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.
Convection does not scale linearly with surface area, it's a fluid dynamics problem rife with Nusselt numbers and Rayleigh coefficients. Since it's not my field, and since I'm more likely to use CFD than first principles in such a situation anyway, I'll point to a calculator that might allow some feel for the problem here (assuming it's actually implemented correctly, of course).
If I stick in reasonable numbers like 100°C for the wire 1, 2, 10mm for the wire diameter then I get total heat loss scaling with roughly the diameter (and thus the surface area) to the power of 2/3.
Preece was looking at fusing current, which would be at a much higher temperature than typical operation of an electrical wire.
Best Answer
Your question comes down to a basic application of Ohm's Law. It says that the voltage drop across a resistor is the current thru the resistor times the resistance.
You apparently have wire with a resistance of 3.688 mΩ/foot. To get the total actual resistance, you need to know the length of the wire. Remember that current has to run out one wire and back on another, so the wire length the current sees is twice the distance from your power supply to your device.
Since you didn't give the distance, I'll work it backwards to find the maximum distance your setup can support.
It seems the power supply puts out 24 V, and the device at the other end of the cable can operate down to 17 V. That means the cable can drop 7 V. You also say the current is 400 mA. Using Ohm's law, (7 V)/(400 mA) = 17.5 Ω.
Now we can find the wire length that results in that much resistance.
(17.5 Ω)/(3.688 mΩ/foot) = 4745 ft
Again, it takes two wires, so the distance from the power supply to the device can be at most half that, which is 2373 feet or 723 meters.