Component getting too hot and I dont know why

[Schlogdawg]

Hey all,
Starting off with a thanks in advance!

My project is to use an arduino uno to light 2 LED strips that will be turning on based on the high voltage of my turn signals in my car. I have all the logic coded for that all to work. I am taking the Accessory power fuse (12V) and putting that through my arduino. The arduino is only powering the board itself. The LEDs are powered off a separate power supply that is taking positive/ground into a 7805 and outputting to the power for the LEDs. All grounds are conntected (car,arduino,LEDs). Data line for the LEDs connected up to the arduino.

The Problem:
(Picture attached)
The issue im having is with the 7805 getting suuuuper hot, where its causing the LED strips to pulse and change weird colors. I burnt my fingers grabbing it. That power supply is just taking the 12V from the car, passing it through the 7805 and sending the power at ~4.8 V to power the LEDs. Its worth noting that the wires going out to the LEDs are pretty long so I have a 1k ohm resistor connecting ground to data (I was having issues with the length of wire acting as an antenna and messing with the colors of the LEDs). The 4.8V is powering 2 LED strips with only 9 LEDs per strip.
Ive heard the common current draw from these is about 30mA per so in total im expecting to draw 540mA. In reality, i measured with my multimeter connecting the output of the 7805 screw terminal to each input power to the LED. Due to the different lengths of wire, one side is pulling 112mA and the other is 130mA. That puts the max current draw from the 7805 to be 242mA. The max current draw for the 7805 is 1.5 A, so why is this thing getting so hot.

Am I missing something when measuring out the current? Or know why the 7805 is getting so freaking hot?! Any idea on what other measurements I could make to get a better idea with the issues going on? Thanks again!

 

[davenn]

The issue im having is with the 7805 getting suuuuper hot,

That puts the max current draw from the 7805 to be 242mA

I'm not surprised it's getting hot

The max current draw for the 7805 is 1.5 A, so why is this thing getting so hot.

no, they are really rated at around 1 to 1.2 Amp and that is with a VERY GOOD heat sink

 

[hevans1944]

Do the math. You are dropping 9 V (13.8 V - 4.8 V) across the 7805 at 0.242 A, so it must dissipate over 2 W. I don't see a heat-sink attached in your photograph. Bolt a good heat sink to that puppy and see what its temperature rise is. Learn how to obey the laws of thermodynamics. Wishful thinking doesn't help.

 

[davenn]

Do the math. You are dropping 9 V (13.8 V - 4.8 V) across the 7805 at 0.242 A, so it must dissipate over 2 W.

I didn't get a chance to add that bit, I had to head out

 

[Terry01]

That's why I love to read the random posts....always something new to be had knowledge wise. These kinds of posts are invaluable for beginners. Also...now I know this, the same principles can be adapted for other things or problems I am sure to come across.

That's why I love EP!

Sorry for the off topic but thought it was worth mentioning.

 

[Schlogdawg]

I thought I had done my research in looking at the max current draw through the 7805 and I guess the source I looked at was incorrect. I had not considered the power dissipation and honestly didnt think that would make for “enough” heat to overheat the component. I looked up heatsinks and cant really find a great explanation on what to add. Obviously the bigger the better, but most places say theyre not worth it without being able to blow some air at it, i.e a fan of some sort. Is there somewhere to look where I can get some more info about how well heatsinks work? or how to pick something like that out, such that itd solve my problem? Resources would be helpful as I am not sure what I am looking for.

 

[hevans1944]

Heat sinks involve a huge amount of theory to "get it right" but often just "ballpark estimates" and experience will suffice. Look for some basic thermodynamics courses first to get a firm grasp on what heat is, how it is generated, and how it is conducted from one location to another. You need to know what temperature rise you can expect to experience as a result of dissipating electrical power as heat.

For semiconductor devices, heat is usually generated in a very small area of the "chip" and it has to be conducted out through the case by bonding the chip to the case and out through the lead wires internally connecting the chip to terminals on the case. Thermal resistance will determine the temperature rise for a given power input, but there are several thermal resistances that appear in series, from the semiconductor chip to the package case and to whatever the case is mounted on that acts as a heat sink. Often the thermal path through the wires bonded to the chip can be ignored because their thermal resistance is much higher than the thermal resistance between the chip and the case. These thermal resistances are sometimes specified by the semiconductor manufacturer, so check the datasheet.

The most important spec is the thermal resistance from chip to case because this will determine the chip temperature as a function of power input when the case is held at a constant temperature. Allowing the 7505 to "hang in air" with its mounting tab sticking up severely limits the allowable power dissipation before the chip temperature exceeds its maximum working value. And, yes, blowing air across that bare tab would help, but it isn't very efficient. A small finned heat sink with vertically oriented fins could allow air to circulate over the heat sink by convection without the need for a fan.

You cannot depend on heat sink mass to control the temperature rise. All a massive heat sink does is slow down the rate of temperature rise, which could be a "good thing" if the power dissipation required varies with time.

Heat is basically transported by three mechanisms: conduction, convection, and radiation with different thermal resistances associated with each mechanism. The temperature rise as a function of power input is non-linear for each of the three transport mechanisms, but conduction is fairly easy to approximate over a small range of power and temperature if you can assume the conducting surface is a heat sink held at a constant temperature, say by blowing ambient air over it, or circulating a cooled liquid through it.

 

[(*steve*)]

In the resources section of this site there is a resource about heatsinks. It is incomplete, but it might help.

 

[Schlogdawg]

In the resources section of this site there is a resource about heatsinks. It is incomplete, but it might help.

Thank you, this led me to some other websites, if others care:
https://www.sparkfun.com/tutorials/314

I wanted to calculate the total power dissipation for the 7805 and ran through the following:

Looking at the 7805 datasheet: https://www.sparkfun.com/datasheets/Components/LM7805.pdf
We have the Tj = 150 degC (the max temp for the 7805)
ThetaJC = 3degC/W
ThetaJA = 19degC/W

To calculate the max Power dissipation, assuming 20deg C as ambient temp:
Pd = (Tj(max)-Ta)/ThetaJA = (150-20)/19 = 6.8 W

Considering that I am only dumping 2W this should be well within the operating range. Is this an issue of theoretical vs. actual? or am I missing something else?

 

[Audioguru]

Doesn't the ambient temperature reach 30 degrees C sometimes? Isn't the 7805 tab temperature that is higher than 60 degrees C "freaking hot"?

 

[hevans1944]

am I missing something else?

You are neglecting the thermal resistance between the 7505 tab and the ambient air. This is NOT zero or you would not be able to hold your finger on the tab and declare it "suuuuuper hot".

From the TI datasheet:

NOTE 1: Maximum power dissipation is a function of TJ(max), θJA, and TA. The maximum allowable power dissipation at any allowable ambient temperature is PD = (TJ(max)–TA)/θJA. Operating at the absolute maximum TJ of 150°C can affect reliability.

I added the bold, italic, red font selection. You need to "design in" a margin of safety for reliability.

Notice that TA is defined as ambient temperature of the case (tab). How to keep TA constant (or least not increasing) when dissipating power is what you are neglecting to consider. With the tab sticking straight up in the air on your prototype, there is a very limited opportunity for passive convective heat transfer, which requires a temperature difference between the tab and ambient air to work. Stagnant (non-moving) air has a huge thermal resistance, so the tab will get quite warm with only a little power dissipated. That is why many (not all) heat sinks use forced-air cooling or water cooling to remove their heat input to the ambient air temperature.

 

[Schlogdawg]

Notice that TA is defined as ambient temperature of the case (tab). How to keep TA constant (or least not increasing) when dissipating power is what you are neglecting to consider. With the tab sticking straight up in the air on your prototype, there is a very limited opportunity for passive convective heat transfer, which requires a temperature difference between the tab and ambient air to work. Stagnant (non-moving) air has a huge thermal resistance, so the tab will get quite warm with only a little power dissipated. That is why many (not all) heat sinks use forced-air cooling or water cooling to remove their heat input to the ambient air temperature.
[/QUOTE]

Okay great, I'll look at adding a heatsink to the 7805. Thanks for the information!

 

[hevans1944]

Okay, just don't get carried away with the size. There are several varieties that are made for directly mounting on the tab. Your task is to calculate which one is best suited for your application. Assume 2 W dissipation, TJ = 125 C, TA = 25 C. A heat sink having a sink-to-ambient air heat resistance of less than 47 C/W, such as this one, should be fine. But do the math yourself.

 

[Schlogdawg]

I was doing some more research on the heat sinks and saw the different methods of attachments. Based on what I was reading, it seems more efficient to use a thermal paste than to use bolt ons. But this got me thinking, whats the "efficiency" of both. Ideally, I would think that with the thermal paste you have the ability to get a larger contact surface thus giving you better conduction. Whereas a bolt only provides conduction through the bolt. I cant find a comparison between the two in terms of number comparisons. Is there a rule of thumb for how much of that sink-to-ambient air heat resistance you get based on the attachment?

Lets say a heat sink is spec'd for 50 C/W, using a bolt do you get ~45 C/W whereas thermal paste may get you ~49 C/W?

 

[hevans1944]

Don't think I have ever seen a comparison such as you describe. The heat transfer is from the flat case (tab) of the semiconductor device to a flat area on the heat sink. The bolt securing the two has very little influence on the heat path, ideally none in the case of plastic (Nylon) hardware. The only reason to use thermal paste is to fill in the microscope valleys between peaks that all machined surfaces, even polished surfaces, exhibit under sufficient magnification. Excellent thermal conduction occurs only where the metal peaks of opposite surfaces meet, with the valleys representing conduction only through trapped air between opposite surfaces.

Thermal paste is not a very good conductor of heat, but it is a heck of lot better than air. So, filling in the valleys with thermal paste is an improvement that lowers the thermal resistance between the two surfaces. OTOH, putting on too much thermal paste means some of it lies between the metal peaks, detracting from their metal-to-metal contact and decreasing their heat conductivity by increasing the thermal resistance between the two surfaces. This problem is usually avoided by the attaching bolt(s) squeezing out the excess thermal paste and not using too much thermal paste to begin with.