12VDC Power Supply Help Needed

Discussion in 'The Projects Forum' started by Iggy, Feb 2, 2009.

  1. Iggy

    Thread Starter New Member

    Feb 2, 2009
    I'm a newbie and want to make a 12VDC regulated power supply to experiment with various LED's combinations. I calculated that for 2.5% ripple, C1 should be a 4700uF 25V capacitor. However, I don't know how to determine what the value of C2 should be or if it is needed at all.

    My calculations so far:
    Transformer is rated 12vrms @ 500ma 60Hz
    VP = (1.414 x 12) - 1.4 = (16.968) - 1.4 = 15.568
    Vrip = 0.3892V X 2.828 = 1.10 V
    C1 = [ ( 0.5A X 0.00833) / 1.10V] X 1000000 = 3786uF

    Any help appreciated.
  2. SgtWookie


    Jul 17, 2007
    4700uF should be fine for C1, but 25v is marginal. 12.6vac when rectified will become around 17v. The "rule of thumb" is to double the voltage rating for electrolytic caps, so you'd want a cap rated for 35v (standard value). This helps to minimize "leakage"; if leakage becomes excessive the capacitor will overheat and blow it's top.

    Use a 0.1uF and 10uF on the output, just to suppress any oscillation tendencies. You won't have any real transients to speak of with LEDs.

    Note that to ensure guaranteed regulation, a 7812 requires a minimum load of 5mA. You could provide that by adding a 2.2k resistor from the output terminal to ground, or simply ignore it if you have more than a 5mA load. Just don't be surprised if the output isn't close to 12v when you do not have a load.
  3. bertus


    Apr 5, 2008

    Just take a look at the datasheet of the 7812.
    There are two smal capacitors on input and output to prevent the 7812 from oscillating.

  4. beenthere

    Retired Moderator

    Apr 20, 2004
    C2 is oversized. 1 uF would be plenty. In fact, you would do well with just a .1 uF ceramic close to the 7812's pins on the output. C1 is probably larger than necessary, but capacitors are inexpensive. The load will not exhibit dynamic characteristics.
  5. Audioguru


    Dec 20, 2007
    An LM317 regulator has this problem because its idle current (current for its internal circuits) goes through the output so its ADJ pin's current is extremely low.
    But the idle current of a 78xx regulator goes to ground through its ground pin so it does not have a minimum load current.
  6. SgtWookie


    Jul 17, 2007
    Please examine a datasheet from ST Microelectronics' L7800 Series or Fairchild's LM7800 Series positive regulators. You'll find that output voltage regulation is specified with a load of 5mA to 1A. To me, this indicates that a minimum load of 5mA is required to achieve guaranteed regulation. This is also consistent with other manufacturer's datasheets.

    With an LM317, if a 120 Ohm resistor is used for R1, there is no additional minimum load required, as 10mA will flow from the OUT to ADJ terminal, and then to ground via R2.

    With an LM317L, R1 can be 240 Ohms, as the minimum load requirement is 5mA.
  7. Iggy

    Thread Starter New Member

    Feb 2, 2009
    Thanks! I'll have at least a 20mA load.

    Thanks, I think I'll use a 2200uF 35V for C1 (5% ripple).
  8. mik3

    Senior Member

    Feb 4, 2008

    Have you considered that such a big capacitor (C1) will cause a large in rush current during power up which may blow the diodes?
    This current must stay below the maximum current the diodes can conduct.

    Also, diodes can withstand a surge current for an amount of time. If this surge current lasts more the diodes are gone.