Converted PC PSU to a bench - why 10Ω 10W is used ?

Thread Starter

2Hexornot2Hex

Joined Apr 16, 2020
52
Hi,
During a PC PSU conversion to a bench lab, in most cases a dummy load is needed to the 5V rail.
Why in a lot of articles, and also here on the forum, a resistor 10Ω 10W is chosen as a dummy load or even some heat sink is applied to this resistor ?

Ohms law + formula for a power... we can take a regular/simplest 1KΩ resistor (for instance), so it will put through 5V / 1KΩ= 5mA and dissipate just 25mW

Why not do this instead of using a heatsink and/or purchasing some 10watts components/sandbar.
Am I right ?
What do I miss ?

BR
 

Martin_R

Joined Aug 28, 2019
101
On many pc psu's the +5v is the only voltage that's regulated, the rest are just 'about right' due to the turns ratio of the transformer. The problem occurs when the +5v has no load on it, and you draw current from (say) the +12v output. It causes the +5v rail to increase voltage, and at 5.25v approx the PSU will shut down to protect the computer. It will then reset and the process reoccures. Loading the +5v with the low value resistor stop this from occurring.
 

Thread Starter

2Hexornot2Hex

Joined Apr 16, 2020
52
The problem occurs when the +5v has no load on it, and you draw current from (say) the +12v output. It causes the +5v rail to increase voltage, and at 5.25v approx the PSU will shut down to protect the computer. It will then reset and the process reoccures. Loading the +5v with the low value resistor stop this from occurring.
It doesn't explain why a higher resistor value isn't good enough.

The resistor needs to draw enough current to keep the supply happy.
Some supplies will be fine without the resistor but some won't be.
This sound like an actual reason

A question, out of curiosity, why PSUs were designed in such a way (to have a load on 5V to be operational) ?
As a safety measure or some other engineering reasons ?
 

crutschow

Joined Mar 14, 2008
27,734
A question, out of curiosity, why PSUs were designed in such a way (to have a load on 5V to be operational) ?
You have the question backwards.
Designing them to require a load generally gives the simplest design.
The supply would require some addition circuitry (and cost) if you want it to properly regulate with no load.
Since a PC power supply always has a load, there's no reason to add the additional circuitry.
 
Last edited:

Thread Starter

2Hexornot2Hex

Joined Apr 16, 2020
52
You have the question backwards.
Designing them to require a load generally gives the simplest design.
The supply would require some addition circuitry (and cost) if you want it to properly regulate with no load.
Since a PC power supply always has a load, there's no reason to add the additional circuitry.
Ok, got it (I mean, the general idea...)
 
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