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When working with Ethernet, whether copper or optical fiber, why are the termination resistors always in a few ohms and never in kilo-ohms?

Example: 49.9R on transmit & receive pair of PHY in copper Ethernet, 130R & 82R divider on transmit & receive pair of PHY in fiber Ethernet.

Reference: Page 62 and 63 of this PHY datasheet: http://www.ti.com/lit/ds/symlink/dp83620.pdf

The only reason I can think of is the driving/sourcing current, which would be less in case of high valued resistors causing issues in high speed communication.

Any thoughts?

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    \$\begingroup\$ google "impedance matching". The characteristic impedance of wires etc. is in the 10's of Ohms range. \$\endgroup\$ Dec 3, 2014 at 9:40
  • \$\begingroup\$ "feeling dumb"; thanks. The second sentence itself answers it. \$\endgroup\$
    – LoveEnigma
    Dec 3, 2014 at 10:20

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For example, using a twisted pair, the formula that determines characteristic impedance, Z0 is: -

\$Z_0 = \dfrac{120}{\sqrt{\epsilon_r}}\cdot ln(\dfrac{2S}{D}) \$

enter image description here

If permittivity (\$\epsilon_r\$) is held constant at 4 then you will find that you have to get a a very large "S" distance before the impedance is in the kohm range.

For D = 0.5mm, an S value of 8 mm is still only 208 ohms. An S value of 1000 mm yields an impedance of 498 ohms. A 10m gap is still only 636 ohms.

Try this for your self here

If you don't terminate cables in their correct characteristic impedance you'll get reflections, standing waves, nulls, over-voltage, data errors etc..

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  • \$\begingroup\$ Thanks a lot, Andy. That clarifies it for now. Actually, I am not very strong in transmission lines and often find myself confused. \$\endgroup\$
    – LoveEnigma
    Dec 3, 2014 at 10:24

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