[SI-LIST] Re: Effective impedance of a uniformly loaded bus
- From: "Dr. Howard Johnson" <howie03@xxxxxxxxxx>
- To: <si-list@xxxxxxxxxxxxx>
- Date: Mon, 28 Nov 2005 11:18:04 -0800
Dear Edi,
Your question relates to the effect of loading a bus with a
number of uniformly spaced capacitors.
Assume each trace of the bus begins with a certain
capacitance (C0) in units of pF per inch. To that trace
capacitance I wish to add the capacitance of the loads. The
trace capacitance is in units of pF per inch, so I must
convert the load capacitance to the same units before
adding.
Assume there are (N) loads and each load has a capacitance
of (CL) picofarads. The total load capacitance is:
N*CL (pF)
This capacitance is spread across the total length of the
bus in inches (assuming uniform loading). The effective
load capacitive per inch added to the bus is, therefore:
N*CL/length (pF/inch)
The total effective capacitance distributed across the
structure (in units of pF per inch) will be:
C0 + N*CL/length (pF/inch)
As long as the spacing between loads is small compared to
the rise/fall time of your signals, the loaded bus will
behave much like a transmission line with extra capacitance
(that is, a transmission line with a lowered impedance and
increased delay).
Lenghtening the transmission line (spreading the loads
across a greater measure of transmission structure) does
indeed reduce the capacitance PER INCH, and raise the
impedance.
The reason this is important is that your driver can only
put out a certain amount of current. If you want full-sized
waveforms, there is always a certain minimum impedance below
which your driver cannot successfully drive the transmission
structure.
If you build a line so heavily loaded (and so low in
impedance) that your driver cannot produce full-sized
waveforms on that line, the formula above suggests that you
might do better to spread the loads further apart. PROVIDED
that the loads still remain within a small fraction of a
risetime of each other, so that the distributed loading
effect applies, lengthening the structure sometimes raises
the impedance into a range that your driver can handle.
The driver must still drive the same total capacitance (even
more, because the line itself is now longer), but by
lengthening the structure you give the driver more TIME in
which to accomplish its job. By spreading the same total
work across a longer span of TIME the driver can do its job
with a lower peak current.
I hope these brief comments are helpful to you.
Best regards,
Dr. Howard Johnson, Signal Consulting Inc.,
tel +1 509-997-0505, howie03@xxxxxxxxxx
http:\\sigcon.com -- High-Speed Digital Design seminars,
publications and films
P.S. -- I'll be in San Jose doing public classes Jan 30 -
Feb. 3, and right after that attending DesignCon showing my
latest films, one made for Xilinx about crosstalk in BGA
packages and one for Sigrity about power system analysis.
See you there.
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx
[mailto:si-list-bounce@xxxxxxxxxxxxx]On Behalf Of Edi
Fraiman
Sent: Sunday, November 27, 2005 11:01 PM
To: si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Effective impedance of a uniformly loaded
bus
Hello experts,
I have question regarding formula C'=3D Cline +
NCload/length and Z' =3D
(L/C')^1/2 ("A Handbook of Black Magic" Paragraph 4.4.3.1).
If I design trace with several loads than according to the
formula I can
conclude next statement:
Longest trace cause of reducing in total capacitance C' and
increase the
Z'.
So, long trace help to driver drive loads (impedance is
larger).
I fill confuse, everyone knows that long trace is not good
for SI.
=20
Could you please clear me this issue?
=20
Best Regards,
Edi Fraiman
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