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Power supply part design: AC DC convert problem

Discussion in 'Electronic Basics' started by [email protected], Mar 4, 2007.

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  1. Guest

    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.
     
  2. 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
     
  3. kell

    kell Guest

    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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