Voltage Regulator 7805 (5v) Overheats Without Heatsink
Mahonroy
Posts: 175
Hey guys,
I have a circuit that is powered from a 13.5v - 15v power source. I am running this into a 12v voltage regulator, and a 5v voltage regulator. After roughly 30 seconds to a minute, the 5v voltage regulator overheats and the voltage drops down to 2.5v or so.
I put a heat sink on the 5v voltage regulator, and the problem is 100% fixed, but this thing still gets really hot! It has not overheated and malfunctioned since though.
So my question is, is this a reliable design? I was considering putting a 9v voltage regulator in front of the 5v voltage regulator, so it does not have to dissipate so much in the form of heat... but then my circuit would have 4x voltage regulators in it (12v, 9v, 5v, 3.3v). Would this be the proper way of doing it though or is there something I am missing? Should I get a 5v voltage regulator that has a higher heat rating, and just stick with the heatsink as well? Thanks and any help/advice is greatly appreciated!
I have a circuit that is powered from a 13.5v - 15v power source. I am running this into a 12v voltage regulator, and a 5v voltage regulator. After roughly 30 seconds to a minute, the 5v voltage regulator overheats and the voltage drops down to 2.5v or so.
I put a heat sink on the 5v voltage regulator, and the problem is 100% fixed, but this thing still gets really hot! It has not overheated and malfunctioned since though.
So my question is, is this a reliable design? I was considering putting a 9v voltage regulator in front of the 5v voltage regulator, so it does not have to dissipate so much in the form of heat... but then my circuit would have 4x voltage regulators in it (12v, 9v, 5v, 3.3v). Would this be the proper way of doing it though or is there something I am missing? Should I get a 5v voltage regulator that has a higher heat rating, and just stick with the heatsink as well? Thanks and any help/advice is greatly appreciated!
Comments
If you have to drop 15 volts down to 3.3 volts at one amp that's 11.7 * 1 = 11.7 watts of power that must be dissipated in the regulator.
That will make them very hot, and fail, unless you have a huge heat sink.
That is assuming your regulators are linear devices like the 7805. Very inefficient.
At this point you will want to look at switched mode regulators. Which are much more efficient.
There are many such switched mode regulators you can find on the net or if you are feeling brave make your own with a device like the MC34063. It's not so hard.
Thanks for the reply! I measured it at 16 mA.
Its actually dropping down to the 5v voltage regulator, the 3.3 voltage regulators input is connected to the 5v output.
May I ask how you measured the current?
I broke the "power in" wire, and ran one end in to the negative terminal on the volt meter, and the other end to the positive terminal on the volt meter. I connected the volt meter terminals to the common and mA reading, then clicked it over to read mA, and it read 16mA.
I'm a bit of a newb when it comes to electronics (I'm a software engineer), so let me know if I didn't do this correctly haha.
EDIT:
Yep sure enough, I didn't measure it right... broke the wrong wire.
Ok I got 180mA now, it should be right.
Still something like 2 watts.
I would have thought a reasonable heat sink would be enough to handle that and everything would be OK.
What kind of heat sink do you have? Do you have any thermal compound in there?
Does efficiency matter to you?
Yes you are correct, I am using this heatsink: http://i01.i.aliimg.com/img/pb/949/985/512/512985949_485.jpg
And everything is working fine with this heatsink. Without the heat sink is when it overheats. When I'm assembling these things, I will use thermal compound for a better thermal connection, but not currently using any.
Efficiency does not matter to me, I just want it to work and be reliable. For instance after this device is enclosed inside a metal box, is in a warm room, etc. I don't want it to have any failures... I just am not sure if this is an acceptable design or not. But if this is all perfectly normal and if this is how its done, then I'm fine with that. I just don't know any better is all.
That's a tiny heatsink.
Also, remember, if you mount all this in a closed case, where is the heat to go? The case will require ventilation of some sort.
Here's a slick little device that will drop your 15V down to 5V --- it's a switching regulator with all the parts packed into small package.
http://www.adafruit.com/product/1065
I haven't tried it but at 1A output it should handle your 180ma okay. At 90% efficiency it shouldn't even get warm; ie no heatsink.
Another option is to put a 50 ohm 2 watt resistor in series with the linear regulator. That way the resistor bears the brunt of the heat. Or ten 500 ohm 1/4 watt resistors connected in parallel (instead of the 50 ohm) would also do the trick. Kludgy approach, but it'll get the job done cheap and make your LDO reliable. ---:zombie:
1st approach = $14 + shipping, 2nd = $1
Dom..
Input 7-28V, output 5V 500mA max.
Its similar to the adafruit but way cheaper. I use it on a commercial board.
http://www.digikey.com/product-detail/en/R-78E5.0-0.5/945-1648-5-ND/2834904
ditto that, this is the part to use because it plugs in the same way as a 7805 AND it's cheap and I use it commercial product all the time now (used to do my own).
Mahonroy, once you use a switching regulator the 5V load will still be 180ma = 0.9W but the 15V current will probably be around 70ma.
The linear solution while fine for testing is not something you want to build into a box and leave running as the heat will eventually cause something to fail, either the regulator or in other cases (not yours) it drys out the electrolytic capacitor on it's input (from an AC supply).
You lost me.
The power consumed by your circuit is given by the current it consumes multiplied by the voltage it runs at:
P = I * V
Where P is the power in Watts, I the current in Amps and V the voltage in Volts.
For example if it sucks 500ma at 3.3v the power consumption is 0.5 * 3.3 = 1.65 Watts.
But now, if that voltage is provided by a regulator dropping down some higher input voltage what would the to total circuit look like from the outside?
Say we had a 15 volt input. The power consumed by the entire circuit, assuming a 100% efficient a regulator, is still P = I * V but now the V is 15 volts. So we have P = I * 15 watts.
But that P is the same P we had before, 1.65 watts. So we have:
1.65 = I * 15
or
I = 1.65 / 15 = 0.11 = 110 ma.
Much less than the current drawn by our target circuit.
Point is that energy is conserved here. If volts go down current must go up and vice versa.
Of course if your regulator is not 100% efficient this does not work out exactly.
I suppose I should clarify that the above post deals with a lossless regulator. Switching regulators can be very efficient so this is a good start.
As opposed to the good old fashioned linear regulators which can be thought of as just a resistor that magically changes it's resistance so that the output voltage remains constant.
For a linear regulator the current at the input to the regulator is the same as the current through your end circuit.
That in turn means the regulator is consuming power P = I * V. Where the V is the voltage dropped across the regulator Vin - Vout.
This can get big when input voltages are large, output voltages are small and circuit current consumption goes up. Hence the need for heat sinks. Ans why a switching regulator is preferred here.
Do you recommend I have the output of my 12v linear regulator connected to the input of this 5V switching regulator? Otherwise, the main power is 13V to 15V, I could run this to the 5v switching regulator instead if it still plays nice with a non constant input?
Why do you have a 12V regulator, is it that you need exactly 12V because if that is the case then that's not guaranteed if you only have 13V supply, it needs more "headroom". The +5V switching regulator is fine for up to 28VDC so you don't need to feed it from another regulator. As they are designed as regulators they respond both to line variations (input input) and load variations, that's what they do, otherwise what's the point of that 12V regulator? I suspect what you really need is to feed the nominal 12V supply direct into the 5V switcher.
The heat is inevitable. The 7805's are linear regulators which means they typically convert voltage to heat.
For example, I had a Sony home theater receiver once that used a 7805 to convert 12V to 5V. It had a heatsink but not enough because after 10 years the heat killed the regulator.
If you can run it off a lower voltage, you'd avoid the heat problem. Or, you can stick a big heatsink on the 7805 and it'll probably be fine. TO220's have pretty excellent heat dissipation.
You might look at Pololu for switch mode power supplies -- buck regulators -- that efficiently convert between voltages. They're inexpensive and work well.
I've been playing with surface mount LM2940's which are automotive grade regulators. Apparently they have a circuit that reduces heat dissipation when the regulator is operating at a dropout voltage in excess of the LDO range.
Michael
Heater - I guess I didn't ask my question clearly.
I was confused by Peter's inference that two in-series regulators would each have different currents flowing.
The basic concepts you succinctly explained, I'm familiar with.
Surely one can put any number of switching regulators in series and the currents will be different between each stage as the voltages are boosted or bucked. After all the power (I*V) flowing between each stage must be the same. Assuming 100% efficiency of course.
Or do you mean something else?
...gonna have to go draw some diagrams to help in the discussion. It'll be awhile as I need to spend the rest of the day hanging kitchen cabinet doors.
Interesting thread. Can someone verify that this would be a drop in for our old favorite 7805?? If the pin spacing is the same as the 7805 I think I'll replace it on my latest project given the issue of the 5 vdc regulator on the Prop Proto board (the one no longer available...) ooops, my bad, I meant it would be a replacement on my SpinStudio Prop Proto board (no longer available) .
This application would put the device in an enclosure close to a HF communication receiver. Switching noise??
Mike B.
/QUOTE=Mahonroy;1297785]Yes you are correct, I am using this heatsink: http://i01.i.aliimg.com/img/pb/949/985/512/512985949_485.jpg
And everything is working fine with this heatsink. Without the heat sink is when it overheats. When I'm assembling these things, I will use thermal compound for a better thermal connection, but not currently using any.
Efficiency does not matter to me, I just want it to work and be reliable. For instance after this device is enclosed inside a metal box, is in a warm room, etc. I don't want it to have any failures... I just am not sure if this is an acceptable design or not. But if this is all perfectly normal and if this is how its done, then I'm fine with that. I just don't know any better is all.[/QUOTE]
As for EMI, I can only say that these pass EMI for FCC and CE with our 3 propeller design. The only issue we had was the PWM output which was driving a transistor for the backlight of a 4x20 LCD display failed EMI. But since the backlight was less than 30mA we could remove the transistor and all passed.
They are a great block device when you have large input voltages.
Mike B.
The current doesn't change no matter what it goes through. If one amp enters a string of regulators, one amp comes out the other end.
Linear regulators yes, but switching regulators convert by storing energy in an inductor. You are familiar with transformers where the secondary supplied 12VAC @2A for instance yet the primary that was connected to 120VAC only drew 200ma essentially. So if the output was supplying 24W (V*I) then 24W (power, not current) is what was consumed on the primary (but not counting losses).
The losses inside a switching regulator depend upon the design itself, the components, the load current etc but are typically around 85% efficient. The losses are what generate all that heat in linear regulators so a linear regulator with 15V in and 5V out @1A would be trying to dissipate 10W which is way way too much even with a decent heatsink (that's a whole subject in itself too). Even though a 7805 might be rated for 1A that doesn't mean it can magically handle that no matter what and if you put in 15V and added a decent heatsink then you might find that you could squeeze 100ma out of the regulator before it overheats.
You are talking through your hat, go read up on the subject.
Power "loss" was mentioned simply to show that energy is lost in any conversion.
I said that the current in a DC circuit is the same everywhere. If there is one amp going into a node, there is one amp coming out. Unlike voltage drops, there is no such thing as a current drop.
So if I put a switching regulator in front of you with 15V in and we are drawing 5V @1A on the output you could show me the 1A on the input??? If you somehow managed to show me that then I would ask you where did the other 10W of energy go (5V @1A = 5W) because it is not even hot.
But you will find that with 1A out you will probably only measure around 330ma or so on the input. Do you grok it now?