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The isolation transformer I'm looking at: http://www.mouser.com/ds/2/336/HX1188NL-515471.pdf

My question is why is it there? Is it not enough to have just the transformer on the left?

why does the part in blue exist?

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That "transformer" is a common mode choke.

It's used to suppress EMI (either being induced onto the line and affecting the circuit or being transmitted from the circuit out over the line).

It's called "common mode" because it's very effective in suppressing HF currents that are common to both lines.

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  • \$\begingroup\$ The key here is not just that they are common but extremely well matched for much higher CMRR's than can be achieved with differential amplifiers 1MHz to as high as 300MHz in some cases. but never more than 3 decades for a given application. One such application is a phone, Another is Ethernet. Why so many points for such a simple answer above? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 28 '17 at 7:04
  • \$\begingroup\$ @Tony_Stewart Perhaps because somebody who has never seen this circuit diagram before will go "Oh. That's beautiful!" when it's explained. I know I did. \$\endgroup\$ – nigel222 Jun 28 '17 at 8:25
  • \$\begingroup\$ it's also a 2 way filter for ingress and egress. I guess they don't teach CM chokes in university . pity they have special properties like a 3dB hybrid which look almost similar aka DC-3 or directional couplers a.k.a. TV cable splitters \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 28 '17 at 16:06
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That is what's known as a common-mode choke. It presents a high impedance to any common-mode currents which might flow in the + and - pairs. Any common-mode currents which flow in the attached ethernet cable will have a strong tendency to radiate at levels in excess of legal EMC limits.

An ideal common-mode choke will present no impedance to differential currents as their developed fluxes will tend to cancel due to being equal in magnitude and opposite in direction.

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