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How to calculate power ripple, and then to add the sutable Capacitor?

Discussion in 'Electronic Design' started by Boki, Sep 30, 2005.

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  1. Boki

    Boki Guest

    Hi All,
    My audio amplifier has a lot of noise, after I connect a VCC to
    GND capacitor, the noise is gone.

    But, I did a boring thing, ==> try and error to find the value.

    How can I calculate the value directly?

    Do I have to use scope to see the ripple first? and then calculate the
    RC time?

    Best regards,
    Boki.
     
  2. Tim Shoppa

    Tim Shoppa Guest

    My audio amplifier has a lot of noise, after I
    Noise? Does this mean hum, or does it mean motorboating, or does it
    mean broadband hash, or ?
    Nothing wrong with using a scope, but this is pretty much chapter 1 of
    any power supply design tutorial.

    Tim.
     
  3. Jon

    Jon Guest

    Boki,

    If you're using a switching supply, your best source of equations would
    be the app notes for the regulator chip that you're using. For a
    simple full wave rectifier-capacitor input filter type design, here is
    a rule of thumb that I have found useful:
    ~
    For ripple < 10%, the following rule applies:
    .. Let R = the load resistance (Ohms)
    .. Let C = the filter capacitance (Farads)
    .. Let f = Frequency (Hz)
    .. Pi = 3.14159......
    (2PifRC) = 10 yields (roughly) 10% ripple and 10% load regulation.
    This rule scales (roughly) linearly, so, for example:
    .. (2PifRC) = 20 would yield roughly 5% ripple and 5% load
    .. regulation
    Remember this is just a rough rule of thumb, but it gets you close.
    Regards,
    Jon
     
  4. Boki

    Boki Guest

    ya, I have to review them recently.

    basically, I think I can measure the ripple first, and then try to
    calculate how larger RC time do I need, am I right?

    but it seems that only a capacitor between VCC and GND, so what I have
    to consider about R is total circuit?

    Best regards,
    Boki.
     
  5. ehsjr

    ehsjr Guest

    This site may help:
    http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/rectct.html#c2

    Ed
     
  6. Pooh Bear

    Pooh Bear Guest

    You're asking the wrong question.

    Your 'noise problem' is unlikely to be related to any ripple current
    requirement. The reservoir capacitor(s) in the power supply do this.

    It would help if you explained what the noise sounds like though !

    Graham
     
  7. Given the context of prior quotes, I think you're using an LM386.

    They are prone to motorboating (that is low-frequency instability) without
    good supply isolation. If the noise you're hearing sounds like a pulse
    train (tick-tick-tick at 5 to 50 ticks/second), that's probably what's
    happening. If it sounds like hum, then it's more likely a problem of supply
    filtering, that is, ripple. If it sounds like hiss, it's either noise on
    the supply lines, or high-frequency instability, the latter again an
    isolation problem. Ripple reduction is a matter of reading an intro text on
    power supply design (it's probably covered in AoE, I don't remember
    offhand). Isolation is harder to solve theoretically, because you don't
    have enough information about the supply impedance or about the internals of
    the LM386; trial-and-error is the best approach there.

    In either case, looking at the output and at the supply line (near the chip)
    with an oscilloscope can tell you a lot. Especially if the problem goes
    away when you touch it with the scope probe :)
     
  8. Pooh Bear

    Pooh Bear Guest

    Or your finger !

    That kind of instability was very popular in the 70s / 80s.

    If Boki used a more suitable amplifier chip the problem would likely vanish.

    Graham
     
  9. Mac

    Mac Guest

    It's not RC. It's I=cdV/dt. I is the max supply current, dV is the ripple
    Voltage you are willing to accept, and dt is the period of the waveform on
    VCC.

    For example if you build a power supply from fully rectified 50 Hz, then
    your waveform is 100 Hz. So you want to use 10ms as your dt. Now you can
    solve for c:

    c=Idt/dV

    HTH

    For high-current power supplies you then you have to calculate the ripple
    current for the capacitor and select a part (or parts) which can handle
    the ripple current.

    --Mac
     

  10. The rule is that an 8300 microfarad capacitor has its volts of ripple
    equal to its amps of current at full wave 60 Hertz.

    Everything simply scales from there.



    --
    Many thanks,

    Don Lancaster voice phone: (928)428-4073
    Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
    rss: http://www.tinaja.com/whtnu.xml email:

    Please visit my GURU's LAIR web site at http://www.tinaja.com
     
  11. I read in sci.electronics.design that Don Lancaster <>
    10 000 uF for full-wave 50 Hz. Even simpler.
     
  12. Pooh Bear

    Pooh Bear Guest

    Actually about 8000 uF @ 50Hz and 6700uF @ 60 Hz

    Don incorrectly assumed a discharge time = 1/2 cycle.

    Graham
     
  13. I read in sci.electronics.design that Pooh Bear
    Unfortunately for the harmonic levels on the mains, the charge time for
    high-efficiency SMPS is so short now that the half-cycle assumption is
    almost true.
     
  14. Pooh Bear

    Pooh Bear Guest

    Uh ?

    Peak loading results in flat-topping which actually extends the charging time ! The
    ac supply's way of 'compenating'. ;-)

    Graham
     
  15. I read in sci.electronics.design that Pooh Bear
    But I've done the measurements.
     
  16. Pooh Bear

    Pooh Bear Guest

    What were you measuring ?

    Just doesn't sound right to me.

    I'll take a look at an smps I'm working on right now. It certainly isn't the case
    for classic line frequency transformer based psus where the charging time is as
    much as 2.5ms at lower powers ( lots of DC R from the transformer ) and typically
    around 2 ms at medium / high power.


    Graham
     
  17. I read in sci.electronics.design that Pooh Bear
    SMPS between 30 W and 75 W. Short conduction angles, like 18 degrees, or
    1 ms in your terms. And it wasn't only me: all the members of the IEC
    Task Force saw it done. Some, but not all, eyebrows were raised.
    More; you can get to the magic 65 degree conduction angle (3.6 ms) that
    meets the IEC/EN 61000-3-2 Class D limits. Not so easy with a toroidal
    transformer, but a few ohms in series helps.
    Transformer/rectifier supplies do have lower harmonic emissions.
     
  18. Pooh Bear

    Pooh Bear Guest

    Ok, I have a 33W smps I can compare with.

    I'll bet the relatively small reservoir cap has something to do with that. I can't
    recall offhand if we use 68uF or 100 uF.

    Was there any R in the way at all btw ?
    I'll also take a look at an R-core based supply we have. The DC R in those is very
    high indeed. The R-cores have the lowest stray flux I've ever met btw. Haven't got
    my hands on an O-core yet but they should be even better.

    Graham
     
  19. I read in sci.electronics.design that Pooh Bear
    Probably not enough if you have to meet the immunity requirements
    against voltage dips and interruptions. People are using much the same
    capacitors as for 200 W supplies. They must be, otherwise we wouldn't
    get such small conduction angles.

    The only impedance in series with the rectifier is to limit the inrush
    current to 50 to 80 A or so, to avoid things going BANG!
     
  20. Pooh Bear

    Pooh Bear Guest

    I just did a quick simulation using EWB. I'm blowed if I can tweak anything much
    away from a 2.5ms conduction period.

    I had a small resistance in series with the supply that I varied from 0R1 to 5R.
    Still didn't change much though.

    Do you recall any more details ?

    Graham
     
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