MDI interface common mode termination resistor

[vignesh_M]

Hello, 

I have question about the TJA1100 data sheet. 

I would like to have more information about the common mode termination function.

In MDI interface the common mode termination network is used, here the 1K ohm resistor, 2 no's is used on 100 ohm impedance line. Please tell me how this resistor pair will give 100 ohm termination on connector side?

here, i attached the datasheet link for your reference:

page no : 45 

fig no : 17

https://www.nxp.com/docs/en/data-sheet/TJA1100.pdf 

[JozefKozon]

Hello Vignesh,

please see below an answer from an application engineer I have contacted.

DESCRIPTION

The termination network is intended to prevent having a floating cable. This is something the OEMs require to prevent the cable from being charged in a certain direction (e.g. by RF disturbance). There is a very high ohmic (100K) DC part and a relative low ohmic (1K) AC path to GND for this purpose.

Now on the question how this fits to the MDI spec. The 2x 1K are parallel to the 100 Ohms in the PHY. This results in 1/((1/100)+(1/2K))= 95,23 Ohms. This is well in the spec of 100+/-10% (= 90 to 100 Ohms). This small mismatch is the price to be paid for getting a terminated channel.

With Best Regards,

Jozef