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Unregulated DC to a voltage regulator - how much heatsink?

Discussion in 'Electronic Basics' started by Jim, Feb 12, 2004.

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  1. Jim

    Jim Guest

    I'm using a 9V unregulated DC wall-wart to supply a 5V regulator. The 5V
    line is going to provide 250mA approx.

    I'm using a standard TO220 voltage regulator (1.7V drop IIRC)

    What spec of heatsink do I need in order to keep the regulator at reasonable
    temperature, say below 50degC/120degF? I guess it depends on the maximum
    voltage output of the unregulated supply under the load - is there a way to
    work out what this is?

    Many thanks,

    Jim
     
  2. Jim

    Jim Guest

    Thanks for the reply, Jamie.

    Do I need to cope with the fact that the *unregulated* 9V supply could be
    supplying more than 9V, e.g. 12V or whatever?

    Jim
     
  3. bg

    bg Guest

    A heatsink is rated in degrees per watt. That means that the temperature
    will rise so many degrees above ambient due to the watts of power flowing
    thru it. Your regulator will need to dissapate
    its wattage thru the heatsink.
    watts = volts times amps
    12 volts - 5 volts = 7 volts across the regulator
    amps = .25 amps thru the regulator
    watts = 7 times .25 = almost 2 watts dissapated by the regulator
    example ---
    If the heatsink is rated at 10 degrees per watt, the heatsink will rise 20
    degrees above ambient, which would be about 90 degrees at room temerature.
    Can you get to where you want to be with this info?
    bg

    Jim wrote in message ...
     
  4. Jamie

    Jamie Guest

    not much at 250 ma's
    a little u-channel peice of alliminimum will do it.
    WDf = 5*0.250 is good enough for selecting the heat sink.
     
  5. Jim

    Jim Guest

    Hi BG,

    Thanks very much for a very clear and helpful answer.

    The 12V max from the supply was a guess though - I still don't know what the
    unregulated supply could put out in terms of voltage with the 250mA load. I
    think I read that an unregulated supply with no load could output up to 1.7
    * nominal = 15.3V in this case. But in that case there will be no current
    anyway, so no heat.

    Is there a way of estimating the voltage from the unregulated supply when
    the 250mA load is in place, or do I need to cope with the possiblity of
    15.3V at 250mA? BTW, I want it to work with pretty much any decent 9V
    unregulated supply, so I can't just measure the one I have here.

    Thanks,

    Jim
     
  6. bg

    bg Guest

    Jim wrote in message ...
    All supplies will have an output resistance. It is the output resistance
    that drops the supply voltage as current is pulled from the supply. There is
    no standard value for a 9 volt wart or any other supply for that matter.
    Many of the three terminal regulators have protection built in to shut them
    down if an over current or over power condition exists. If you heatsink the
    regulator so it can handle it's full rated power dissapation, you will
    probably be safe with just about any typical 9 volt wall wart.
    bg
     
  7. John Fields

    John Fields Guest

     
  8. John Fields

    John Fields Guest

    ---
    Yes. Either get a copy of the spec's for the wall-wart or load it down
    with 250mA and measure its output. It would also be nice to know what
    the output voltage ripple under load looks like, or at least the size of
    the filter cap in the wall-wart.

    Having said that, though, we can make an educated guess that the
    wall-wart's output won't rise above 15V when it's unloaded on nominal
    mains, and with mains high, it'll rise to 16.5V, tops. Assuming that
    the wall-wart's current output capability is much greater than 250mA
    allows us to say that its maximum output voltage will be 16.5V, no
    matter what, and we can use that for the rest of our calculations.

    With 16.5V on the input of the regulator and the output at 5V with 250mA
    going into a load, the regualtor will be dissipating the difference
    between the two voltages multiplied by the current into the load, or

    P = (Vin-Vout)Iload = 11.5V*0.25A = 2.875W ~ 3 watts

    The reason for using a heat sink in the first place is to keep the
    junction of the regulator's pass transistor from rising to a temperature
    which could damage it, but most regulators now (certainly the 7805?
    you'll be using) have internal overtemp protection which will shut the
    regulator down if it gets too hot, but this gets to be a PITA if you
    don't want your circuit to stop operating when the regulator shuts down.

    So, the job then becomes one of selecting a heat sink which will keep
    the regulator from shutting down under worst case conditions. That means
    under maximum load with high mains and a high ambient temperature.

    The 7805 has a thermal resistance of 5°C/W from junction to case, so its
    junction temp will be 15°C hotter than its case when it's dissipating 3
    watts.

    That also means that its case temperature needs to stay below 135°C when
    the regulator is dissipating 3 watts, so in order to do that you need to
    find a heat sink with a thermal resistance which will keep the case
    below 135°C when the ambient temperature rises to its highest. Assuming
    50°C ambient, that means the heat sink has to get rid of 3 watts worth
    of heat when the ambient temp is 50°C without letting the regulator's
    case get to 135°C. That's an 85°C difference, so the heat sink's
    thermal resistance needs to be 85°C/3W = 28.33°C/W from the surface of
    the heat sink to ambient.

    You can check:

    http://www.aavidthermalloy.com/bin/standarda.pl?Pnum=&PgNum=1&NumPerPage=v3&Device=10&SortBy=V4

    For what you need. The higher the page number, the higher the thermal
    resistance, and you'll find you won't need much of a heat sink at all.

    Maybe none at all, depending on what your wall-wart's output looks like
    with a 250mA load on it.

    Also check:

    http://www.aavidthermalloy.com/products/standard/pdf/Page4.pdf

    for more information on how to choose a heat sink.
     
  9. John Fields

    John Fields Guest

    ---
    It'll need to be a little better than that, since the space between the
    regulator and the heat sink will impose its own penalty.

    Check
    http://www.aavidthermalloy.com/products/standard/pdf/Page4.pdf

    for more details.
     
  10. Jim

    Jim Guest

    Many thanks John for taking the time to write that comprehensive reply. I
    think I understand fully now. The links are very useful too.

    Jim
     
  11. Jamie

    Jamie Guest

    i assume your using a 7805 type reg.
    to be more clear most heat sinks that you get
    made have a Tempiature per watt rating..
    if you are using this type of reg then most of the
    heating you should be concerned about is in the IC it
    self and like i said, if you want to contruct one you can
    get a peice of small U-channel from your local hardware store
    and that will work just fine..
    to get proper wattage calculations one would need to subtract
    the out from the in.
    for example
    W= 9V-5V=4*0.250= 1;
    now lets assume a 16 volts in.
    W= 16V-5V=11*0.250= 2.75
    the above are just theory answers and not exact because a
    7805 type reg like many other circuits do have a minimum of input
    to concider.
    so using these rough calculations a piece of Small Gauge U-Channel
    still fits the bill!
     
  12. Jim wrote:

    If in doubt, don't estimate - measure instead. You can use a power
    transistor (2N3055 or so) or a power MOSFET as variable load. Measure
    the voltage across the output of the supply unit as a function of
    current flowing.
     
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