[SI-LIST] Re: pcb board capacitors
- From: "Lee Ritchey" <leeritchey@xxxxxxxxxxxxx>
- To: "Fred Townsend" <fred@xxxxxxxxxxxxx>, dan1_st@xxxxxxxxx
- Date: Mon, 5 Dec 2005 17:14:14 -0800
These things we call bypass capacitors are actually coulomb buckets that
supply local charge to support switching events. When they are not large
enough to supply the charge their terminal voltage goes down and we get
ripple. Fix the ripple by putting large enough coulomb buckets nearby that
will operate at the frequencies of interest.
Wish we could erase the name decoupling capacitor from our vocabulary, it
misleads rather than enlightens as to the purpose of the capacitors.
Lee W. Ritchey
Speeding Edge
P. O. Box 2194
Glen Ellen, CA 95442
Phone- 707-568-3983
FAX- 707-568-3504
I just used the energy it took to be angry to write some blues.
Count Basie
> [Original Message]
> From: Fred Townsend <fred@xxxxxxxxxxxxx>
> To: <dan1_st@xxxxxxxxx>
> Cc: <si-list@xxxxxxxxxxxxx>
> Date: 12/4/2005 10:48:14 PM
> Subject: [SI-LIST] Re: pcb board capacitors
>
> Wouldn't it be great if there was a chart you could just look up 3.3 V @
> 5 GHz and get the proper value. Then all the SI engineers could turn in
> their badges and go home.
>
> I think you have a mistaken idea what a bypass cap does. Can you look up
> from this text and answer the question? I give you a clue. It doesn't
> have much to do with filtering, at least in the frequency domain sense.
> That's done back in the power supply.
>
>
>
>
> Answer: At 5 GHz there are really two functions. Let's consider the more
> common. For every trace or wire, particularly power and ground, there is
> a certain amount of inductance. For every di/dt in the chip there is a
> corresponding -v= L(di/dt). This voltage opposes Vs causing the voltage
> at the chip to increase or decrease according to sign of di. If you are
> dealing with an analog chip it will introduce a common mode signal to
> the chip. If you are dealing with digital (time domain) it will extend
> rise and fall times.
>
> Let's look at the variables again. -v=L(di/dt) Three variables you
> haven't given us a clue as to their value. We would need a wegee board
> to give you a WAG.
>
> Now let me shock you. Capacitors have inductance as do leads to and from
> capacitors. Therefore capacitors have a self resonate frequency at which
> they switch from being capacitors to inductors. This means the capacitor
> package and capacitor leads are two more variables you interject into
> the equation.
>
> What happens if you can't get enough capacitance, below the self
> resonate frequency, you need. Inductance can be reduced by paralleling
> capacitors. For critical applications it is common to see a capacitor
> network of different packages and values. Yet another variable.
>
> At 5 GHz it is possible to see another problem. The pins of an IC form a
> matrix of capacitive values. If the pins happen to furnish an unwanted
> positive feedback path oscillation may occur. When this happens it is
> necessary to null out or swamp out the unwanted capacitance. Such
> capacitors are known as neutralizing caps.
>
> I hope I have explained why you will not be able to use a chart or rule
> of thumb to determine what capacitor value to use. At 5 GHz the PCB
> layout person controls lot more of of the design than the circuit design
> engineer.
>
> Fred Townsend
>
>
>
>
> david stern wrote:
>
> >hi
> >
> >I am designing a board working at 5GHz. The input
> >voltage is 3.3v.
> >What values of capacitors I have to use in order to
> >filter the input voltage how I calculate the capacitor
> >values where can i find material about it?
> >
> >THANKS
> >
> >DAVID
> >
> >
> >
> >
> >
> >
> >
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