# how does bypass capacitor work?

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

1. ### [email protected]!!!Guest

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....

2. ### Lord GarthGuest

The noise spikes are high frequency. Does it help you to think of the cap
as a small
local power supply?

3. ### Stefan ArentzGuest

I just found this with Google:

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

Excellent piece of info!

S.

4. ### tempus fugitGuest

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. ### John PopelishGuest

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. ### WongGuest

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 GriseGuest

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. ### Robert MonsenGuest

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. ### John PopelishGuest

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. ### Robert MonsenGuest

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. ### John PopelishGuest

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.