Engineering Lexicon
"A" "B" "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M",
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Voltage Droop

Voltage Droop, Output: defines the loss in output voltage from a device as it tries to drive a [capacitive] load. The reduction out voltage level may or may not be an issue depending on the circuit in question. For example if the receiving IC is a digital logic IC the level reduction might not even be noticed if the final value never fall below a transition point.

Not supplying sufficient current to an IC that needs to drive a heavy load is one example that could cause voltage droop. Increasing the value of the by-pass capacitor used by the IC is a possible fix. The value of the Decoupling Capacitor that should be used with an IC depends on the load the IC has to drive. To some extent, the larger the decoupling capacitor used with the IC, the larger the load the IC will be able to drive; because the decoupling capacitor supplies current to the IC as it switches.

If device A has to drive two separate inputs at 3.3 volts then the current demand depends on the load of both inputs, taking into consideration the rise time of the signal. The input capacitance for a given device is parameter Ci in the device data sheet. If the receiver [B] has a load of 12pF [Ci] and the output driver [A] has a rise time of 1nS then the current required is: I = dV / dt or I = 12pF*(3.3v)/1nS. The current required = 39.6mA. If we keep the voltage droop to 3.0 volts, or a reduction of 300mV. The required by-pass capacitor then equals: C = I * dt / dV. C = 39.6mA * 1nS/300mV = 22pF.

Pulse Waveform
Voltage Droop on Pulse

The example above uses the input capacitance of an IC to explain voltage droop or loss in voltage level due to an excessive load. However the capacitance could just as well be the capacitance of a long cable, or even a cable in addition to the destination IC. Either way; if the IC is able to drive the load, than a quick fix is to add a larger by-pass capacitor to supply the needed current.

PC motherboard

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