Switching regulator (boost) voltage ripple

Discussion in 'The Projects Forum' started by Tyco, Mar 20, 2007.

  1. Tyco

    Thread Starter New Member

    Mar 20, 2007
    2
    0
    I've been working on a project that requires a 12V and 3.3V supply. The 12V is used for various sensors and an OLED main power while the 3.3V is utilized by an AVR microcontroller and associated logic for the OLED/User input. Currently the power source is 4xAA batteries in series for a total 6V power.

    The issue I'm running into is a very large ripple (2Vp-p) on the 'regulated' switching boost regulator. I'm utilizing a LM2733 to boost from the AAs to 12V, following the specifications in the datasheet. With no load, the regulator produces a fairly clean 12V output, but at full load the ripple emerges.

    While exploring the problem I've tried setting up a zener diode shunt regulator to limit the noisy 12V signal to 11V which produces a clean 11V unloaded voltage, but upon loading the voltage drops to 7.5V steady.

    This same effect is seen when I wired together an Op-amp peak detector to the output; clean 12V with no noise unloaded, and ~7V loaded.

    These voltage drop-off lead me to believe the load current required exceeds that which the 4xAA batteries can supply. However, with just the LM2733 switching regulator a similar voltage drop is not seen, just the large voltage ripple.

    I do not, unfortunately, have a variable power supply to test the circuit on to see if it relates to the batteries. Also, due to the complexity of the circuit predicting equivalent load resistance to arrive at current requirements would be a difficult task.

    Any ideas/recommendations?
     
  2. thechief

    New Member

    Mar 19, 2007
    3
    0
    A shunt regulator has a series resistor, which will drop voltage when you apply a load. That's the nature of the beast. On your switching regulator, what do you have for the output capacitor? You can calculate the ripple by (load I (A) / switching frequency(Hz)) / capacitance (uF) = ripple (Vp-p). Maybe you just need a larger capacitor. If you still need lower ripple, you may need to use an LC filter on the output.
     
  3. Tyco

    Thread Starter New Member

    Mar 20, 2007
    2
    0
    I have a 4.7uF cap on the output of the switching regulator which was resulting in the 2Vp-p ripple. In efforts to decrease the ripple I've since added 2x 33uF caps on the output as well, which has decreased the ripple to a more manageable level.

    Ideally I would like to eliminate the ripple entirely. It seems that one option would be to use the LM2733 to boost the voltage to 15V and then use a 7812 linear regulator to bring the voltage back down to 12V, but the loss in efficiency in that combination is something I'm trying to avoid.
     
  4. ashokcp

    Active Member

    Mar 8, 2007
    50
    2
    You could try an LDO regulator, Low dropout voltage regulator, and, keep the input to the LDO just above the minimum Vdropout. These require a dropout of less than 1 V to provide regulation. So for your app, you could keep the input to LDO regulator at about 13, which would minimise your losses compared to 15V for 7812.
     
  5. nomurphy

    AAC Fanatic!

    Aug 8, 2005
    567
    12
    You need more output capacitance. Try a 330uF - 470uF, low ESR tantalum. With a 12V output, you should use a tantalum rated at 25V.
     
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