Electronic – How does “bidirectional” transmission on gigabit ethernet work 2

One pair is used to transmit, the other - to receive.

If you connect two computers without a switch or a hub, in the past you had to use a cross-over cable, where one pair is connected to pins 1,2 on one end and pins 3,6 on the other end. Hubs had reversed pinout of the socket, so you can use a straight cable to connect a PC to a hub, but would need a cross-over cable to connect two hubs (unless one hub had an "uplink" port).

Modern Ethernet devices can be connected with any cable - they will figure out whether the cable is straight or crossed and will reconfigure themselves accordingly. Gigabit Ethernet works a bit differently - it uses all pairs (instead of just two) and can reconfigure each pair as "transmit" or "receive" as needed.

Now, as to why pairs are used instead of single wires:

To transfer data, you need to be able to get some current to the receiving device. As we know current only flows when there is a closed circuit, so you need at least two wires connecting the devices. Your "scheme 2" will not work as you drew the "batteries" not connected. This can be done in one of two ways - easier is to have one or more data wires and one ground wire (called Single Ended system). Here ground is shared among all the signals and you need less wires. However, this system does not work well for long distances - noise can get in the cable quite easily and the receiving device may not be able to understand the transmission. One solution is to use a coaxial cable (it shields the data wire from noise), but they are expensive and you would need one cable for each data pin. Still, multiple coax cables are used, say, for connecting a VGA monitor to the computer (at least in the better monitor cables). It is also true for analog audio.

A better way to do things is to have two wires for each signal. Now you send the signal in both wires, but invert one of them, that is, if you send "1" in one wire, you send "0" on the other - so the voltage between those two wires is always non-zero. You also use a twisted pair cable. This is called differential signaling. Now, the noise affects both wires in a pair equally and the receiver can cancel it out (by measuring the voltage between the wires instead of each wire to ground). This allows the signal to be sent further using cheaper twisted pair cables. Professional analog audio also uses differential signalling for, say, microphones etc (the XLR connectors have three pins - positive signal, negative signal and ground), so that longer cables can be used without noise affecting the signal.

An example of differential signalling:

As you see, in this case what matters is the polarity of the received voltage, so if whatever noise affects both wires the same, the polarity will not change and the information will still be transmitted.

To transmit in both directions (but not at the same time, so-called "half-duplex") over the same pair of wires you can do it like this:

Now when any switch is closed, both lamps light up, so any end can transmit taking turns. This arrangement is called "open collector".

If you don't need compatibility with ethernet, it doesn't matter which pair you use. It doesn't even really matter much that you use pairs.

Your main concern in this endeavor will be the resistance of the cables. You can find a chart of wire resistance and from that, calculate the total resistance of your cable. For an example, cat-5 cable is typically 24-26 AWG. From a table, I see that 100 feet of 23 AWG wire has a resistance of about \$3.2\Omega\$.

Say you want to move 500mA on this wire. We can calculate the power lost in the cable:

\$ P = I^2 R = (0.5A)^2 3.2\Omega = 0.8 W \$

or the voltage you will lose on this cable:

\$ V = IR = 0.5A \cdot 3.2\Omega = 1.6 V \$

Keep in mind, this is for 100 feet of wire, but since you need a supply and a return, the total length of the wire is twice the length of the cable.

This 0.8W is not only power not doing useful work, and less power available at your load, but also heat generated in the cable. A longer cable has more area to radiate heat, so can dissapate more power. This is why most tables also include a maximum current. \$500mA\$ is just a bit too much for 23 AWG cable.

Also, the voltage drop will reduce the ability of a voltage regulator to regulate the voltage on the other end. The voltage regulator has no way to know what the voltage on the other end is, so it can't regulate it.

Another concern is the maximum voltage the insulators in the cable can withstand. Power over ethernet provides 48V, so anything below this should be safe. Anything above that, consult the datasheet for your cable.