common mode choke over damped? - Page 1
Forum section: electronics / beginners

author: 3roomlab
author_rank: Frequent Contributor
timestamp: November 22, 2015, 02:19:03 pm
content: i was reading this articlehttp://www.exergia.info/Lightning/lightning7.htmvery nice calculation guide on common mode choke.but what will happen if the AC is overdamped? more power loss in the choke? or phase delay in filtered AC?say if the wriggly symbol (damping) is over 8?

author: f5r5e5d
author_rank: Frequent Contributor
timestamp: November 22, 2015, 06:51:36 pm
content: that page doesn't appear to model choke losses - which will be frequency dependent - AC mains frequency may only see the wire winding Rand the common mode choke connection cancels most effects for differential current - which how you are supposed to be drawing power from the mainsthe equation of Rload with filter components to a Technical ground is not a good approximation for most electronics

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 22, 2015, 10:10:53 pm
content: That's "zeta", by the way.A peak in the response corresponds to a possible peak in your EMI response. Both susceptibility (noise introduced from the mains) and emissions (noise produced by your circuit).It's noteworthy that, although EMI filters are almost always measured at 50 ohm conditions, this is at best only partially true of the mains side (which is officially tested as 50 ohms per conductor, but obviously could be anything in practice). The load side is typically a capacitive load to ground, due to stray capacitance in the switching transformer, and Y type capacitors. This is rather different from a matched source and load filter, so you aren't going to pull the values for it out of a table of filter designs.As mentioned, CMCs are typically lossy, in the 10kHz to 1MHz range (depending on type, with ferrite cores being lossy over a narrower range and at higher frequencies, and tapewound (Metglas, Vitroperm, etc.) cores being lossy to lower frequencies and over wider ranges). They are also capacitive (or complex -- multiple winding resonances) at higher frequencies. The capacitance limits the asymptotic attenuation (i.e., the circuit looks like a capacitive divider with constant attenuation, rather than e.g. -40dB/dec forever), and the resistance can be used to help dampen the filter network, if component values can be dimensioned correctly. If not, you may want to add R+C networks in strategic places, or ferrite beads, to provide the damping necessary.Tim

author: 3roomlab
author_rank: Frequent Contributor
timestamp: November 23, 2015, 01:15:21 am
content: for the first example (1st order, only using CMC)say we have a 100mH CMC and a load of only 10wusing the equation on the website ... inductance=load/(2*pi*F)the corner frequency of setup = 15.9Hz (hmmm is this right?)at such low load, does it mean the inductor will also filter off AC mains voltage? (ie: voltage will sag? and the load doesnt really see full AC?)

author: 3roomlab
author_rank: Frequent Contributor
timestamp: November 23, 2015, 06:07:11 am
content: i tried to look up more calculations. ZL= 2*pi*FLif just by 50Hz AC, the AC is looking at a 31ohm impedance? or by common mode operation, this impedance doesnt exist? ( i am guessing it doesnt)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 23, 2015, 09:40:57 am
content: The equivalent circuit looks like this:The filter choke is ultimately working into the 50 ohm source impedance, but that's not where the interesting stuff happens. The interesting part is the impedance it's working against: a capacitively coupled noise source, and capacitance across the isolation barrier, with very little loss otherwise.R1 and R3 represent the LISN, which in turn is used to represent the AC line during testing. (In general, a real AC line will not be a pair of resistors, but a complex, lossy and random transmission line. But that doesn't really help, so we blindly go on assuming it's okay to use resistors. Applying some RF theorems, it's not actually as bad an assumption as it sounds.)C3 represents the winding-to-winding and isolation capacitance in a typical SMPS, and V1 represents the switching voltage which appears on those windings. The voltage seen by the isolation barrier is roughly the average of the voltage across the winding. So for a flyback supply that's driving 400Vpp across the primary (and has a low voltage secondary), you might expect V1 is 200Vpp.C1, C2, C4, C5 and L1 comprise the familiar EMI filter. C5-C7 may be used in various permutations, depending. (C4 may or may not be used. Despite how it looks, it actually has little impact on what L1 is doing.)In any case, the equivalent sum of C3 and C5-C7, is the capacitance which resonates with L1. You can simply draw the equivalent circuit: GND--25 ohms (they act in parallel)--L1--Ceq--GND.Ceq is C5 + ((C7 + C3) || C6). That is, C3 adds to C7, so we have 2.3nF, in series with 2.2nF is 1.12nF. And this adds with 2.2nF, giving 3.32nF total equivalent.Now that we know the capacitance and inductance, we know the impedance and frequency!If that impedance is very different from the mains resistance, then we will have a problem: the filtering will either be very sloppy (Z < R), or very peaky (Z > R).So we have, Zo = sqrt(5mH / 3.32nF) = 1227 ohms. Which is much larger than 25 ohms, so there will be a substantial peak at some frequency, where V1 is able to transmit energy to the line resistors (or vice versa). If everything is lossless, the Q factor is 49, so we can expect the peak to be quite strong indeed!And, the frequency of that peak will be very close to F = 1 / (2*pi*sqrt(L*C)) ~= 39kHz. The simulation gives 39.21kHz:Now, -40dB doesn't sound very thrilling, but the fact is, you're starting with hundreds of volts. And the regulations say it has to be millivolts or less. So this is only getting you down to single volts, and you have a whopping 60dB yet to go!Since the Q factor is so high, we can add some resistance to that coil, worsening the attenuation at very high frequencies, but dropping the peak substantially. This looks like so:with 5kohms strapped across one of the windings (doesn't much matter which one). Now the peak is -59dB, an improvement of 19dB.I don't have a realistic CMC model handy for a part of this size, but the effective parallel resistance can be found in the datasheet: on a graph of Zcm vs. F, it is the peak value of the impedance. At low frequencies, Z is rising (inductive characteristic), and at high frequencies, Z is falling (capacitive characteristic). Where those ranges meet, L and C cancel, and Z is real (resistive), and equal to Rpar. Most times, it will be much too high (e.g., >10k for a part of this size), so you might consider damping the capacitors instead (e.g., replace C5 with 1nF, and put (2.2nF + 1kohm) in parallel with it).Lastly, an SMPS really isn't perfect white noise. If you can place the transmission peak at a frequency below the switching frequency (or subharmonics, just in case), you don't need to worry about it, because there's so little power being transmitted at that frequency. That leaves plenty of time for the lowpass filtering to kick in, before the fundamental, which might be at 100kHz or so, and harmonics beyond there.Tim

author: 3roomlab
author_rank: Frequent Contributor
timestamp: November 23, 2015, 11:01:58 am
content: thanks for taking time to give the above example/calculationi think i learnt something new, but my level of understanding is shallow (n slow)if i understand correctly, shouldnt a larger C5 promote smaller Zo (so instead of 2.2n - 1n, it becomes 10n)? and dull the Q? or that wont do it?hmmm my bad, it is smaller for higher zeta, = dull Q

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 23, 2015, 08:14:29 pm
content: Yes, in Z = sqrt(L/C), higher C gives lower Z, and the closer Z matches to R, the lower the Q. But you can't go much more than 10n, because of other restrictions (ground leakage current), and shouldn't need to use much more than 4.7nF total for most situations.You can also reduce L, but this raises the cutoff frequency, and it will be uselessly too high by the time it's a closer match, even with C at the limit.Any kind of damping works, so we can reduce Q by shunting R across the L or C. If we still need a largely reactive component, we can use an R+C or R+L where the time constant of the combo is near the time constant of the filter we're damping (plus or minus a bit, so it's a fairly good loss at the cutoff frequency, without causing too much trouble otherwise).Tim