EEVblog Electronics Community Forum
Electronics => Projects, Designs, and Technical Stuff => Topic started by: tree on June 11, 2014, 05:31:15 pm
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What would be the ideal placement/routing/decoupling strategy for a board that has only ONE voltage regulator meaning that both the analog circuitry and digital circuitry are supplied power from the same supply?
I've seen articles that suggest connecting power to the analog circuits first, then connecting it to the digital circuits with some sort of series decoupling, whether a low value resistor or a ferrite bead. I've also seen articles that suggest that the digital circuits are supplied power first and then the analog circuits with a ferrite bead. This is of course assuming a series power connection and not a star power connection.
Would the scenario change depending on whether the supply is linear or switching?
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Highly depends on your particular circuits (digital and analog). General rule - keep return ground path's separate, minimize interference between digital and analog section. Linear supplies have lower noise (usually), but have much less flexibility in terms of conversion options.
So not knowing what you actually trying to do, it's hard to make any guesses, not even talking "ideal" strategy :)
Post your schematics and placement, than much more can tell. :)
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Take a look at this datasheet on page 89. http://www.analog.com/static/imported-files/data_sheets/ADUC845_847_848.pdf (http://www.analog.com/static/imported-files/data_sheets/ADUC845_847_848.pdf)
Most Analog/Digital IC actually designed with separation in mind and datasheets often have recommended layout. When I designed board with ADuC845 I used one LDO for digital and one for analog supply.
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Highly depends on your particular circuits (digital and analog). General rule - keep return ground path's separate, minimize interference between digital and analog section. Linear supplies have lower noise (usually), but have much less flexibility in terms of conversion options.
So not knowing what you actually trying to do, it's hard to make any guesses, not even talking "ideal" strategy :)
Post your schematics and placement, than much more can tell. :)
Sorry, it's not an application, just a general question.
Take a look at this datasheet on page 89. http://www.analog.com/static/imported-files/data_sheets/ADUC845_847_848.pdf (http://www.analog.com/static/imported-files/data_sheets/ADUC845_847_848.pdf)
Most Analog/Digital IC actually designed with separation in mind and datasheets often have recommended layout. When I designed board with ADuC845 I used one LDO for digital and one for analog supply.
I know that the best strategy is to use separate regulators, but that's not what I was asking. I specifically asked for the best method if you only had ONE regulator.
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Well in that case like you already mentioned it yourself - ferrite followed by filter cap (>10uF).
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Well in that case like you already mentioned it yourself - ferrite followed by filter cap (>10uF).
Yes, but how? Suppose there's one power trace and you connect both analog and digital components to the power trace, do you place the ferrite bead on the trace to the digital line, to the analog or both? Instead of a cascaded connection would it better to have separate traces to the circuits straight from the power supply, i.e. a star power connection?
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Well in that case like you already mentioned it yourself - ferrite followed by filter cap (>10uF).
Yes, but how? Suppose there's one power trace and you connect both analog and digital components to the power trace, do you place the ferrite bead on the trace to the digital line, to the analog or both? Instead of a cascaded connection would it better to have separate traces to the circuits straight from the power supply, i.e. a star power connection?
Yes, separate traces would be better as inductance and resistance in the copper traces will convert the pulsating digital supply current into voltage ripple. The ferrite bead is inserted just before the decoupling capacitors for the analog circuitry.
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Doesn't matter very much.
Consider the power supply network impedances. Both will have bypass capacitors in their respective domains (analog and digital), so we can assume whatever noise is present on one acts as a low impedance source.
The only thing different is which side the regulator is connected to. So let's consider the regulator's impedance. A 7805 or similar has fractional ohms impedance at signal frequencies, rising slightly at higher frequencies (roughly corresponding to the internal emitter resistance of the pass device). See Fig. 32, p.36:
http://www.st.com/web/en/resource/technical/document/datasheet/CD00000444.pdf (http://www.st.com/web/en/resource/technical/document/datasheet/CD00000444.pdf)
At such frequencies, bypass caps take over (~20m to 20 ohms depending on type).
So the question is, which side should get slightly more bypass -- because, the regulator itself acts as a dynamic bypass, as can be seen from the impedance curve. Not a great one: a pretty good voltage source at low frequencies, but inductive at higher frequencies.
Which is why you'll get a "blippy" sort of step response when paralleled with a small capacitor, but it needn't be unstable (i.e., an oscillator). Sufficiently large, and optimally lossy, caps can be chosen to effectively bring that resonance down to a frequency where it just goes "thud" instead, and then you have a well bypassed supply. (Roughly, 50mohm and 3kHz would suggest 1000uF, quite a lot for a linear regulator, but hey, if you're desperate, it should be optimal. Just beware of its peak current capacity, eh?)
In short, not very much difference. One way, the regulator shunts digital noise directly, then the ferrite bead and bypassing filters it.
The other way, the bypassing shunts digital noise directly, then the ferrite bead, and regulator and bypassing, filters it.
BTW, ferrite beads are probably useless here: they saturate at extremely low currents (compare the recent video on ceramic capacitors, but just switch everywhere "voltage" with "current"). Look for real inductors, and design the supply network so it remains well damped. (FWIW, a ferrite bead is roughly 0.1 to 2uH depending on value and size.) Usually, a bunch of small, low ESR caps, with one or a few bulk caps with optimal ESR to dampen any resonances (between the small caps and trace inductances, and between filter inductances and whatnot).
Tim
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For low speed analog circuits i try and break away the ground and power rails right at the regulation point (the feedback point in your local regulation) as that means feedback action will reduce any of low frequency noise most effectively for that point,
for digital segments i try and box the noise in by using LC filters on its positive supply rails, as for its ground supply, respect both path of least resistance and path of least impedance, any fast edges current will want to flow on the ground plane directly underneath the signal trace, but some of it will still follow the path of least resistance, which if it overlaps something sensitive may require some careful thinking
the other golden bit of layout i near religiously try and do is group by function, it makes things prettier, but in most cases it lets you shrink the size of the current loop, if you ever look into switch mode converter layout, one of the goals is to keep your current loops small to reduce the inductance and reduce the amplitude of any radiated noise, similar things should be observed if any of your digital signals have an appreciable current behind it,
to take this a little further thinking of every trace ground plane included as a resistance, and every current as having both a signal and a return path gets you in the right mind set for how to best possibly layout your circuits for minimal complications,