Circuit traces can literally support amperes of current flow, but it takes time for the current to reach these levels. To supply high step currents in nanoseconds, the characteristic impedance of the transmission path must be very low. As an example, if the characteristic impedance is 0.1 ohm and the voltage is 5 V, a load of 5 ohm will draw a current of 0.97 A with essentially zero rise time.

In an integrated circuit, many output logic traces can be connected to the source of voltage at any one time. If 10 lines are connected then the characteristic impedance at the logic junction is about 5 ohm. If an 11th logic trace is connected at clock time, the initial energy supplied to the connected line comes from 10 other lines in parallel plus the trace connected to power. In the other extreme, if no logic is connected and a logic trace is connected to power and if the source impedance is 50 ohm, the voltage will drop to 50%. This assumes that no energy storage is on the die. This example shows that the voltage levels at the die will vary depending on the number of traces that are connected and the number of traces being connected.

Consider a logic system where the number of drivers that demand current at any one clock time is fixed. Under these conditions the current demand will also be a constant. This arrangement does not require decoupling capacitors as there are no step demands for current. This approach would require dummy transmission lines ...

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