3 A test load

Hi team, I went to the local store and grabbed some 1N5404 silicon diodes and a 5 W 1 ohm resistor. Since I'm building a 2 W 445nm laser there is a need to set the current as high as ~1.8 A and my old dummy load would get extremely hot and the diodes would fail to stand so much load. Anyway, I don't know how many of those diodes to use so I can simulate a 445 at 1.8 Amps. Here is the datasheet for those who are not familiar with these high diodes. http://www.datasheetcatalog.org/datasheet/WINGS/1N5406.pdf

I figured that I can determine the vdrop on the diodes on FIG.3 but I'm not sure, since somebody claimed that they would drop 1.1 Volts even at ~2 A. But that's @ 3 A isnt it? I haven't taken any electronics classes yet and I only stuck to the practical side of things without really understand how they work. Please I would appreciate it if you can enlighten me on this.

6 diodes would be enough

Each one is 0.7 volts so 0.7 x 6 == 4.2

LaZeRz said:

6 diodes would be enough

Each one is 0.7 volts so 0.7 x 6 == 4.2

Click to expand...

Right, are you certain about it my friend? Remember these are not 1N4004's Have you used them before? Also, are they 0.7 at all currents or close to 2 A ?

Well, according to the datasheet

they all have the same specs

http://www.datasheetcatalog.org/datasheet/WINGS/1N5406.pdf

So Fig.3 gives us the number we want Thanks! Do you know of any good method to heatsink these diodes? Should I place some brick of aluminum on them using arctic silver?

I've got some of those diodes. They're absolutely huge and thick. You'll need to consult the datasheet for the general forward voltage. Different manufacturers will have different forward voltage drops. ON semiconductors' have 1V, Fairchild's are 1.2Vf @3A, etc. Measuring the drop with the multimeter shows only half a volt.

I wouldn't worry about it too much, because the diodes are only there to get the dummy load "within range" of the voltage that will be dropped by your laser diode. Being off by even a volt probably won't make much difference to the driver's reference voltage you're using to set the current.

For heatsinking, I wouldn't worry too much about it unless you're working at the rails. You can leave the leads longer so the wires themselves act as the heatsink and suspend the diode in the air somewhat so there is more air movement. Look at ON semiconductor's datasheet for the 1N5400 for some suggested mounting styles.

Silvershot said:

So Fig.3 gives us the number we want Thanks! Do you know of any good method to heatsink these diodes? Should I place some brick of aluminum on them using arctic silver?

Click to expand...

If you can scavenge an old CPU heatsink you can stick them to that using some thermal paste.

"FIG. 3 - TYPICAL INSTANTANEOUS FORWARD VOLTAGE"

They drop ~0.9V at 3A, and ~0.85V at 1.8A. It looks to me like you only need 3 or 4 of them.

Your resistor will only handle 2.2A though (P = I²R). At 1.8A, it will drop 1.8V (V=IR)

This test load is not necessary for your application. Since you know what current you want to set it at, you only need to find a power resistor that drops a similar amount of voltage. Again, use V=IR and P = I²R to solve for resistance and power rating. 2.5 ohms at 10W+ would work.

If you didn't want to simulate the loading on the driver, you could even use just your 1ohm resistor.

Cyparagon said:

"FIG. 3 - TYPICAL INSTANTANEOUS FORWARD VOLTAGE"

They drop ~0.9V at 3A, and ~0.85V at 1.8A. It looks to me like you only need 3 or 4 of them.

Your resistor will only handle 2.2A though (P = I²R). At 1.8A, it will drop 1.8V (V=IR)

This test load is not necessary for your application. Since you know what current you want to set it at, you only need to find a power resistor that drops a similar amount of voltage. Again, use V=IR and P = I²R to solve for resistance and power rating. 2.5 ohms at 10W+ would work.

If you didn't want to simulate the loading on the driver, you could even use just your 1ohm resistor.

Click to expand...

Thanks Cyparagon.

As for the test load way, in case I use 4, wouldn't that be 3,4 V? While with a 2,5 Ohm 10 W resistor I would simulate a vdrop of ~4,5 Volts. Did I miss something here? Maybe I didn't mind the vdrop on the 1 Ohm resistor

Does anybody have a link to a power chart for a 445nm diode?

Is that the exact datasheet for those diodes? Some manufacturers will have different voltage drops at different currents, but it shouldn't matter too much if its still within range.

Here's the PIV plot by IgorT:
http://laserpointerforums.com/f65/3x-445nm-ld-piv-plot-53927.html

Depending on what current you want the resistor will add a noticeable voltage drop as well at higher currents. You could measure the voltage going through the resistor at your desired current and add that along with the voltage drop of the 1N5404 diodes.