Welcome to Laser Pointer Forums - discuss green laser pointers, blue laser pointers, and all types of lasers



Laser Pointer Store

Dead M140?

IVO

New member
Joined
Jun 2, 2017
Messages
15
Likes
1
Points
0
Well, that's why I don't use lm317 circuits, because of the efficiency. With this MOSFET design, you can get a very low dropdown voltage on the transistor. For example, I'm using an irfz44n, wich has a Rds(on) of 0.0175R, and with a current of 1.8A the voltage drop will be V = 0.0175 * 1.8 = 0.0315V. Plus the drop of the resistor in series, 0.0315 + 0.55 = 0,5815V. Of course, that is providing that the Vgs is enough. So that means that you'll need an input voltage that is around 0.6V more than the diode (at that current). If the diode is a M140, you'll need 4.8 + 0.6 = 5.4V minimum input voltage. That's why it doesn't work with the 5V power bank. The 9V battery was enough but it drains too fast. The power supply was the only thing left.

Switchmode regulators can do a tad better, but add a lot of complexity to the point where you cannot build them from scratch at home.
I don't think so... You can make a switchmode regulator at home. With a little bit of practice, you can make very nice homemade PCBs. But I get the point, it's a lot easier to solder a resistor onto a lm317 or just buy one. But the MOSFET design isn't that complicated to make at home. I guess it just depends on the will of people to learn about driver circuits and DIY some PCBs.

This topic is very interesting, but can we please go back to the dead diode? XD I've provided details. If you need more information just ask. Any ideas of what could have killed it?

I know that investigating the death of a diode might be boring, but I think it's a good way of learning from the expensive mistakes and it'll be useful for me and other newbies like me that try to start with lasers.
 
Last edited:

Benm

Well-known member
Joined
Aug 16, 2007
Messages
8,004
Likes
665
Points
113
I think Aaron got my point of linear regulators versus voltage regulators repurposed to do so and the voltage drop required :)

As for -proper- linear constant current sources, i still like them and would recommend them for a situation where the battery voltage is just a bit higher than the required diode voltage. As mentioned the dropout voltage can be quite small using the proper mosfets, opamps and such.

As for your dead diode: it is difficult to determine what caused this.

As i understand it it never worked properly, so i could have arrived dead at your doorstep. This could be caused by static electricity, or just be a manufacturing defect, no way to tell.

Often there is no way to tell at all - even if it has no visible damage of any kind a diode could be killed by numerous factors. There are some failure modes that you could see when looking at it under a microscope such as a fractured facet, but the absence of that does not prove anything.

Microscopic inspection could prove over-current operation if there are any bond wires that evaporated, but that's about it.

There is a chance that the damage occured during shipping, with the vendor sending off a tested good laesr diode amd you receiving a led'ed one on the other end. It's nearly impossible to prove either way around - if you buy something off a reputable vendor it could arrive dead. If you google for videos about how badly packages are handled in airmail you'll get an idea of how even good packaging could (not) protect something.
 

IVO

New member
Joined
Jun 2, 2017
Messages
15
Likes
1
Points
0
Ok, I understand.

I was just asking because I don't know if I was doing something terribly wrong that could kill the diode (as newbies do). But from what all of you are saying, the circuit is probably not the problem, and there are a lot of other factors (and a little bit of luck) involved.
So I'll try again. This time I'm going to buy the diode from another vendor (just in case), I'm going to buy an antistatic wrist strap and antistatic mat (I'm not willing to take that chance again), and while I'm waiting, I'm going to continue with my boost converter design. I'll make a couple of models (350mA and 1.8A) and I'll make one at home (DIY style), just to prove how you can make one switching mode power supply at home. XD

Thanks all for the help and advices.
When I have news I'll make a new post.
 

paul1598419

Well-known member
Joined
Sep 20, 2013
Messages
13,233
Likes
1,738
Points
113
I'm not sure if it has been mentioned here or not, but you should at least test your drivers with a test load, or dummy load, before using them on a LD. I have made several for different power LDs, but the theory is the same for all. You need 10 silicon rectifiers connected anode to cathode, and on the last cathode you put a 1 ohm 5 watt resistor in the series. I use header pins and shunts to be able to put any number of rectifiers in the series as you are trying to approximate the LD's load on the driver to measure the current it is supplying to the LD. The positive lead from the driver is connected to the anode of the last diode in the series and the negative side is connected to the free end of the 1 ohm resistor. If you measure the voltage drop across the resistor, it will give you the current. 1000 mV=1000 mA. There are many diagrams of these dummy, or test loads in threads on here. If you need a schematic, one can be found by searching for test loads. :D
 
Last edited:

IVO

New member
Joined
Jun 2, 2017
Messages
15
Likes
1
Points
0
Yes, I know that they have to be tested with a load. Since I don't have a dummy load yet, I use high power LEDs with similar voltage drop. I know it's not the best... but it does the job. The dummy load is much more useful and accurate. I'll make one as you described as soon as I finish my university access exams (tomorrow, Wednesday and Thursday).
Thanks!
 

Benm

Well-known member
Joined
Aug 16, 2007
Messages
8,004
Likes
665
Points
113
Using power LEDs as dummy loads is not -that- bad, although their voltage drop is quite a bit lower than that for 445 nm diodes. Besides that they behave pretty similarly to laser diodes as a load, and you can use a scope to check for spikes on power-up and similar problems.

If you use a linear driver based on something like a opamp with mosfet power element it's not likely to kill your diode, but check with a shunt resistor and scope just to be sure there is no dramatic overshoot.

As for static electricity: don't be overly concerned with that unless you get zapped when touching taps and other grounded metal pieces around your house. The wrist straps are usually not needed if you work on an anti-static (i.e. slightly conductive) mat as a work surface. Just touch the mat before handling anything on it and you'll be fine.

Common sense helps a lot with static problems: you can lift a cpu from a computer perfectly safely without any wrist straps or such, just grab hold of the frame before you do.
 
Joined
Jun 20, 2015
Messages
67
Likes
38
Points
18
Ok, I understand.

I was just asking because I don't know if I was doing something terribly wrong that could kill the diode (as newbies do).
I had mentioned before under a selection of your text I had quoted of which you had stated that you connected the diode to an already powered on regulator, my response to that (paraphrasing) was "this was probably what did it" or some such thing.

I'm pretty sure that's the closest you can get to what most likely killed it, assuming that it arrived to you in working condition which it may or may not have.. and regardless of the cause, not connecting diodes to already powered regulators waiting to begin regulation based on the load presented is the one thing that absolutely should be taken as a learning experience from this. Connecting diodes to already on regulators is connecting a diode to a not regulator for exactly the amount of time in the response curve for that regulator.. turning on a regulator with a load already connected will have an entirely different set of events than connecting a load to a powered non loaded regulator, one notable difference being the setup with the diode already connected will be the setup that doesn't have a diode receiving 99% of the unregulated supply voltage and current for a few nanoseconds before the regulator can react to the load.
 
Last edited:

IVO

New member
Joined
Jun 2, 2017
Messages
15
Likes
1
Points
0
I had mentioned before under a selection of your text I had quoted of which you had stated that you connected the diode to an already powered on regulator, my response to that (paraphrasing) was "this was probably what did it" or some such thing.
Yes, I got that. And it's something I'll never do again. Thanks. I was just asking if there's something else.
So the procedure is the following: 1- connect the driver with a dummy load to test current. 2- disconnect driver. 3- discharge capacitors. 4- connect diode. 5- Power the driver with the diode already connected.

But I have one question about the dummy load... Why do you measure the current with the voltage drop of the resistor? I know that according to the ohm's law it should work because the resistor is 1 ohm, so V=A, but that will be 100% true only in a perfect world. What I want to say is that because of the tolerance of the resistor, wouldn't this be less accurate than just using a multimeter set to current?

As for static electricity: don't be overly concerned with that unless you get zapped when touching taps and other grounded metal pieces around your house. The wrist straps are usually not needed if you work on an anti-static (i.e. slightly conductive) mat as a work surface. Just touch the mat before handling anything on it and you'll be fine.
Well... I don't know because the block that I live in is very old and doesn't have a proper ground connection. The power outlets have ground pins, but the are not connected to anywhere. So I'll have a problem connecting the mat to ground. But I have something in mind for my workshop/lab..
And I know it's not that big of a deal, but this will be something very useful for the place where I do my stuff with electronics, not only for laser diodes.
 
Last edited:

paul1598419

Well-known member
Joined
Sep 20, 2013
Messages
13,233
Likes
1,738
Points
113
I use 1 ohm 1% and 0.1 ohm 1% resistors in all my dummy loads. They aren't that much more expensive, either.
 

WizardG

Active member
LPF Site Supporter
Joined
May 9, 2011
Messages
639
Likes
243
Points
43
This:

"But now I remember that I adjusted the current with the diode still on. I placed another resistor in parallel, raising the current up to 900mA."

Your regulator built with a MOSFET and bipolar transistor will not respond kindly to the above behaviour. Without a capacitor on the output I can easily see adding another resistor in parallel while the circuit is running creating a nasty, if brief, spike at the output. It's quite possible that if you were holding the leads of the resistor you added that a tiny amount of ESD could get amplified by the driver circuit into an output spike.

If you have an oscilloscope you should hook up a test load to your driver and play with the circuit a bit and compare the output of your driver with and without a small (1uF say) capacitor smoothing the output of your driver circuit. Without the capacitor I'd bet you will see some nasty spikes.
 
Joined
Jun 20, 2015
Messages
67
Likes
38
Points
18
But I have one question about the dummy load... Why do you measure the current with the voltage drop of the resistor? I know that according to the ohm's law it should work because the resistor is 1 ohm, so V=A, but that will be 100% true only in a perfect world. What I want to say is that because of the tolerance of the resistor, wouldn't this be less accurate than just using a multimeter set to current?
There's probably a very good reason to use a resistor but I don't know what it is, I'm guessing it is under the assumption that the regulator will trip overload or that the fraction of an ohm load it presents will give a non-compliant output in some which way or another.

That being said-

I just flick the DMM to current and drop a dead short over that sucker and I've had nothing but spot on expected, calculated current as the result... which is really quite impressive that (in my analog linear world) a dirt cheap off the shelf part can precisely and with rock solid stability regulate a sizable chunk of current in to a dead short. I reason that if it kills the regulator then I shouldn't have used it in the first place and if the output drifts or is non-compliant to the specification for that of which "Lord 'Set Resistor'" hath commanded then I'll toss it's substandard, flawed @ss in a cold, callous yet decisive and confident manner, lacking of respect, devoid of any recognizable basic human empathy, without even a second thought.. like it's nothing to me, worthless, trash. Sometimes I smoke a cigarette while performing this callous tossing, for added effect. Other times I'll do it while wearing a black leather jacket while also smoking a cigarette to show how serious this stuff is and that I'm not messing around. And to what is the aim of these tossings for which they are tossed? There is only one logical choice and that is the Trash Pit of Despair which will leave no doubt to the tossage the certainty of it's inescapable doom. This doom is shared with peers of other flawed, useless, broken, crumpled, crusty, stuck together or perhaps just downright ugly things of various types, all previously deemed unworthy and unfit for physical existence, sentenced to certain doom all of which was tossed prior in a similar cruel, cold and disrespectful manner and, as before, without even a second thought as if it means nothing at all to me, nothing at all, like it's just trash.

Woah.

I mean yeah, I use the DMM on current and it works fine.
...
 

paul1598419

Well-known member
Joined
Sep 20, 2013
Messages
13,233
Likes
1,738
Points
113
It is still better to have a load that simulates the LD as you are adjusting or checking the current of whichever regulator of current you have decided to use. An ammeter in series with your LD is an LD waiting to enter your trash bin. I have made a point to use 1% resistors in all my dummy loads and to keep the leads large and short as to not add appreciable resistance to the circuit. Also, I use large traces that are kept short for the same reason.
 

Benm

Well-known member
Joined
Aug 16, 2007
Messages
8,004
Likes
665
Points
113
There are two reasons to construct a dummy load from a bunch of silicon diodes and a resistor in general:

- it actually represents how a laser diode behaves, the current-voltage curve is not as steep as that of many silicon diodes.

- the resistor gives you a convenient point to measure the voltage and determine resistance.

1 ohm is sort of handy as voltage will equal current, but also pretty realistic to simulate 445 nm diodes. The tolerance on the resistor is not that problematic. You could go for a 1% or better version, but even ones labeled 5% are usually within 1% of the declared value. If in doubt you could measure the resistance too, though your average multimeter will not do this very well for such low resistances.

All that said it probably was connecting the laser diode to the already running power supply - especially if it has any output capacitors (which help to stabilize things but kill laser diodes on dodgy connections or just connecting them after powerup).
 
Joined
Jul 1, 2016
Messages
526
Likes
152
Points
0
M140 are fairly cheap atleast.. when I first got here I did a DIY LM317 circuit because there was alot of guys who had used them succesfully.
 

Benm

Well-known member
Joined
Aug 16, 2007
Messages
8,004
Likes
665
Points
113
Yeah, it works well if you have enough voltage difference between power supply and diode, and don't mind the resistor dropping 1.25 volts and dissipating power according to that either.

The circuit worked -really- well in the case of having 2 lithium cells powering a red diode from a dvd writer, at the time at 200 mA or so. The voltage difference was big enough for it to work (though not technically within spec) and the resistor would dissipate 250 mW, which is what most standard through-hole ones are rated for (400 mW for metal film, but not in a tight space i reckon).

Problem is it has gone to powering things like 445 nm lasers that have a higher voltage drop and also higher current to get full power, in this case the lm317-type drivers really don't perform well. It's fine if you just want to test something from a 12v / bench power supply and can but a big heatsink on the lm317 and use a power resistor, but far from ideal.
 

IVO

New member
Joined
Jun 2, 2017
Messages
15
Likes
1
Points
0
Ok guys... I finished my exams, now is time to get some work done.

I'll make a dummy load and I'll post. Any specific diode model recommendations? The common 1N4001 only takes 1A...

If you have an oscilloscope you should hook up a test load to your driver and play with the circuit a bit and compare the output of your driver with and without a small (1uF say) capacitor smoothing the output of your driver circuit. Without the capacitor I'd bet you will see some nasty spikes.
It will be useful to see that. I'll try to get my hands on an oscilloscope as soon as possible. That's one of my goals for this summer, buying an oscilloscope because I need one. But I'll have to repair a lot of smartphones and laptops to save for an oscilloscope...

M140 are fairly cheap atleast.. when I first got here I did a DIY LM317 circuit because there was alot of guys who had used them succesfully.
Yes, compared to other diodes... That's why I thought it would be good to start with it.
By the way, a new M140 is already on its way...
 
Last edited:




Top