Hello All!
I want to calculate how much capacitance I need for my design's switching
currents.
Q1. Is it possible to obtain all of your bypass capacitance from two power
planes where I have
+3.3V
dielectric
GND
?
Q2. If I calculate the total bypass capacitance, does this value have be
divided into bulk (surface mount bulk capacitance) and the capacitance of
the planes(see last section below)?
--------------------
Section 1: Background info:
I calculated the total load capacitance. I then used the equations in
"High-Speed Digital Design" book by Johnson and Graham (Example 8.1 p273)
dI = NC dV/dt where d represents the Greek letter Delta
N =147 outputs
V = 3.3Volts
t = 6 ns (rise time and fall time required at the outputs)
= 4.453 Amps (worst case peak while charging all loads)
dV = 0.050V (from noise margin budget)
X_max = dV/dI The maximum common-path impedance we can tolerate
= 11.2m Ohms
(Embedded image moved to file: pic15447.pcx)
C_bypass = 1/(2pi*f*X_max) where f = 33 MHz
= 429nF
------------------
Section 2. The equation for parallel plane capacitance is:
C_p = .225 * er * A /h pF
where
er = dielectric constant (dimensionless)
A = Are of the parallel plates
h = thickness of the dielectric
A = 57.2 square inches
h = 5 mils
er= 4.1
C_p = 10.5nF
So, does this mean that I have to have 418.5 nF being provided by SM bulk
capacitors (i.e. four 0.1uF caps)and the rest (10.5nf) from the board
capacitance?
Note that the most I can get from this board is ~30nF if the thickness of
the dielectric is 2mils.
---------
Sal Aguinaga
(847) 222-2833
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