Connect with us

how does bypass capacitor work?

Discussion in 'Electronic Basics' started by [email protected]!!!, Oct 20, 2004.

Scroll to continue with content
  1. Hi:
    how dose bypass capacitor filter out the noise on the signal?
    say i have a 74HC74 chip... with a bypass capacitor connecting the Vcc and
    Gnd, i get a really nice wave form. but w/o one, i got noice all over the
    place! can't even get a waveform on the scope...
    it's like magic!
    i know how a cap can filter out high freq. signal. but just can't figure out
    how a bypass cap work....

    thank you for any answers...
     
  2. Lord Garth

    Lord Garth Guest

    The noise spikes are high frequency. Does it help you to think of the cap
    as a small
    local power supply?
     
  3. I just found this with Google:

    http://www.seattlerobotics.org/encoder/jun97/basics.html

    Excellent piece of info!

    S.
     
  4. tempus fugit

    tempus fugit Guest

    Basically, the cap at the Vcc of the chip diverts any noise spikes to
    ground. Think of it as water flowing downhill instead of up. The cap
    provides a low impedance path for any noise spikes to ground, so they go to
    ground rather than into the chip.
     
  5. Wires have inductance. Circuit board traces have inductance. It
    takes time for current to make its way from a power supply to a chip
    that has a sudden drop in resistance across its supply lines (and
    modern chips can do this extremely quickly, compared to the time it
    takes light to make its way to the supply and back). Inductance takes
    time for current to change by the differential equation:
    V=L*(di/dt). A capacitor is defined by the differential equation:
    I=C*(dv/dt). This means that the current through a capacitor is
    proportional to the time rate of change of the voltage across it.
    (Whew!)

    So taking all that into account at the same time (imagination is
    faster than the speed of light) when a chip passes a sudden pulse of
    current, instead of that pulse having to make its way through the
    circuit board trace inductance and supply wiring inductance (including
    the time it takes for the current to change through those inductance
    and not even worrying about the speed of light over that distance) and
    having to put up with all the voltage sag it takes to drive that
    current pulse through all that inductance, if you put a capacitor very
    close to the chip, and if the capacitor is big enough, a small sag in
    the capacitor voltage allows it to supply a very quick pulse of
    current to the chip.

    Then the inductive supply distribution system charges the cap back up
    before the next pulse.

    Isn't that easy?

    ;-)
     
  6. Wong

    Wong Guest

    You may apply the Xc formula. What's the frequency of a DC and AC voltage ?
     
  7. Guest

    When the chip makes a sudden demand for energy it pulls the +ve supply
    downwards and the -ve supply upwards (as sort of momentary short
    circuit).

    Why? Because although the supply tracks are "just wires" they still
    have a little inductance and if you try to push a fast enough *change*
    in current though them, they will "react" to that change and "delay"
    it leaving a temporary "vacuum" at the demand end, i.e. a drop in
    voltage at the +ve (and a rise at the -ve end).

    The "bypass" cap, as the above poster explains, is a temporary battery
    that makes good the temporary deficit more or less provided you keep
    its legs as short as possible and connect *both* ends as close as
    possible to the supply pins of the IC.

    But evidently all things suffer from this effect, that is why you
    might see as many as three bypass capacitors, all connected in
    parallel e.g. 100uF, 10nf, 100pf because each have limitations
    frequency-wise: the smaller caps "clean" up the faster transients.

    Cheers
    Robin
     
  8. Rich Grise

    Rich Grise Guest

    Well, the gag straight answer is, a bypass wroks by filtering out the high
    frequency signals from the power supply leads. ;-)

    When a totem-pole-output chip switches, both output devices are on
    simultaneously, briefly. During this short transition period (you'd
    have to look up the spec - probably on the order of 1 ns), they
    draw a heavy current pulse. The charge on the capacitor provides
    charge to power that current pulse, and then recharges at its
    leisure. So it filters out the high-frequency component of the
    signal.

    This is also known as a "decoupling" capacitor, which they generally
    use instead of "bypass" when you're doing digital. Technically,
    a "bypass" simply presents a low impedance to the signal at the
    frequency in question - notably, to complete an RF circuit.

    HTH!
    Rich
     
  9. How much inductance does a trace generally have? If you pick the wrong
    capacitor, can it resonate with this inductance, given a particular
    frequency for a chip's pulses of current?

    I recall a T1 card that was designed at a former company (not by me, I'm
    a software guy) that had issues with the 32 bit address/data bus having
    too much inductance. There were data corruption issues, and when they
    clipped on logic analyzer probes, the corruption went away. They fixed
    it by hacking on tiny smt capacitors. Is that kind of thing easy to
    predict? (It was between a QED MIPS core and a Galileo memory/pci
    controller.)

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  10. That is difficult to nail down without modeling the surroundings, but
    here is an article about the inductance of loops (and the supply
    connections to a bypass capacitor form a loop):
    http://members.aol.com/marctt/CV/Abstracts/inductance.htm

    If you pick the wrong
    Yes. Somewhat lossy capacitors are not necessarily bad for bypassing
    for this reason. I have connected an ohm in series with a .1 uf on a
    board near the connection points for the supply wiring to damp such
    ringing, in addition to the distributed bypass capacitors.
    Bus drivers are especially bad actors, because they have high current
    capability per bit line, and are switched on and off in large groups.
    Big knocks.
    Not for me, maybe for others. I try to select the adjacent bypass
    values based on peak current drawn and duration of pulse, to limit the
    sag during the pulses, neglecting the current supplied by the rest of
    the system. If no part of the system can generate a significant
    instantaneous voltage swing across the supply rails, there is little
    likelihood that any large resonance will occur. But I always check
    and have provision for ringing dampers as I described above, as cheap
    insurance.
     
  11. This guy claims that you can get between 0.5uH/m and 1uH/m of
    inductance. That's amazing. It implies that with a 10cm run, and a
    bypass cap of 0.1uF, you can get resonance at 2.25MHz. There are clearly
    write or read sequences that can generate lots of harmonics near there
    given a 66MHz data rate.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  12. This shows why big capacitors become pretty silly at high frequencies,
    and you better get that capacitance a lot closer than 10 cm to the
    die. With surface mount caps, it even makes a difference whether you
    punch to the power plane under the cap or beside it or at the end of
    it. And two vias on opposite sides of the cap are better than one.
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-