Voltage Regulator.

Discussion in 'General Electronics Chat' started by NM2008, Jan 27, 2012.

  1. NM2008

    Thread Starter Senior Member

    Feb 9, 2008
    135
    0
    Hi there,

    I have a 30DC supply that I have dropped to 12V.
    At 12V I need a current of 1A.

    The regulator I have been using is a KA350, 3A regulator. The problem is I have roughly an 18V drop, and it does not take long to heat up a 100x60x3mm aluminium plate heat sink.

    Is there another regulator better suited to an application like this?

    Regards NM
     
  2. w2aew

    Member

    Jan 3, 2012
    219
    64
    Simple ohm's law. 12v * 1A is 12W. No matter what linear regulator you use, the result is the same - 12W of power dissipation in the regulator. Power dissipation creates heat. How much heat depends of the thermal resistance of the package and heat sink.

    The only way around this really, is to use a switching regulator. A lot more complex to design and build, but will have better efficiency (thus, less dissipation in the regulator).

    If you don't care about efficiency, the only other thing you could do is drop some of the voltage on other devices (diodes, etc.) in series with the input to the regulator. But, in the end, you're still dissipating the same 12W in total - just spreading it out over several devices. You'll have to worry about dealing with the heat generated by each of these parts.
     
  3. russ_hensel

    Well-Known Member

    Jan 11, 2009
    818
    47
    The cheap device to drop it with is a resistor. This is fine if a not very efficient supply is ok. A scope to look at the loaded ripple in the circuit would be useful. Power resistors are not expensive I would use about 20 watts of them.

    In fact if the load is constant and the 30 volt supply is regulated all you need is a resistor, this is not likely.
     
  4. praondevou

    AAC Fanatic!

    Jul 9, 2011
    2,936
    488
    It's (Vin - Vout) * Iin = 18W.

    I agree.
    The easy way is to buy a DC/DC converter drop-in replacement for your linear regulator but they are not so cheap. http://search.digikey.com/us/en/products/V7812-1000/102-1718-ND/1828611

    The other way is to build a switching regulator yourself.

    You can also verify if it is not possible to lower the input voltage.. in order to have less voltage drop over the linear regulator.
     
  5. NM2008

    Thread Starter Senior Member

    Feb 9, 2008
    135
    0
    Thanks for the help everyone, will be changing to a switching regulator for this one.
    Regards NM
     
  6. w2aew

    Member

    Jan 3, 2012
    219
    64
    "It's (Vin - Vout) * Iin = 18W."

    Yeah - duh! I read it too fast and assumed it was 12V across the regulator. Note to self - read posts more carefully before posting!
     
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