When creating a barebones arduino (or use a regular one), a 100nF capacitor is already connected from +5V and +3V3 to GND.
Many IC's recommend a 100nF capacitor to be connected from their Vcc to GND.
My question is, since such a capacitor is already connected on that line (in the arduino circuit), do we still need to bypass the IC's Vcc and GND lines with another one?
Take as an example this 9-axis sensor or this temperature sensor. If we hook these up to an arduino, isn't the built in capacitor on the breakout boards obsolete? Could or should it be omitted in such a circuit?
123Splat:
and locate them as close as possible to the IC...
This is why you need one for each individual chip. It's not just the capacitance itself, but its physical proximity that is important. CMOS digital circuits do not draw a steady current, they take their current in great gulps when they switch state. The parasitic inductance and resistance in the traces can cause the power supply to brown out for just that chip, even while the power supply is functioning properly.
A small, nearby capacitor provides a low-inductance reservoir of charge that the chip can pull from during the surge, which is refilled relatively more slowly by the power supply.
Its only the parasitic inductance that matters, the supply traces resistance is a few milliohms, not an issue
(unless drawing very large currents). Low parasitic inductance means wide short traces between chips supply
pins and the decoupling cap, and its usually to use ground-plane (or both ground and supply planes in a 4 layer board).
On a timescale of a nanosecond the trace inductances have reactances of the order of tens of ohms per cm - this completely dominates the impedance of the supply where fast changing signals are involved, and is why the
cap needs to be as close as possible to the chip - the cap holds its voltage at these high speeds.