7805 So hot!

Go To Last Post
43 posts / 0 new
Author
Message
#1
  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Hi all,
I am using a 7805 and a LM3940 to supply 5V and 3V to my embedded system, which draws only 100mA.

The 7805 is so hot (after just 1 min or so) until I cant touch it (burning hot) and have to shut down to avoid damage. The DC input is 12V from a power adapter (which rated 1A). I know 7805 is dissipating (12-5)x.1A= 0.7W, but that should not be that hot right?

Attached is my schematic. Is there any problem with my power supply circuit.

I know I should add a diode between 7805 and LM3940 to prevent current flowback to 7805, but is that the reason of why 7805 is so hot? Also, the output cap of LM2940 is tantalum to reduce oscillation.

Thanks.

cs

I'm happy ytd, today, and tmr :)

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Hook up a 200 ohm resistor to that 12 volt supply and see how hot it gets over a minute. That's about the same power dissipation. The rule of thumb is if your thumb can tolerate it for three seconds the silicon can handle it for as long as you are likely to care. If energy consumption is a consideration use a buck converter.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Energy consumption is not my concern at all...

200 ohm needs to dissipate .1x.1x200=2W, so I need to use a 5W resistor.

Does that mean it is not important to add a diode between the 7805 and LM2940?

cs

I'm happy ytd, today, and tmr :)

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

First, is your 12V power supply an unregulated "wall wart" ? If it is, then it's output might be considerably higher than 12V. Especially if it's rated for 1amp, and you are only pulling 100mA. Measure the input to the 7805 to see your actual input voltage.

Second, a 7805 (TO-220) mounted in free air with no heat sink you should expect the tab temperature to be about 46C hotter than ambient at 0.7W. At 25C ambient, that 71C. That's pretty warm to the touch, but not hot enough to burn you as you describe.

Check your input voltage.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Yup, the power supply is unregulated...Sometimes it go as high as 15V!

Attaching the heat sink to 7805 ease the heat a bit, but still hot when touching (not burning though).

cs

I'm happy ytd, today, and tmr :)

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

You are basically ignoring the critical fact about linear regulators. Power dissipation is the product of the current and the voltage drop (Gee, same rule as a resistor! How strange!) But, this WAS recognized in the original post - just not carried to its logical conclusion. So, lets see the gory numbers

Vin = 15V, Vout = 5V, Drop = 15V - 5V = 10V
Current = 0.1A
Power = 10V * 0.1A = 1W

One whole watt. Thats not so much you say? Well a TO220 package (like a 7805) has a thermal resistance of 65C/W from the internal chip to ambient (free air).

So, what does this mean? It means that 1W of power dissipation will make the junction about 65C hotter than local air temperature. At normal room temperature (25C), that means that the chip inside will be 25C + 65C = 90C. But, you say, thats only inside? True, BUT, without a heat sink to remove heat, the tab and body will be ALMOST the same temperature as the chip, say 80C or so. I'd call that pretty hot.

This long story tells us that, for THIS component, 1W (or even 0.7W) really is quite a bit of power dissipation.

You have several choices: (1) add a heat sink, (2) use a lower supply voltage, or (3) reduce the current, or (4) switch to a buck converter. Adding a series resistor (or diodes) as some may suggest, will reduce the voltage at the regulator input, reducing ITs power dissipation. But, this strategy does not reduce the total power dissipation. It only shifts its location a little. If its all going to be on a board or in a box, then you have gained very little, at all.

Jim

Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

What I usually do: Add a heatsink (as suggested)
and dont worry if the heatsink has a temperature
of 40C or even 50C.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

I also add a heatsink, it feel a hot but still ok to touch, I have tried to run the board for few hours continously without problem (touch wood)...

My question is, with this heat problem, can the board run 24/7 in a commercial environment where it is put inside a metal casing and in isolated area with no fan (not much air-circulation) like that on the roof...for 365 days non-stop?

cs

I'm happy ytd, today, and tmr :)

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Why not just do the job properly, and stop wasting electricity/warming the planet.
Use a switcher - I believe it's fairly easy these days as companies like Linear Technology have on-line tools that work out what components you need, order them for you and then send round someone to build a prototype.

BTW, what happens on the 366th day?

Four legs good, two legs bad, three legs stable.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

If you use a switcher in a commercial environment one
expects that you fulfill the EMI regulatons.
With a switcher you may run into trouble with
emissions if you have no experience in filtering.

So I still vote for a linear regulator. What is the
maximum ambient temperature for you equipment ?
How large is the "box" your equipment is in ?
If the surface of the box is big enough and the
ambient temperature is low enough there will
be no trouble in dissipating 1 or 2 Watts.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Suggestion, use an LDO such as LM2940. It will help.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

chartman wrote:
Suggestion, use an LDO such as LM2940. It will help.
Brilliant idea! Apply for a Nobel Prize immediately! :D:D:D

Warning: Grumpy Old Chuff. Reading this post may severely damage your mental health.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Why will a LDO (linear) regulator help (genuine question)

LM2575 5v switching regulator would be my choice - one inductor, one schottky diode and a couple of caps and you'll get virtually no heat from it at those supply voltages and current consumptions

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Quote:

My question is, with this heat problem, can the board run 24/7 in a commercial environment where it is put inside a metal casing and in isolated area with no fan (not much air-circulation) like that on the roof...for 365 days non-stop?

Things in direct sunlight can get to over 70c easily, in a roof cavity maybe a little less. If your temp rise is 40c over ambient, then we could be over 100c. As mentioned, waste less or get a bigger heatsink. Recom have some nice 7805 drop in replacement switching regs that will solve your heat problem, but these are 20 times the cost of a 7805.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

[Mods, please move this thread to General Electronics forum.]

chanseng738,

What you really need is something like this:

Attachment(s): 

Warning: Grumpy Old Chuff. Reading this post may severely damage your mental health.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

MBedder wrote:
chartman wrote:
Suggestion, use an LDO such as LM2940. It will help.
Brilliant idea! Apply for a Nobel Prize immediately! :D:D:D

I'm not sure absolutely sure that one can apply. I believe you have to be nominated.
To be honest, I doubt that suggesting an LDO is sufficient grounds anyway.

Four legs good, two legs bad, three legs stable.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

chanseng738 wrote:
The 7805 is so hot (after just 1 min or so) until I cant touch it (burning hot) and have to shut down to avoid damage.

Don't panic.

Measure the temperature and check the datasheet for permitted junction temperature range. For example, ON Semi's 7805 operating temperature range is from chilly -65 up to boiling +150 degrees Celsius (+302 in Fahrenheit scale).

Cool heat sink is a symptom of bad engineering.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

dak664 wrote:
... The rule of thumb is if your thumb can tolerate it for three seconds the silicon can handle it for as long as you are likely to care. ...
The rule of thumb where we work is if it is hot enough to burn the co-ops thumb then it needs a heat sink.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Quote:
Don't panic.

The only thing you shouldn't do is not to exceed parameters specified in Absolute Maximum Ratings section.
I am not sure about that but do all LM7805 have internal over-temperature and over-currect shutdown circuitry?
Use LM317 (NS,ON,TI) which has this feature and it is virtually impossible to damage it or the circuit it is powering. When it gets too hot or shorted, it simply shuts down and restarts after a while.

Anyway, linear regulators work up to 125*C, some even up to 150*C, Zeners to 175*C, so do not keep (Si of) these at 70*C or they get cold :)

Comparing similar solutions of linear and switching regulators, it seems the former are about 15x cheaper, if the energy is not a concern.

Does anybody know a reasonable price switching solution (0,1-0,2A), at a price even close to LM317 (TI costs about 0,13$)?

No RSTDISBL, no fun!

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

What no one has said about heat sinks is that one needs to use one of the correct size and attach the part correctly. For the T0-220 package used by the 780x series parts there is quite a range of heat sinks sold ranging from small clip on types to massive finned blocks of metal. In any case you need a thin film of thermal grease between the back of the IC and the heat sink. Too much grease is not a good thing. The heat sink surface must be flat and the IC torqued down with the right sized screw. (I've seen pop rivets used, not good!). The 780x series does not normally need an insulating washer between the IC and the heatsink, unless you need to float the DC ground for some reason.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Seems to me that if you don't want to redesign your power supply to utilize a switching core, the easiest way to reduce power dissipation is to choose a better wallwart that is more closely matched with your requirements.

Given you have a 7805 that is not an LDO, you could probably use a REGULATED 9V wallwart and reduce power dissipation in your circuit.

With 15V in at 5V@100mA out you have 1W being dissipated in your 7805 right now. With 9V in at 5V@100mA out you would have 0.4W being dissipated instead.

Toss out the 7805 and replace with an LDO and you could use a 6V regulated wallwart and now you would be at 0.1W dissipation.

cheers,
george.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Or you could use a pre-regulator such as a LM117 set for 9v output (or find a 7809) ahead of the 7805. Yes, this just puts two hot IC's on the board, but each one will run cooler that just the 7805 alone and you might get by with smaller heat sinks. Switcher regulators work their magic by not drawing power all the time, they use PWM to regulate the voltage and draw less total power. However this requires storing energy in an inductor (magnetic storage) or a capacitor (electrostatic storage). NSTAAFL

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

georges80 wrote:

Given you have a 7805 that is not an LDO, you could probably use a REGULATED 9V wallwart and reduce power dissipation in your circuit.
Given the regulated wallmart is allowed, using the 5V one eliminates the 7805 at all and leaves only a 3.3V LDO. So your suggestion to use a 9V regulated wallmart to regulate it further to 5V is just another gibberish.

Warning: Grumpy Old Chuff. Reading this post may severely damage your mental health.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

I cant use 9V supply because I am using the 12V unregulated supply too to power other components. So, I need 12V and 3V only.

I also cant redesign my PCB because it is already printed. Can I use a big heat sink to dissipate the heat? I am also adding a power resistor to share the heat with 7805, that will hopefully reduce the heat to a manageable level...

cs

I'm happy ytd, today, and tmr :)

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

MBedder wrote:
georges80 wrote:

Given you have a 7805 that is not an LDO, you could probably use a REGULATED 9V wallwart and reduce power dissipation in your circuit.
Given the regulated wallmart is allowed, using the 5V one eliminates the 7805 at all and leaves only a 3.3V LDO. So your suggestion to use a 9V regulated wallmart to regulate it further to 5V is just another gibberish.

Gibberish eh... very technical...

Using a 5V wallwart is not the same as having a regulator onboard. The onboard regulator assures 5V on the board at a specific noise level (linear regulator) and regulation level under various loads.

A wallwart with 'some' length of wiring will introduce voltage drops that are current draw specific. It also means if someone plugs a 12V wallwart etc in the board will be destroyed. At least a 9V wallwart would reduce his power dissipation while still maintaining reverse polarity protection on point of load regulation in his existing design.

Anyhow, as I wrote "IF HE DOES NOT WANT TO REDESIGN HIS BOARD" using a 9V regulated wallwart is a good option.

cheers,
george.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

chanseng738 wrote:
I cant use 9V supply because I am using the 12V unregulated supply too to power other components. So, I need 12V and 3V only.

I also cant redesign my PCB because it is already printed. Can I use a big heat sink to dissipate the heat? I am also adding a power resistor to share the heat with 7805, that will hopefully reduce the heat to a manageable level...

Sounds like you've designed yourself into a corner.

Anyhow, yes, just use an APPROPRIATE sized heatsink for your dissipation levels AND maximum ambient temperatures Wouldn't hurt to put a thermocouple on the regulator once you have the heatsink in place and see what kind of delta you get from the 7805 and ambient.

For future projects you will hopefully consider input voltage and regulator choice more carefully :)

cheers,
george.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Though I prefer NOT to admit it, I had a commercial design much like chanseng describes. A 7805 was mounted on a circuit board in a small metal enclosure. It was commercial (all the more shameful). The regulator was simply bolted to the board where there was some copper to "improve" heat removal. Yes, the regulator was hot, but I could hold my finger on it with a little discomfort.

Then, after a year or so, they started coming back. The regulator still worked. Sort of. BUT, the ECB was a total mess. The resin had "evaported", leaving only the woven fiber glass. Everything wggled. The board traces were hanging there in mid air. Even some of the solder had migrated away from the pads used by the regulator.

I HAD to redesign the device. The regulator was attached to the inside of the metal box to give it better heat removal. It still was not good, but it was acceptable. It cost us a LOT of money.

So, my recommendation? Hang the fact that you already have a board laid out. Do it right, or it will cost you, far more, in the end. And, it won't be "just" money!

Jim

Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Using the copper on the board as a heat spreader isn't the worst idea (certain SMD regulators HAVE to be done this way). However you can't use the standard thickness of copper, you'd have to specify PC board material with thicker copper, as well as a substrate material that is rated for the temperature.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

There are drop-in switching regulators that will fit in the 7805 footprint. They aren't cheap, but it's one alternative. Another choice would be to make your own dropin using one of the "simple switcher" style regulators on a daughterboard.
/mike

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

in your original post you have a drawing that states : 12V DC power supply, and you are using a 12V AC power supply, so you violate your own design parameters.

Also in that drawing you have a 12V node, do you use the 12V on other parts of the board, or is it just to tell yourself that there should be 12V there? Note then again that you just measured yourself that this is not the case. There also will be a extremely large ripply on the 12V anyway. You only use half the sinewave making the 7805 really work for the output to be regulated.
why not really start using a DC 12V ( or if you can lower) power supply. Or add a full bridge rectifier with large cap in your design to at least have a more stable 12V supply.
But like others said, best option is to use an smps to get the 12V(peak probably about 18V) down to 5V. This will save a tremendous amount of current, and thus you can use a lower power supply.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Quote:
What no one has said about heat sinks is that one needs to use one of the correct size and attach the part correctly. For the T0-220 package used by the 780x series parts there is quite a range of heat sinks sold ranging from small clip on types to massive finned blocks of metal. In any case you need a thin film of thermal grease between the back of the IC and the heat sink. Too much grease is not a good thing. The heat sink surface must be flat and the IC torqued down with the right sized screw. (I've seen pop rivets used, not good!). The 780x series does not normally need an insulating washer between the IC and the heatsink, unless you need to float the DC ground for some reason.

In case nothing change and used heatsink.

Used thermal grease and mica plate(but didn't important).

With thinner plate of Aluminium(not Aluminium foil) better than bold plate.It dissipated the temperature faster to air than bolt plate of aluminium.
With lattice grid(thin plate) is sufficient.

There's somewhere at electronics store for standard heatsink for TO-220.....

Quote:
The regulator was attached to the inside of the metal box to give it better heat removal.

Place the 7805 at the edge of PCB.

Jeckson

.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Quote:
With 15V in at 5V@100mA out you have 1W being dissipated in your 7805 right now. With 9V in at 5V@100mA out you would have 0.4W being dissipated instead.

And if you provide additional series resitor, you can cut that down to 1,9V*0.1A=190mW. Even SO-8 can dissipate 600mW @25*C, your TO-220 will get bored.
With linear regulator you can achieve 5/12~40% energy efficiency. With high end smps that is perhaps ~80% with at least 15 times higher cost.

No RSTDISBL, no fun!

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

This is a switching drop in replacement for the 7805:
http://www.mouser.com/ProductDet...

The cost is $3.93 USD for one. It should run much cooler.
Matt

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

meslomp wrote:
in your original post you have a drawing that states : 12V DC power supply, and you are using a 12V AC power supply, so you violate your own design parameters.

Also in that drawing you have a 12V node, do you use the 12V on other parts of the board, or is it just to tell yourself that there should be 12V there? Note then again that you just measured yourself that this is not the case. There also will be a extremely large ripply on the 12V anyway. You only use half the sinewave making the 7805 really work for the output to be regulated.
why not really start using a DC 12V ( or if you can lower) power supply. Or add a full bridge rectifier with large cap in your design to at least have a more stable 12V supply.
But like others said, best option is to use an smps to get the 12V(peak probably about 18V) down to 5V. This will save a tremendous amount of current, and thus you can use a lower power supply.

No, I never use 12AC. I only use 12VDC and 3VDC.Yes, I use 12VDC in other components. I dont have to worry on the ripples in 12VDC because the components has its own filtering.

cs

I'm happy ytd, today, and tmr :)

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

matt6ft9 wrote:
This is a switching drop in replacement for the 7805:
http://www.mouser.com/ProductDet...

The cost is $3.93 USD for one. It should run much cooler.
Matt

Looks like a great substitute with such a reasonable price! Do I need heat sink for this one?

cs

I'm happy ytd, today, and tmr :)

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Wouldn't need a heat sink.

Cool part.

The largest known prime number: 282589933-1

Without adult supervision.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Read the datasheet right there on the Mouser page "“No heat sink” direct replacement for 3-terminal
78xx-series linear regulators"

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

The inductor is on the front side of the PCB, like in the picture on Mouser. On the back is an 8 pin SOIC, a diode and some other height variable small surface mount parts. (some are maybe 0402, very small)

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Why OKI-78SR-5/1.5-W36-C is so cheap as compared to other switching regulator like
Recom 5V/1A Switching Voltage Regulator USD12.45! One Recom can buy 4 OKI-78SR-5/1.5-W36-C! Moreover the Reciom is rated at 1A only and OKI-78SR-5/1.5-W36-C is rated at 1.5A. What is the catch?

http://www.nghobbies.com/cart/in...

This one DE-SW050 from Dimension Engineering is rated at 5V/1A too but cost USD15!
http://www.dimensionengineering....

Why the price difference is so big?

cs

I'm happy ytd, today, and tmr :)

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Guys, I think 100mA regulator is needed, not a 1,5A monster. Do you know about some smaller calibre switching devices of aceeptable price (at least +-about the price of linear ones)?

No RSTDISBL, no fun!

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

If 1.5A cheaper than 100mA regulator, why not?

cs

I'm happy ytd, today, and tmr :)

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Murata OKI-78SR-5/1.5-W36-C, 5V/1.5A $3.93 (Mouser)
Murata 7805SR-C 5V/0.5A $11.78 (Digikey)

Wonder why a 1.5A rated switching voltage regulator is 3 times cheaper than a 0.5A rated similar regulator from the same manufacturer!

cs

I'm happy ytd, today, and tmr :)

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Quote:
Wonder why a 1.5A rated switching voltage regulator is 3 times cheaper than a 0.5A rated similar regulator from the same manufacturer!

Three times higher current, three times lower price. Seems (inverse) proportional. Then 100mA one would cost 60$ :)

No RSTDISBL, no fun!