Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs
Community Proposals
Community Proposals
tag:snake search within a tag
answers:0 unanswered questions
user:xxxx search by author id
score:0.5 posts with 0.5+ score
"snake oil" exact phrase
votes:4 posts with 4+ votes
created:<1w created < 1 week ago
post_type:xxxx type of post
Search help
Notifications
Mark all as read See all your notifications »
Q&A

Post History

75%
+4 −0
Q&A Power Line Filter. Location of Cy and common mode choke.

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 ...

posted 6mo ago by Olin Lathrop‭

Answer
#1: Initial revision by user avatar Olin Lathrop‭ · 2024-05-04T12:45:04Z (6 months ago)
For easy reference and to protect against possible edits to the question, here is the circuit you are asking about:

<img src="https://electrical.codidact.com/uploads/aizp4lvhg7z5uvnb314e9n6ck51i">

In order to attenuate a signal, you need two impedances.  Imagine you had a 0 &Omega; 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 k&Omega; impedance.  If you load that with 3 k&Omega;, you'll get 2.5 V.  With a 2 k&Omega; 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 k&Omega; 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 &Omega; 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 k&Omega; 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 <i>common mode choke</i> also sometimes referred to as a <i>balun</i> (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 <i>common mode</i> signals, and low for <i>differential mode</i> signals.  The power being delivered is all differential mode, but much of the high frequency noise you want to attenuate is common mode.