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LM78XX imput/output capacitors

Discussion in 'Electronic Design' started by jmariano, Sep 13, 2007.

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  1. jmariano

    jmariano Guest

    Dear All,

    Sorry if this is a basic question. I deed search the web for an
    answer, but could find one.

    I'm designing a power supply with various outputs (+24, +15, +10, -15,
    -5), using LM78/79XX line regulators. The datasheet states that a .1
    uF capacitor should be placed at the output to improve transient
    response, and a .22 uF at the input, if the regulator is far from the
    filter capacitor. But when I look at designs published in journal and
    such, I found all kind of variations, including different values for
    the input cap., even if
    the regulator is close to the filter capacitor (in a PCB), and usually
    an additional electrolytic capacitor at output.

    So I guess my question are: 1 - do I need a capacitor at input and
    what is is value? (and is purpose).
    2 - Do I need an additional electrolytic capacitor at output, besides
    the 0.1uF, and how do I calculate is value.

    If anyone knows of a bibliographic reference ware this things are
    explained, I would appreciate.

    Regards

    José Mariano
     
  2. Hello,

    these are minimum values.

    the input capacitor doesn't replace the filter capacitor.

    for the filter capacitor, take 1000uF per Ampere for 1 volt input variation.

    for the output, put big value if you want low impedance...you can add a 220
    nF plastic capacitor cause it is better at high frequencies.

    Regards

    Vincent


    "jmariano" <> a écrit dans le message de

    Dear All,

    Sorry if this is a basic question. I deed search the web for an
    answer, but could find one.

    I'm designing a power supply with various outputs (+24, +15, +10, -15,
    -5), using LM78/79XX line regulators. The datasheet states that a .1
    uF capacitor should be placed at the output to improve transient
    response, and a .22 uF at the input, if the regulator is far from the
    filter capacitor. But when I look at designs published in journal and
    such, I found all kind of variations, including different values for
    the input cap., even if
    the regulator is close to the filter capacitor (in a PCB), and usually
    an additional electrolytic capacitor at output.

    So I guess my question are: 1 - do I need a capacitor at input and
    what is is value? (and is purpose).
    2 - Do I need an additional electrolytic capacitor at output, besides
    the 0.1uF, and how do I calculate is value.

    If anyone knows of a bibliographic reference ware this things are
    explained, I would appreciate.

    Regards

    José Mariano
     
  3. Eeyore

    Eeyore Guest

    I take it you mean ripple ? Your numbers are out by about 7-8:1

    You need ~ 8000uF for 1V (pk-pk) of ripple @ 1A load and 50Hz.

    Graham
     
  4. JeffM

    JeffM Guest

  5. MooseFET

    MooseFET Guest

    If the LM78XX regulator can see an inductive input line it is likely
    to oscillate. You want a low impedance capacitor right at its input
    terminals. The values in the data sheets are minimums based on
    several assumptions about how the rest of the system looks. They are
    usually enough.

    Another reason to put a larger capacitor is that logic circuits often
    draw current in narrow pulses. A capacitor to ground and some
    impedance in the input wire can stop these pulses from making their
    way up the power wiring.

    One thing that people often overlook is that nearly all capacitors
    cost about the same by time they are installed. Using a low number of
    different values can often save you more than the price differences.
    I normally put added capacitance at some small distance from the
    output. The purpose is to provide a low impedance path to ground for
    the supply. The output impedance of a 78XX rises at high
    frequencies. I prefer the impedance to fall.
     
  6. Yes, you should have the capacitors. Their purpose is to keep the
    regulator stable and stop it from oscillating.
    Not to keep the regulator stable, no.
    You may need additional capacitors on the output located elsewhere on
    the board to provide localised decoupling to those parts of the
    circuit.

    Be aware that "low dropout" voltage regulators are a different beast
    and have fairly critical input and output decoupling requirement, this
    will be explained in the datasheet for each device. Standard
    regulators like the 78xx series are much more tolerant to this.

    You might see extra caps in some designs because people tend to get
    overzeleous in use of decoupling caps. Nothing wrong with that apart
    from extra cost and board space.
    This might get you started:
    http://en.wikipedia.org/wiki/Decoupling_capacitor

    Dave.
     
  7. Oh, that's right !!! Sorry, I apologize for my mistaque....
     
  8. No, noone told me....surprised ? Anyway I'll take a look at this, thanks !!!
     
  9. Eeyore

    Eeyore Guest

    From which manufacturer ?

    More than a couple of inches typically. It's to counter the input inductance
    caused by the long pcb trace.

    The designers are being cautious. An extra cap doesn't cost much. A piece of
    returned equipment costs a lot.

    Negative regulators require this IIRC.

    I think TI or Nat Semi once published a 'Voltage Regulator handbook' - I expect
    it would be available as a download these days. The one I'm thinking of is TI -
    I have the 1977 copy.

    Anyway, IIRC, negative voltage regulators like a slightly larger cap than just
    0.1uF on the output pin. I use a 10uF electrolytic in parallel with a 0.1uF
    plastic film cap.

    Graham
     
  10. Eeyore

    Eeyore Guest

    I think you do with a 79xx especially. I'm sure I once forgot it and it gave
    trouble.

    Graham
     
  11. Eeyore

    Eeyore Guest

    No problem.

    Don't forget to add the transformer loading when calculating 'voltage variation'
    too.

    Graham
     
  12. John Fields

    John Fields Guest

     
  13. Eeyore

    Eeyore Guest

    Let's see your calculation then if you think my number is wrong. Or did you want me
    to suggest an E6 value ?

    Graham
     
  14. John Fields

    John Fields Guest

    ---

    Idt 1A * 0.01s
    C = ----- = ------------ = 1E-2F = 10000µF
    dV 1V


    For a standard +/- 20% part,


    Cnom 10000µF
    Cmin = ------- = --------- = 12500µF
    0.8 0.8
     
  15. Eeyore

    Eeyore Guest

    Where the hell do you get 10ms from ? I suggest you go and look at a ripple waveform.

    Graham
     
  16. John Fields

    John Fields Guest

     
  17. Eeyore

    Eeyore Guest

    That's not the relevant answer.

    What's relevant is the discharge period. Hint - NOT 10ms. I suggest you either measure or
    simulate.

    You're not as smart as you claim to be.

    Graham
     
  18. neon

    neon

    1,325
    0
    Oct 21, 2006
    To begin with if you intend to put a big cap on the output then you will also need a to put a diode across input to output. That is to save the regulator when power is shut off and the output is present because of the cap, For a sytem where there are many boards and circuitry it much beneficial to instal small caps across each device as opposed to a big across the rail,
     
  19. John Fields

    John Fields Guest

    ---
    I've never claimed that I was smart, just less stupid than most.
    Including you.

    However, in one of your rare moments of lucidity, your "What's
    relevant is the discharge period." brought me up short since that's
    correct and I hadn't earlier considered that the recharge time was
    irrelevant.

    Thanks for that. :)

    Moreover, including the recharge time in the waveform's period,
    'dt',

    Idt
    C = -----
    dV

    makes the solution inaccurate because it doesn't consider the
    discharge time boundary created when the rise of voltage from the
    rectifier causes the reservoir capacitor's losses to cease.

    So what's the right way to do it?

    As I see it, first determine the peak DC needed into the regulator.
    For a 7824 that would be:

    Vin = Vout + Vdo + Vrpl = 24V + 2.5V + 1.0V = 27.5V

    Where Vin is the peak input voltage to the regulator,
    Vout is the output voltage of the regulator,
    Vdo is the regulator's dropout voltage, and
    Vrpl is the allowable peak-to-peak ripple voltage on the
    input of the regulator.

    Next, determine the angle which corresponds to the valley voltage of
    the ripple:

    Vin - Vrpl 26.5V
    arcsin = ------------ = arcsin ------- = 74.5°
    Vin 27.5V

    Then, determine the length of the capacitor's discharge time.

    For a 50Hz sinewave, we're dealing with:

    1 0.02s
    t = ------ = -------
    50Hz 360°

    which is also:

    0.02s
    t = ------- ~ 55.5µs/°
    360°

    Since the cap is fully charged at 90° and discharges to 26.5V in the
    time it takes to go from 90° to 0° and then back up to 74.5°, that's
    a total of:

    n = 90° + 74.5° = 164.5°

    To get the discharge time, then, we multiply the total excursion,
    164.5° by 55.5µs/°, and wind up with:

    164.5° 55.5µs
    Td = ------- * -------- ~ 9.13E-3s = 9.13ms
    1 °

    Now, since we're feeding a fixed voltage (the output of the
    regulator) into a fixed resistance, (the load) we can say that the
    input current into the regulator will be what the load dissipates
    and we can finally say that the value of the reservoir capacitor
    should be, at least,


    Idt 1A * 9.l3E-3s
    C = ----- = -------------- = 9.13E3 ~ 9100µF
    dv 1V
     
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