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Help Building Test Load

ChrisNL

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

How many watts can your resistors handle?

Problem is that i have only 1 ohm resistors @ 0.25 watt each. i think they are too light for testing isn't it?
 



AUTO XX

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The 5400 series can handle 3 amps instead of 1 (1N4000 series)
The resistors are only 2ohm resistors, 1 watt I think.
Wired in parallel, the resistance is 1/2 because basically the current has two paths to travel instead of one.
The 4 resistors in mine are: 2 parallel, series wired to 2 the other parallel.
So it works out to 1ohm @ 2W resistance....

*edit* Just something to keep in mind, I started by trying to find the cheapest 1N5400 diodes. A few weeks ago I found 1N5409 diodes for ~1/4 the price of the 1N5400.
The second two numbers are the voltage rating 00 is 50V, 01 is 100V, 02 is 200V, etc. They have no real relevance to what you will need them for here.
 
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Leodahsan

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fast question, what if I measure a driver with 3 diodes and with 5 diodes? how much the results will vary? :)
 

rhd

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Fast answer (might need more input from other users):

- Might not vary at all if your driver is capable of supplying the current at both levels. IE, if your driver is set to output 500mA, and is more than capable of supplying this for both a ~3V red and ~6V blue diode, then your measurement results might be exactly the same (500mA) for both 3 and 5 diode test loads.


AUTO XX:

- I could be wrong, but I think you need MORE than 2W resistors. I might be applying Ohm's law incorrectly. Do we use:

V x C = W
or
(C^2) x R = W

I'm not real clear on that. The first approach might suggest around 8W for the resistor (depending on the current etc), the second would only require around 2.5W. Now that I think of it, it's probably the second formula we'd follow, but I don't know now.
 

Leodahsan

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@rhd
Makes sense. Maybe the diodes heat more when low voltage :p

LM317 and FLX drivers are designed to supply many voltage levels..

+1
 

SHIN

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4 of 2W 1ohm equals one of 8W 1ohm.
P = I*V = I*I*R
8 = I*I*1
I = 2.8
So this can be used for 2.8A test load.

I heard that 5W/10W resistor like "coil in cement" may act like inductor.
The test load with this resistor may not show the exact current of some drivers.

SHIN
 
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AUTO XX

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I had it figured out before but have forgotten the numbers on it.
All I know is, built the way it is, that test load will handle 1.5A @~4.5V for more than 2 minutes (the longest I have ever taken to set a flexdrive) without getting too hot.
I could put my fingers on any part of it, it was warm yes, but not hot.
An 8W resistor might be overkill ;)
You will also need to heatsink your driver at these currents, some people forget about that.
 

rhd

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4 of 2W 1ohm equals one of 8W 1ohm.
No.....

If it's series: 4x 2W 1 Ohm = 2W 4 Ohm
If it's parallel: 4x 2W 1 Ohm = 8W 0.25 Ohm

If it's a mix of the two (2S2P): 4W 0.5Ohm
 

SHIN

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No.....

If it's series: 4x 2W 1 Ohm = 2W 4 Ohm
If it's parallel: 4x 2W 1 Ohm = 8W 0.25 Ohm

If it's a mix of the two (2S2P): 4W 0.5Ohm
Hi. rhd.

I think when I built 4(2s2p), 9(3s3p) and 16(4s4p) with 1ohm resistors, all of them are 1 ohm.

Please correct me if I was wrong...:yh:

SHIN

edit] And I think:

4S of 2W 1ohm = one of 8W 4ohm
4P of 2W 1ohm = one of 8W 1/4ohm
2P2S = one of 8W 1ohm
 
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rhd

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If you parallel equal resistances, it halves the Ohm rating.

If you series equal resistances, it doubles the Ohm rating.

So if you have 1 Ohm 1W resistors, and you need 1 Ohm 2W resistors, then you'll need to use 4 of the resistors to accomplish your goal. 2 in parallel to double the wattage (halving the resistance) and then another set in series to bring the resistance back up.

Cheers
 

AUTO XX

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Yeah, I think that's right... I need to go find my receipts or look up the code on the resistors. There is only one problem with that: I'm lazy.
I usually don't bother with the color code anyway, I just use the DMM to test resistance if it is a random resistor or get it out of my marked bags if I need a specific value.
I might have to trade something important if I memorized the resistor code tables :na:
 

ChrisNL

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i bought a resistor of 1 ohm, 3 watt. 3 watt should be enough for the test load right?
 

SHIN

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i bought a resistor of 1 ohm, 3 watt. 3 watt should be enough for the test load right?
Hi. ChrisNL.

3W 1ohm resistor can permit 1.7A.

If you want a 2A test load, the wattage of 1ohm resistor is minimum 4W.
If you want a 3A test load, the wattage of 1ohm resistor is minimum 9W.

SHIN
 

chipdouglas

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one thing that isn't metinoned. you can save time and accuracy.. if you are testing a linear driver (rkcstr, lm317) you can hook up your multimeter directly to it. just make sure it is set to the correct setting 10amp or 2 amp.

michael.
 

AUTO XX

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Also to note, lower rated resistors should work unless you leave your test load hooked up for a while. Technically mine are not good enough for the current I throw at them, they still go for more than a few minutes without getting hot.
 

ChrisNL

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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..
So when i need 300mA with 4 diodes, its 3.1V that i measure? (300mV + 2.8V for the diodes?)
 
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