# calcuting test load for a modular power supply

#### liamObsolete

Joined Jul 5, 2021
8
I was recently asked to repair 3 modular power supplies. I usually mainly work on Audio systems but I do have a reasonable understanding of linear PSU and SMPS design and I have a good amount of general fault finding experience. All of the units were the same model (ACE450F by Cosel). They have 4 otuput rails (5v, 15v, 24v and 48v). Two of the units had no 48v outputs and the third unit was dead altogether. I have managed to get all three units up and running correctly with no loads but now I need to test them under suitable loads. the labels on the unit state that the output power rating is: 400 / 450W MAX and then the individual outputs state the following: -

> 5V - 10A
> 15V - 20A (27A)
> 24V - 6.5A (8A)
> 48V - 3.2A (4A)

I'm looking at buying some wire wound load resistors to build a dummy load but I'm a little unsure of what values I should be using.
Can i just calculate resistance using V / I for each output or do I have to calculate the power first? I'm guessing where it says OUTPUT: 400 / 450W MAX is referring to the maximum combined load of all outputs... is this correct?

I understand that I might be completely wrong here so I thoughts it's definitely best I obtained a little advice so as to not blow up the PSU's which I have just spent a day fixing.

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#### ericgibbs

Joined Jan 29, 2010
15,353
Hi Liam,
I would individually test the voltages, one at a time, using these limits.
Allow +50% for the resistive power dissipation load capability.

5V - 10A = 50W
> 15V - 20A (27A) = 300W
> 24V - 6.5A (8A) = 156W
> 48V - 3.2A (4A) = 153W

E

#### liamObsolete

Joined Jul 5, 2021
8
Thank you Eric,
So would I then just calculate the resistor values using V(squared) / P

so for example; for the 48V output I would use 48(squared)... which is 2304

and divide it by the power: 2304 / 153 = 17.06

so I would need a resistance value of around 17 ohms?

#### MrChips

Joined Oct 2, 2009
25,930
Use one of the three variants of Ohm’s Law.
R = V / I

#### ericgibbs

Joined Jan 29, 2010
15,353
hi Liam
This method is the easy option
E
BTW: to buy these resistors is not going to be cheap, or easy to acquire.

5V - 10A = 50W................. 5V/10A = 0.5R
15V - 20A (27A) = 300W................ 15V/20A = 0.75R
24V - 6.5A (8A) = 156W..................... 24V/6.5A =3.7R
48V - 3.2A (4A) = 153W.................... 48V/3.2A =15R

#### liamObsolete

Joined Jul 5, 2021
8
Use one of the three variants of Ohm’s Law.
R = V / I
Thank you... yeah I thought this was ythe case but I just wanted to make sure there was nothing else i needed to account for before I ordered the resistors.

Much appreciated

#### liamObsolete

Joined Jul 5, 2021
8
hi Liam
This method is the easy option
E
BTW: to buy these resistors is not going to be cheap, or easy to acquire.

5V - 10A = 50W................. 5V/10A = 0.5R
15V - 20A (27A) = 300W................ 15V/20A = 0.75R
24V - 6.5A (8A) = 156W..................... 24V/6.5A =3.7R
48V - 3.2A (4A) = 153W.................... 48V/3.2A =15R
Thanks again Eric. I'll have a look and see what I can find.

Do you know of another way that I can test these units under load then?

#### ericgibbs

Joined Jan 29, 2010
15,353
Hi Liam,
For testing 12V and 24V supplies, I use Car/lorry headlamp bulbs.

Look at this link for ideas, you will need series or parallel combinations for 12V and 24V testing.

The 5V and 15V will need resistor loads.
The 5V @10A ie: 50W will require 5 off 2.5R at 10W in parallel

The 15V at 20A is a problem.!

I would suggest contacting a local test centre to get the PSU tested.

E
https://www.autobulbsdirect.co.uk/241-24v-truckbulb-ba15s.html

#### MrChips

Joined Oct 2, 2009
25,930
Thank you... yeah I thought this was ythe case but I just wanted to make sure there was nothing else i needed to account for before I ordered the resistors.

Much appreciated
You still need to meet the power rating!
Where will you find a 1Ω 500W resistor?
500W is a hell of a lot of power to dissipate!