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35v dropout - LDO?

Discussion in 'Electronic Design' started by Valentin Tihomirov, Aug 11, 2003.

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  1. I was using LF50 to regulate my 5v power. Source was 7.5v. The chip was
    increadibly hot. Specification tells that input can be up to 40v. A low
    droput regulator?
     
  2. A E

    A E Guest

    Power dissipation is dropout voltage times load current. Doesn't matter what
    kind of regulator. You probably need a heat sink.
     
  3. 60 degrees celcius is also incredibly hot, but still okay, although
    I like them cooler myself too. Linear regulators tend to get hot
    rather easily, in your case each amp is 2.5 watts of heat. You need
    a heatsink, no doubt.

    LF50?
     
  4. Brian

    Brian Guest

    The maximum current the LF50 is rated for is 1 amp. If you are running it
    at max. current, that would give you 2.5 watts dissipation. The junction to
    case temperature rise for the LF50 is 50 to 60 degrees per watt (depending
    on what case you have). Without a heat sink, that would make the junction
    case about 125 to 150 degrees centigrade plus what ever the ambient
    temperature is. The LF50 junction temperature maximum rating, is 125
    degrees centigrade. If you use a heatsink, the junction to case temperature
    rise is 3 to 5 degrees centigrade per watt. You need to use a heatsink, or
    lower the source voltage. The dropout voltage for the LF50, is about .7
    volts, at maximum temperature.
    Hope this helps,
    Brian
     
  5. Brian

    Brian Guest

    With no heatsink and a junction temperature of 125 degrees Centigrade, the case
    would be about 117 degrees above ambient. Let's add 25 degrees Centigrade
    (ambient) to that, which would make a case temperature of 142 deg. Centigrade,
    which would be 287.6 degrees Fahrenheit. As you can see, that is pretty hot to
    the touch.

    If we want:
    (1) 100 deg. C. max. junction temperature
    (2) 2.5 watts to dissipate
    (3) Junction to case thermal impedance of 3 deg. C. per watt.
    (4) Case to sink thermal impedance of 1 deg. C. per watt.
    (5) Work in an ambient temperature of 50 deg. C

    The heatsink would need a Zsa (surface to air impedance) of about 16. This
    equates to a heatsink of about 1.5 square inches of surface area. A larger
    heatsink would be even better.
    Brian
     
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