7805 & heat sink

[Adhith]

hello everyone..
   for my project (12v  1A input)  I'm using 7812 & 7805 regulator ics. but due to space limitation i can only place a small heat sink with it. so the 7805 is getting heated up very quickly, is there any way to reduce the heat rather than placing a bigger heat sink?? also, if i provide 7805 a regulated 12v input( from 7812) does it make any difference in heating??

[pelule]

How about replacing the line 7805 by a 7805 similar switching converter solution - less losses, thus less heat generation?
What is the current via the 7805?

[wraper]

Give or take, with linear regulators you cannot reduce dissipated power without decreasing the input voltage or current consumption. Dissipated energy won't disappear miraculously.
https://en.wikipedia.org/wiki/Conservation_of_energy
You would need to use step-down converter to improve efficiency and therefore reduce dissipated power.

also, if i provide 7805 a regulated 12v input( from 7812) does it make any difference in heating??

Then 7805 will heat less but 7812 will heat more. Total heat won't change.

[XFDDesign]

Add a large box-fan with a funnel down to your small heatsink.

[Brumby]

Give or take, with linear regulators you cannot reduce dissipated power without decreasing the input voltage or current consumption. Dissipated energy won't disappear miraculously.
https://en.wikipedia.org/wiki/Conservation_of_energy
You would need to use step-down converter to improve efficiency and therefore reduce dissipated power.

also, if i provide 7805 a regulated 12v input( from 7812) does it make any difference in heating??

Then 7805 will heat less but 7812 will heat more. Total heat won't change.

This  ^  ^  ^

[Audioguru]

Add a power resistor between the 12V output and the 5V regulator input to share its heat.

[vulturebetrayer]

Have you considered a buck converter?

They are a bit more expensive but may solve your problem.

https://www.adafruit.com/products/1065?gclid=COry_6eg2c8CFQ5Efgod0ncMQg

https://www.amazon.com/DROK-synchronous-Converter-6-5-24V-Replace/dp/B00CCXDSRU

[DimitriP]

Add a power resistor between the 12V output and the 5V regulator input to share its heat.

I like the resistor idea...or
7812-->7808-->7805 , maybe because I'm allergic to buck converters

[wraper]

Add a power resistor between the 12V output and the 5V regulator input to share its heat.

I like the resistor idea...or
7812-->7808-->7805 , maybe because I'm allergic to buck converters

All of that is useless, because:

but due to space limitation i can only place a small heat sink with it.

That still would be the same more than 12W dissipated in the device (depending on the input voltage). It would still run as hot as it was.
 

[DimitriP]

Add a power resistor between the 12V output and the 5V regulator input to share its heat.

I like the resistor idea...or
7812-->7808-->7805 , maybe because I'm allergic to buck converters

All of that is useless, because:

but due to space limitation i can only place a small heat sink with it.

That still would be the same more than 12W dissipated in the device (depending on the input voltage). It would still run as hot as it was.

Assuming worse case , the 5V regultor is drawing the full 1A  with 12V  fed into the 5V regulator we have :
12v x 1A = 12W Pin to the 7805
5V  x 1A  = 5W Pout
12-5 = 7 W  to  get rid off of the regulator as heat

With 8V feeding the regulator:
8V x 1A = 8W Pin
5V x 1A = 5 W Pout
8-5 = 3W  to get rid off of the 5V regulator as heat 

So I dont' see where it's "useless"

[wraper]

Add a power resistor between the 12V output and the 5V regulator input to share its heat.

I like the resistor idea...or
7812-->7808-->7805 , maybe because I'm allergic to buck converters

All of that is useless, because:

but due to space limitation i can only place a small heat sink with it.

That still would be the same more than 12W dissipated in the device (depending on the input voltage). It would still run as hot as it was.

Assuming worse case , the 5V regultor is drawing the full 1A  with 12V  fed into the 5V regulator we have :
12v x 1A = 12W Pin to the 7805
5V  x 1A  = 5W Pout
12-5 = 7 W  to  get rid off of the regulator as heat

With 8V feeding the regulator:
8V x 1A = 8W Pin
5V x 1A = 5 W Pout
8-5 = 3W  to get rid off of the 5V regulator as heat 

So I dont' see where it's "useless"

Then check the size of that resistor, and figure out you could increase the size of heatsink as well instead, taking space required for that resistor. And that heat still remains in that very same hot spot, slowly frying everything around.

[rob77]

Add a power resistor between the 12V output and the 5V regulator input to share its heat.

I like the resistor idea...or
7812-->7808-->7805 , maybe because I'm allergic to buck converters

All of that is useless, because:

but due to space limitation i can only place a small heat sink with it.

That still would be the same more than 12W dissipated in the device (depending on the input voltage). It would still run as hot as it was.

Assuming worse case , the 5V regultor is drawing the full 1A  with 12V  fed into the 5V regulator we have :
12v x 1A = 12W Pin to the 7805
5V  x 1A  = 5W Pout
12-5 = 7 W  to  get rid off of the regulator as heat

With 8V feeding the regulator:
8V x 1A = 8W Pin
5V x 1A = 5 W Pout
8-5 = 3W  to get rid off of the 5V regulator as heat 

So I dont' see where it's "useless"

and where is the 12V -> 8V step producint  4W  of heat ? the total heat dissipation will be the SAME no matter what. if you have  a small box where you have to dissipate 7W of power- then it doesn't matter in how many stages you do it... it's still 7W in the same volume.

the only solution is buck converter - it's called converter because it's converting energy... for example it can do output 5V 1A from 12V 0.6A if the efficiency is ~70% and will dissipate only ~ 2.2W compared to the 7W with linear regulator.

[DimitriP]

The 4Watts are on the 8V regulator keeping it warm. You end up with two warm regulators instead of one very hot 5V regulator

[WaveyDipole]

I get the impression that your circuit is requiring both a 12v and a 5v supply? Bear in mind that if you place the 7805 in series with the 7812 (via the 7808), then you will need to add the current you are drawing from the 7805 to the current being drawn from the 7812. The extra load on the 7812 will result in additional power being given off as heat.

If you are drawing 1A at 5V at the 7805, which required 5W of power, then at 8V (i.e output of the 7808), this will be 5W/8v=0.625A. At the output of the 7812 will be an additional 5W/12v=0.416...A.

So assuming that you are drawing a maximum of 1A at 12v and 1A at 5v, then I would expect a current draw and power consumption at each regulator as follows:

7812 : (12v * 1A) + (12v * 0.4166A) = 12W + 5W = 17W
7808 : 8v * 0.625 = 5W
7805 : 5v * 1A = 5W

So in terms of power lost as heat:

At the 7805: (8v - 5v) * 1A = 3W
At the 7808: (12v - 8v) * 0.625A = 2.5W
The power lost as heat at the 7812 will depend on the unregulated source voltage.

Since you might potentially be drawing 1.4166A at the 7812, then you will require something like the 7812CV which is rated at 1.5A rather than 1A. If you drive the 7808 directly from the source rather than via the 7812, then the heat loss at the 7808 will increase depending on the source voltage.

So yes, you will have more than halved the excess power to be dissipated as heat by the 7805, although this will now be spread across the other two regulators, but the total heat generated and needing to be dissipated by the heatsink will be the same.

You don't mention what the source voltage is derived from, but if you are using an AC source and deriving DC via a rectifier bridge, then one option might be to use half wave rectification to feed the 7805. It only requires one extra diode (or perhaps 2  in series for extra reassurance) but you end up with half the DC voltage and therefore less heat to dissipate. A somewhat larger smoothing capacitor might be beneficial.

[KL27x]

Power resistor can certainly help.

Sticking a 7805 next a 5W lightbulb is better than turning it into a 5W lightbulb. You can get greater output and longevity by dissipating the heat away from the regulator IC.

AND because power resistor can operate at higher temp, ambient temperature produces a higher gradient. Meaning you don't need to put same size heatsink on the power resistor... you can run it at 90C butt nekkid.

[wraper]

The 4Watts are on the 8V regulator keeping it warm. You end up with two warm regulators instead of one very hot 5V regulator

But you take the space where you could put larger heatsink (as space is premium in this case). On overall it runs just as hot.

[rob77]

The 4Watts are on the 8V regulator keeping it warm. You end up with two warm regulators instead of one very hot 5V regulator

unless you use that 8V regulator as butt-hole warmer it will dissipate into the same confined space as the 5V regulator => you will still have the 4W + 3W = 7W dissipated into the same volume.

furthermore... there is something called thermal resistance - it's in kelvin or Celsius per watt. it says how many degrees C (or kelvins) the temperature will rise when 1W dissipated.

a small TO220 heatsink like this which is barely bigger than the TO220 footprint has a thermal resistance of ~ 30K/W - so your 5V regulator dissipating 3W will be 90degrees C hotter than the surrounding air. and your 8V regulator will be 120degree C hotter than the surroundings. with this setup you are out of scpecs right away.

let's say you use a bigger heatsinks like this it's 40x23x16mm so it takes some space but also it's better it has ~ 12K/W - so you'll end up with temp differences 36celsius and 48 celsius above the surroundings which might work if the case is ventilated.. if the ambient is 30celsius the heatsinks will be 66C for the 5V and 78C for the 9V and the silicon itself will be slightly under 100C. btw.. 78degree celsius is definitely not just warm

furthermore if the case will be not ventilated well - the ambient in that case will rise (thermodynamics is a bitch) and the temperature of the heatsinks will raise too... and things will get out of spec again.. (you have to watch the max temperature of the sillicon).

i bet you will find many tutorials on thermal design, so invest some of your time and have look at those

[DimitriP]

If you feel  lowering the power dissipated from the 7805 by more than half doesn't answer the OP question and you have issue with the overall thermal design of the project, feel free to take it up with the OP

[KL27x]

@ Rob and Wraper:

Heat inside the enclosure is not causing nearby components to fail. It is causing failure of the parts creating the heat, in this case the 7805, because of reduced temp gradient to cool it.

This why power resistor is better solution where applicable. Despite inside of case same temp as before it is still a huge improvement. Resistor can run at much higher temp with zero decrease to regulator max output.Two regulators helps, but not as much.

Despite same internal temp in enclosure, the parts dissipating heat are the hottest. Make hottest part a passive resistor with positivery coefficient of resistance to temp and can operate at tremendous temp without longterm failure is most foolproof with built-in safety/limiting in case of change in ambient temp.

Two regulator also is improvement.

Bigger heatsink, alone, is for sucker. Not best solution. Least efficient for space and cost per equivalent improvement in output and longterm reliability. You will never match improvement of series power resistor with heatsink, alone, until heatsink is huge and costly. And using smaller heatsink with power resistor, and pow, you are destroying heatsink-only solution by a huge margin that can't be touched. If overhead voltage was closer to 3 or 4 volts, then heatsink is fine. With over 7V overhead, heatsink alone is dumb brute-force method that is quite limited. In fact, OP never stated supply voltage, which is obviously at least 15V if he run 7812 on it. So at least 10V of overhead; at this point, bigger heatsink should not even be considered a solution, by itself. Even if current draw is very, very low, a small power resistor is more compact and cheaper than small heatsink. In such case, you do not need heatsink on the 7805, at all.

OP. To use series power resistor, start with max current draw and use kirchoff law to calculate value of power resistor. Leaving about 2V of overhead for the 7805 Regulator. So the 7805 needs minimum of 7V supply voltage. Rest can be dissipated by the series resistor when current is at max. At less than max draw, the series resistor won't drop as much share of the voltage, but that won't matter.

[darrellg]

You need something like this pin-compatible, drop-in switchmode replacement for the 7805:
https://www.tindie.com/products/ddebeer/5v-1a-switch-mode-voltage-regulator-40v-max-input/?pt=full_prod_search

[wasedadoc]

You don't mention what the source voltage is derived from, but if you are using an AC source and deriving DC via a rectifier bridge, then one option might be to use half wave rectification to feed the 7805. It only requires one extra diode (or perhaps 2  in series for extra reassurance) but you end up with half the DC voltage and therefore less heat to dissipate. A somewhat larger smoothing capacitor might be beneficial.

You do NOT end up with half the DC voltage. It remains about the same. You do end up with a net direct current taken from the ac source. Which if that source is the seondary of a conventional mains ransformer can be a bad thing to do.