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test load

Wallydraigle

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Cause the jumper have virtually zero resistance, like a piece of wire, and the current always flow in the less resistance path (basical electronic laws) ..... if there was a resistor, in place of the jumper, the current flow in both the branches, proportionally, but if one of the brabches (the jumper) have resistance = zero, all the current flows there.
That makes sense. Thanks for explaining.
 



bhank

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I seem to have lost the loads I made for BR and red, and now that I'm getting ready to build my first 445 I think I'm going to go ahead and build one of these loads to use with all my drivers. And because it's more substantial than my last 2 loads I'll be less likely to lose it haha
 

HIMNL9

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I made one for 445nm drivers too ..... seen that now we need something that hold more current, i thoughd it can be helpful .....

I took an old heatsink, 8 diodes BY225 and a 0,33 ohm 5W resistor (wanted a 0,2, but not found one of 5W, so i had to adapt that what i've found :p) ..... glued the diodes side by side on the heatsink, placing them with 2mm of space between them, and aligned them ..... and glued the same way the ceramic resistor on the heatsink too ..... once it hardened, filled all the space between them and around with the same thermal silicone glue, and placed a strip of metal hooked with 2 screws over them for keep them better (i removed it in the pics, for show the diodes) ..... once completely hardened, i bent the leads and soldered them making "arcs", so for change the number of diodes in the load, i just need to hook the clip on the right "arc" :p ..... the two thin wires brown and green just goes to other 2 clips for hook the DMM leads .....

It born this way as "provvisory setup" for make a quick test, but seeing that is easy to use, robust enough and simple as it can be, it becomed the definitive version (also cause i'm too lazy for rebuild it with a PCB :p :D)

The only "problem" that it have (if you really want to call it problem :p), is that you need to divide the reading of the DMM for 0,33, or multiply it for 3,03 , for have the right value, and this is a bit boring ..... i was searching a 0,2 ohm so it was a bit more easy to do "at mind", cause with 0,2 ohm, you need to divide for 0,2, or multiply for 5, and multiplying a number for 5 is more "immediate" at mind ..... but, also this way, with an electronic calculator it's working good :p :D
 

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qumefox

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Wouldn't it have been easier just to use a 1Ω shunt resistor instead? then mv=ma and no multiplication is needed.
 

HIMNL9

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Yes, but with 1 ohm, the dropout rise too, and, overall, the dissipated power, for high currents ..... as example, if i use it for set a 2A load, the dropout on the resistor is 2V, with 1 ohm resistor, and if i'm using a critical schematic driver, or a boost driver near its maximum output voltage, it can give problems in the setting (or make a wrong setting) .....

Also, the resistor only, at 2A, dissipate 2W, that can seem not too much, but you have to add also the heat dissipated from the diodes ..... even as example, for a 5V LD simulation (7 diodes), to the 2W from the resistor, in this case, you have to add at least others (0,7*2)*7=9,8W ..... i can ensure you that the assembly become very hot, with high currents (i used it for a "burn-in" test on a current sink module, and after some hours, it was so hot that i can't hold a hand on it .....)

So i preferred to use a lower value .....
 

SHIN

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Jul 25, 2009
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Hi. HIMNL9.

Nice work for high current test load.;)
And I like heatsink.

At high current as 2A, is there more voltage-drop at test load diodes more than 0.7V?
Like 0.8V to 1.0V.

SHIN
 

HIMNL9

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@ SHIN: yes, as any other diode, the dropout increase with the current ..... just a bit less than normal 1A diodes like the 1N400x serie ..... with BY255, i measured 0,72V at 100mA, 0,79V at 1A, and 0,86V at 2A ..... tried 2 of them and the values was the same .....

Also tried at 3A, but only for short time, cause it's the limit current for BY255, and measured 0,89V.

Better diodes for a similar use can be glass-passivated ones, or 5 / 6A ones, like SF15, or also diodes in TO220 case, i know ..... but i had at home only these ones, when i needed the load :p



@ Kevlar: Yes, it's just a resistor with different body from the "squared" ceramic ones that i have used, but is still 5W, so it must work (also, 0,2 ohm need to dissipate less heat than my 0,33 ohm).
 

kylemiller1994

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Hey, if I wanna hook my dummy test load to a o-like bluray driver, do I do it like on this pic I added stuff to ?

Anode towards negative output or cathode towards negative output ?
 
Last edited:

lasersbee

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Your Anode and Cathode identification of the Rectifier
Diode is correct...
Your points for measuring the Voltage drop across the
1 Ohm resistor is correct..
But your (+) and (-) identification of the Laser Driver
output is Wrong... It needs to be reversed...


Jerry
 

Kookapeli

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swap the +ve and -ve signs around on the diagram :)

*edit* got there before me Laserbee :D
 

HIMNL9

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Yes, 6 diodes will emulate a 12X decently without problems (in fact they're a little bit higher in dropout than a 12X, but this just mean that if the driver works with this load, it also work good with the LD ;))
 

ped

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Many thanks big fella :thanks:
 

millirad

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Yes, HIMNL9 has helped many builders since posting a diagram of the load circuit. :)
 




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