Trying to work out correct current for driver...

[nzoomed]

Im wanting to use an M140 diode with this driver here:

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Specs for the M140 are 2W max output and max current is 1.3A
After reading the specs for the driver it says 5.5V max boost circuit, and the working voltage is 3-5V

Since with Ohms law, Voltage * Current = power, if we go by the lowest permissible voltage 3V * 13.A, it comes to 3.9 Watts, which is much higher than the diodes maximum output.

Going by these calculations, 1.54V would be needed to run this diode at its maximum 2A output.

Im assuming the driver therefore would have to lower the voltage in turn to make the diode draw less current, or am i missing something here?
5.5V seems to high for the specs, so im assuming the voltage can be adjusted, or am i going to need to find a different driver? I had seen some good talk about this driver on here so thats why i ordered it.

 

[paul1598419]

This driver will not drive an M140 diode to 2 watts because the driver current necessary to get that out is 1.8 amps. This driver is limited to 1.3 amps total input current. Boost drivers generally pull more current than they supply because they also have to supply that current at a higher voltage for the diode's Vf. You would do better with a buck driver instead of a boost, but you will need to use two batteries instead of just the one. If you do use a boost driver it will be necessary to heat sink the driver too. A buck driver would likely not need this.

 

[nzoomed]

This driver will not drive an M140 diode to 2 watts because the driver current necessary to get that out is 1.8 amps. This driver is limited to 1.3 amps total input current. Boost drivers generally pull more current than they supply because they also have to supply that current at a higher voltage for the diode's Vf. You would do better with a buck driver instead of a boost, but you will need to use two batteries instead of just the one. If you do use a boost driver it will be necessary to heat sink the driver too. A buck driver would likely not need this.

Yes i was not planning to run it at 2 watts anyway, as i would probably need extensive heatsinking to cool it at that power.
I had about 1.6W in mind to run it at.
My host can take 2x18650 cells if i keep the extension tube on, so perhaps a buck driver is better as you say.

One other question is how do you actually set the current for a diode? Do you turn the current pot all the way down to zero, hook up an ammeter in series with the diode, power it on and measure the current draw and adjust until its where you want it set?

 

[paul1598419]

You will need to get a dummy load to setup these drivers with. They use a sense resistor to measure the voltage drop across and this will give you the current without the need of an ammeter. There are several threads on the LPF concerning their use and how to build one, if you don't wish to buy one.

 

[nzoomed]

You will need to get a dummy load to setup these drivers with. They use a sense resistor to measure the voltage drop across and this will give you the current without the need of an ammeter. There are several threads on the LPF concerning their use and how to build one, if you don't wish to buy one.

ok thanks. will take a look.

 

[paul1598419]

This particular driver is a boost one. This would not be my first choice for this diode. It has the advantage of only using one battery, but at the current you wish to run it it will need heat sinking. It will need to pull more current from the battery you use than what you set it to deliver to the laser diode. That will necessitate the use of a large high drain battery.

 

[nzoomed]

yeah im looking at buck drivers now. I see some fixed ones at 1.6W which might be all i need?

 

[nzoomed]

Would this fixed buck driver be OK? its rated at 2W

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or the 1.6W version might be a bit safer on the diode.

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[smallfreak]

yeah im looking at buck drivers now. I see some fixed ones at 1.6W which might be all i need?

No. You see drivers capable to deliver „up to“ 1.6W. These are all fixed CURRENT. The voltage is adjusted to feed exactly this amount of current through the diode. It is depends on the type (and individual sample) of diode you use and changes with temperature. Semi conductors lower their resistance with rising temperature, which would draw more current the hotter they get, when you keep the voltage fixed.

The Laser driver instead keeps the current at the specified level, suitable for your diode.

What you should look for is „how many Milliamperes does my diode need?“ and „how do I set the driver to this value?“.

The first value can you read from the data sheet. Look for the term „maximum current“ and take the term „maximum“ as it was meant. Stay below that value with some safety left.

Your battery must deliver higher voltage than the nominal „forward voltage“ of the diode plus some working overhead when using a buck driver. The rest of the magic is done by the driver itself.

So, depending on the voltage that your diode will settle when fed xxxmA it will draw yyy mW of power that is converted into zzz mW of Laser light (depending on the efficiency) So these Watt numbers on the driver may mean different things than you might expect.

 

[Alaskan]

I used this driver with a M140 laser diode, it's big, but if you are not trying to put the diode in a small pointer, worked great for me. I built a big beacon driving sixteen of those diodes, each with their own driver, never a failure.

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[paul1598419]

Would this fixed buck driver be OK? its rated at 2W

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or the 1.6W version might be a bit safer on the diode.

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Unfortunately, they don't say what current either of these drivers are set to deliver to the laser diode. If I had either one, I would set it up on a dummy load to see what current it delivers. It is entirely possible that either driver would work work for your purposes, but there is no way of knowing that ahead of time.

 

[nzoomed]

Is this driver any good?
Reviews seem OK, and its a buck driver.
My calcs say that it should easily power up to 2W if I want.

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[paul1598419]

That is not a buck driver. It is .linear. Being liunear it won't be as efficient as a buck driver and is also limited to 500 mA of current. You won't get anywhere near 2 watts out at that current.

If you want 2 watts out, figure 1.8 amps of drive current for an M140 diode.

 

[nzoomed]

That is not a buck driver. It is .linear. Being liunear it won't be as efficient as a buck driver and is also limited to 500 mA of current. You won't get anywhere near 2 watts out at that current.

If you want 2 watts out, figure 1.8 amps of drive current for an M140 diode.

I thought it was a buck driver because it cuts the voltage down and regulates it at around 5V im assuming.
500mA of current should be more than enough for a 2W module I would have thought?
.5A * 5V = 2.5W so this means that this would over power the diode if i had it running at its maximum current.

I only need 400mA to run this at 2W on 5V am i not mistaken?

 

[paul1598419]

Don't confuse drive current with dissipated power. The current used by the driver to apply a current of 1.8 amps to the diode is much higher than 1.8 amps. All the excess power is dissipated as heat. You have not factored in the heat into any of your equations.

 

[nzoomed]

OK, how much current does an M140 typically draw at 2W?
I probably am happy to run this thing around 1.6W anyway.