# Terminating Resistors on Ethernet Port

I'm using Ethernet (CAT5/5e/6) cables to communicate with the PCA9615 Differential I2C buffer. These cables roughly have a characteristic impedance of 100 ohms, but I have a few questions about choices of terminating resistors.

simulate this circuit – Schematic created using CircuitLab

The datasheet for the IC gives an example of how to terminate one of the twisted pair ends. In there, they chose a combination of 600 ohms and 120 ohms for their resistors with the claim that the parallel combination yields a termination of 100 ohms at each end of the twisted pairs. Sparkfun also made a breakout board, but their choices were 390 and 100 ohms, so there's a slight difference. From what I've researched, the resistor between the two differential pins is the one that terminates the network while the other resistors act as pull-up resistors to VCC and/or ground. If we have the same terminating resistor on both sides of the cable, wouldn't the driving end see a parallel combination of the two (so in the image, the IC would see 60 ohms)? Wouldn't it be better to double it to match the impedance of the CAT5 cable?

Also, if some twisted pairs are not being used, it seems to be good practice to terminate them. However, if I'm using them for DC power and ground and not for signals, is it fine to not terminate them with a resistor?

• the image shows 120 ohms in parallel with 1200 ohms Feb 12, 2020 at 16:24
• Note that for an ideal theoretical transmission line, you want your termination to match the impedance of the line. But, transmission lines are not perfect, they can have losses and dispersion. For the best performance you might use a terminator with slightly higher or lower impedance than the transmission line. Sometimes a slightly underdamped (larger value) terminator is best. Feb 12, 2020 at 17:39
• @Mattman944 Communication is bi-directional, but it feels like having terminations at both ends would require larger resistor values overall. Take the Sparkfun breakout boards. Their resistor choices come to ~88 ohms (100 || 780), so if you use two of these modules, the line sees ~44 ohms with the two networks in parallel. Doesn't this value seem too low in this case? Feb 13, 2020 at 14:19
• Any parallel equivalent impedance at the source has no effect on the pulse travelling down the transmission line. The end needs to be terminated with the characteristic impedance of the transmission line (with the caveats in my other comment). The driver, of course, needs to drive both loads (after reflections have settled). Feb 13, 2020 at 14:28

The whole network of resistors is the termination, not just the 120 ohm resistor. So the differential pair is terminated with approximately 109 ohms, which is close enough to 100 ohms. When both ends of a differential pair are terminated properly to 109 ohms, indeed the total termination is 54 ohms. If you use some pairs for power, they are not unused any more and don't need termination, as voltage source drives then with almost zero impedance. Completely unused pairs can be terminated, sometimes there a 50 ohm resistor to ground on each unused wire.

• 120 || (600 + 600) = 109 ohms, not 100 ohms Feb 12, 2020 at 17:27
• Yes it is 109 instead of exactly 100; I blindly believed the datasheet without checking. Elsewhere they use 54 ohms as load which would equal the given resistances, doubly terminated. Feb 12, 2020 at 19:01
• Thanks for your reply. I'm a bit confused though. Are you looking at the network from the perspective of one end of the differential signal? If so, say from DSCL+, wouldn't that be (120+600) || 600? So, like you said, if both ends of the pair are properly terminated, you get 54 ohms as seen by the transmission line. Wouldn't that cause the line to be mismatched given CAT5's characteristic impedance? Feb 13, 2020 at 13:56
• The perspective of differential signaling always means differential termination between signal pair. Differential impedance is 120||(600+600) = 109. Common mode impedance is 600||600 = 300. Single mode impedance (one wire driven only, one floating) is 600||(600+120) = 327. Each end must be terminated with the characteristic impedance to make it look like infinitely long. So 100 (109) ohms on each end is what the transmission line must see. Total load seen by a driver at any point in wire is 50 (54) ohms. Feb 13, 2020 at 14:39
• @user101402 that example of split termination does not apply here - that would be the case of no CM termination at all versus 50 ohm CM termination biased to ground. Just stick to what the datasheet suggests, the 300 ohms CM termination should be enough and it also works as the biasing. Datasheet does not suggest anything else, like AC-coupled split termination to reduce CM impedance, like sometimes seen on CAN or RS-485. Feb 13, 2020 at 18:30