adding resistor power ratings together?

[willbanks]

hi all. i'm turning an atx psu into a lab psu. sources online tell me to use a 10ohm 10watt resistor between the 5V wire and ground.
I don't have a resistor of that rating, but i have 2 5watt .22ohm resistors and 1 1/4 watt resistor. if i put them in series would that add the power ratings together to get around 10 watts?

[tszaboo]

No.

[kalel]

hi all. i'm turning an atx psu into a lab psu. sources online tell me to use a 10ohm 10watt resistor between the 5V wire and ground.
I don't have a resistor of that rating, but i have 2 5watt .22ohm resistors and 1 1/4 watt resistor. if i put them in series would that add the power ratings together to get around 10 watts?

if the 2 5 watt resistors have 0.22 ohm each (so 0.44 ohm combined), what about that 1/4 watt resistor?

[Zero999]

Why do you need to connect a 10 Ohm resistor to the output? I understand that some ATX PSUs have a minimum load current but don't ever recall it being has high as 500mA.

Firstly measure the voltage, with nothing connected to the output. If it's too high or oscillating then, try something more sensible such as a 50mA load first. The required resistor value and minimum power rating can be calculated using Ohm's law, which can be found using a search engine.

[alm]

Putting two 0.22 Ohm 5 W resistors in series gives you a 0.44 Ohm 10 W resistor, assuming the resistors are far enough apart to receive sufficient airflow, does not work if you put them side by side. If you put a 2 Ohm 1/4 W resistor in series, your new resistor is 2.44 Ohm, but the voltage across the 1/4 W resistor is limited to \$\sqrt{2 \Omega \cdot {1\over 4} W} = 0.7 V\$, or 0.7 V / 2 Ohm = 0.35 A. So the total current through the resistor is also limited to 0.35 A. 0.35 A through a 2.44 Ohm resistor means a maximum power dissipation of \$(0.35 A)^2\cdot 2.44\Omega = 0.3 W\$. You can do the math for any other resistance value.

Maybe you could put 40 390 Ohm 1/4 W resistors in parallel instead .

[bson]

Why does it need 10ohm on the output?  Maybe it should be 10k to sense when there's something plugged in?  If it has a minimum load, surely that's already part of the internal design; who would design a PSU that exhibits runaway voltage if the motherboard connector comes loose?  (But, I don't know anything about PC power supplies.  I just screw it into a case and plug in the connectors and forget it exists.)  Be aware that illustrative schematics frequently have a token load, sometimes designated R_L that isn't part of the supply and is just there to show where the load (the thing powered) goes.

5W resistors are smaller, cheaper, and easier to find than 10W ones, so if you REALLY need about half an amp of load, use a slightly bigger value, like 12ohm, this will keep power dissipation to about 5W even with 10% overvoltage.  And if you really do need a 10W 10ohm resistor, just bite the bullet and buy the right thing.

If you need a current sense resistor, then it should be some fraction of an ohm.

[willbanks]

the 1/4 is 10 ohms

[tszaboo]

Please, just try to solve it with Kirchhoff's voltage law and Ohm's law. Read them on Wikipedia. It should not be too hard to solve it, I think I learned it in 6th grade elementary school. It is very hard to do electronics if you dont understand the bare basics.

[alm]

If you have a 10 Ohm resistor in series with two 0.22 Ohm resistors (you can count this as one 0.44 Ohm / 10 W resistor), and put 5 V across that set. How much voltage is dropped across the 0.22 Ohm resistors, and how much across the 10 Ohm resistor? How does this split the power between them? How much power would the 10 Ohm resistor dissipate if the total dissipation was 2.5 W?

This can all be determined from Ohm's law and a basic understanding of resistive voltage dividers (also based on Ohm's law). The power dissipation of a resistor R is the current through it multiplied by the voltage across it, or \$V\cdot I = {V^2\over R} = I^2\cdot R\$ (substitute either I or V using Ohm's law).

[kalel]

You could also use a simple simulator like this to check:
http://tinyurl.com/y8gwt3bs

If I got the values wrong, you can change them with right click > edit. If you hover your mouse over a resistor, you can see how much power it is dissipating.
You can also change the voltage on the left.

Note that in real world, there are various things that this simulator doesn't account. It should still give a rough estimate.