Ohm's law and Overheating LEDs

I have a bunch of your standard 2-pin led lighting strips that I have connected to my camper's 12v power source (shore power converted to DC via camper's power center).

The problem is they run really, really hot. I have these setup in two applications and the same problem seems to exist.

In my first application I pulled out all the hardware for my fluorescent bulbs and put in LED strip inside the lighting cover and connected it directly to the campers 12 volts. my strips always seem to go out at my wire connectors. And I think it has something to do with heat damaging the connection points because there seems to be brown burnt-like residue in the vicinity.

I'm my second application, I have an Arduino controlling 15amp mosfets through which I relay the camper's 12 volts to the LED strip. My mosfets are ridiculously hot and my camper's power center will often kick the cooling fan whenever I have these lights on. Admittedly, this was done a through a breadboard while prototyping, but if it was just heat dissipation through the low amp capacity breadboard, would the power center of the RV still be working so hard it needed to cool itself? I connected my little multimeter between the LED strip and the power source and it closed the circuit to turn the lights on, but gave me an open-loop message on the readout...

If the 12-volt supply voltage is fixed, and there are resistors built into the LED strips, then doesn't that just leave amps to be determined by the rest of ohm's law? I'm not sure why so much heat is being built up in the LED strips in either of these contexts...

I guess my question is, what am I missing? Is there a relationship between resistance and heat dissipation? When I add a potentiometer to the circuit, why does it burn-out instead of just limiting the amount of current that passes through (I thought this would prevent heat build-up).

Are you sure it’s 12V? If the voltage is higher than 12V, that could be the problem. Also, it would help to know what LED strips you’re using. Are you sure they are 12V?

And, regarding power, and power dissipation, there’s another formula:

[b]P=IE[/b]

Easy to remember because it spells “PIE” :wink:

And, when you combine the two formulas – Ohm’s Law and this power formula – along with a bit of Algebra, you get the following suite of formulas, that can be used to figure out just about everything you need for DC electronics:

** **E = IR** **

** **P = IE** **

** **P = I[sup]2[/sup]R** **

** **P = E[sup]2[/sup]/R** **

So, for instance, if you want to come up with how much power a resistor is going to dissipate, for a particular voltage across that resistor: P = E2/R … so, if your LED strip has, for instance, 150Ω resistors, and there are 3 LEDs in series with one of those 150Ω resistors, and each LED is running at, say, 3.2V, then

[b]P = (12V - 3 * 3.2V)[sup]2[/sup]/150Ω = 38mW [/b]– [Thus: 38mW is dissipated by each 150Ω resistor]

If you want to know why that potentiometer is burning up: assuming your strip has 20 sets of 3 LEDs, and, again, the LEDs are running at 3.2V, then each 3 LED set is drawing the following amount of current:

[b]I = (12 - 3 * 3.2V)/150Ω = 16mA[/b]

and, the total power, for the strip, is:

[b]P = 10 * IE = 10 * 16mA * 12V = 1.9W[/b]

So, if you put that in series with a 1/4W potentiometer, worst case, you’re probably going to dissipate around a watt in that poor thing, and thus all the smoke and fire!

Now, some "LED"s used in LED strips, are actually little mini-COBs [“COB” = “Chip On Board”], made up of THREE LEDs! So, our above calculation would, then, become:

[b]I = 3 * (12 - 3 * 3.2V)/150Ω = 48mA[/b]
[b]P = 10 * IE = 10 * 48mA * 12V = 5.8W!!![/b]

So, unless you’re using one of those big-ol high wattage potentiometers, it probably doesn’t have a chance in hell!!

BTW: Your heating problem could also have to do with putting the LED strip in a closed box. You might need to provide some sort of ventilation – especially if you are using a strip with those 3-LED COBs.

1 Like

Details matter!

What LED strips? What is their voltage and power rating? What 15A MOSFETs? What circuit did you use to connect to the Arduino? Have you measured the voltage output of your "camper power center" so you know that it is 12V?

When you "connected your little multimeter" are you sure it was on the current measuring range? What EXACTLY did the display say?

Steve

ReverseEMF:
Are you sure it’s 12V? If the voltage is higher than 12V, that could be the problem. Also, it would help to know what LED strips you’re using. Are you sure they are 12V?

And, regarding power, and power dissipation, there’s another formula:

[b]P=IE[/b]

Easy to remember because it spells “PIE” :wink:

And, when you combine the two formulas – Ohm’s Law and this power formula – along with a bit of Algebra, you get the following suite of formulas, that can be used to figure out just about everything you need for DC electronics:

** **E = IR** **

** **P = IE** **

** **P = I[sup]2[/sup]R** **

** **P = E[sup]2[/sup]/R** **

So, for instance, if you want to come up with how much power a resistor is going to dissipate, for a particular voltage across that resistor: P = E2/R … so, if your LED strip has, for instance, 150Ω resistors, and there are 3 LEDs in series with one of those 150Ω resistors, and each LED is running at, say, 3.2V, then

[b]P = (12V - 3 * 3.2V)[sup]2[/sup]/150Ω = 38mW [/b]– [Thus: 38mW is dissipated by each 150Ω resistor]

If you want to know why that potentiometer is burning up: assuming your strip has 20 sets of 3 LEDs, and, again, the LEDs are running at 3.2V, then each 3 LED set is drawing the following amount of current:

[b]I = (12 - 3 * 3.2V)/150Ω = 16mA[/b]

and, the total power, for the strip, is:

[b]P = 10 * IE = 10 * 16mA * 12V = 1.9W[/b]

So, if you put that in series with a 1/4W potentiometer, worst case, you’re probably going to dissipate around a watt in that poor thing, and thus all the smoke and fire!

Now, some "LED"s used in LED strips, are actually little mini-COBs [“COB” = “Chip On Board”], made up of THREE LEDs! So, our above calculation would, then, become:

[b]I = 3 * (12 - 3 * 3.2V)/150Ω = 48mA[/b]
[b]P = 10 * IE = 10 * 48mA * 12V = 5.8W!!![/b]

So, unless you’re using one of those big-ol high wattage potentiometers, it probably doesn’t have a chance in hell!!

BTW: Your heating problem could also have to do with putting the LED strip in a closed box. You might need to provide some sort of ventilation – especially if you are using a strip with those 3-LED COBs.

wow thank you so much! This is exactly what I’ve been needing… but didn’t know I needed! I’m going to sit down and really digest this. Thanks for the time you put into the explanation!

@ReverseEMF

I want to thank you for taking the time to break these formulas out. This is perhaps the cleanest and best explanation I have seen.

Not covered in above discussion is why the FETs are getting so hot; I'd expect that for a BJT (which is an obsolete technology as far as this sort of use case is concerned), but a modern logic level MOSFET left on continuously shouldn't be getting hot unless you're pushing the specs, or this string of lights is a lot bigger than I'm imagining.

What MOSFETs are you actually using? Are you PWMing them?

What voltage are you actually getting from the "12v" camper supply? My guess is more than 12v (which is not unusual).

Those LED strips are expected to run warm - you can run them off a slightly lower voltage if you need to reduce the power consumption and heat (ie, with adjustable DC-DC buck converter rated for more than the total current that the LEDs consume - plenty more if you get it on ebay, because the ebay-from-china modules inflate the headline specs. For finding DC-DC converter on ebay, search "dc-dc step down" or "dc-dc step up" - don't use buck or boost, because all the sellers of both kind of converters put both keywords in the listing title).

A potentiometer dissipates the extra energy as heat. You can only use small pots for very low current, tens of mA total - high current pots suitable for that sort of dimming are expensive and bulky, and would generate a lot of heat.

1 Like

Hi,

Can you please post a copy of your circuit, in CAD or a picture of a hand drawn circuit in jpg, png?

Do you have a DMM?

Thanks.. Tom.. :slight_smile:

I don’t think those led strips are supposed to run on 12V.
It would have to dissipate 12V-VF = 12-2.0 =10V (for VF) = 2.0V
in the form of heat.

slipstick:
Details matter!

What LED strips? What is their voltage and power rating? What 15A MOSFETs? What circuit did you use to connect to the Arduino? Have you measured the voltage output of your "camper power center" so you know that it is 12V?

When you "connected your little multimeter" are you sure it was on the current measuring range? What EXACTLY did the display say?

Steve

Yes of course. I only figured that my problem was conceptual and not an minor voltage issue.

Here are the links to my parts

WeiMeet RFP30N06LE 30A 60V N-Channel Power Mosfet TO-220 ESD Rated for Arduino(10 Pieces) https://www.amazon.com/dp/B07CTF1JVD/ref=cm_sw_r_cp_apap_2ToLdstiHxnIJ

LE 12V LED Light Strip, Flexible,... Amazon.com: LE 16.4ft LED Strip Light, Super Bright, 300 LEDs SMD 5050, Non-Waterproof LED Tape, Flexible Rope Light for Home, Kitchen, Under Cabinet, Bedroom, 12V Power Supply Not Included, Warm White: Home Improvement

The LEDs are spec'd for 12 volts, but I assume that these products are able to handle some variability of voltage input. I'm just not sure what decimal place that range occurs in. Is 1 volt extra too many, or is .1 volts too many?

I've seen voltage readings in my camper electrical system upwards of 13.6 volts. But I haven't measured it at the light source directly, I'll do that and report back today.

======

yes, I'm using an Arduino to send a pwm signal to the mosfets in order to control the brightness of the LEDs.

【Safe to Use】: The working voltage is 12V, extremely low heat, which is safe to touch. But please do not use power adapter that is higher than 12V to supply. No worry. Just enjoy. The wattage is 40W, Please be kindly informed that the total wattage of LED strip lights should not exceed the max wattage of power adaptor.

I believe you said you are not using the optional power supply because your are running off DC ?

I connected my little multimeter between the LED strip and the power source and it closed the circuit to turn the lights on, but gave me an open-loop message on the readout...

Do you know how to use a meter ?

You didn't even tell us what you were measuring or what range the meter was on .

"really hot" is subjective.
Don't you have a thermometer ?

If not, why not buy an DS18B20 temp sensor, follow the online instructions and measure the temperature ?

You say the led strips are running hot but the vendor information above says :

extremely low heat, which is safe to touch

so who are we to believe ?

I have a bunch of your standard 2-pin led lighting strips that I have connected to my camper's 12v power source (shore power converted to DC via camper's power center).

What does this mean ? What is the Main power source ? (AC or DC)

If

converted to DC via camper's power center

Have you tried testing the led strips running straight off the 12V battery so see if they run cooler when
not running of DC converted by the power center ?

Maybe the power center 12V is not filtered enough.

rooshio:
The LEDs are spec'd for 12 volts, but I assume that these products are able to handle some variability of voltage input. I'm just not sure what decimal place that range occurs in. Is 1 volt extra too many, or is .1 volts too many?

Strips like this have three LEDs and one current limiting resistor in series.
Three white LEDs in series normally drop about 10volt, leaving 2volt across the current limiting resistor.
Rated power depends on that 2volt drop and the resistor values on the strip.
Increasing 12volt to 13.8volt (charging voltage) almost doubles the voltage across the current limiting resistor,
resulting in almost twice the power used by the LED strip.
Probably not an issue for the strip, but more power is ofcourse more heat.
12volt/40watt is already more than 3Amp, and increasing that to 6Amp might be too much for the connectors. Maybe you should be powering the strip at both ends.
Leo..

Wawa:
Strips like this have three LEDs and one current limiting resistor in series.
Three white LEDs in series normally drop about 10volt, leaving 2volt across the current limiting resistor.

There are also strips that have 3 LED "COB"s. In that case, there are three COBs and three resistors. And it's wired similar to what Wawa described, only each LED, in the COB, is like the one LED in the "Wawa string". So, if we number the LEDs in the COB as LED1, LED2 & LED3, then all the LED1s are connected in series [e.g. 3 LEDs in series], plus one resistor, also in series. All the LED2s are in series with yet another resistor, etc., thus making three series circuits. And three times the power dissipation.

The strip is advertised as 300 LEDs and 40watt@12volt.
That works out to 40mA per string of three LEDs (@10volt Vf).
Most likely 60mA LEDs (3*20mA chips inside).
Leo..

Here’s the thing. Based on the Amazon photographs, the series resistors are 180Ω, which is higher than usual [typically around 150Ω], which takes us in the opposite direction of “hot”. But, at 13.6V, they could be getting hot.

These are the 3 LED per chip "COB"s that I was talking about, so they are going to generate more heat per inch than the single LED chips stings, and if they are running at, say 3.2V each, that’s (13.6V - 3 * 3.2V)/180Ω = 22mA per LED, and thus 3 * 22mA = 66mA per COB, so that’s a bit higher than the reccomended 60mA per COB, but not much.

It is possible they’re running at something like 3.0V, which would shift the math thusly: 3 * (13.6V - 3 * 3.0V)/180Ω = 77mA, which is rather excessive. So, maybe that’s your problem!

Anyway, they are designed to run at 12V and they predict 40W, so the math says:

40W/12V = 3.33A
3 * 300/3 = 300 actual series LED paths
3.33/300 = 11mA
180Ω * 11mA = 1.98V so… 12V - 1.98V leaves 10V for the LEDs, so that means 3.33V each LED, which is plausable. But, shoot…that means these LEDs are being run really cool!

And, if we plug 3.33V into the 13.6V equation:

3 * (13.6V - 3 * 3.33V)/180Ω = 60mA

So, heck, running at the nominal current. Especially considering the LED forward voltages are only going to [slightly] increase [with increased drive voltage], which puts their running current a tad lower – and thus a tad less than nominal. But, that’s all assuming the specs are not marketing lies [or something].

But, confining LED strips in a cramped, poorly ventilated space is going to cause them to heat up, so that’s the only straw I have left.

Well, anyway, hopefully I taught you to fish [a little bit].

In my first application I pulled out all the hardware for my fluorescent bulbs and put in LED strip inside the lighting cover

I'd guess that's the problem... 40W from a straight 5 meter LED strip should remain cool.

But if you concentrate that heat into a regular 40W light bulb it gets hot. If you concentrate 40W at the end of a soldering iron you can melt solder. If you cram 5 meters of LED strip into a smaller container with no air circulation (or leave it in a roll) it's going to get warm. (The total heat energy is the same, but it's concentrated.)

I don't know what the problem is with the MOSFET, but you might need a heatsink.

My experience with white LED strips (the metal kind; in the 15-22W/m range) is that they run hot, really hot, as in too hot to touch. That's when exposed to the air on all sides. I suppose those strips are meant to be mounted onto a metal surface that can act as heat sink; if placed in an armature the heat of course goes up. This is normal for such strips, after all even with LEDs some 80% of the power supplied is directly converted into heat.
Your MOSFET running hot implies wrong type for the job and/or too low gate voltage.