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#### Last One Left

##### Member
Hey everyone i know this is a question that has been asked many many times, but i cant quite find a definite answer..i want to build my own test load, here is what i am thinking about using..

Diodes

Resistor

Im hoping to build a test load that would test up to 3Amp, so i can do all my testing using this..
Im thinking of following this design posted on the forum, using the diodes and resistor from ebay to achieve the 3Amp test load..

Please Let me know if this will work the way i am thinking, thank you..

#### Coherent Light

##### New member
Those parts will be fine to do a 3 amp test load.

#### Last One Left

##### Member
Thanks a lot man! That is all i need to know..Ordering parts now..

#### SHIN

##### Member
Hi.

Max. current of 1N400x is 1A.
Max. current of 1N540x is 3A.

If you choose 1 ohm resistor at 3A, it should be min. 9W.
[ P = I * V = I * (I * R) = 9W ]
You can measure I = V
[ V = I * R = I * 1]

If you choose 0.2 ohm at 3A, it should be min. 1.8W.
You can measure I = 5 * V

Good luck! :yh:

#### Last One Left

##### Member
Dude, you lost me a bit lol..can you please explain this a little to me..i dont understand the equation..
Which resistor should i use with the 1N540x Diode to get V=I?

#### Coherent Light

##### New member
He's just giving you the formula to calculate resistor wattage, and a 10W resistor is big enough for your project.

#### Last One Left

##### Member
Ok kool i understand, thank everyone..

#### ChrisNL

##### New member

I want to build my own also, but i'm new in this category and don't want to mess it up the first time

#### bluuray

##### New member
voltage requirement for each diode
and each of the 1N diodes have a 0.7v voltage drop(if thats what its called)
bluray = 4.2-4.5v so ( 6 * 0.7 = 4.2v )
red= 3v so ( 4 * 0.7 = 2.8v )

#### rhd

##### New member
For what it's worth - I've found that bluray (405) and kasEO (445) diodes behave a bit differently (now in reality, this may just have to do with the fact that I tent to test 445s at higher currents, so the diodes eat more voltage each). But this generally works best for me:

405: 6 diodes
445: 5 diodes
650: 4 diodes

#### ChrisNL

##### New member
In a local shop nearby me they have only the 1N4007 diodes (Diode 1000Volt 1Amp).

Can i use that one also?

#### Cyparagon

##### Well-known member
Yes, as long as you keep the current below 1A. For more, you'd need another string of them in parallel.

#### ChrisNL

##### New member
They have the 1N5408 @ 3A i saw later on, that's ideal for my future 450nm i've spotted on ebay, thanks for the info, i'm building a testload soon

#### AUTO XX

##### Active member
I used 540* series diodes on mine, works great for a 3A test load

#### LuxorLasers747

##### Member
For what it's worth - I've found that bluray (405) and kasEO (445) diodes behave a bit differently (now in reality, this may just have to do with the fact that I tent to test 445s at higher currents, so the diodes eat more voltage each). But this generally works best for me:

405: 6 diodes
445: 5 diodes
650: 4 diodes
So you say that using 5x 1N4001 diodes is the best for testing a 445 diode right?
In modwerx, there is a test load circuit wich has 6 diodes, do i need to remove one or it doesnt really matter if i use 6?

#### AUTO XX

##### Active member
The 405nm diodes eat a lot higher voltage when you get into the higher currents.
I use 6 diodes to reproduce the characteristics of a high current 445nm diode build but I have never understood the need for exact precision that way.
If the drivers regulate voltage based on the diode's demand, then why is it required to be so close to the specific voltage with the test load?
I get that you might get skewed results if you only used a 2 diode test load (1.4-2V) for a 405nm (6V) laser build but is it that important?
I have never pushed a laser diode to its maximum so I guess a little variation in the drivers' output wouldn't be as much of an issue as if you wanted to max the diode out.