it just generates more heat to be dissipated by both the load (diode) and the regulator, but it doesn't go away. The reason it creates more heat is there is now more power going through the circuit. Power = volts x amps. So if you have 1.3V extra at 250mA you have 325mW of extra heat being generated, but the 1.3V is still there. This is a common misconception with linear regulators. Boost/Buck and BuckBoost are the only regulators which actually remove any excess voltage and turn it directly in to heat without the load actually seeing the extra voltage.

And no, 4.4V is not enough to kill the diode.

Great information, thanks again!

if you use an op amp you can vastly improve efficiency