jonnyGURU Forums

jonnyGURU Forums (
-   Electronic Component Discussion (
-   -   Heating up a couple of Resistors! (

frogwash 09-20-2012 12:02 PM

Heating up a couple of Resistors!
I have the need to attach a 9 volt battery to a couple resistors to heat them up just a tad to test an IR thermometer. I assume I can wire these in parallel, 1 end to + and the other to -...? I would like to heat the resistor up to about 100/105 degrees just to show a bit of heat over ambient room temp. I am a bit concerned how long this 9 volt batt is going to last also, but will find out if I can get an idea of what value resistors to start with for experimentation. I am a mechanical guy and have no idea where to begin as far as selecting the values. I am real familiar with electronic components and their uses but when it come to designing any kind of circuit... forget about it!!! Can someone give me an idea where to start? :confused:


mariush 09-20-2012 02:08 PM

Resistors have a power rating... the basic leaded resistors usually have 0.25 watt , 0.5 watt

Resistors in parallel :

1 / R total = 1 / R1 + 1 / R2 +.... 1 / Rn

So for example, if you have two 100 ohm resistors with a power rating of 0.25 watt, the resistors behave like a single resistor with a value of 50 ohm, but they can dissipate twice the amount of power, 0.25 watt x 2 = 0.5 watt.

So you want to put resistors in parallel only to increase the maximum power a resistor can dissipate (transform energy to heat)

The power dissipated in a resistor is calculated with the formula I x I X R where I is the current value and R the value of the resistor.

You have Ohm's law, which says Voltage = Current x Resistance or V = I x R

So if you want to determine the I , you have I = V / R

Let's say you want the limit the current to 0.1 A... you have 0.1 = 9V / R = > 0.1 R = 9 => R = 90 ... so let's pick 100 ohm resistor.

V = I x R = > I = V/R = 9 / 100 = 0.09 A

So now we can go back to the previous formula and calculate I x I x R = 0.09 x 0.09 x 100 = 0.81 watts

So this tells me that if I want to discharge 0.1 A in a circuit, my resistor has to be rated for something about 0.81W (that would be 1 W) otherwise it will overheat

[ - battery + ]--------- /\/\/\/\ R /\/\/\/\ ------ (to - at battery)

If you'll pick a 0.25w or 0.5w one, it will get hot fast.

You can repeat this with other values ... for example let's say a 47 ohm resistor.

V = I x R => I = V/R = 9/47 = 0.19A

Power dissipated in resistor = I x I x R = 0.19*0.19 * 47 = 1.7 watts -> you need a 2w resistor to prevent it from overheating. If you use a 0.5w - 1w rated one, it will be hot.

PS. How much they heat from normal temperature, it says usually on the datasheet, it varies from resistor to resistor .. for example on these 7w resistors, they go up to 150C when they dissipate 5w of power constantly (see page 2, graph on right) :

frogwash 09-20-2012 02:52 PM

Thanks! (I think!!!) Like I said I am a mechanical guy and totally lost when it comes to volts, ohms, watts etc.... I leave that stuff up to Sparky. Looks like I might want to start with 1 - 160 ohm, 1/2 watt res. Or 2 - 320 ohm, 1/4 watt res. This was a suggestion from someone who semi calculated what it was were are trying to achieve. If these will overheat, then obviously this is not what I want. I only want maybe 30 degrees above ambient temp (ideal room temp of say, 70 degrees). Am I to understand is it possible to stop the temperature rise at any given (+/- 5 to 10 degrees) point or will the resistor keep on building in temp? So 100 degrees will be the max I am looking for. ughh the link didn't work.. thanks anyway!

mariush 09-20-2012 04:03 PM

A 9V can not output a lot of current constantly - most can output between 15 and 100 mA. If your device tries to demand more power, they just heat up and eventually die or leak the contents and the voltage drops fast.

For example, on this website the guy tested a 9v rechargable battery and found the voltage dropped to 8.5v in less than an hour when the drain was about 50 mA:

The cheapo 9v batteries will drop the voltage even faster than that. So stay on the safe side and plan to drain up to 30 mA from the battery constantly... that's 0.03 A.

With AA batteries, these have more capacity and can handle higher drain current ... for example see this website :

As you can see, one AA battery can output 200mA constantly for 2 hours until they drop down to about 1.3v (you look at the red line, alkaline are the most best non-rechargeable AA batteries):

So you would be better off using a pack of 4 AA batteries or even just one battery.... total voltage is not really important, current and how much a battery can "output" matters.

But let's get back to the 9v battery. I picked 30 mA or 0.03 A

With this current, the power dissipated for a 160 ohm would be 0.03 x 0.03 x 100 = 144 mW or 0.14 Watts

So it's obvious, if you want the battery to last a long time (more than 10-20 minutes), it won't be able to heat up much.

If you push it to 100mA, or you switch to AA batteries, you have another story:

I x I x 160 = 0.1x0.1x160 = 1.6 watts

Now we're getting somewhere ... you need at least a 2 watt resistor for that. For example this 3w one would do:

Now this resistor will heat up, but how much, i can't tell you.

What you can do though, is to use some kind of circuit that would connect and disconnect the battery from the battery several times a second... if you want it to heat up just a bit, keep it connected 10% of a second, if you want it to heat more keep it connected 50% of the time, if you want to get it to get a hot a lot, keep it connected all the time.

Here's such a simple circuit :

You just put the resistor instead of the motor... depending on how much you turn the 100k potentiometer, the resistor is going to be connected in circuit with the battery.. you power the circuit from a 9v battery, and the resistor from a simple AA battery or a second 9V battery.

Here's the modified circuit from that page:

Digerati 09-21-2012 08:45 AM


to test an IR thermometer
That sure seems like an extreme method of testing an IR thermometer.

Why not point it at your kid's forehead (avoiding laser pointer in eyes)? Or a hot pan on the stove, or an ice cube?

frogwash 09-21-2012 11:15 AM

Thanks Mariush! It sounds like I need to switch to AA's for better performance.

Digerati... I don't think we could get a kid to travel around trade shows to show how this operates. This excersize is for a client of ours. They want 2000 portable (hand carried in a small case) units with their tester in it. This case has a "mock" panel, whether it be a circuit breaker box or some other electrical visual. This device will detect heat behind the panel showing a possible loose wire or something that will heat up due to electrical issues. Would be nice if we could just send them out the door with a kid, stove and hot pan or an ice cube...!!!

Digerati 09-21-2012 01:09 PM

lol I understand. Plus, the truant officers might get after you. But still. I think there should be something better than shorting, or nearly shorting out a battery with a couple resistors. A lightbulb perhaps. It is just for demonstration, right? Even hot water will work, or a match.

frogwash 09-21-2012 04:22 PM

I actually mispoke... it's actually a thermal imager. This again is a carry around Pelican Case type deal all contained, foam cutout for the imager, inside lid has a graphic that will look like a circuit breaker panel, behind the graphic is going to be a couple of these resistors, battery pak, wiring etc... Switch it on, grab the imager and see how hot spots will look with this tool

I am now testing a 1/2 watt 220 ohm resistor in a room at about 70 degrees. It has held fairly steady for 2 hours @ 91/92 degrees. I am now using 4 - AA batteries. After 2 hours it appears to have lost about 1 degree.

Digerati 09-22-2012 08:54 AM

Okay - I see what you are doing now. Thanks for clarifying.

You might have to come up with something that will get a bit hotter as ambient temperatures in some facilities without air conditioning on a hot summer day may approach 90F. Or the beating sun on an outside wall may obscure the image.

allikat 09-22-2012 10:50 AM


Originally Posted by Digerati (Post 92261)
Okay - I see what you are doing now. Thanks for clarifying.

You might have to come up with something that will get a bit hotter as ambient temperatures in some facilities without air conditioning on a hot summer day may approach 90F. Or the beating sun on an outside wall may obscure the image.

Direct sun on a hot day will wipe out almost all useful data from a thermal imager anyway.

All times are GMT -4. The time now is 11:31 PM.

Powered by vBulletin® Version 3.8.7
Copyright ©2000 - 2018, vBulletin Solutions, Inc.
Copyright © 2000 -