Power Line Filter. Location of Cy and common mode choke.
A circuit under question:
Why are CY capacitors on the source side, and not on the load side?
Henry Ott describes this with:
Because filter attenuation is a function of impedance mismatch, the role of the power line filter is to maximize the mismatch between the source and the load impedance
Can someone explain the reasoning behind this?
Why for maximum attenuation the high impedance filter element shall face low impedance LISN, and the high impedance noise source shall face low impedance Cy?
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Why are CY capacitors on the source side, and not on the load side?
You may not be grasping what the image shows. The left-side port connects to the AC power-line. It supplies power to the right-side port via the filter. The right-side port (called power supply) is in fact a switch mode supply (SMPS) like a flyback converter.
What may be confusing you is that the right-side port is labelled "source" and that the left-side port is labelled "load". Those terms are used because the image is showing how interference (generated by the SMPS) can be inflicted back on to the AC power line.
Hence, the source of interference is the SMPS (the aggressor) and, the left-side port is regarded as a potential "victim" of that interference.
The Y capacitors attempt to equalize both lines on the right-side port to the same level of interference with respect to earth/ground. This then means that the CMC is dealing with mainly common-mode interference from the SMPS.
In other words, if both lines on the right are made to have the same interfering voltage (with respect to earth) then the CMC does a much better job of reducing the interference coupled back to the port on the left.
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For easy reference and to protect against possible edits to the question, here is the circuit you are asking about:
In order to attenuate a signal, you need two impedances. Imagine you had a 0 Ω source at 5 V and wanted to make a 2 V signal from it. How would you reduce the voltage?
Putting a single resistor across the source isn't going to change the voltage, just increase the current the source must supply. That's because the source has 0 impedance.
Now imagine a 5 V source with 3 kΩ impedance. If you load that with 3 kΩ, you'll get 2.5 V. With a 2 kΩ load, you get the 2.0 V you wanted.
The reason the second case worked and the first didn't is because in the second case the source had a known finite impedance. It takes two impedances to make an attenuated voltage. Think of a simple resistor divider. Even though you only used a single resistor in the second case, the top resistance of the divider was effectively still there because the source had 3 kΩ impedance. The first case didn't work because there was no top resistance for the divider.
So what if you do have a really stiff regulated supply that has nearly 0 Ω impedance for practical purposes? You can add the top resistance of the divider yourself. You can take the ideal 5 V supply of the first case and put a 3 kΩ resistor in series with it to get the second case. Both are equivalent Thevenin sources. (If you don't know about Thevenin and Norton sources, look those up. They are not complicated, but worth understanding.)
The power line has a very low and unpredictable impedance. Therefore you can't rely on what you get by just putting a capacitor across the power line. To be guaranteed to attenuate high frequencies, you have to add some series impedance somehow. The trick is you don't want that series impedance to get in the way of the power line delivering power.
You then notice that the signals you need to attenuate are much higher frequency than the 50 or 60 Hz of the power the power line is delivering. You can therefore use frequency-dependent components to attenuate high frequencies but leave the power frequency largely alone. This is done by putting inductance in series and capacitance as a shunt. This is like a resistor divider except using complex impedances instead of simple resistances. Dividers actually worked with impedances in general all along, just that limiting them to pure resistances makes things simpler and frequency-independent, which is what you want to just attenuate a voltage by a fixed scale factor.
Since the impedance magnitude of an inductor goes up with frequency, and that of a capacitor goes down with frequency, a divider with an inductance at top and capacitor at bottom will attenuate ever more as the frequency goes up. This is exactly what you want in the case of trying to remove high frequencies from the power line.
In your circuit, L1 provides the inductance that is the top impedance of the divider, and the capacitors the bottom impedances of the divider. This circuit is more complicated than a simple divider because it is intended to work in both directions.
Imagine the power line with high frequency crap on it connected to the left end. C3 doesn't do much, at least nothing we can rely on, because we have to assume the power line has low impedance. L1 adds inductance in series with the power line. L1 is adjusted so that it has little impedance at the power line frequency so it doesn't get in the way of delivering power. However, it has substantial impedance at high frequencies. C1 and C2 have low impedance at those high frequencies, so heavily attenuate them. However, they have high impedance at the power frequency, so leave it alone.
The works flipped around for high frequencies coming from the right end. Again, you can't rely on C1 and C2 doing much since the impedance of those high frequencies is unknown. L1 then adds a known impedance, with C3 then being able to attenuate those high frequencies.
Another wrinkle is that L1 isn't just a simple inductor in each power lead. It is a common mode choke also sometimes referred to as a balun (balanced, unbalanced). This makes use of the fact that to deliver the power, the current must flow in one direction in one lead, and the opposite direction in the other lead. To the extent the magnetic coupling between the two windings works, the impedances of the windings subtract from each other to signals that flow in opposite directions, and add to signals flowing in the same direction. Put another way, the inductance is high for common mode signals, and low for differential mode signals. The power being delivered is all differential mode, but much of the high frequency noise you want to attenuate is common mode.
One additional wrinkle not (at present) mentioned is that Class X and Class Y capacitors are used in those specific locations for safety reasons, having to do with the capacitors being connected directly to the line supply.
Class X capacitors are rated for the use shown - across the line. If they fail short, it's expected that a fuse will blow or a breaker will trip.
Class Y capacitors are rated for the use shown - line to ground, where failing short has the nasty unfortunate side effect of applying line voltage to a ground. This is particularly problematic where the ground in question is an audio ground not connected to powerline ground, as in vintage guitar amps (which may well predate Class Y capacitors, as well; I have not been successful in determining when those classifications were defined and implemented), or gear where someone has cut off the power ground prong to accomodate outdated receptacles. So class Y capacitors are designed to fail open when they fail.
Some "Safety Capacitors" are dual rated for both X & Y uses.
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