26
Decoupling and Layout Recommendations
Mobile AMD-K6
Processor Power Supply Design
22495C/0—May 1999
inductances between the capacitor and the processor as shown
in Figure 15 on page 28. The dotted line shows that connecting
the capacitor directly to the processor eliminates two series
inductances. However, this trace also has inductance—if it is
too long or too narrow it can be worse than the vias.
Figure 14 on page 28 shows the effect of the inductance at
higher frequencies. (The numbers outside the X and Y axis
indicate the minimum and maximum values plotted). The
inductance used is 1.8nH (0.7nH for each of the two vias and
0.4nH for the capacitor itself). The capacitor is a 0.1-
μ
F X7R
multilayer Ceramic MLC. The inductance of a capacitor is a
function of the case type. An 0612 case is assumed here.
The following steps show how the number of required
capacitors is calculated:
1. Decide what to allow as a ripple voltage budget. In this
example calculation the ripple-voltage budget (dv) is 30 mV.
2. The measured AC transient current is 0.75A. This transient
current has a typical duration (dt) of 2.5 nsec. The amount
of capacitance required can now be determined using the
following equation:
I = C (dv/dt)
C = I (dt/dv) = 0.75A (2.5nsec/30mV) = 0.625
μ
F
This equation indicates that if the capacitors didn't have
inductance, only six 0.1-
μ
F capacitors would be required.
3. Determine the number of capacitors required based on the
inductance of the capacitor. Use the following formula:
V = L (di/dt) = L
(0.75A/2.5nsec) = 30mV
Solving for L, the allowed budget is 100pH.
4. The inductance of the capacitor and via = 1.8nH (0.7nH for
each of the two vias and 0.4nH from the capacitor itself).
Because each capacitor usually has two vias (one on each
end), the effective via inductance must be:
2
0.7nH + 0.4nH = 1.8nH