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# Help with driver math calculations

#### Amply

##### New member
I am working on making a laser. I bought the 2W 445nm M140 Blue Diode in Copper Module laser diode from DTR's laser shop. I am now wanting to make a driver for the laser.

Here is what i know, (or think i know..?)

I need 2 Watts to power the laser diode
The laser diode can not exceed 1.8 amperes
I know that i want my power supply to be close to the power needed to power the diode while calculating for voltage drop out (for heat dissipation purposes)

((Keep in mind i have pretty much no idea what i am doing and i'm going completely off random research i have done online so most of this could be wrong just need to be pointed in the right direction))

I am going to be using a LM1085 regulator (my goal is to have a potentiometer so i can regulate the power from 500mA all the way up to 1,700mA (1.7A) -I don't want to go to 1.8A for fear of blowing the diode.-
( http://www.ti.com/lit/ds/symlink/lm1085.pdf) data sheet
The data sheet says that the max Voltage drop out is 1.5volts

So using Ohms law i should be able to figure out the resistance needed
(I = V/R)

v = Voltage drop out
I = Amperes
R = resistance in Ohms
E = Voltage

So re arranging the equation I should be able to figure out how much resistance i will need to get 1.7 amps with a 1.5 voltage drop out
R = V/I
So I = 1.7 , V = 1.5 , R = unknown
1.13 Ohms resistance

Using the equation W/A = V (v being volts this time not drop out voltage)
W = watts and A = amperes
I can figure out the least amount of volts needed to power the diode at 1.7A and determine what kind of power supply i need to run the diode.
W = 2 (as classified as a 2W Laser diode)
A = 1.7 (how much amperes i need for maximum output of the laser diode)
2W/1.7A = 1.176V
so this means i need
1.176v + 1.5v (I read somewhere that dropout voltage needs to be added to the voltage needed to operate something?)

Looking at these calculations I have a feeling i'm doing things completely wrong and I'm hoping someone can steer me in the right direction...

I was thinking the circuit would look something like this..?

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#### diachi

##### Well-known member
I am working on making a laser. I bought the 2W 445nm M140 Blue Diode in Copper Module laser diode from DTR's laser shop. I am now wanting to make a driver for the laser.

Here is what i know, (or think i know..?)

I need 2 Watts to power the laser diode
Nope - You need more than 2 watts - diodes aren't 100% efficient. Nothing is.
The laser diode can not exceed 1.8 amperes
That sounds correct!
I know that i want my power supply to be close to the power needed to power the diode while calculating for voltage drop out (for heat dissipation purposes)
Yes, although less of an issue with switching drivers, definitely something to consider with a linear driver (which is what you are using!).

((Keep in mind i have pretty much no idea what i am doing and i'm going completely off random research i have done online so most of this could be wrong just need to be pointed in the right direction))

I am going to be using a LM1085 regulator (my goal is to have a potentiometer so i can regulate the power from 500mA all the way up to 1,700mA (1.7A) -I don't want to go to 1.8A for fear of blowing the diode.-
( http://www.ti.com/lit/ds/symlink/lm1085.pdf) data sheet
The data sheet says that the max Voltage drop out is 1.5volts

That component looks fine to me.

So using Ohms law i should be able to figure out the resistance needed
(I = V/R)

v = Voltage drop out
I = Amperes
R = resistance in Ohms
E = Voltage

So re arranging the equation I should be able to figure out how much resistance i will need to get 1.7 amps with a 1.5 voltage drop out
R = V/I
So I = 1.7 , V = 1.5 , R = unknown
1.13 Ohms resistance

That calculation looks good to me - remember to take into account the power being dissipated by the resistor. In this case 2.55W - So you'd need a resistor rated larger than 2.55W.

Using the equation W/A = V (v being volts this time not drop out voltage)
W = watts and A = amperes
I can figure out the least amount of volts needed to power the diode at 1.7A and determine what kind of power supply i need to run the diode.
W = 2 (as classified as a 2W Laser diode)
A = 1.7 (how much amperes i need for maximum output of the laser diode)
2W/1.7A = 1.176V
so this means i need

Your biggest problem is assuming that the diode consumes 2W electrical power. Laser diodes are measured by OUTPUT power, not INPUT like LEDs are. It needs more than 2W input, it'll be more like ~10W. DTR should have the voltage/current/power OUTPUT pictures available on the product page. There will be graphs around if you search too.

Looking at these calculations I have a feeling i'm doing things completely wrong and I'm hoping someone can steer me in the right direction...

I was thinking the circuit would look something like this..?

Need some extra characters...

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#### Amply

##### New member
So I'm assuming that this is where it shows that information? https://sites.google.com/site/dtrlpf/home/diodes/445-m140-didoes

It shows 1.7A and 4.6V (Is this what i'm looking for?)

So that would mean that using the equation W = V x A
W = 4.6 x 1.7
W = 7.82
but i need to add 1.5 volts to the 4.6 to account for voltage dropout right?
so W = 6.1 x 1.7
W = 10.37
So i will need at least a 6.1V power supply to obtain 1.7amps running at 10.37W Output
Then using R = V/I
R(resistance) = 6.1/1.7
R = 3.588 Ohms?
Also would the potentiometer account as a 3rd resistor? in my circuit image (i think i drew it correctly..?) Imgur: The most awesome images on the Internet

#### DashApple

##### Well-known member
In your picture you have it setup as a voltage source if you want to do it this way , current wont be regulated and will wander with temperature of the diode ,

but if you want to know the resistors to use for a 1.25V - 4.6V output its the formula under the first picture in the data sheet . R1 Would be 121 Ohms and R2 would be 320 Ohms and will give a range from 1.25V - 4.56V

Assuming you do it that way then yes the minimum input voltage to the circuit would be 4.6+1.5 = 6.1V

///

Id recommend you need to set the regulator up as a constant current source such as people do with the common LM371 instead of a voltage source

I think you'll find the 1.5V dropout is the regulators maximum expected > Minimum < Dropout

I know this isn't variable as you want but for reference with it setup as a CC source see below ( ideal conditions )

Its feedback voltage is 1.25V , so for a current of 1.7A you would need a resistor of -

1.25/1.7 = 0.735 Ohms .

Resistor power rating

1.25*1.7 = 2.1W .

Minimum input voltage to circuit .

1.5V regulator drop , 1.25V resistor drop , 4.6V Laser diode drop = 7.35V minimum

Power lost in regulator

1.7*1.5 = 2.55W

Diode input power

1.7*4.6 = 7.82W

7.82W - 2.27W Optical = 5.55W

Total power into circuit -

7.35*1.7 = 12.495W

////

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#### paul1598419

##### Well-known member
You could also go with a 1.8 amp buck driver which would have far less loss from heat in the driver and also regulates better. DTR sells these as well, and so does lazeerer who made the driver. If you are set on building your own, Dash has given you a good linear diver.

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#### paul1598419

##### Well-known member
Trying to control the power in your laser by adjusting the current might not be the best idea. If the pot has any noise it could kill your diode. Is this going to be a handheld pointer or a desk/ lab style?

#### Amply

##### New member
It's going to be a desk/lab laser
I have proper heat sinks for the LM1085 and the laser diode module itself as well as fans to keep things cool.

Eventually I'm going to be hooking this up to a Array of these modules that will focus in at one point depending on certain computer inputs and basically move in order to change the distance where all the lasers beams converge using Arduino and various stepper motors and among other things.

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#### paul1598419

##### Well-known member
Amply, you are missing the point. That is a voltage regulator and you MUST have a current regulator. I don't actually know of any lasers that allow you to adjust the output power of your laser diode while it is operating. You would be much better off with a 1.8 amp buck driver which would give you an output of over 2 watts with a typical single element aspheric lens.

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#### Amply

##### New member
If i pitch the pentiometer idea and just have it fixed will i be ok?
I don't know if i'm skilled enough to build a buck driver yet i know very little about them. I went with linear because it has decent efficiency and pretty simple. In the future i would like to explore other driver options as i learn more.

Imgur: The most awesome images on the Internet

#### paul1598419

##### Well-known member
Yeah, you have it set up as a voltage regulator in this diagram. It won't work because the voltage will remain constant while the current will increase with the laser diodes heat, and as it is doing so, it will go into thermal runaway and destroy your laser diode. If you just have to use this IC and no others, I will take some time and see if I can draw one out for you that will actually regulate the current. As long as you keep the current constant, it doesn't matter what the forward voltage on the laser diode is. It will change as the diode heats up and drop to keep the current constant.

#### paul1598419

##### Well-known member
Sorry, I had to take care of something, so I'm getting back to later than I normally would. I checked the data sheet on this and it looks like the easiest way is to make R1 between 0.7 and 0.82 ohms at three watts. This will give an output between 2 watts and 1.5 watts. You needn't use R2. I'd heat sink the IC as it will run hot. This output power is dependent on your using a G1 or G2 lens as a compound lens will give you about 30% less.

#### OVNI

##### 2
If i pitch the pentiometer idea and just have it fixed will i be ok?
I don't know if i'm skilled enough to build a buck driver yet i know very little about them. I went with linear because it has decent efficiency and pretty simple. In the future i would like to explore other driver options as i learn more. Imgur: The most awesome images on the Internet
OP's schematic in the above link (and the first one) doesn't show up in my browser for some reason ... :thinking:

...You needn't use R2...
Agree with paul1598419's comment, configure as a constant current source instead of a voltage source. And if this part works like the LM317 configured as a constant current source, then make sure to change the output to be after R1 (i.e. the ADJ pin) and not before (i.e. the OUTPUT pin) as currently shown in your schematic.

EDIT-1: LM317 configured as a constant current source.

EDIT-2: Found a thread on using an LM1085 here on LPF.

EDIT-3: OP's schematic marked up with changes to work as a constant current source.

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#### DashApple

##### Well-known member
Most 3 pin regulators that use a method of feedback like this can be used for a constant current source in the same manner as the LM317

I'm my first post it gives you the values you need for a current source including power dissipations based on minimum dropout

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#### Isaac Clarke

##### Banned
and add in those capacitor to smooth out those spike!!!

#### OVNI

##### 2
:bumpit:

Is this allowed to get Edits to an existing post (#13) to cause the thread to reappear with added info?

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