Assistance on voltage regulators

Discussion in 'The Projects Forum' started by ElectricalNewb, Aug 14, 2013.

  1. ElectricalNewb

    Thread Starter New Member

    Aug 14, 2013
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    Im in the process of building a project variable power supply and was wondering is anyone could help me with the voltage regulator. I'm really wanting to use a linear power supply instead of a switching power supply, its just a personal preference, but Im stuck between the LM350 and the LM317. My transformer only puts out 20V/2A. The ready made power supplies I have seen similar to my application run an LM317 with a small heatsink in an enclosed box.

    Where Im concerned is with the size of the heatsink that I will need to run either of these ICs. So I was hoping that someone would be able to help me out.
     
  2. #12

    Expert

    Nov 30, 2010
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    For the TO-3 package, theta Jc = 1.5
    For a full wave bridge with capacitor filter, I = .62 X the transformer rating.
    The transformer can deliver 1.24 amps at 28.28 volts peak.
    Subtract the loss of voltage in the rectifier assembly to arrive at about 26.7 volts.
    If the output voltage is adjusted to 1 volt, the regulator chip will have 25.7 volts across it at 1.24 amps maximum.
    That makes 31.84 watts, minus the effect of ripple voltage at full load, which isn't much.

    For the chip limit of 125C and a room temperature of 25C, you are allowed a 100C temperature rise at full load (but you should try not to push the chip to its melting point. Better to use 100C as the high limit.)

    The thermal resistance of the TO-3 package is 1.5C/W which means the chip will heat up by 1.5 C per watt if it is connected to a perfect, infinite heat sink. There is no such a thing, so you calculate the minimum heat sink with these numbers.

    P X resistance to heat flow = 100C
    3.14 is the limit for heat flow resistance. The chip has already used up 1.5 of those, and the goo between the TO-3 package and the heat sink will use up another .1 C/W
    3.14 - 1.6 = The Heat Sink Rating
    That's 1.54 C/W which is a label you will find when trying to buy a heat sink.
     
  3. ElectricalNewb

    Thread Starter New Member

    Aug 14, 2013
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    So a heatsink that measures 117x61x37mm with a thermal resistance of 1.3C/W should be sufficient for the LM350? Or would I be safer dropping down to an LM317 and running the same heatsink?
     
  4. #12

    Expert

    Nov 30, 2010
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    LM350 and LM317 do not say which package you are using. You have to declare which package and use the thermal resistance of the package the chip is in. Read the datasheets.

    But, yes, 1.3C/W will heat up less than 1.54 C/W
     
  5. ElectricalNewb

    Thread Starter New Member

    Aug 14, 2013
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    Yeah, I apologize, I forgot to mention that the ones Im looking at are TO-220 instead of TO-3. I just ask because I suck at the math and understanding the formulas.
     
    Last edited: Aug 14, 2013
  6. #12

    Expert

    Nov 30, 2010
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    The TO-220 case has 4 to 5 C/W and won't work.
    It doesn't require math to look up the thermal resistance of the package by looking at the datasheet, especially after I did the math for you.
     
  7. crutschow

    Expert

    Mar 14, 2008
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    The power dissipated is independent of the regulator you use. It strictly depends upon the voltage drop across the regulator and the current.
     
  8. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
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    Correct. But the temperature rise of the die caused by the power is very much dependent on the package selected: a typical TO-220 mounted to a heatsink has a rise of about 4C/W PLUS THE RISE OF THE HEATSINK, while a TO-3 can be closer to 1C/W depending on die size.

    In this case with a 20VAC/2A transformer, the design could be upwards of maybe 22V/1A which would put as much as 20W or more of power in the reg so the 4C/W number would not be viable for use. You would have more than 100C rise above ambient even with a pretty good heatsink.
     
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