# Resistor Wattage

1. May 17, 2006

### mrjeffy321

I am looking for a low resistance resistor to put in my circuit to lower the current slightly.
My circuit is/will be running off a 12 power supply and normally will draw somewhere between 11 and 13 amps under normal operation. I would like to add in an additional maybe .5 to 1 ohm resistor so as to make sure not to over load the power supply (it currently is if I leave it unattended for too long).

Obviously, when I stick in an extra resistor into my circuit and run that many amps through it, after a while, it is going to get very hot.
I want to make sure I get a resistor which will resist getting too hot. I know that resistors have a power rating, but I am not sure whether I am looking for a high power rating or a low power rating.
Does the wattage on a resistor tell you its heat dissipation or accumulation?
In other words, for a given resistance, would a higher or lower wattage resistor stays cooler?

2. May 17, 2006

### Averagesupernova

The wattage rating of a resistor tells you how many watts of power it is able to dissipate without damaging itself. The wattage is determined by muliplying the voltage across the resistor by the current through it. A good rule of thumb is to use a resistor with twice the wattage rating that will be dissipated in it. What are you powering?

3. May 17, 2006

### Staff: Mentor

Sounds like you need to use a real power resistor. You can get them in little heat sink packages, so you can bolt them to your metal package to get good power dissipation. The power rating for resistors corresponds to some temperature rise, given some sort of heat dissipation situation. For small resistors, it's assumed that they are soldered to a PCB, and the only heat sinking is just to the air. For the heat sink packaged power resistors, they assume some heat sink attachment. Just look at some power resistor datasheets to see what delta-T corresponds to the power rating. As Supernova says, the delta-T at rated power is intended to keep the resistor from being damaged. Also keep in mind that the ambient temperature range of your product comes into play.

4. May 17, 2006

### mrjeffy321

So I need to be looking for a higher power rated resistor, OK.

That is what I indended to do. I was going to try to help conduct the heat away by strapping the resistor to a metal object and possible use some of that heat transfer gel as well.

Now that tha Radio Shack near my house is closing down, I can get all sorts of stuff (like resistors) dirt cheap [as if they were expensive before].

5. May 17, 2006

### Suerbatica

This goes for ALL Electrical components. They are rated for WHAT THEY CAN DO, not what they can't.

Suer

6. May 18, 2006

### Cliff_J

Look halfway down this page, the resistor has an aluminum heatsink already attached in the diagram.

http://www.mouser.com/catalog/626/500.pdf [Broken]

current^2 * resistance is power dissapated in resistor.

13^2 * 1 = 169W!!

But when you insert that resistor, it becomes part of the circuit and therefore changes the parameters. It obviously would only dissapate that kind of power if it absorbed all the power the supply could deliver.

So instead, assuming the load you have currently is about 1 ohm, adding another 1 ohm in series will cut the voltage in half, and the power to 1/4!!!

Instead, maybe a .1 ohm resistor would be more logical, or maybe a larger power supply!

Last edited by a moderator: May 2, 2017
7. May 18, 2006

### mrjeffy321

I bought a .47 ohm, 5 watt, wireround, resistor to see just how hot it would get.
The new current through the circuit was about 6.2 amps, and within a few seconds, the resistor was so hot, I burnt my find touching it. I cooled it down somewhat by dripping water on it, but the water just boiled off.
As a temporary solution, I immersed the resistor in a can off, mostly ice, water. It does not take long for all the ice to melt and the resulting water to become quite hot too.

Actually, even at only ".47 ohms", the current is a little too low for my taste. A smaller resistor might still be needed, perhaps a .1 , .2 or .3 ohm resistor with a MUCH higher wattage rating. It looks like they do have a nice selection of small resistance resistors on that page with the Aluminum heat sink, but at only 10 or 12 watts, will that be enough?

I couldnt find any of that heat sink gel one would sometimes place on hot electrical components in order to better conduct the heat into its heat sink when I went to the store. Would this something to look into do you think, or useless for my application?

8. May 18, 2006

### Staff: Mentor

Can you describe exactly what the load is? How much can it vary? What voltage variations can it tolerate? You said something about using the resistor to ensure that you don't "overload" the 12V power supply -- what do you mean by that? There is probably a better solution than just a power resistor.

9. May 18, 2006

### mrjeffy321

The load in my circuit is an electrolytic cell which has very little, built in internal resistance. The resistance through my cell can vary a little bit depending on the internal conditions, but for all intensive purposes, it does not vary more than about +/- 1 ohm usually. The cell does have a minimum voltage needed to operate (less than 2 volts), but other than that, it can (in theory) handle any voltage higher than this.

I mentioned overloading the power supply because the power supply I am using is a computer ATX power supply. I am using two of its voltage outputs [not at the same time], either +5 volts or +12 volts. The +5 volts has a very high maximum current (~22 amps I think) which it will allow before it "overloads"/trips its circuit breaker and turns itself off. The +12 volt output of the power supply has a lower max. current (~14 amps) before it turns itself off.
I want to maximize current through the cell in order to maximize reaction speed. If I use the +5 v. output on my cell, I get a current somewhere between 2 and 3 amps [depending on cell conditions]. If I use the +12 v. output, I get a current which starts out at about 11 amps and slowly increases higher and higher, until it overloads the power supply and it turns off. Allowing the power supply to turn itself off if not good for several reasons, but at the same time, I want to get a high current.

When using a small resistor (like my .47 ohm resistor), I get a much more constant current (with a couple tenths of an amp rather and +/- 1 or 2 full amps), but the trade off is a lower current. I am willing to accept a current somewhere within the 5 to 10 amp range as long as I am sure it will not overload the power supply and it will not pose a fire hazard when I am not around to watch it.

Someone proposed to me the idea of using two ATX power supplies, wired in parallel, both using the +5 volt output. Using this method, I should be able to combine both of their currents, but at the same time, have the higher max. current protection on the power supply.

10. May 18, 2006

### Staff: Mentor

If you want to stick with a standard source power supply, then it seems like you would want to figure out what V-I range you want to be able to run your cell in, and add a post-regulation component between the power supply and the cell. Your use of a resistor is a simple first cut at the post-regulation component. If you can show that a resistor value will give you the range of current and voltage that you want at the cell, then it's the best solution. But if it is too limited in its ability to set stable V-I points for the cell that allow the power supply output to always be within its output current specification, then you'll need to use an active post-regulation component.

For example, If your calculations for your cell show that you want to always have it running at 10A, and at voltages from 3-10V, then you could make a heat-sinked 7V active voltage drop circuit out of a number of TO-3 heat-sinked NPN transistors configured for 10x Vbe as the Vce drop (with some small emitter resistors to ensure that all the transistors share the overall current well and nobody runs away and fries). 10A at 7V is 70W, and with a fan and good heat sinking you can probably to about 10-20W per transistor. If the constant voltage drop is still not sufficient for your cell's operating area, then you can do a voltage regulator version of it with multiple TO-3 heat-sinked voltage regulators, and small sharing resistors.

BTW, if you can use just a plain power resistor as the dropping element, you can experiment some with doing a square of resistors to share the power dissipation. If you want a value of R overall, and you have a number of R resistors, you can parallel up series connections of the R resistors and still get R overall. Like, (R+R) // (R+R) = R. Oh, and still use a fan to keep the square resistor assembly cooler.

Last edited: May 18, 2006
11. May 18, 2006

### mrjeffy321

Hey, now that is an idea. That just might work.
I could go out and get a couple 1 ohm, 10 watt resistors and wire them up in parellel to get my desired resistance. Since I would be using about multiple resistors with double the wattage of what I am currently using, that will work pretty well.
It is just so simple, I love it.

What you were describing with the whole constant current, transitor, ideal sounds like it would be the ideal solution to what I want to do, but since I didnt understand a lot of what you said, I probably couldnt build it.

12. May 18, 2006

### Staff: Mentor

Have you worked out the equations to figure out what value of overall resistance will do the job? It seems like it needs to be small enough to pass enough current to the cell at a voltage that the cell supports, and it also needs to be big enough to limit the current to something the power supply can support (less than the 12V output Imax spec).

You should be able to write two equations for the resistor value, to determine the Rmax and Rmin values that will work with both the power supply constraint (Imax) and the cell constraints (Imin, Vmin).

13. May 18, 2006

### mrjeffy321

I am not too sure how I would do that, I didnt plan on taking it that far.

What I had planed to do is choose the extra amount of resistance I want to add to the circuit when running off the +12 v. output, then work out how many, and of type, resistors I would need or order to achieve this equivilent resistance.

For a start, say I choose to add (to make the math easy), .33333 ohms to the circuit and I wanted to do this while still being able to dissipate 70 watts of heat (meaning I would need 7, 10 watt resistors).

1 / r1 + 1/r2 + 1/r3 + ... 1/r7 = 1/.3333 = 3
assuming all resistors have the same resistance value,
7 / r = 3
r = 3/7 ohms or about .42
But since it might be hard for me to find 7, .42 ohm, 10 watt resistors, I would just play around with the math a little more until I found something close to what RadioShack would stock.

Even though I am using the +12 volt output on the power supply, I really am not getting a true 12 volt potential across my cell, I measure it to be more like 9.something the way everything is arranged right now. This drop in the voltage makes it a little more difficult to set up equations to give me the exact current I would like to have.

Last edited: May 18, 2006