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Is it normal for a 7805 regulator to warm up with 6V power supply?

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mik3ca

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I'm running a circuit where a microcontroller is used with a 22.1184Mhz crystal along with a radio module (HM-TRP) that continuously transmits data at 38.4kbps.

I hook it up to a wall wart as my power supply. Its specs are 6VDC/900mA.

When I run the circuit, the overall functionality is good, but after 10 mins of continuous power, when I put my finger on the part (identified as L7805CV), the part is warmer than the rest of the board, but not hot.

Some people say up the input voltage to the regulator (to 7+V instead of 6V) but I don't want to create more heat. So is this regulator going to last for a long time at warmer temperatures, or should I replace it with one that has a longer lifespan? I will consider a heat sink if t helps. In the future, I expect to have the circuit run for a minimum of 12 hours every day.
 

Externet

Well-Known Member
The heat sink is necessary, and the part better be a 78T05. The supply voltage better be at least 8 V. Measure when working.
 

mik3ca

Member
It is in a T0-220 style package and is rated to handle 1A. I did add a heatsink and didnt notice much in temperature change
 

MikeMl

Well-Known Member
Most Helpful Member
It is in a T0-220 style package and is rated to handle 1A....
But only if is bolted to a Huge heat sink (>20cm x 20cm). The postage stamp size thing you put on it is not much of a heatsink.

Here is the test that counts: spit on the T0-220. If the spit instantly boils off, it might be too hot! Otherwise, don't worry about it...

You have a much more serious problem. Look at the attached snippet from the L7805 data sheet. Look at what I circled. Do you know what it means?

l7805.png
 
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audioguru

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Most Helpful Member
The datasheet for a 7805 shows that its minimum input is 7.0V and a 0.33uF capacitor must be mounted at its input and ground pins and a 0.1uF capacitor must be mounted at its output and ground pins. Why did you feed it only 6V?

Heat is produced by the voltage across it times the current in it. You did not say the current of your load but if it is 400mA and if the input is 9V then it dissipates (9V - 5V) x 400mA= 1.6W. Its datasheet says its thermal resistance from its chip to ambient air is 19 degrees C per Watt so if the ambient is 30 degrees C its chip will be 30 degrees C + (19 x 1.6)= 60.4 degrees C which is fine since its recommended maximum allowed chip temperature is 125 degrees C.

A heatsink bolted to the metal tab needs a thin layer of "thermal compound" between them to provide good thermal conduction. The heatsink will be rated with its own thermal resistance. If the heatsink is small with a high thermal resistance then it and the 7805 will not be cooled much. The 7805 and its heatsink should be in open air, not enclosed in a small enclosure. If it is enclosed then you will probably need a fan to suck cool air in and blow hot air out.
 

mik3ca

Member
Here is the test that counts: spit on the T0-220. If the spit instantly boils off, it might be too hot! Otherwise, don't worry about it...
My spit is bad. I don't want to create shorts. Maybe I'll try a tiny raindrop of water.

But only if is bolted to a Huge heat sink (>20cm x 20cm). The postage stamp size thing you put on it is not much of a heatsink.
I used one a bit bigger than a postage stamp. Mine looks similar to the one below. It's about 2cm high by 1cm high and 1cm out.


You have a much more serious problem. Look at the attached snippet from the L7805 data sheet. Look at what I circled. Do you know what it means?
It means I can't expect 5V but that's ok. I measured with a voltmeter an output of roughly 4.2V and the circuit runs fine, simply because all parts connected to it are not fussy with the voltage. I'd be in serious trouble if I used old-fashioned 7400 series chips like the 74xx or 74LSxx series.

Its minimum input is 7.0V and a 0.33uF capacitor must be mounted at its input and ground pins and a 0.1uF capacitor must be mounted at its output and ground pins. Why did you feed it only 6V?
I upped the capacitors to smooth out ripple. I used 2.2uF minimum (with 50V rating). I was trying to make things efficient, and I noticed the 6V adapter was at a low price in the store (about 75% cheaper than the rest of the adapters), so I decided to use it. Heck, when I bought the second adapter, they were nearly sold out. and this is another reason as you stated:

Heat is produced by the voltage across it times the current in it.
So by using 6 instead of 7 volts with the same current requirements, the heat should be about 14% less?

A heatsink bolted to the metal tab needs a thin layer of "thermal compound" between them to provide good thermal conduction.
If my project continues to be too warm then I'll look into the compound. For now, I'll say I'm over-reacting.
 

audioguru

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Most Helpful Member
The 7805 regulator with a 6V input does nothing and does not regulate anything. It might reduce the voltage a little like a couple of series diodes (6V in, 4.8V out). A voltage regulator with the input always above its minimum recommended input voltage is an excellent ripple eliminator, much better than a simple capacitor. Maybe you did not have a large filter capacitor at the output of the bridge rectifier so then the voltage regulator dropped out (stopped regulating) on each half cycle from the rectifier? No, with your 6V input the 5V regulator did not regulate the voltage.
 

mik3ca

Member
Oh. Well the funny thing is that's the only thing on my board that's warm of all things. I guess in any event, I'll be safe since I haven't exceeded its absolute maximum ratings of well over 30 or so volts at input. and ok, I'll call it a funny diode then.
 

crutschow

Well-Known Member
Most Helpful Member
If you want 5V regulation with 6V input then you need to use a low dropout type such as one of these.

Since it will be dropping less voltage at 5V out as compared to the 4.2V you are getting with the 7805, it will be dissipating less power also.
 

Colin

Active Member
"Since it will be dropping less voltage at 5V out as compared to the 4.2V you are getting with the 7805, it will be dissipating less power also."

Not necessarily so . . . . . Not necessarily so . . . . .

At 5v, the circuit will be taking more current.
 
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