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How to connect the common mode choke in a circuit?
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sbdada:
Hello, I'm confused to use a common mode choke in an smps primary input line. The first image shows the winding structure, and I did exactly same. The second image shows the schematic of the circuit.
So, which lead will be connected with which point of the circuit? Assume I've connected Lead A to point 1, and C to point 2. Then where B and D will be connected? B -> 3 and D -> 4 ? Or the other way?
Thanks in advance.
helius:
The two dots in the schematic are called "phase dots" and indicate the phase relationship or polarity between the two windings. If the dots are at the same end, the windings are in parallel. The windings in the first diagram are also in parallel: If you wrap the fingers of your right hand around the core, the winding from B to D goes like your fingers when your thumb is pointed clockwise thru the core. If you then turn your hand to the top, the winding from A to C goes like your fingers as your thumb points in the opposite direction (counter-clockwise). This means that B-D is parallel to A-C.
The "inputs" or left connections to the choke schematic are 1: A and 3: B, and the "outputs" are 2: C and 4: D.
See https://en.wikipedia.org/wiki/Polarity_(mutual_inductance)
sbdada:
--- Quote from: helius on January 21, 2019, 07:33:35 pm ---If you then turn your hand to the top, the winding from A to C goes like your fingers as your thumb points in the opposite direction (counter-clockwise).
--- End quote ---
This line is confusing. Isn't the thumb pointing same direction (clockwise) as before when the fingers wraped from A to C?
T3sl4co1l:
A and B both enter the same face of the core, therefore they are both starts. Or both ends, but just pick one direction as convention for argument's sake.
Another way to think of it: slide the turns around, until they overlap. You can reverse the position of B and D by loosening the winding, and pushing the turns over and under each other, without ever passing an end of the winding through the core. This does not invert the phase of B and D, B is still the start and D is still the end. Now, while the turns are still loose, rotate them around the core until on top of A-C. You will find that B and A line up perfectly, and the two windings follow identical paths (once you snug up the wires), all the way to C/D. You can merge them together mentally as a single cable (this is called bifilar construction, when done intentionally), or connect them together and see that the inductance is the same as the single winding alone.
The key factor is, how many loops are made around/through the core, and in which direction, front to back or back to front. (A starts at the front, and C ends at the back.)
Incidentally, a "necklace" (a single toroid core on an otherwise seemingly straight wire) is one turn. How? Well, sooner or later, if we are to measure a current through that wire, it has to complete a loop. It doesn't matter that that loop is the size of the universe, or wrapped tightly around the core -- it still makes one complete turn.
Ultimately this connects with topology and winding number (knot theory). :)
Tim
T3sl4co1l:
--- Quote from: sbdada on January 21, 2019, 08:15:29 pm ---
--- Quote from: helius on January 21, 2019, 07:33:35 pm ---If you then turn your hand to the top, the winding from A to C goes like your fingers as your thumb points in the opposite direction (counter-clockwise).
--- End quote ---
This line is confusing. Isn't the thumb pointing same direction (clockwise) as before when the fingers wraped from A to C?
--- End quote ---
Correct, the right-hand test shows them in the same phase. (The description I gave above is just the more abstract way, in the event you don't have a hand...handy.)
Tim
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