Why is my resistive heater smoking when connected to a car battery?

[Genthree]

I am constructing a resistive heater that I plan to operate using a standard car battery. Resistance has been measured at about 1 kilohm, but whenever I hook up the battery, the wire begins to smoke at the source, but the element does not actually see any temperature rise.

With resistive heating is there supposed to be a gradient in temperature between the source and drain? Are my voltage or current to high? Does anyone have any experience with this type of issue?

Thanks in advance.

 

[berkeman]

I am constructing a resistive heater that I plan to operate using a standard car battery. Resistance has been measured at about 1 kilohm, but whenever I hook up the battery, the wire begins to smoke at the source, but the element does not actually see any temperature rise.

With resistive heating is there supposed to be a gradient in temperature between the source and drain? Are my voltage or current to high? Does anyone have any experience with this type of issue?

Thanks in advance.

Welcome to the PF.

12V across 1kOhm will not generate much heat. Can you post a picture showing the setup and the problem?

 

[Genthree]

I'm afraid I can't post a picture as the heating element itself is proprietary, but I'm using a sourcemeter alternatively set to 12V or 1A. This is connected to a thin sheet resistor which serves as the heating element. When the sourcemeter is turned on, the resistor begins to smolder where the source enters. The resistor itself does not appear to heat at all.

I was under the impression that resistive heating was dependent on the current, not so much the voltage. (q = I^2 * R * t) Does voltage factor into this issue? Also, I can tweak the resistance pretty easily, what kind of values should I be aiming for to get effective heating out of a 12V car battery?

I apologize as I know this is a somewhat trivial question. I am a materials engineer, not an EE, so this is a little outside my usual field.

 

[vk6kro]

The power used by a resistor is equal to the product of the voltage and the current.

So, power = voltage * current.

If you knew the resistance and the voltage you could calculate the power like this:

Power = voltage * (voltage / resistance)... because current = (voltage / resistance)

or Power = (voltage squared ) / resistance.

In your case, the voltage is 12 volts and the resistance is 1000 ohms,
so the power = 144 / 1000 ohms or 0.144 watts.

This should not be causing any obvious heating, so maybe you have it connected wrongly or there is a short circuit somewhere.

 

[berkeman]

I'm afraid I can't post a picture as the heating element itself is proprietary, but I'm using a sourcemeter alternatively set to 12V or 1A. This is connected to a thin sheet resistor which serves as the heating element. When the sourcemeter is turned on, the resistor begins to smolder where the source enters. The resistor itself does not appear to heat at all.

I was under the impression that resistive heating was dependent on the current, not so much the voltage. (q = I^2 * R * t) Does voltage factor into this issue? Also, I can tweak the resistance pretty easily, what kind of values should I be aiming for to get effective heating out of a 12V car battery?

I apologize as I know this is a somewhat trivial question. I am a materials engineer, not an EE, so this is a little outside my usual field.

Well something strange is going on, so I'd recommend a couple of things.

1) Put a fuse in series with the output of the battery. Size it larger than the expected current. This should help to prevent catching things on fire.

2) Put a current meter in series with the output of the battery, and put a voltmeter across your resistive heating element. V = I*R is always true in a situation like your setup, so use the meters to be sure you are getting what you are expecting (based on the measured resistance of your heating element).

The heating will depend on the power absorbed by the heating element. The power is related to both voltage and current and resistance:

P = V * I = V^2 / R = I^2 * R

If you have a voltage source, the current is determined by the voltage and the resistance. If you have a current source, the voltage is determined by the current and the resistance.

My initial guess is that there is something different between the configurations for measuring the element's resistance and hooking it up to the battery, but that's just a guess at this point.

 

[berkeman]

Dagnabbit. vk6kro beat me to the punch again. By a nanosecond!

 

[Genthree]

I'm dealing with sheet resistance which changes rapidly based on how far apart the source and drain are and I am afraid berkeman might be correct about me measuring resistance under different conditions. I won't be able to play with it again until Monday, but I'll tinker with it then and report back to you guys.

Unfortunately I am more familiar with viscoelasticity than circuit design. Thanks for all your help so far.

 

[berkeman]

I'm dealing with sheet resistance which changes rapidly based on how far apart the source and drain are and I am afraid berkeman might be correct about me measuring resistance under different conditions. I won't be able to play with it again until Monday, but I'll tinker with it then and report back to you guys.

Unfortunately I am more familiar with viscoelasticity than circuit design. Thanks for all your help so far.

BTW, one other note. Current meters will typically contain their own overcurrent fuse, so check that out before connecting it up in your circuit. Be sure that the fuse you put on your battery is smaller than the fuse in your current meter, or you risk blowing the current meter's fuse. And at least for my Fluke DVMs, the current measurement fuses are a weird type, and hard to replace. Not that I've ever managed to blow one up, mind you...