# 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 Commented 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. Commented 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? Commented 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). Commented Feb 13, 2020 at 14:28