Heating problems on regulator with low current

Thread Starter

Zeno53

Joined Sep 23, 2020
91
Hello guys,

i need your help if possible. I created a dual voltage power supply using LM317 and a LM337 both in TO220. The problem is that when empty everything is ok but when I connect the load they heat up within two minutes also with heatsinks. I give a brief description of the circuit. The transformer generates me 17-0-17 Volt AC (150mA + 150mA I know it's small but it's just for testing) which rectified and stabilized become +22.8 0 -22.8 Volt DC. This is the input voltage to the voltage regulators. There are 6 of which three are negative and three positive where each one generates a different output voltage which is:
+/- 12 Volt
+/- 5 Volts
+/- 1.5 Volt

Let's leave out the 12 Volt that for the moment I don't have the possibility to test it, the measured currents are:

For the +1.5V and -1.5V 77mA.
For the + 5V I read 90mA and for the -5V 34mA.

The difference in current from + and - is due to the fact that the + V has to feed one more component.

I did some calculations so :

(22.8-5) * 0.090 = 1.6Watt for +5V
(22.8-5) * 0.034 = 0.6Watt for -5V
(22.8-1.5) * 0.077 = 1.6Watt both for + and - 1.5V

What i ask :

1) the correctness of the calculations and if exact, why the regulators heat with such low absorption?.

2) possible solutions.

As I said above these dual stages are 3 in parallel to 22.8Volt . If I wanted, I could intervene on every single entry of the regulators if the fact that they are in parallel does not generate further problems.

I attach the diagram of one of the three dual stages and the image of heatsink used.

Thanks.
 

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crutschow

Joined Mar 14, 2008
28,216
Yes, your power calculations are correct.
they heat up within two minutes also with heatsinks
How much do they "heat up".
You need to know the thermal resistance of the heat-sink (temperature rise versus power dissipated).
With such a small heat-sink, I would expect 1.6W to give perhaps a 30-40 °C rise in the heat-sink temperature, which should not be a problem.
 
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Thread Starter

Zeno53

Joined Sep 23, 2020
91
To answer all:

I would prefer not to use fans but I agree that ventilation helps a lot. Unfortunately, I do not know the dissipation because it is a missing data (taken on amazon). I can tell you that they are made of aluminum and are also mounted upside down for space reasons and this probably makes the situation worse. In this case, however, I can solve by finding a heatsink that develops in height with greater dissipating power. As for the temperature in about 3 minutes in the room at 26 ° C I measured 55 ° C with a laser thermometer but it probably would have risen even more if I had continued the measurement. Not very accurate due to the small surface but rather probable. I didn't think it could generate so much heat and I didn't give much space to the heatsink on the PCB. Now I don't have much experience with the fate of heat. What I would like to understand is:
Since the calculation is correct, is it normal to expect these temperatures at these conditions and with such low currents?
Because if they are normal it means that the circuit responds normally to the current conditions and I can simply solve it by lowering the voltage of the transformer and changing the heatsinks. I attach photos of the heatsink.

Thanks

hjk.jpg
 

dcbingaman

Joined Jun 30, 2021
498
I would drop the transformer to a lower voltage even 12V out instead of 22.8V will make a significant difference.

From your calculations (which are correct) I calculated efficiency PowerToLoad/PowerIn
(22.8-5) * 0.090 = 1.6Watt------ for +5V - 22% power to load 78% wasted as heat
(22.8-5) * 0.034 = 0.6Watt for---- -5V - 22% power to load 78% wasted as heat
(22.8-1.5) * 0.077 = 1.6Watt---- both for + and - 1.5V : 7% power to load, 93% wasted as heat

I try to shoot for at least 75% efficiency, reducing the transformer voltage will make a lot of difference and you will not be wasting so much power. Because the currents cancel power efficiency is nothing more than output voltage over input voltage for most linear regulators as usually most of the input current becomes the output current only a small fraction is consumed by the linear regulator itself and can usually be ignored when determining efficiencies.
 

Thread Starter

Zeno53

Joined Sep 23, 2020
91
Thanks to everyone for the answers, I will make the necessary changes and repeat the measurements once completed. I've found the thermal characteristic of the heatsink which is 15 °C/W (air @ 1 m/s) .

I have a further question for @BobTPH . What alternatives could I have used instead of linear regulators?
 

BobTPH

Joined Jun 5, 2013
4,053
As audioduck said, a switch mode regulator is how you get efficiency. If you need less noisy output, switch mode down to 3V followed by a linear down to 1.5.

Bob
 

crutschow

Joined Mar 14, 2008
28,216
A 12Vac transformer, as dcbingaman suggested, will significantly reduce the amount of heat dissipated in the linear regulators, and you won't have the noise and complexity of a switching regulator.
A switching regulator seems overkill for such low currents.
 

Thread Starter

Zeno53

Joined Sep 23, 2020
91
I've substituted transformer from 15 0 15 AC with 12 0 12 AC. Heat lowered but 60°c after 4 minutes. Alternative: For testing = use a fan when powered on. What I would like most is to recreate the same circuit with the K version of the regulators that are in TO3 with a fairly large one like the one in the picture. But I don't know if the TO3 can increase the noise.

1.jpg

2.gif
 

ElectricSpidey

Joined Dec 2, 2017
1,972
legitimate TO3 versions of those regulators will set you back a pretty penny, but please don't be tempted to buy the "less expensive" versions available on eBay and such unless you can get some NOS from a reliable dealer.
 

Thread Starter

Zeno53

Joined Sep 23, 2020
91
I've seen the price of K version and WTF... Perhaps i must really reconsider a bigger heatsink a new PCB layout because the present doesn't have more space.
 
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