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7805 high input problem

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eldermalagon

New Member
Hi everybody.

I´m trying to supply a circuit that requires only 500ma and I have a 24v switching power supply as input to the 7805. The thing is that I have a couple of detectors that are supplied with 24v and they have a single 7805 without heatsink and the 7805 does not get hot at all!!! but when I try to power mine with 24v it gets very hot and over time the output decreases. I would like to know if any of you have a suggestion or idea for preventing this and proper supply the 7805 with 24v.

I´ve noticed that some circuits have a small value resistance in series with the input of the IC, I´ve tried this also but the resistance gets hot too! if any of you have a circuit will be fine, o an idea to do this.
 

MikeMl

Well-Known Member
Most Helpful Member
Do a little basic math:
P=I*E
P=0.5*(24-5)= 9.5W



A 7805 (the big one in TO220 package) will require a large heatsink, a good quality thermal mounting washer, and white heat sink compound...
 

Torben

Well-Known Member
Hi everybody.

I´m trying to supply a circuit that requires only 500ma and I have a 24v switching power supply as input to the 7805. The thing is that I have a couple of detectors that are supplied with 24v and they have a single 7805 without heatsink and the 7805 does not get hot at all!!! but when I try to power mine with 24v it gets very hot and over time the output decreases. I would like to know if any of you have a suggestion or idea for preventing this and proper supply the 7805 with 24v.

I´ve noticed that some circuits have a small value resistance in series with the input of the IC, I´ve tried this also but the resistance gets hot too! if any of you have a circuit will be fine, o an idea to do this.
Hello, and welcome to the forum!

24VDC is way too high for input into a 7805 with no heat sink which is expected to supply 500mA. I don't have the formula in my head right now (It's sadly been a while since I've had to do this calculation so I hope I'm corrected if I mess up the actual numbers here) but I think that's something like 12 Watts [Edit: should be 9.5W] of wasted power which the 7805 must dissipate as heat. The 78xx series have a thermal resistance of 65°C/W above ambient from the junction to air, which IIRC means that with no heatsink, the junction's temperature will rise by 65°C above ambient for each Watt of power which must be dissipated. The maximum operating temperature is 125°C, so you can see that this is clearly being exceeded unless you're running the thing in a space which is -492.5°C. The 7805 won't like that and will go into thermal shutdown (it tries to protect itself by shutting down instead of just frying).

I think that the reason you don't see this with your other "detectors" (whatever they are) is that they draw almost no current through the 7805s. The more current you draw through it, the more power you need to dissipate as heat.

You need to bring down the input voltage to the 7805. The resistor you sometimes see on the input of a 7805 is for this purpose: to drop some voltage, and thereby shift some of the heat dissipation off the 7805 and onto the resistor. That resistor would need to be a power resistor and would probably also need to be heat sinked, at least with a 24VDC input. What I *do* know is that with a 24VDC input, you are not going to get 500mA out of a 7805. If you feed the 7805 with 9VDC instead of 24VDC you can do it.

I'd suggest using a more appropriate power source than your 24VDC. Or you could use one or more power resistors to drop the voltage, but then you're just moving the heat dissipation off the 7805 and you will need to dissipate it from the resistors (which may be a better idea). You could also put some other regulators (say, maybe a 7818 and 7812 in series) between the 24VDC power supply and the 7805, but that's a messy solution IMHO. [Edit: No matter what, you're going to need a good heat sink and heat sink compound.]

Also--you're including the input and output capacitors on the 7805, right? Without them, your 7805 might oscillate and heat up no matter what voltage you feed the thing with.

At any rate, good luck! I still suggest using a more appropriate power supply than the 24VDC one. Using that means that you have to do more work to keep things cool enough to work.


Regards,

Torben

[Edit: MikeMI got his numbers right and I got mine wrong, I think! You still have to dissipate way too much power, though.]
 
Last edited:

eldermalagon

New Member
Do a little basic math:
P=I*E
P=0.5*(24-5)= 9.5W



A 7805 (the big one in TO220 package) will require a large heatsink, a good quality thermal mounting washer, and white heat sink compound...
I´ve seen circuits having a 7805 being supplied with 24v, and the regulator doesn`t have any heat sink, in fact, it doesn`t get hot at all. Any other Ideas? ´cause is very strange that these circuits doesn´t blow. I´ve tried to use the same components but it Doesn´t work I think is some calculation for the series resistance at the imput of the 7805 but I can´t figure it out.
 

Torben

Well-Known Member
I´ve seen circuits having a 7805 being supplied with 24v, and the regulator doesn`t have any heat sink, in fact, it doesn`t get hot at all. Any other Ideas? ´cause is very strange that these circuits doesn´t blow. I´ve tried to use the same components but it Doesn´t work I think is some calculation for the series resistance at the imput of the 7805 but I can´t figure it out.
MikeMI is actually quite good at this, from what I've read of his posts. :) Trust him. I suspect that either A) the circuits you have seen drew almost no current through the 7805 (which means cooler running) or B) the circuits you have seen were drawn by people who don't know what they're doing and haven't actually built the circuits.


Torben
 

Gayan Soyza

Active Member
I grabbed this reply from another forum.

Code:
A much better way to go from 24V to 5V would be a switching voltage regulator. 
You can get complete modules for little money these days.Look at Mouser or Digikey 
for DC/DC converter modules. There are thousands of choices.

If you need to be linear because of switching noise issues, a good way of dealing 
with the voltage drop of (24V-5V)=19V is to use a pre-regulator. If you know your 
load current, a resistor will do fine. Just make it large enough that it drops 10-15V 
at the desired current. Diodes in series before the regular or a power zener diode 
(rare and expensive) will also do the job and take "most of the heat".

You can also use two regulators in series, which also has the advantage that input 
noise is suppressed much better than with a single regulator. In your case, a 12V or 15V 
regulator followed by the 5V regulator is a good idea.

No matter what you do with pre-regulators/cascaded regulators, you have to get rid of the heat. 
If your load current is 100mA, the total power dissipation on these parts is 1.9W. In a good 
design you will never stress the parts with that much without a heat sink that can bring the 
package temperature to below 50 degrees Celsius (that is the highest temperature on a metal 
surface you can touch for more than a short time before it feels very uncomfortable). You have 
to keep in mind that the silicon (junction) temperature is always MUCH higher than the case temperature. 
How much higher can be calculated from the junction-to-case thermal resistance and the total dissipated 
power. Whenever you design a power supply or any electronics that can get hot, YOU ABSOLUTELY HAVE TO DO 
THIS CALCULATION. Never get the junction beyond 125 degrees C. Most silicon will work as high as 
175 degrees, but it will degrade much, much faster once you get close, especially if you also have 
high voltage across the parts or are running high currents through them. Keeping the temperature low is your 
best bet to build well working, reliable electronics.

And then, please do not forget to put decoupling caps directly on input and output. 78xx parts are generally 
unforgiving if you don't and will oscillate. You might not notice or you may be marginally lucky but in a good 
design both caps are ALWAYS included. If you have two regulators, you will need three caps, of course, one on the 
input of the 12V regulator, the next one in between and the third one at the output of the 5V regulator.

Good Luck!
 
Last edited:

eldermalagon

New Member
Hello, and welcome to the forum!

24VDC is way too high for input into a 7805 with no heat sink which is expected to supply 500mA. I don't have the formula in my head right now (It's sadly been a while since I've had to do this calculation so I hope I'm corrected if I mess up the actual numbers here) but I think that's something like 12 Watts [Edit: should be 9.5W] of wasted power which the 7805 must dissipate as heat. The 78xx series have a thermal resistance of 65°C/W above ambient from the junction to air, which IIRC means that with no heatsink, the junction's temperature will rise by 65°C above ambient for each Watt of power which must be dissipated. The maximum operating temperature is 125°C, so you can see that this is clearly being exceeded unless you're running the thing in a space which is -492.5°C. The 7805 won't like that and will go into thermal shutdown (it tries to protect itself by shutting down instead of just frying).

I think that the reason you don't see this with your other "detectors" (whatever they are) is that they draw almost no current through the 7805s. The more current you draw through it, the more power you need to dissipate as heat.

You need to bring down the input voltage to the 7805. The resistor you sometimes see on the input of a 7805 is for this purpose: to drop some voltage, and thereby shift some of the heat dissipation off the 7805 and onto the resistor. That resistor would need to be a power resistor and would probably also need to be heat sinked, at least with a 24VDC input. What I *do* know is that with a 24VDC input, you are not going to get 500mA out of a 7805. If you feed the 7805 with 9VDC instead of 24VDC you can do it.

I'd suggest using a more appropriate power source than your 24VDC. Or you could use one or more power resistors to drop the voltage, but then you're just moving the heat dissipation off the 7805 and you will need to dissipate it from the resistors (which may be a better idea). You could also put some other regulators (say, maybe a 7818 and 7812 in series) between the 24VDC power supply and the 7805, but that's a messy solution IMHO. [Edit: No matter what, you're going to need a good heat sink and heat sink compound.]

Also--you're including the input and output capacitors on the 7805, right? Without them, your 7805 might oscillate and heat up no matter what voltage you feed the thing with.

At any rate, good luck! I still suggest using a more appropriate power supply than the 24VDC one. Using that means that you have to do more work to keep things cool enough to work.


Regards,

Torben

[Edit: MikeMI got his numbers right and I got mine wrong, I think! You still have to dissipate way too much power, though.]
well thanks a lot, that will be useful!!

Also talking about linear regulatos, another matter with them is that I´m trying to supply the back light of a LCD module with a 7809. Ií trying to feed all the circuit with 13v and bring it down to 9 with the 09 but when I connect the back light the input voltage drops to 7v aprox and the output of the 7809 drops to 4v aprox and there is no heating just voltage drop. It´s a pain because I don´t know either what happen with the 7809 and I cannot supply my LCD´s backlight!! Any suggestion? I really need to know why these regulators do this cause they are driving me crazy
 

eldermalagon

New Member
You are right

Thanks, in fact the circuits supplied with 24v draw almost no current, so that´s why they doesn´t heat up. I think it´s better with a switching power supply, is painless
 

eldermalagon

New Member
Ok the thing is that I have a switching power supply that is 24v output but I want to power the digital section with it to reduce cost and avoid having more supplies than project. I need the 24 volt for motors and a thermal printer and buying another switching power supply is expensive. That´s why I´m trying to do the whole thing with 24v
 

Torben

Well-Known Member
well thanks a lot, that will be useful!!

Also talking about linear regulatos, another matter with them is that I´m trying to supply the back light of a LCD module with a 7809. Ií trying to feed all the circuit with 13v and bring it down to 9 with the 09 but when I connect the back light the input voltage drops to 7v aprox and the output of the 7809 drops to 4v aprox and there is no heating just voltage drop. It´s a pain because I don´t know either what happen with the 7809 and I cannot supply my LCD´s backlight!! Any suggestion? I really need to know why these regulators do this cause they are driving me crazy
In that case I'd suspect a flaky power supply, or else that what you think is a 13VDC power supply is actually something more like a 9VDC wall wart adaptor when measured with no load on it. When you measure the output of a typical cheap wall adaptor, you will see a voltage well above what it's rated at. When you actually pull current from it, the voltage sags down closer to the number stamped on the back.

So if you're using a 9VDC wall wart I could see that reading as close to 13V when measured with no load. (Your meter presents almost no load to the supply.) But once you connect the load (the backlight), power is drawn and the voltage sags. Since the 7809 needs the input to be at least 11V, once the supply dags below 11V (including ripple from the supply, which if it's linear will increase with current draw) then the 7809 will be unable to regulate and the output voltage is no longer guaranteed.

Also, are you using the input and output caps on the 7809?


Regards,

Torben
 

eldermalagon

New Member
Yep I´m usin the caps. but I think the problem really is that indeed I´m usin a cheap multi-output wall wart adaptor. I suspected on it but I wasn´t really sure, now I know for sure that is that cheap thing!
 

Thunderchild

New Member
just use a 10 W resistor in series of about 30 ohms or the huge heat sink better still a combination of both a heat sink of some sort is recommended anyhow
 

Torben

Well-Known Member
Yep I´m usin the caps. but I think the problem really is that indeed I´m usin a cheap multi-output wall wart adaptor. I suspected on it but I wasn´t really sure, now I know for sure that is that cheap thing!
OK, cool. I wasn't sure but supply sounded like it might be the problem.

Good luck. Post the results after trying another supply. :)


Cheers,

Torben
 

SPDCHK

Member
... unless you're running the thing in a space which is -492.5°C...
Torben, may I be so bold as to point out that absolute zero is −459.67°F (which is -273.15°C, which is 0 Kelvin).

You were close though :D:D
 

Torben

Well-Known Member
Torben, may I be so bold as to point out that absolute zero is −459.67°F (which is -273.15°C, which is 0 Kelvin).
Certainly. :) That's partially my point.

You were close though :D:D
hehe

I suspect that my number may be off somewhat--like I said, it's been a while since I've done this and my notes are buried. But still, there's a lot of heat to account for. :)


Torben
 

audioguru

Well-Known Member
Most Helpful Member
Maybe the heating in the 7805 is caused by very antique high current TTL logic chips instead of Cmos that draws almost no current.
 
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