Linear Power Supply - Heat Dissipation Problem

What type of PSU is better?

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EM Fields

Joined Jun 8, 2016
583
Hi, I have build an Linear Power Supply similar like that on the attached picture.
I use and 230:6x2 transformer that is insulated in epoxy.
Line (socket) that is connected is about 235V-245, 50Hz (EU).
Instead the C1 470u of the pic use 2200u, and instead C3 100n block, I use 100u electrolytic.
The load that is connected is an led strip, that makes less than 200mA.
The transformer is rated for 1A output like the LM7812 and the Diode Bridge.
My problem is that the Transformer and the Volt Regulator (LM7812) get very hot (60-80Celsius).
Transformers in normal condition get warm. But mines get very hot.
My question is: Why the transformer and 7812 get very hot even the load is less then 200mA, and transformer and 7812 are rated for 1A.
Thanks in Forward.

If you're using a transformer which outputs 16 volts RMS when the transformer is under a full 2 ampere load, then its output voltage will be higher than that when it's loaded with only one ampere. Small transformers usually exhibit something like 30% regulation from no-load to full load, so just for the sake of argument let's say that with a 1 ampere load the transformer's output is up 15% from 16 volts, which will put it at about 18.4 volts, RMS, which is about 26 volts peak.
After that you'll get about a 1.6 volt drop across the bridge, so the reservoir cap will charge up to 24.4 volts on each rectified half-cycle and present that voltage to the input of the 7812, and if the 7812 is holding its output steady at 12 volts, there'll be 12.4 volts across the 7812 during those peaks.
If your load current is 1 ampere and the regulator is dropping 12.4 volts, then it'll be dissipating 12.4 watts while its input is at 24 volts.
That won't be true all the time of course, because the rectified input to the cap will vary, but the larger you make the reservoir cap the closer it'll stay to 24 volts while the load is discharging the cap through the regulator and the hotter the regulator will get. It's no picnic for the transformer either, since it has to power the load AND charge the cap on its peaks, so it'll get hot too.

If you have to use a linear supply, you should use a transformer sized to run the load and charge the cap so that the cap's voltage never falls below the regulator's dropout voltage, and get a cap which will hold enough charge from cycle-to-cycle so that its ripple valleys never allow the regulator's input voltage to fall below the regulator's threshold voltage, but get close.
 
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EM Fields

Joined Jun 8, 2016
583
Well, I would just use a low drop-out regulator to regulate the 12V, but if you are happy with such a kludge circuit, than that's all that matters. :)
If you use a 12 volt source to feed an LDO, you won't get 12 volts out of it, so it makes sense that if you have a string of LEDs and you're feeding the string with something like a 7812, that you give the 7812 enough headroom to work by setting its input voltage to about 14 to 15 volts.
 

Thread Starter

StephanG

Joined May 21, 2017
22
Hi,
I did the following changes: removed C2 (0.1uF) and C3 (100uF elec) -> C4 (100nF block).
Now the transformer and 7812 less heating when the LEDs are on. When the LEDs are off the 7812 is cold and the ttransfoemer is very little warm.
The circuit attached is the actual.
 

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R!f@@

Joined Apr 2, 2009
10,004
You need to have 100nf decoupling caps on Input and Output, as close as possible to the regulator
 
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