Driving Capacitive Loads

"...consider a 74ACxx chip (5V output swing in 3ns) driving a total load capacitance of 25pF". I = C(dV/dt) = 25pF * (5V/3ns) = 41.667mA, and this current returns through ground (HIGH to LOW transition) or the 5V line (LOW to HIGH transition), producing little spikes at the receiving end. Consider that wiring inductance is roughly 5nH/cm, so 1 inch of ground wire (5nH * 2.54cm = 12.7nH) carrying this logic transition current would have a spike of V = L(dI/dt) = 12.7nH * (41.667mA/3ns) = 0.176V. "If the chip happened to be an octal buffer, with simultaneous transitions on a half dozen outputs, the ground spike would be over a volt" (from The Art Of Electronics, Paul Horowitz, Winfield Hill, ISBN: 0-521-37095-7, see Chapter 9.11 for more details).