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Power supply part design: AC DC convert problem

I have a PSU parrt design problem.
It should receive 24V AC input to get 12V DC.
It start with a snub circuit, which is a varistor. The second is the
bridge rectifier, which should gives out full wave rectified DC. Then
the bridge rectifier's V+ output connect to a 220uF capacitor, whose
negative pin connect to the V-.
Following the 220uF capacitor, is the 12V regulator.

The problem is when I supply the circuit with the 24V AC, or actually
it is 27V AC, the regulator will enable the thermal shut down
protection. No DC output.

But if I only give the 24V DC supply input, the PSU part is fine.

I figure out the problem is the AC/DC part problem. But don't know the
exact answer.


The current consumption is 120mA. The 220uF is oversized, a 22uF might
do the job, which I had a test and it seemed fine.
 
A

Anthony Fremont

Jan 1, 1970
0
I have a PSU parrt design problem.
It should receive 24V AC input to get 12V DC.
It start with a snub circuit, which is a varistor. The second is the
bridge rectifier, which should gives out full wave rectified DC. Then
the bridge rectifier's V+ output connect to a 220uF capacitor, whose
negative pin connect to the V-.
Following the 220uF capacitor, is the 12V regulator.

What is the part number for the regulator? Is it 7812? Does it have a
heatsink?
The problem is when I supply the circuit with the 24V AC, or actually
it is 27V AC, the regulator will enable the thermal shut down
protection. No DC output.

What is the voltage of the AC after the rectifier? If you are feeding 27VAC
to a full wave rectifier, the output of the rectifier will be something
around 35V.
But if I only give the 24V DC supply input, the PSU part is fine.
I figure out the problem is the AC/DC part problem. But don't know the
exact answer.

The current consumption is 120mA. The 220uF is oversized, a 22uF might
do the job, which I had a test and it seemed fine.

The recitifed AC is at least several volts higher than the DC, depending
upon your circuit. This increases the power dissipation of the regulator.
This causes more heat. If the regulator is seeing 35V on its input, it is
having to dissipate (35 - 12) * .120A or almosts 3W. That's going to get
hot pretty fast without a heatsink.

Also, you should have a .1uF capacitor on the output of the regulator to
ground. This will help prevent oscillations.

http://www.national.com/pf/LM/LM78M12.html
 
K

kell

Jan 1, 1970
0
What is the part number for the regulator? Is it 7812? Does it have a
heatsink?


What is the voltage of the AC after the rectifier? If you are feeding 27VAC
to a full wave rectifier, the output of the rectifier will be something
around 35V.



The recitifed AC is at least several volts higher than the DC, depending
upon your circuit. This increases the power dissipation of the regulator.
This causes more heat. If the regulator is seeing 35V on its input, it is
having to dissipate (35 - 12) * .120A or almosts 3W. That's going to get
hot pretty fast without a heatsink.

Also, you should have a .1uF capacitor on the output of the regulator to
ground. This will help prevent oscillations.

http://www.national.com/pf/LM/LM78M12.html

The OP could put a 3 or 5 watt power resistor of 150 ohms at the input
of the regulator to take the heat. 150 ohms at 120mA would drop 18
volts, bringing the voltage down to about 17, for a 5 volt drop across
the regulator. at 120mA that's about 0.6 watts heating the
regulator. Still might need a heatsink, but only a small one.
Another possibility is to use a transformer that puts out less voltage.
 
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