Filtering is a good idea, yup.
What is a filter, here? An RLC network with well behaved frequency response and impedance. Note the impedance is both inseparable and necessary. Impedance is a ratio of voltage to current. It might represent the peak change in voltage caused by a step change in load current. It's also equivalent to sqrt(L/C), for a suitable combination of L and C.
Exactly which L and C to combine, depends on the network, and what you're asking of it; effectively, the inductor in a CLC network is shared between both C's, or the C in an LCL -- so a factor of 2 or 1/2 commonly shows up. That's simple enough. If you want to know the asymptotic LF or HF equivalent instead, you might be after the total L and C in the network; or perhaps just the last L and/or C.
I bring this up, because one should not throw around components willy nilly and expect it to work (of course!). A filter is a part of its environment, so it matters how you measure it, and it matters what it's made of. Do you test it with a bench supply on long wire leads? With a CDN/LISN? With a big fat cap hanging off it? All the above are used at some point or another; it depends. Of course, a good power supply filter won't care about any of these; which actually tells us a bit about how it needs to be designed, too.
Ferrite beads. They aren't generally a great idea in filters, because they saturate easily.
If we define saturation as a 30% loss of inductance or impedance, typical ferrite beads saturate by anywhere from 5 to 20% of rated DC current. (It could be even worse than this; I just don't think I've noticed many that are.)
The DC rating is only a thermal rating, nothing else.Incidentally, ceramic capacitors (type 2 dielectric, X7R and the like) are the exact same way, and this is no coincidence; one is ferromagnetic, the other ferroelectric. The physics is actually amazingly similar. Always check that you have the capacitance
at DC bias that you intended.
Ferrite beads are different from inductors in three important ways:
1. Ferrite beads are typically sold by impedance at frequency, not inductance.
2. Inductors almost always are sold by saturation current, ferrite beads almost never.
3. Ferrite beads are most often used to dampen EMI resonances, i.e. have high losses over a wide frequency range. Inductors are intended to handle useful reactive power (i.e., lower losses).
In a filter, losses cause a soft transition band (i.e., for frequencies near cutoff). Say you pick values for a nice crisp Butterworth prototype. Put in a lossy inductor and you end up with some soggy thing that looks more like a Bessel, or even worse. The stopband (cutoff asymptote) might not even be the e.g. -60dB/dec (for a CLC) you were expecting. What's more, transmitted plus reflected power don't add up, i.e. there's appreciable insertion loss there. Obviously, it's going into...losses, this is rather redundant right? Well the nice thing about losses is, if we don't have many to begin with -- or we have an element that's actively working against us, trying to cause feedback -- we can dampen it out, and still get a well-enough behaved filter response (one without peaking; who cares how sharp it is anymore).
And what would cause feedback? Consider the input resistance of a switching regulator: for constant load, as supply voltage rises, supply current drops. It has negative incremental resistance. Pair that with too little damping resistance, and a low enough resonant frequency, and you've got an oscillator instead of a filter!
So losses aren't necessarily a bad thing for PS filtering. But the saturation is a problem. That means we can't control the cutoff frequency, or the filter impedance: both depend on L, but L depends on current draw.
And all this again applies to capacitors, where we have to be careful applying ceramic capacitors (they're usually well enough rated to still be able to design with -- ferrite beads only rarely have DC bias curves; which, the Laird catalog is the most complete of any brand I've seen), and can make use of the notable ESR of tantalum and electrolytic capacitors for example.
So let's do a design example. I've still not covered nearly enough to actually do this, I'll just walk through the process and hopefully not lose you too much along the way...
Consider the case with a bench supply and wire leads: say the wires are a meter long, loose (not really twisted together or in a cable or anything). Wires have inductance, whether they're wound into springs and other shapes or not. A good ballpark for this situation is to take about the inductivity of free space (\$\mu_0 \approx 1.257 \mu H/m\$) multiplied by the length. So, times one meter length is about 1.2µH, neat.
Why start here? Well, the bench supply probably has a fat ass capacitor across its output; it has a very small AC impedance. We can model that as a short circuit. AC power isn't going anywhere, it's bouncing straight back off of that. So we should consider the filter network, including this short, the connecting wires (which aren't just dead conductor, they have reactance), the filter, and the filter's load. Then we'll have a good idea of how the circuit will respond.
So, say our filter is CLC, 2 x 10uF (and we'll say they're ceramic and properly sized, so it's 10uF at voltage, and very low ESR) and 1uH (a proper inductor, not a ferrite bead). We have an additional 1.2uH in front, and a DC-DC converter after.
Which means the first 10uF is kinda split between the two ~1uH inductors, so it kind of counts for half; but also the middle inductor is split between the two 10uF's, so kind of counts for half as well. So we can get the impedance of this network as more or less Zo = sqrt(1uH / 10uF) and the halfs cancel out, simple enough. (You can see where this might not be true: if the filter were a single C, it might be split between the input wiring, and whatever impedance the converter has.) So the impedance is about 0.32Ω, and the cutoff frequency is 1 / (2 pi sqrt(L C)) or 50kHz.
Say the converter is buck type, for sake of illustration. Uh, might as well put some units to it, too; say 12V 1A input, 5V 2A output (give or take efficiency). And it's running at 500kHz.
Then, the converter's switching frequency is higher than the filter's 50kHz cutoff, which will probably then be within its control bandwidth. That is to say: it will be able to respond to changes in the source voltage or load current, at frequencies around 50kHz. And so the converter will have some negative resistance characteristic. Probably not a whole lot -- its current draw can't respond instantly; it probably has an inductive characteristic as well. So, negative R in series with some L. R is the incremental resistance: say we increase the supply by 1V, we expect current to drop from 1A to 0.92A, or -0.08A; R = 1/(-0.08) = -12.5Ω. And probably L is... around the value of the output filter choke divided by the average duty cycle?
So let's look at the Q factor of the filter. If the capacitors have 10 maybe 20mΩ ESR, then that's 1/30th to 1/15th of Zo. ESR < Zo, so Q > 1, indeed by about this factor. Loss can't be any lower than this (we haven't included all losses yet, we can only add more from here), so this gives a maximum Q factor.
Q factor is many things, for example the ratio of resonant impedance to characteristic (Zo) impedance. If we get a peak or notch in the impedance, this is (approximately) the ratio from the peak/valley to Zo. (So if we have Q < 1, the impedance is very flat, and any voltage/current peaks in the response will be comparable to the plain old impedance: which means both that the step response will be flatter, and also that it won't ring to much high peaks.)
Except I kind of lied, because the converter's negative resistance does cancel out real (loss) resistance; so we do need to do our due diligence there. If we compare filter Zo to incremental resistance, we get (0.32) / (12.5) = 1/39, which is a good sign -- effectively it would give a Q of -39 with a lossless filter. But our filter is already much lossier than this (i.e., Q = 15 is more lossy than a Q of -39 is "gainy"), so we should be okay -- stable.
Indeed even if we halve the supply voltage (which should about double the current draw, and quarter the negative resistance), we're still at least marginal; and the converter can't operate much lower (a buck can't make 5V from less than 5V input!) so this about the worst case we'll ever need to worry about.
It does seem that a bit of extra loss would be a good idea. If we put a few ohms in parallel with the inductor, we lower its Q to single digits (say with 2.2Ω, we'd expect Q < (2.2) / (0.32) ~= 7). Notice what we're doing to the filter, though: at lower frequencies, it's still L straight through (this loss mechanism does not affect DC performance!), but at high frequencies (when X_L >> 2.2Ω) it's just a plain old CRC filter, which has -40dB/dec response -- not the -60dB/dec we might've been hoping for, from the CLC filter we thought we built.
And hey, it's still not bad, that's a fair amount of filtering just going from the 50kHz cutoff, up to the 500kHz switching fundamental -- a factor of 100 give or take. We don't usually put full differential antennas on DC inputs, it's not likely to radiate much of anything anywhere. (The loose wires going to the bench supply notwithstanding, but they're also much shorter than the wavelengths in question, so aren't going to radiate very far.)
Alternately, we can make the capacitors lossy. As for parallel resistance with the inductor, so it is for series resistance with the capacitor. If we stick in ESR, we can dampen the filter in much the same way. We should probably put it on the outside facing capacitor, since we want low losses to the converter's bypass cap (it's switching hard, we don't want its ~2A current pulses to dissipate excess power). Here, anything up to 0.32Ω would be fine (just as anything in parallel with the inductor, down to 0.32, would be fine), of course "fine" being in terms of just damping, while we're making the filter's high-frequency response worse by a proportional amount.
Can we do better? Sort of. There's a minimization law underlying these systems, which says you can't have arbitrarily sharp cutoffs
and a stable response as source/load impedance vary. (And we're at one extreme condition here, with the shorted-inductor input.) If we don't mind that our filter isn't sharp (and we don't), we can make a crappy capacitor by putting ESR in parallel with it instead.
Say we change the input capacitor to 10uF || (22uF + 0.3Ω). Just consider this network by itself: towards DC, Xc >> ESR, and the capacitors act in parallel: 32uF. At high frequency, the 10uF dominates (Xc << ESR). Inbetween -- coincidentally, near the 50kHz transition band of our filter -- the ESR dominates. The impedance curve of an ideal capacitor is a straight diagonal line, but we've put a kink in it, where it's falling from DC to about 16kHz, then flattens out, then continues on diagonally, above 53kHz. That flat region is resistive rather than capacitive, and it acts to dampen -- or terminate -- our filter.
This will give us a nice low Q factor, maybe 1 or 2. It could be improved, but it's not a bad start, and it's definitely not unstable with respect to the converter.
Note where this
won't help us: if the input is shorted by an even lower impedance, the ESR will be swamped, and our filter may misbehave again. This might happen if our converter is a module in a larger project, tied to a low-impedance rail. Effectively, if there were no lead length to the bench supply (the assumed AC short). We've just defined a source condition: we're happy while the input impedance is greater than some minimum. Neat.
Suppose we go through the other source conditions:
- A LISN (Line Impedance Stabilization Network) is a network with relatively large series inductors, then R+Cs shunting to ground. A DC source connects to the inductors on the open side, and the EUT (Equipment Under Test) connects to the shunted side. If the source is again a low impedance, or some stray wiring (some ~uH), those series inductors aren't much different than the above case. The total inductance rises a bit, which raises Zo, but not by much (no more than half, I think, because of the input capacitor being shared by the two inductances). Our damped capacitor still does its job. In fact, we've even got a little more R in parallel with it, which can only help! (The R is actually the input impedance of a spectrum analyzer, typically 50Ω. It's not much compared to the ~0.3Ω filter impedance, though, not enough to account for the greater L.)
- A big chunky capacitor might be used to test the ripple of a power supply. This is... kind of disingenuous, I never liked it as a test.* But however it happened, it's either, a de-facto standard in the industry, or maybe it actually comes from a publication somewhere, I don't know. Anyway, this puts a low impedance near the supply (typically over a short cable, which will have lower inductance than loose wires -- maybe 1/2 to 1/4 of µ
0 times length, so in the 0.2uH ballpark). This may be a concern: both Fo and Zo are reduced by the added capacitance, and the L is mismatched to the rest of the filter, so it's at least not going to be a sharp cutoff CLCLC. We might address this by adding a small inductance in front, which sets our minimum impedance; which can also be lossy, maybe we use R || L here. Honestly it's not a bad place for a stupid ferrite bead, just because we don't really mind what the inductance is at this point (as long as it's doing anything at all), and it saves us the trouble of selecting a proper inductor plus another resistor.
*It's understandable, because you'll most likely be using a power supply with your own on-board filtering/bypassing, and that's roughly what they're showing, some capacitance in parallel with the output of the supply. (See
RS-15-5 for example, output ripple and note 2.) The worst though is the 20MHz bandwidth, while supplies generate considerable noise up beyond 100MHz. It would be nice to know about
that part. And a good measurement of that would use a LISN, and a standard setup with ground planes. Those frequencies don't like to stay in wires, they couple into various things quite easily: the measurement should include normal mode (each wire independently), or common mode (both together), not just differential (one with respect to the other). We expect the voltage difference across a beefy cap to be very small of course, but that doesn't mean there isn't voltage on the cap with respect to ground.
The same logic of course applies to the converter's output side. There, we have an LCLC network, with a square wave applied to the front, so we know exactly how much attenuation (in terms of voltage gain) we need. (Just to clarify, the input network has a square wave current applied to it, and if we're worried about source voltage ripple then we're actually asking about the resistance...through the network, the trans-resistance as it were. The voltage and current are in different places so it's not a literal impedance anymore, hence the "trans".)
On the output side, we are concerned with control loop stability and output impedance. Stability is helped by damping (which may not depend on the switching inductor -- that can be isolated by current mode control). Output impedance is proportional to characteristic impedance Zo, with a peak in the transition band (around Fc), and dropping off to either side (the DC resistance depends on the controller, and the HF impedance is largely the output cap, and everything else connected to it).
Tim
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