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Different Voltage, same power. Is it ok?

ron

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Hey All,

Ok. Here is the situation. I am about to purchase a laser diode from Stonetek. If I am not mistaken, its the Sony SLD1236VL. I have read the data sheet and it states that:

Operating Voltage: 2.5 Volts
Operating Current: 130 mA

So I am assuming that the total power output is 0.325 Watts (Power = Voltage X Current). The optical output is 250mW, so I guess the remaining energy is released as heat or something.

Now I have an LED flashlight that uses 3AAA batteries. So this means a total output of 4.5 Volts. If I am able to regulate the voltage, I fear that the current wont reach 130mA, and vice versa. Does this matter? Or it does not matter that I have a different voltage (or different current output) as long as the power still is 0.325 Watts.

I am planning to use that famous driver circuit that everyone is talking about (the one with the LM317T regulator). I know that it requires a minimum of 6 Volts of current. I remember reading somewhere online that the LM317T needs a minimum of 1.5 Volts of difference between the input and output in order to work. So if I use the 3 AAA batteries with a total of 4.5 Volts, will that work? Or do I have to build a different type of circuit? If so, got any suggestions?

Many thanks for your help. This forum really is a good resource. Keep it up fellas.

Ron
 

chido

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Hey ron welcome to the forum. To get things started the diodes that senkat has been shipping lately are not the SLD1236VL, here's a thread about that:
http://www.laserpointerforums.com/forums/YaBB.pl?num=1199990106
And here's the thread on how to build the LM317 circuit:
http://www.laserpointerforums.com/forums/YaBB.pl?num=1185701612
The circuit needs a minimum of 6v to work, but it's recomended to use a 7.2v input so that it doesn't start dropping out that quickly.
You can also use this driver if you can't fit a 7.2v power source in your host:
http://www.laserpointerforums.com/forums/YaBB.pl?num=1191898669/0
 

Gazoo

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ron said:
The optical output is 250mW, so I guess the remaining energy is released as heat or something.

Ron
The 250mw rating is if the diode is pulsed. It is rated at 80mw's cw running with 130ma's. We don't apply pulsed currents to the diodes. Refer to the links chido posted..and welcome to the forum.. ;)
 

ron

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Thanks Chido and Gazoo.

Appreciate your help.

Wow. Luckily you informed me early about the different LD model. Otherwise, I would have destroyed my diode by wiring it up to a more powerful circuit.

So for the AMC 7135, I have to install some resistors in serial in order to reduce the amount of current to the optimum level for the diode. But what about the voltage? Does it matter if e.g. I use 3 AAA batteries in my LED flashlight, which means that the total voltage is 4.5 Volts. Won't that mean that the power going into the diode will be much higher, and thus, fry it?

Again, thanks for your help. :)

Ron
 

Gazoo

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The AMC7135 is a completely different regulator than the 317. One drawback to using the 7135 is the laser module must be electrically isolated from the rest of the flashlight body, if the case is being used for the ground side. The reason is the ground wire coming from the 7135 is not the same as battery ground. Don't know if you read my review of it here:

http://www.laserpointerforums.com/forums/YaBB.pl?num=1191898669/0
 

ron

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Hey Gazoo,

Ok. So in that case, where do I connect the ground wire from the 7135 to?

Sorry, but maybe I did not catch the answer, but what about the voltage and current mismatch issue?

:) Take care.

Ron
 

ron

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Hey Gazoo,

This might be a silly question, but where in the world is the ground wire of the laser diode? Is it the same as the negative terminal? My AAA flashlight's negative terminal is connected to the body of the flashlight. So in this case, where am I going to plug the ground wire from the 7135 to?

Ok. Maybe I explained it wrongly earlier. So here goes. Hopefully it makes more sense now:
Let's say I am using a Laser Diode Sony SLD 1236VL, and a flashlight housing that uses 3 AAA batteries.

My flashlight uses 3 AAA batteries, which means its 4.5Volts of input voltage. Now using the 7135, this means that the output from the 7135 will also be 4.5 volts, and the current will be reduced to (let's just say for example) 400mA. This means that the current is too much for the diode and I will fry it. Let's say I add some resistors in series to it to reduce the current to 130mA, which is the operating current for that diode (according to the data sheet). But the voltage will still be at 4.5 Volts, and according to the data sheet, the diode only requires 2.5 Volts. So even though 150mA is perfect, the voltage is too much. So my diode will be fried right?
Now what I was thinking about was this:
Power = Current X Voltage
So the power required by the laser diode (if I multply the voltage and current values in the data sheet) is 0.325 Watts.
So since my batteries give out 4.5 Volts, I would have to add enough resistors to reduce the current to 72mA, so that the total power would be the same (0.325 Watts). Will this work?

Am I approaching this issue in the right manner? Or have I got it all wrong?

Thank's Gazoo for your knowledge. :)
 

Gazoo

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Approach it looking at only the current you want to drive your diode with. We know the 7135 puts out 350ma's. We also know that is too much current for the diode. The only way to reduce the current going to the diode if using the 7135 is to put a resistor in parallel with the diode. If you put one in series, the current going to the diode will not change, it will be 350ma's. I have been recommending putting 33ohm resistors in parallel with the diode. The 33 ohm resistor will absorb appx. 100ma's. Therefore the diode will be taking up the remaining 250ma's. The voltage going to the diode will be appx. 3 volts.

The ground wire of the diode is the same as negative. The case of the diode is negative. So what do you think will happen if you use the 7135 in a metal flashlight that has a negative polarity case? The regulator will be bypassed and the diode will in effect be connected to the batteries. The only way to prevent this from happening is to isolate the diode. Since the diode is mounted in a module the module would have to be isolated. And if the module is mounted in a heatsink then you would have to isolate the heatsink from the body of the flashlight.

As far as the regulator, it will provide a constant 350ma's with an input voltage ranging from 4.5 down to 3.2 volts. Below 3.2 volts it will quickly drop out. This makes it ideal because it is efficient, and also ideal for 3 nimh batteries, or 1 li-ion battery. When the regulator drops out is is obvious so you know it's time to charge the batteries, which will also prevent over discharging the batteries.
 

ron

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Ok Gazoo. I get you. Just two things that puzzles me:

"Therefore the diode will be taking up the remaining 250ma's. The voltage going to the diode will be appx. 3 volts."

1.) But the datasheet says that you should be feeding 150mA and 2.5 Volts. Isnt this too much for the laser diode?

2.) How come the voltage reduced to 3 volts, and not 4.5 volts as it was when it came from the batteries? Does the 7135 work as a voltage regulator too?

Sorry if my questions seem a bit stupid. Hopefully others can benefit from this too. :)
Thanks man.
 

Gazoo

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250ma's is overdriving the diode but that is what many of us do. I have not tried the new batch of diodes in question, but it's starting to look like they should not be driven with more than 160ma's. Therefore, you would require a resistor less than 33ohms. I don't know how much less, but it would probably need to be a 1 watt resistor as opposed to a 1/2 watt resistor. I might play around with this next week.

Since the new batch of diodes are not lasting as long as the old batch there is another option and that is to use an open can diode harvested from a DVD burner. There are lots of threads discussing it in the experiments section of the forum. They can easily be driven with ~400ma's. I am running one with the 7135 at 350ma's, and the voltage of the diode is 2.8 volts. The only other component needed is the capacitor that gets soldered across the diode.

The voltage is determined by the resistance of the load. That is why you will see the output of voltage drop with the 7135 or the LM317T when it is connected to a load, in our case the diode.
 

chido

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Hey Gazoo, when did the new batch start to ship out, I ordered my diode pre installed in an aixiz husing right before SenKat had his surgery and it shipped out on the 1st of December, I can't tell the code on it since it's in the housing, is it one of the bad ones, or am I safe? :-?
 

Gazoo

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chido said:
Hey Gazoo, when did the new batch start to ship out, I ordered my diode pre installed in an aixiz husing right before SenKat had his surgery and it shipped out on the 1st of December, I can't tell the code on it since it's in the housing, is it one of the bad ones, or am I safe? :-?
I have no idea. The only way to know is by looking at the codes...
 

ron

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Hey Gazoo,

Thanks a lot for your input. Really appreciate it. :)

Thanks for your help too Chido.
 

ron

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Hey Gazoo, just realised something.

The 7135 is a current regulator, whereas the 317 is a voltage regulator.

Which should I be more concerned with? The voltage regulation for the diode, or the current regulation for the diode? :)
 




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