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Calculating out put on a M140

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Hello everyone! Ok I have a question I simply have been unable to answer using my google-fu. I am a electronics engineer but I am just now branching out into lasers above a class 2. Here is my question. I have a M140 diode with a G2 lens coming in shortly. I am using a custom constant source driver that uses a SDcard and a actual programming file where I can control specifically how may mA I can drive to the M140. My voltage source is 7.5-7.75 volts depending on battery charge. And I can push up to 2A on my primary channel.

I have a potential input of up to basically 15W but I'm trying to figure out how to calculate the out put of the M140. I have found many sights with loads of data. Looking at thermal variance and in put to out put seams leaner around 25-30deg C. Everywhere I read says that the M140 is not a voltage dependent device, and hooking it up straight to a power source would cause thermal runaway and kill it. I also know that I should stay under 1.8A. And my threshold value should be around 140ma (the G2) might lower this, I am unclear on this, according to the limited info I could find online.

My question is, is there a way to get a rough calculation on out put vs input with out a actual measuring device. Using good old math. I keep running into conflicting data and this is why I am posting here. This seems to be the best resource on the net for Laser hobbyist.

The board has the capability of switching between 1-3 files each file I can set to a different mA level. So does the input voltage affect the out put of the M140 in calculations because if so. P=AxV so the get 2w theoretically I only need around .250mA but I feel like I am missing some crucial information on how to figure out what I want to know. Any help would be appreciated.
 

123splat

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Thresholh current is the level at which the diode will START LASEing, at or above the Voltage drop reguired by the diode structure. The ABSOLUTE Max.s are the stopping point, beyond whech the manufacturer does not recommend going, and will not garantee the product (go boom,boom).
you are calculating electrical power, not optical power. There are complex equations to approximate optical power, but they require optical and physical variables, as well.
best you get a hold of an optical power meter.
 
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Ideally i will, right now it's not in the budget. So what your saying though is the voltage in won't affect the optical out as long as the mA are regulated?
 

ChaosLord

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Ideally i will, right now it's not in the budget. So what your saying though is the voltage in won't affect the optical out as long as the mA are regulated?
Sounds right. Amperage is what affects optical output, assuming enough voltage is there for the diode. The m140 drops between 4-5V, just how much depends on how much current you feed it.

As long as your constant current driver can supply that voltage while regulating current, then the current will determine optical output.
 
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Hi Ageless. The pictures in this DTR OP should give you a ball park idea. I have the same diode and lens in handheld. At 1.75 amps, I'm measuring about 2.25 watts. Pretty close to the what DTR measured with that diode.

Best wishes on your build.
 
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I think I saw this post when googling last night. Deff thanks to pointing me back to it though! I was mainly concerned about the voltage affecting output. But as I have already had that cleared up no I am just waiting on parts. Again thanks for all the info. This community is a great resource. Once I get my build finished I'll post her up.
 
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Right now I am designing the heat sink. Is there any info put there on how much heat the m140 puts off near the 1.5A mark?
 
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Ok I have a question about DTRs OP on the m140 using a g2. Looking at the pics you can see a wattage increase that's fairly in sink with the mA increase I get that. Bit my question is this. There is also a voltage increase. Is that the diode drawing more voltage as the wats go up or were was the voltage manually increased for example

You see 259mW @ 300mA/3.8v
724mW @ 600mA/4.0v

But at the optimal 1.7A it's

2.272W @ 1700mA/4.9v

Is the spies just naturally pulling more volts thus regulating it's self or was the voltage increased independently to deliver the correct mA? I ask because this goes back to my op and the fact that the main circuit has a max potential of 7.5V do I need to step this down or is that safe for the M140? I guess what I am trying to ask my driver is using 7.4v because of other components and the main output channel can go up to this max so does the diode pull the voltage it needs for the mA being fed to it or does the voltage need to be limited as well?
 
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I don't have the electrical knowledge to directly answer your questions. I'll tell you what I know and hopefully a more knowledgeable member will drop in.

First, the heat generated at 1.5 amps. It's very easy to calculate if you know the equation. Something like watts x junction temp (on data sheet for the particular diode), but it's too late for me to look up.

Second, regulating voltage or current. Current is what is controlled. Adequate voltage needs to be available for the current to be used. DTR probably set voltage for his power supply slightly higher than he knew the diode would demand at full power. The then probably increased current throughout the test. The power supply is reporting the voltage that the diode is consuming. The LED's in lasers need current regulation to be properly controlled. Most drivers somehow provide the voltage needed by the diode.

For example, I like to make sure I have more than enough voltage to ensure regulated output. So I use buck drivers and 2S or more cells. The driver can provide all the way up to the full voltage of the cells (less the overhead it needs). I set the driver to the current I desire. I could connect 1 more diodes in series as long as the total voltage required by the diodes at the desired current is less then the total voltage of the cells less voltage sag and driver overhead requirements. The driver will provide the same current whether I have 1 diode, or more diodes in series. It will adjust the voltage provided automatically. I wish I know how it did this.
 
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Besides the heat you answered the question perfectly thank you. Yes my set up is for control of current. And will provide 7.5-.7.75 max volts so I shouldn't have any issues, according to what your saying. The m140 should act no different than the class two 2s I've been using the demand around 3.0-3.5 @ around 700mA and when attached to the primary circuit Chanel they only pull the voltage they need as long as you limit the current.

The M140 seems to be no different. Other than the fact it cost me about 4x The $ so that's why I'm asking a lot of questions. I have not been able to find any form of actual data sheet on this diode as they are "harvested" typically rather than coming from an actual manufacture. If there is a data sheet put there point me to it. I love hard numbers. It would get me put of your hair. I am sorry if any of my questions seem stupid or redundant. I've tried to educate my self on everything from the tech specs to the dangers of class 4 lasers. Before I even hook this bad boy up for the first time. Murphy likes to live under my work bench so I try to avoid having him knocking on my table when I can.

Again thanks for the responses, each if not into early at least in part has given me or pointed me to some of the relative info I required. I sent DTR a pm as I actually got the m140 from his shop. So hopefully the heat question I'll have an answer to soon as well.
 

DashApple

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As you increase the current though the diode the voltage over the diodes junction will increase aswell ,

As said above , more current the more electrons producing photons , more stimulated emission , more optical power out .

Heat generation is power out taken from power in , but that's if you are measuring optical output power

2.272W out @ 1700mA/4.9v would be 8.33W in , that's 6.058W heat .

Other way would be look at a power / current / voltage chart of the diode .
 
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LDs require a current source (whatever that means electronically). With very high currents voltage losses on wires must be taken into account.

" is there a way to get a rough calculation on out put vs input with out a actual measuring device."

Depends on your required roughness. Get approximate efficiency value W/A (datasheet or from others) and you might be fine.
 
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Ion, thanks for the conformation the class 2s I am used to working with are the same way. There just not as dangerous as a class 4 so again that's why I had the questions. The calculation for heat is perfect and gives me am idea on what I am trying to dissipate. Thanks!

As you increase the current though the diode the voltage over the diodes junction will increase aswell ,

As said above , more current the more electrons producing photons , more stimulated emission , more optical power out .

Heat generation is power out taken from power in , but that's if you are measuring optical output power

2.272W out @ 1700mA/4.9v would be 8.33W in , that's 6.058W heat .

Other way would be look at a power / current / voltage chart of the diode .
 
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Is there a link to a data sheet on the m140 someone can direct me to? I have scoured the web looking for one but was un able to find one through Google.
 




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