Resistor dissipation

[Calibretto]

Hello gents

I've been looking through the site for a couple of days now and done some searches. I can't seem to find anything that describes in enough detail what I need to know, so I'm going to go ahead and make a thread for my super newbie question.

How close can you take the wattage to a resistor's dissipation capabilities before you're cutting it too close?

An explanation of why I'm asking:

I'm making some custom circuitry for the internal power in a computer. Specifically for some 12V cooling fans.
I've found that I want to reduce them to as close to 7V as I can, without getting under and I found these resistors.

Here's some math I then did with the help of a friendly website.

Original fan voltage: 12V

Original fan current: 0.2A

    V=IR (Ohm's Law), so 12 = 0.2R   =>   R = 12 / 0.2   =   60

Fan resistance: 60 Ohms

Target voltage: 7.045V

    V=IR (Ohm's Law), so 7.045 = 60I   =>   I = 7.045 / 60   =  0.117

Target current: 0.117A

Voltage drop across resistor: 12V - 7.045V  =  4.955V

    V=IR (Ohm's Law), so 4.955 = 0.117R    =>    R = 4.955 / 0.117    =  42.2

Target resistance:  42.2 Ohms

Power dissipated by resistor:  4.955V * 0.117A  =  0.582W

So you see. These are 0.6W resistors and I need to dissipate 0.582W
IS that going to be fine or am I cutting it too close? Second, what kind of heat development can I expect from this resistor.

Thanks in advance. I'm honored by whoever took the time to read this.

/Cal

[mariush]

The resistor is rated for MAXIMUM that wattage listed on it.  It's never a good idea to constantly have a resistor working at that wattage.

In addition, resistors have a temperature coefficient, the resistance changes with temperature. So if the resistor gets hot by running at close to its maximum wattage, the resistance will change, and the power dissipation will also change.

Conclusion: I would be comfortable with a 2-3w resistor, and I'll tolerate a 1w resistor if that's the only thing I have.

[SeanB]

First, the fan is definitely not going to be a resistive load, or very linear with lower voltage giving a lower current draw. They tend to draw the same current or reasonably close to it at lower voltage.

Resistors can typically handle the rated power when mounted in the manufacturers test fixture, typically a jig where they have a 5mm lead and then a metal block, with the resistor mounted horizontal in free air at 25C. Then they typically will have a surface temperature of 70C, 90C or 110C, all depending on the manufacturer. If they are in moving air like that from a fan then they run cooler, or conversely can dissipate more power than rated. Power dissipation is determined mostly by the placing, and the temperature of the surrounds.

I would suggest instead of the resistors using a variable speed fan, for a computer you can get variable speed fans with an inline controller, which allows the speed to be varied from full to close to off with no problems.  Otherwise try using a LM317 and the appropriate resistors, or a power transistor like a TIP110 and a variable resistor to give a lower voltage to the fan.

[Fsck]

I'd actually vote for a mosfet-based solution. Many of the high end consumer fan controllers use such a solution.

[andtfoot]

If it's a standard computer power supply, I've heard you can just hook the fan between the 12V and 5V rail to get the 7V.
12 - 5 = 7V

[Calibretto]

If it's a standard computer power supply, I've heard you can just hook the fan between the 12V and 5V rail to get the 7V.
12 - 5 = 7V

Not so good on the power supplies. Some shut down from that trick.

[andtfoot]

If it's a standard computer power supply, I've heard you can just hook the fan between the 12V and 5V rail to get the 7V.
12 - 5 = 7V

Not so good on the power supplies. Some shut down from that trick.

Ah fair enough... I've noticed that's the way a few of the speed-selectable computer fans do it, so figured it might be safe.

[Calibretto]

If it's a standard computer power supply, I've heard you can just hook the fan between the 12V and 5V rail to get the 7V.
12 - 5 = 7V

Not so good on the power supplies. Some shut down from that trick.

Ah fair enough... I've noticed that's the way a few of the speed-selectable computer fans do it, so figured it might be safe.

Yeah it's crazy it's like that when shitty PSUs can't deal with it. Appreciate the idea though.

[Calibretto]

I really appreciate the answers guys. I'll be back at the drawing board for a bit.

[elgonzo]

Ah fair enough... I've noticed that's the way a few of the speed-selectable computer fans do it, so figured it might be safe.

It is safe in the sense that this wouldn't damage your PSU (or computer components).
But yeah, that doesn't mean that the PSU will happily start up with such a dodgy wiring job...

[IanB]

Not so good on the power supplies. Some shut down from that trick.

As long as you already have sufficient other loads on the 5 V rail the power supply won't be able to tell you have done this. To the power supply it will just look like the load on the 12 V rail went up a bit and the load on the 5 V rail went down a bit.

Example:

The 12 V rail has an existing load of 1 A, and the 5 V rail a load of 2 A. You connect your fan drawing 0.2 A between the 12 V and 5 V rails. Now the 12 V rail has a load of 1.2 A and the 5 V rail a load of 1.8 A. As long as these loads are within the operating parameters of the supply, everything will work out fine.

Of course, what you can't do is feed current into the 5 V rail instead of drawing current out of it. The power supply will get very upset if you try to do that. You could always add a dummy load like a 6 V lamp to the 5 V rail if you don't already have a good load on that output to absorb the current.

[wraper]

I have actually done that trick with connecting fan between 12V and 5V rails in ~5 PSUs, all were pretty happy with that  , no problems at all.

[Rerouter]

completly agree, so long as that current on the 5V rail is actually sunk somewhere there is no issue with the 7V method,

[elgonzo]

Most PSUs will not have an issue with a 7V adapter.
Some are unlucky, like this fellow for example.
Anyhow, since you can't do any damage with it, it shouldn't stop you from trying it out...