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Test load for 445nm

Wallydraigle

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When setting driver current, can I use the same test load as for blu-ray, or do I have to do something special?
 



Eudaimonium

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Actually that'll be a very tough question, as these diodes can even survive 1A of current, but ordinary rectifier diodes might not be able to keep up.
We would need a high power resistor (5W or more) that will have the aproximate V drop at the desired current.

That's my reccomendation.
If we , for example , need 5 V drop at 1A, resistance will be 5 ohms, but the resistor needs to be 5W at least , if not more. Huge resistor.

We definetly need some sort of current testing for these high amp settings.
 

HIMNL9

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Less resistor value = less heat dissipated ..... stay on 1 ohm, just get 3 or 5W resistors, or make "grid" mounts (4 elements of 1W, or 9 of 1/2W, as example), if you don't find the high power ones .....

And for the diodes, yes, they are 1A, but also remember that they have an 1W maximum dissipation ..... better switch to some more powerful ones, like BY225, or 1N540x, or similars) ..... almost any silicon 100V 3A or 5A can be good alternatives .....
 
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Last time I used my test load I was setting it at 1.2A and the diodes got so hot it melted the solder and they started sliding off LOL also fried the resistor :p
 

Gryphon

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^haha I fried the resistor once aswell, fun stuff
 

Eudaimonium

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^haha I fried the resistor once aswell, fun stuff
Ahh, resistors smell really bad when fried, it has some typical smell that only resistors smell like - take a while to vent it off your workroom.

But I cannot disagree that it's fun :D
 

Bionic-Badger

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In the future, we might need to use 1N540n diodes instead of 1N400n ones for 3A.
 

ReNNo

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Less resistor value = less heat dissipated ..... stay on 1 ohm, just get 3 or 5W resistors, or make "grid" mounts (4 elements of 1W, or 9 of 1/2W, as example), if you don't find the high power ones .....
True, but we want to simulate diode. If we put just 1ohm resistor then forward voltage would be 1V. 5ohm is closer to Blue diode, it'll make 5V voltage drop on 1A current.

Even resistor is not what we want. We can use it just for certain current.
Changing current forward voltages changes differently than blue diode would act.

What we need are silicon diodes rated to higher current.
Then we can put certain number of diodes in series to simulate blue diode on any current.
 

HIMNL9

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True, but we want to simulate diode. If we put just 1ohm resistor then forward voltage would be 1V. 5ohm is closer to Blue diode, it'll make 5V voltage drop on 1A current.

Even resistor is not what we want. We can use it just for certain current.
Changing current forward voltages changes differently than blue diode would act.

What we need are silicon diodes rated to higher current.
Then we can put certain number of diodes in series to simulate blue diode on any current.
Sorry, but i think you are confusing something .....

The dropout is made with the diodes (that need to be 6 or 7, for give 4,9 / 5,4V approximatively) ..... the diode series is already that what we use for do this ..... the resistor is used simply for read the current in the range "1mA for each mV of drop on the resistor", with an 1 ohm resistor ..... if you use a 5 ohm resistor, you have 5mV of reading for each mA flowing in the circuit ..... ;)
 

ReNNo

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I know that we can use 1ohm resistor for current readings but Eudaimonium mentioned 5ohm resistor to use as a dummy load.
 

HIMNL9

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Yes, i see, but this just rise the confusion another bit ;)

I mean ..... you CAN use a resistor for the dropout, but, other than dissipating more power in heath, it change a lot the dropout related to the current ..... we normally use the diodes for cause the dropout in the dummy load, mainly cause, in this way, the dropout is almost the same (except for the resistor part), when you vary the current, so the driver become tested in a condition that is similar to the real diode one ..... the LD don't change its own dropout so much, when you change the current .....

Suppose you want to set a driver for 500mA and 1A ..... with the diodes and 1 ohm resistor, the Vdrop change from the two currents of 500mV, plus 40 or 50mV for the change in dropout junctions ..... with the resistor alone, your dropout in the dummy load change is 2,5V, from the two currents, so you are not testing the driver in an optimal condition .....
 

Eudaimonium

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I know that we can use 1ohm resistor for current readings but Eudaimonium mentioned 5ohm resistor to use as a dummy load.
drlava used 5.4 ohm or soemthing to test out efficiency of Flex and Micro at 5.4 V drop - well at least for Micro, since it can provice one amp, dunno for Flex.

Is it just me, or you two are arguing wherever you meet?
 

Asherz

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Just out of curiosity, what's the max current the linear rckstr driver hits?
 




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