Twitter Quiz – the answer


NEWPIC

Filtering capacitors routing is very important in digital world. One and same circuit may work or may not work correctly if routed bad or good, although schematic is same and components are same.

In this quiz the correct answer is Right circuit is better routed as it has twice less wire lengths and thus twice better prone for injected noise and emitted noise through the power supply wires.

Explanation is simple:  All digital circuits have power supply spikes when change their state from on to off, inside one microcontroller there are thousands of such digital switches which produce high frequency noise on the power rail lines. They may be simple represented by resistors from Vcc to GND which value changes very quickly and change the power consumption for very small amount of time (nano seconds). For these high frequencies the PCB tracks are not just conductors but behave also as inductors, although with small inductance when frequency is high they start to play significant role and do not allow the power from the power supply to feed the digital circuit immediately, and if you look with scope you can see lot of noise up and down spikes caused by the changed power consumption and track inductance, to clear these spikes capacitor is placed close to the digital IC which is acting as reservoir with energy and feed energy for these short times until the main power supply energy overcome the inductive resistance and come to the digital circuit. In our case left side circuit have two times longer tracks which go to the capacitor and will emit and receive twice more noise than the circuit in the right hand side.

Right side circuit have one disadvantage though – it allow more cross talk between the two VCC power supplies, i.e. Vcc1 will inject more noise to Vcc2 and vice versa, so of course best solution is to place two capacitors, so it’s up to engineer decission which in his case is better to spare one capacitor and allow more noise cross-talk between Vcc1 and Vcc2 or to place two capacitors and rise the BOM🙂

All above is based on the assumption for high frequency digital design as the filtering capacitor is with small value 100 nF, if this was audio circuit with low frequency power noise the left circuit is better, it’s also known as star topology, I guess this confused most of the responses, for low frequency and if the filtering capacitor was with big value 1000uF the left side circuit would behave better as in this case PCB tracks do not behave as inductors, but the cross talk between the different Vcc will be minimal.

Random.org says our winner is @avrnoob, please contact us for your present – OLinuXino (paper) notebook:

notebook

see you next Friday🙂

4 Comments (+add yours?)

  1. Guest
    Oct 20, 2015 @ 00:49:24

    > the correct answer is Right circuit is better routed

    I (an amateur though) dare to object. The main job of these Cs is to provide an extremely low impedance current source and the left layout is better for that.

    > it has twice less wire lengths and thus twice better prone for injected noise
    > and emitted noise through the power supply wires.

    [I assume all tracks are the same width and power is supplied at the capacitor]

    The wire length from each Vcc-pin to the capacitor is the same in both layouts. But the left one has less resistance (two tracks carrying the power vs one) and less inductance (two parallel inductors giving halve inductance). You already mentioned less crosstalk in the left version – important if Vcc1 and Vcc2 power separate blocks in the chip – in the right layout any noise from pin 1 is directly fed into pin 2 and power is first pulled from the other pin before C kicks in.

    For a good decoupling *between* different circuit blocks, you want a high impedance between their power lines (RC/LC/RLC filters, ferrit bead, etc). But these should be between the power supply and the reservoir caps.

    In short: low impedance between reservoir caps and chip, high impedance between reservoir caps of different chips.

    Reply

  2. Javier Vela
    Oct 20, 2015 @ 09:42:15

    It is very educative

    Reply

  3. Lachlan
    Oct 21, 2015 @ 10:16:29

    On a side note, you can reduce noise on the power rails if the circuit your decoupling is operating at a constant Fq, by putting in a series resonant LC close across the power pins, tune to the Fq. Only works for constant Fq’s thou.

    Reply

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