To send data, the sender first turns each 4-bit nibble into a 5-bit
word, which ensures that five straight zeroes is never valid and
indicates signal loss
Not exactly. This encoding does much more than just detecting signal loss. It makes sure that the same number of zeros and ones are sent (a.k.a. DC balanced), does some error detection, and has otherwise useful properties for this type of work.
Now, a change in voltage must propagate through the wire; first the
recipient will see it, and then the sender themselves will see it on
the "undriven" side of the circuit. The sender must see this feedback
in order to ensure continuity (doesn't it?).
No. Ethernet has properly terminated signals (the termination is on the other side of the isolation transformers), and so the signal does not reflect back to the transmitter. In Ethernet there is no concept of continuity, only link. Link is established by a handshake type protocol between the two ends of the cable. If device A can send data to B, and B can send data to A, then there is a good link between the two devices.
So, the limit to the total circuit length, assuming the ideal that
voltage propagates at c, is how far light can travel in 31.25
microseconds. That distance, given a simplistic c = 3*108 m/s, is 9.6m
~= 31.5 ft. Since that's total circuit length from sender to receiver
and back, the actual total cable span is half that, or 4.8m ~=
15.75ft. Beyond this length of Cat5, it is simply impossible for the sender to toggle the voltage fast enough to maintain the fundamental
frequency, so the two parties negotiate a lower frequency, resulting
in a lower maximum bitrate over the longer cable.
No. Since there is no reflections, there is no relationship between bitrate and cable length. To put it differently, a Gigabit Ethernet cable that is 100 meters long can have up to (approximately) 600 bits worth of data "stored" in the cable.
By the time we get out to 182m, the Cat-5 specification's maximum
cable length at which simple resistance of the spec'ed cable will have
reduced signal voltage below the threshold of the receiver's
distinction between the three states, I calculate that this
speed-of-light limitation will also have reduced the maximum
sustainable fundamental frequency to approximately 1.65MHz, for a baud
rate of 6.6Mb/s and a true data rate of only 5.28Mb/s.
Ethernet spec allows for a maximum cable length of 100 meters, not 182 meters. And this has nothing to do with the bitrate or voltage thresholds. It has everything to do with collision detection and minimum packet size.
I do Ethernet all day long and we are able to transmit 900 Mbps of real data over a 100 meter long cable with absolutely no issues with reduced throughput.
if I have any unk-unks in this, it could be completely off.
Yeah, completely off. Sorry.
Any real conductor has an effective resistance per unit length. The voltage drop over that length of conductor, and the power dissipated in that conductor, is dependent on the amount of current you draw through that conductor. For a current draw of I, and a Resistance per Meter R you will experience:
(I^2 * R) watts per meter of power lost as heat in the conductor
and
(I * R) voltage drop per meter of conductor
If your endpoints cannot tolerate the worst case voltage drop, or if your power budget does not account for the losses in the cable, you can expect them not to function as intended.
This PDF is a reference I have used for stranded wire.
Best Answer
14 AWG has 2.525 ohms per 1000 foot so clearly, at 2A, this is not going to work because the volt drop at 2000 foot is twice this and then twice again because 2 wires need to be used for send and return.
You need to push 24 watts to the far end so maybe consider a boost converter. It will step up the dc voltage at the sending end and then, at the receiving end a buck converter can restore it back to 12 volts.
I'm also thinking that to get 24 watts at the far end, it's not unreasonable to push 30 watts into the cable at the sending-end - this means a loss of 6 watts in the cable.
Given 2000 ft of 14 awg cable (doubled) has a resistance of 10.1 ohms, you could argue that to dissipate 6 watts, the current should be: -
Power = I\$^2\$ R therefore I = \$\sqrt{P/ R}\$ = 0.77 amps
Given that you need to feed in 30 watts, the boosted voltage at the sending-end will be: -
30/0.77 = 38.9 volts.
This seems perfectly doable using a booster to raise the DC voltage to maybe 45 volts (gives a little bit of extra power for inefficiencies in the switchers).
I'd use a booster because it's fairly simple, can easily have current limiting to protect the cable and uses a safe voltage range. You could repeat the calcs for 24 AWG wire using the above method to determine sending voltages for other gauges. If you are in EU, a good safe DC voltage limit is 60 volts - this is known as SELV (safety extra low voltage) - paying a little bit more than lip-service to safety specs could save you a law suit!!
Gut feeling is that 24 AWG might need more than 60V.