Electronic – Voltage transmission

This is a really good question and one that is misunderstood by many.

one key is to realize that it is not electrons that are traveling down the line, its actually a CHANGE in the electric FIELD that is traveling down the line. The electrons themselves travel at a much slower rate as the wire is made up of molecules in a crystal structure, and the electrons bump into other molecules and electrons all the time. The field pushes them in a general direction but they do not move at anywhere near the speed that the field moves.

if you remember basic physics, an electric field exists between any two points that have a voltage difference between them. The strength of that field is dependent on the material that fills the space between those two points.

On a transmission line you start with a steady state situation where the voltage is constant, and the field is therefore constant. You then abruptly change that field at one end of the line. The CHANGE in the field propagates down the line at the speed of light, if the material between the wires is free space, or somewhat slower depending on the material (called the dielectric). In a coax cable, its close to the speed of light, in a PCB its about half the speed of light. Rule of thumb for PCB is 2ns per foot, or 180ps per inch. At the speed of light, the wave would propagate at about 1ns per foot.

A good analogy is a wave in water. Start with a smooth water surface. Drop a rock in the middle. A wave propagates from the rock drop point outwards at some speed. The individual molecules of water do not travel with the wave. They mostly just move up and down as the wave passes by. they will probably move a bit in the direction of the wave, but much slower than the wave is actually moving.

There is also an analogy for how the wave in water bounces off when it hits a wall, which is like an open circuit in the electric case. But I'd have to go look up the analogy for a short circuit and a well terminated line!

One more analogy: you can think of a transmission line as a series of capacitors. as the field propagates down the line you are charging each capacitor, one after the other. in the process of charging them, some current must flow from one wire to the other until the capacitor reaches the field voltage. The actual value of the capacitor, and the resistance the current sees is entirely dependent on the physical makeup of the structure: the spacing between the wires, the dielectric material in that spacing, etc. so based on the change in voltage (delta V), you get some current to flow to charge the capacitors (delta I) and therefore you can calculate an equivalent resistance from ohms law: delta R = delta V / delta I. this is called the characteristic impedance of the transmission line, also referred to as Z0.

In free space and with wires that are far apart, you reach a limit of about 120 ohms for Z0. In PCB's it is common to have lines that are between 30 ohms and 70 ohms, based on the width of the PCB traces, the spacing between them, and the PCB dielectric material between them.

I know this is getting long and wordy, but when the field finally gets to the end of the line, it finally sees what is there. This is a boundary condition that the field must react to. if its an open circuit, then no current will flow at the end of the line between the two wires. BUT there is this delta I that's been happening, and because of the inductance of the line (I know i haven;t mentioned it yet but there is inductance there as well) the current can't stop immediately, so it kicks to the opposite voltage which causes a change the electric field, which then starts to propagate back down the line IN THE OTHER DIRECTION!!! When that change in field hits the original source, it sees some impedance and reacts accordingly and changes the field voltage again which propagates back to the end, and so forth and so on.

If there is a resistance across the line at the end which matches the Z0 of the line, then it perfectly matches the delta V and delta I that is coming down the line and no change in field results so there are no reflections. that is what is called a perfectly terminated transmission line.

But what happens if we do not add the resistor? ... The LED drops 2.5V, but the battery's potential is 9V. What about the "missing" 6.5V?

If you have a battery and LED with no current-limiting resistor, then you basically have the same scenario as when you just connect a wire across the two battery leads, but now with only 6.5 V instead of 9 V.

Basically, in this scenario you can no longer neglect the internal resistance of the battery or the forward resistance of the LED (and possibly the resistance of the wire) if you want to determine the actual current. A very large current is produced.

What happens to the battery, and what to the LED?

Like in the short-circuited battery case, the battery heats up. If the LED is only rated for 20 mA, and you put 100 mA or 1 A through it like you might in this situation, then very quickly the light-emitting diode becomes a smoke-emitting diode (at least momentarily) and then it usually becomes an open circuit.

What is the actual current in the circuit?

As I mentioned above, you need to consider the internal resistance of the battery and the equivalent resistance of the LED, and maybe the wire resistance. From these you can estimate the actual current. You probably can't calculate it exactly because the numbers depend on temperature, which will be rapidly changing as the circuit heats up and then cools down after the LED pops.