Power supply dummy load resistors

Thread Starter

Napalm1432

Joined Apr 5, 2019
17
Hi there,
I'm converting a ATX power supply to a "lab power supply" and found i need to have a dummy load to be able to use it.
I know this can be done using a 10ohm/10W power resistor, but i was wondering, since i don't have that laying around, and don't really want to order parts since i have loads of resistors, can i put smaller resistors in parallel and do the same?
I have different values of resistors laying around, the one i was planning to use is a 10ohm/0.6W resistor. this would come to 16.6 resistors, so rounded 17 resistors.
 

BobTPH

Joined Jun 5, 2013
2,055
Those resistors are all in parallel, resulting in 0.6 Ohms.

Ypu could get 10 Ohms by putting 4 strings of 4 resistors in parallel.

Bob
 

Zaishens

Joined May 20, 2019
28
Like Bob said, 4 resistors in series gives 4*10=40 ohm. those 4 strings in parralel gives 1/(1/40+1/40+1/40+1/40) = 10 ohm...
the power rating you can just multiply, so the 4x4 resistor would be able to handle 16x0,6= 9.6 watt, close enough?
 

Thread Starter

Napalm1432

Joined Apr 5, 2019
17
Thanks for all the answers, and ofcourse the link to the parallel circuit guide!
I actually had less 10ohm resistors then i thought, but i do have some 12ohm resistors.

So according to my calculations, if I put 2 * 10ohm + 2 * 12ohm resistors in series, then 4 strings of those in parallel, I would get 11.36 ohm total resistance.
Both resistors use 0.6W, multiplied by 16 I will still get the desired 9.6W
Am I correct in this? If so, I'm starting to understand!
 

Zaishens

Joined May 20, 2019
28
Almost there:
2x10+2x12=44 ohm in series
1/(4/44) = 44/4 = 11.0 ohm exact...
Dunno where you got the .36 ohm from

For the power rating, it's not so easy anymore. Because you are using different resistors, they see a different voltage (V=R×I) and thus a different power (P=V×I) or P=R×I^2

Each individual resistor shouldn't be loaded more than it's power rating. And keep them apart from each other so the can cool in ambient air, more airflow helps aswell.
So the maximum voltage that can be applied to this network of resistors will be:
Sqrt(0.6 W / 12 ohm) × 4 strings × 11 ohm = 9.84 V, but it all depends on the ambient air conditions/cooling
 
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