How much ripple is too much on the input of an LDO?

Discussion in 'General Electronics Chat' started by TheLaw, Jun 4, 2012.

  1. TheLaw

    Thread Starter Member

    Sep 2, 2010
    228
    2
    Hello,

    I'm designing a bench power supply and I trying to build it into a relatively small case and have quite a bit of functionality. One of the problems I'm having is space, as you might assume.

    Whereas before, in my stupid hobby mindset, I'd just get input capacitors that will bring the ripple down to next to nothing, but in this case, I actually need capacitors that are properly suited because I can't waste space.

    Is 1V p-p too much for decent performance? 0.5V? 2V?

    The max current draw will be around 2A.

    --
    And one other question if I may add..I'm trying to figure out how I am going to wire everything up. I'm basing my power supply on this design: http://www.extremecircuits.net/2010/06/fully-adjustable-power-supply_17.html

    When using a pass transistor, does the LM317 need to be heatsinked? Does it 'share' the load with the pass transistor or does all of the current go through the external transistor. Trying to figure out how I am going to mount everything, especially to heatsinks. That's one of my concerns.
    Thanks.
     
    Last edited: Jun 4, 2012
  2. #12

    Expert

    Nov 30, 2010
    16,257
    6,762
    Partial answer:
    The ripple limit is when the drooping supply voltage gets low enough to hit the drop out limit on the regulator and the drooping part gets through to the load.

    radical2 C Er(p-p) F=I

    Figure from your peak voltage the p-p droop of the supply voltage and make sure it doesn't get low enough to violate the drop out limit of the regulator.

    If you have lots of excess voltage, 2 or 5 or 9 volts of ripple won't make any difference. If you don't have much excess voltage, the ripple has to be small. That's why we have math.
     
  3. TheLaw

    Thread Starter Member

    Sep 2, 2010
    228
    2
    Thank you sir. I didn't really understand that concept. Much appreciated.
     
  4. #12

    Expert

    Nov 30, 2010
    16,257
    6,762
    In that circuit, the external transistor starts to take over when the current through the 317 chip times 33 ohms = about .6 volts.
    That's only .018 amps plus the base current of the pass transistor. The 317 will survive without a heat sink.
     
    Last edited: Jun 4, 2012
  5. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    FYI, that circuit is not an LDO (low dropout regulator). The circuit shown will require about 3V or more differential from input to output to regulate.
     
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