[SI-LIST] Re: Estimation of trace capacitance
- From: "Beal, Weston" <weston_beal@xxxxxxxxxxx>
- To: si-list@xxxxxxxxxxxxx
- Date: Wed, 5 Feb 2003 15:53:49 -0800
Prasad,
By calculating the total capacitance of the trace and load you are assuming
that the driver is charging the entire net at the same time. If your driver
transition time is less than the wave propagation delay across the 5 inch trace
(Tr < 180ps * 5) then that assumption can be very wrong. If a fast driver
(e.g. 500ps risetime) drives a 50 ohm trace that's 10 inches long with 1 load
at the end or if it drives a 50 ohm trace that's 20 inches long with 10 loads
at the end the driver will drive only the 50 ohms during its transition.
On the other hand if you are using 7400 TTL chips you are probably OK
calculating circuit capacitance. :)
Later,
Weston
-----Original Message-----
From: Mangipudi, Prasad [mailto:Prasad_Mangipudi@xxxxxxxxxxx]
Sent: Wednesday, February 05, 2003 3:07 PM
To: si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Estimation of trace capacitance
Hi SI experts,
How do you estimate the capacitive load of PCB traces to calculate the total
load on a driver? For example, there is a 50 ohm PCB trace of 5 inches
length (5 mil trace width, dielectric constant of 3.8 with suitable stack
up, propagation delay of 180 pS/inch) connected to a receiver of 3pF load.
Assume the trace is on a ground plane and the frequency of operation is 166
MHz clock for DDR333 operation.
I have the following situations:
1. Calculate the trace capacitance using the formula given in High Speed
Digital Design book by Howard Johnson (page 307, Eq 9-15),
I get C per inch = Td/Z0 where Td is the trace propagation delay and Z0 is
the trace impedance, I get 180/50 = 3.6 pF/inch.
For 5 inches, this is 3.6x5 = 18pF and add receiver capacitance, I get 21pF
load on the driver. This loading seems to be large??
2. Calculate the trace capacitance using the traditional parallel plate
capacitance formula,
C = 8.85 A Er/D pF where A is the area of parallel plates(meter sq), Er
is the relative permittivity of the material between the parallel plates
(3.8 in this case) and D is the separation between the plates in meters( in
this case 5 mils),
I get after conversion to inches, C = 0.84 pF/in or 4.2 pF for 5 inches.
This gives a total load of 4.2+3 = 7.2 pF on the driver. Reasonable??
The difference in both the cases seems to be large and want to know which is
correct. Are there any other methods of calculation? Any thoughts?
Thanks,
Prasad
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