Voltage regulator heat sink question

[Electron Tornado]

I'm building a power supply for the work bench just for testing of pedals. I'm using LM78xx series regulators, and will use one regulator per outlet. Since the tab on the TO-220 case is ground, if I attach them to the chassis, will that be enough heat sink for them?

[aron]

I did that before and it worked fine.

[EBK]

Are you planning on using some thermal paste too?

[ElectricDruid]

Are you planning on using some thermal paste too?

+1 agree. Thermal paste for bonus points and longer-lived regulators.

[antonis]

It depends on chassis area, surface and regulator heat dissipation..

In case of mains grounded chassis, use silicon (no need for thermal paste) or mica heat insulator with proper washer & screw..
(in case of painted enclosure, grind a TO-220 area with a 800 - 1000 grit sandpaper..)

[Rob Strand]

[Antonis posted before me]

You need to take a step back.

There's so many details not filled in.

Assumed output voltage 9V
What DC input voltage to the regulators?
How many channels?
How much current from each?
How much current total?
What size enclosure?

Pulling 10mA from each supply is very different from pulling 1A.
And you might have 5 x 10mA supplies but one day you might want to pull 1A
from only one.

Pulling 1A from a single device needs more care with the heatsinking and mounting.
Pulling 10mA you don't need any heatsinking at all.

As for thermal grease, see fig 5 and fig 6.  Not much advantage for direct mounting to case.
but very significant when using insulating washers
https://www.onsemi.com/pub/Collateral/AN1040-D.PDF

Were you intending on mounting directly to the case without insulating washers?
I suspect this would be OK if you are powering from a wall-wart.  Once in a blue moon you could get a weird grounding issue.

If it's a mains powered supply with a mains ground to chassis it would be best to use insulating washers.

[antonis]

@ Electron Tornado: Unfortunately, not just a M3X8 screw and a M3-0.5X5.5 hex nut cinch..

P.S.
Neither imperial nor metric units make things easier..

[Kevin Mitchell]

Assuming this is only for plugging in a stompbox circuit or two, you likely can get away with no heatsink.

Two factors;
-What's the current draw going to be? (usually it's pretty low, which is why on-board regulators are not usually heat sunk)
-How much voltage are these going to be burning off? (the closer the raw voltage is to the LDO's output voltage the lower the dissipation will be). So if you're running a 12 volt adapter to a 7809 then it's probably not necessary - depending on the current demand of course.

[Electron Tornado]

Thanks for all the replies. Here is some more info:

I do have some heat sink compound I can use.

This supply is just for testing pedals on the bench. It will be a bit easier than a multitude of wall warts.

The supply will be mains powered into about 7 outputs. I do not foresee using more than 2 at a time.

The outputs are: 5V for a USB outlet, three 9V outlets with sag control on one, one or two 12V outlets, and one 18V outlet. I have an LM317 I can use for the 9V with sag. The LM317s tab is not ground, however, so it would need its own heat sink.

I looked at commercial pedal board power supplies available and found these current ratings for their outputs:

9V  100ma, 300mA, most supplies only have one or two 9v outlets rated up to 500mA*
12V  100mA, 300mA
18V 100mA

* This tells me there aren't terribly many pedals that need 500mV. The other current ratings are well below the capability of the individual regulators.

The transformer I have has a secondary of 18VDC at 1A. (Again, I'm not running all of the outlets at once.) A bridge rectifier will give about 24V after the two diode drops. I'm worried less about the current demand through an individual regulator, but more about the heat dissipation requirement due to the amount of voltage being dropped by the regulators, especially the 5V.

[antonis]

The transformer I have has a secondary of 18VDC at 1A. (Again, I'm not running all of the outlets at once.) A bridge rectifier will give about 24V after the two diode drops. I'm worried less about the current demand through an individual regulator, but more about the heat dissipation requirement due to the amount of voltage being dropped by the regulators, especially the 5V.

Taking into account USB 500mA nominal current capacity, you'll have to dissipate about 9.5W max so you need a heatsink of thermal resistance of about 8^oC/W..
Mounting on enclosure should be marginally OK..

https://www.giangrandi.ch/electronics/thcalc/thcalc.shtml

[R.G.]

Math to the rescue!
18V transformer secondary has a peak of 18V*1.414 = 25.42V. This will be minus two diode drops or ... yep, about 24Vdc.
The TO-220 package can dissipate about 2W just sticking up into the air around it if the air is about 25C/room temps.
The power dissipated in a linear regulator is the input voltage minus the output voltage times the output current. So for a 5V output, the regulator dissipates (24V-5V)=19V times any output current. For free air, you can only run 2W/19V = 105ma before the regulator wants to go into thermal shutdown. For 9V out, this becomes 2W/15V = 133ma.

So yes, you're likely to need either a heat sink or some more knowledge about your maximum currents. It is possible to put a power resistor in series with the regulator to eat up some of that excess headroom. If you know you're only going to need, say, 500ma of 5V, then you can calculate:
5V out plus 2V minimum headroom for a 7805, leaving 17V from a 24V source. If you use a resistor of 17V/0.5 = 34 ohms, then the resistor will drop 17V at 500ma. Picking a 33 ohm resistor you can actually buy, then the regulator will only have 2.5V left across it. The resistor will then dissipate 16.5*0.5A = 8.25W and the regulator will have 2.5V *0.5A = 1.25W of dissipation. In free air, this will get hot but not go into thermal shutdown. Connection to a chassis will be plenty extra.
Most linear regulator problems are exercises in Ohm's Law, and adding/subtracting voltages, then computing power. The other regulator dissipations can be calcuated this same way.

[R.G.]

I should probably add that at least during my active career, it was entirely possible to have a successful circuit design career with no higher math than Ohm's Law, quadratic equation solutions, Thevenin, Norton, and Kirchoff's theorems.
I found I'd wasted a lot of time in calculus and differential equation classes. 

[Electron Tornado]

Thanks R.G.!

The datasheet shows the use of a resistor before the regulator to drop the input voltage in Figure 22.  https://www.ti.com/lit/ds/symlink/lm7800.pdf

I only thought to add the 5V outlet as a battery charger for a phone, perhaps, but mostly for this little test gadget https://www.ebay.com/itm/353404276114?chn=ps&_trkparms=ispr%3D1&amdata=enc%3A1lAWp4mLOTfqsXVPYp-HSTg30&norover=1&mkevt=1&mkrid=711-213727-13078-0&mkcid=2&itemid=353404276114&targetid=4580702890871448&device=c&mktype=&googleloc=&poi=&campaignid=418640321&mkgroupid=1233652283797640&rlsatarget=pla-4580702890871448&abcId=9300602&merchantid=51291&msclkid=99a830f3553a19d8010051afb04e1e99

[Rob Strand]

If you know you're only going to need, say, 500ma of 5V, then you can calculate:
5V out plus 2V minimum headroom for a 7805, leaving 17V from a 24V source

Using a resistor to get the DC input  2V above 5V is probably cutting things a bit fine.   The 17V drop isn't well defined.

We actually need the dips of the ripple after the resistor to be at least 7V (2V+5V).   Considering the regulation of transformer and how that affects the DC regulation and the ripple on the filter cap we could easily see down to 20V to 22V at the *bottom of the ripple*.    The resistor is probably going to end up at 22 ohm more than 33 ohm, without allowing any safety margin for uncertainties.     

I didn't check the regulator power dissipation but it will increase.  You would be better trading a good margin on the DC input of the 5V regulator and have a bit more power dissipation and a larger heatsink (or mount to the case with an insulator).

The main point is the resistor will need tweaking to accommodate the uncertainties.

FWIW, an 18VA (18V 1A) transformer should only be loaded with about 12W DC load from the unfiltered input supply.   That's say 0.5A at 24VDC (middle of the ripple).  Not 1A.   Also 12W DC doesn't mean a the 5V rail can pull 2.4A it can only pull 0.5A.  The current is the limit.

I only thought to add the 5V outlet as a battery charger for a phone, perhaps, but mostly for this little test gadget

The battery charger chip will pull about 500mA (perhaps 550mA to 600mA worst case.)

[Electron Tornado]

I didn't check the regulator power dissipation but it will increase.  You would be better trading a good margin on the DC input of the 5V regulator and have a bit more power dissipation and a larger heatsink (or mount to the case with an insulator).

The main point is the resistor will need tweaking to accommodate the uncertainties.

FWIW, an 18VA (18V 1A) transformer will should only be loaded with about 12W DC load from the unfiltered input supply.   That's say 0.5A at 24VDC (middle of the ripple).  Not 1A.   Also 12W DC doesn't mean a the 5V rail can pull 2.4A it can only pull 0.5A.  The current is the limit.

I only thought to add the 5V outlet as a battery charger for a phone, perhaps, but mostly for this little test gadget

The battery charger chip will pull about 500mA (perhaps 550mA to 600mA worst case.)

I plan on mounting the regulator to the case for heat sink.

If i recall from the datasheet, I think it showed a circuit that would allow for current limiting. A lower current limit will still charge batteries, though not as fast.

[antonis]

If i recall from the datasheet, I think it showed a circuit that would allow for current limiting. A lower current limit will still charge batteries, though not as fast.

I think you get "current limiting" mixed up with "constant current"..

Don't know to which circuit you refer on but - in general - current limiting is obtained via voltage reduction..
That might be OK for charging PbO₂ (Lead/Acid) batteries but a major problem for anything else requiring a constant voltage for proper operation..

[Rob Strand]

I plan on mounting the regulator to the case for heat sink.

It's a good idea because it  gives you a chance to lower the resistor if you need it.

If i recall from the datasheet, I think it showed a circuit that would allow for current limiting. A lower current limit will still charge batteries, though not as fast.

Simple consumer products with a rechargeable  Li-Ion battery and built-in USB charger often have a small IC dedicated to charging the battery.   The chips have a number of features which prevent damage to the Li-Ion batteries.

I believe the TC-1 Multi-function Tester uses one of these ICs.  Something like a Top Power TP4057 but maybe not that exact chip.   As far as I am aware these chips have the current limiting built-in.    The datasheet for that device is on-line but it is in Chinese.

A similar device with an English datasheet is this one,

https://www.richtek.com/assets/product_file/RT9502/DS9502-02.pdf

If you look at the chart on page 4 you can see if the input supply drops below some value the device stops charging (UVP = Under-voltage Protection)

The reason I mention that is if you add a current limit circuit to the power supply it can cause the device to stop charging.   A current limit circuit lowers the output voltage until a point it reached where the current is at that limit (as Antonis mentioned).   If the current limit on the add-on current limiter is less than the current limit of the battery charger IC the input to the battery charger IC will be lowered and that may cause the chip to stop charging, as the lower voltage triggers a UVP fault.     Obviously if the current limit of the add-on current limit is higher than the charger chip's current limit the chip will do the limiting and the input voltage will not fall and cause a UVP fault.

[Electron Tornado]

I think you get "current limiting" mixed up with "constant current"..

I think you're right! The datasheet from TI shows a "current regulator" circuit.


Also, thank you, Rob for the charger IC info.  For my application, I think the simplest solution is the series resistor RG mentioned above.

[Electron Tornado]

Math to the rescue!

So yes, you're likely to need either a heat sink or some more knowledge about your maximum currents. It is possible to put a power resistor in series with the regulator to eat up some of that excess headroom. If you know you're only going to need, say, 500ma of 5V, then you can calculate:
5V out plus 2V minimum headroom for a 7805, leaving 17V from a 24V source. If you use a resistor of 17V/0.5 = 34 ohms, then the resistor will drop 17V at 500ma. Picking a 33 ohm resistor you can actually buy, then the regulator will only have 2.5V left across it. The resistor will then dissipate 16.5*0.5A = 8.25W and the regulator will have 2.5V *0.5A = 1.25W of dissipation. In free air, this will get hot but not go into thermal shutdown. Connection to a chassis will be plenty extra.
Most linear regulator problems are exercises in Ohm's Law, and adding/subtracting voltages, then computing power. The other regulator dissipations can be calcuated this same way.

R.G., for a 5V regulator used as a battery charger, is there a way to determine how much a load on the output would drag down the voltage at the input of the regulator? At 2.5V across the regulator, that is close to it's dropout voltage.

[antonis]

Batteries are VERY HEAVY loads..
(consider them as an almost short circuit..)

They can (and will) drag your charging configuration output voltage down to whatever is their own one..
e.g. a 5V regulator used to charge a 3.6V (nominal value) cell with 2.8V current voltage value will exhibit a 2.2V voltage drop plus its Vin-Vout voltage drop..
BUT
the main issue for chargers is the current demand..!!

You can have the same power dissipation by using a LM317HV of 61.25V In and 1.25V Out at 1A current with 7.25V In and 1.25V Out at 10A current..

P.S.
For charging purpose, you always DO need some kind of current limiting configuration..