Why does the voltage double at the end of line in a open transmission line (physical explanation)

Question

I'm seaching a way to explain physically what happens in a wave reflection in a transmission line, if we apply a DC voltage in a open circuit transmission line, the voltage at the doubled.

I saw this video presentation, but I would like to understand physically the reflection that is represented in the video.

1. In the first moment of the signal travel it's charging the capacitors that are in between the conductors, but according to the video (minute 2:08) after a reach a few capacitors the first capacitor started to discharge, why does the voltage decrease?

2. When the current reaches the last capacitor, the voltage in the last capacitor almost doubled its value. I would like to know what happen regarding to the charge in the last capacitor.

Answer 1

I prefer this explanation...

To understand reflections (and reflection coefficient) we can set up a thought experiment: -

• Assume a cable (transmission-line or t-line) of characteristic impedance \$Z_0\$
• At some distance away a resistor (\$Z_L\$) terminates the t-line
• Assume a voltage (\$V_F\$) is applied at the start of the t-line
• The current (\$I_F\$) that initially flows into the t-line equals \$V_F\$ divided by \$Z_0\$ (ohm's law)

When the voltage and accompanying current reach the end of the t-line there may be a violation of ohm's law if \$Z_L\$ does not equal \$Z_0\$. The process of fixing a violation uncovers the meaning of the reflection coefficient.

For instance, if \$Z_L\$ > \$Z_0\$ we have to mathematically: -

• Make the voltage arriving at \$Z_L\$ a bit bigger and, at the same time...
• Make the current arriving at \$Z_L\$ a bit smaller
• Adjust voltage and current to force a ratio of \$Z_L\$ (make ohm's law work)
• But, the adjustments have to "go somewhere" and, indeed they form a reflection
• That reflection travels back up the line to the source

Algebraically we could say: -

$$\dfrac{V_F + \delta V_F}{I_F - \delta I_F} = Z_L$$

I've added a bit of voltage and, subtracted a bit of current. Mathematically it is like this: -

$$\dfrac{V_F}{I_F}\cdot \dfrac{1 + \delta}{1 - \delta} = Z_L\longrightarrow Z_0\cdot \dfrac{1 + \delta}{1 - \delta} = Z_L$$

$$\text{Hence,}\hspace{1cm}\delta Z_0 +\delta Z_L = Z_L - Z_0$$

$$\text{And,}\hspace{1cm}\delta = \dfrac{Z_L-Z_0}{Z_L+Z_0}$$

But, of course, we call \$\delta\$ by it's usual name (reflection coefficient) \$\Gamma\$. The symbol \$\delta\$ is just a device I invented to get through the thought experiment.

However, the important subtlety that prevents an ohm's law violation is the "bit" we add to voltage and the "bit" we subtract from current (\$\delta V_F\$ and \$\delta I_F\$).

If we examined their ratio we would find it is \$Z_0\$. This means that they can naturally flow (together) back into the transmission line because they have the perfectly correct ratio to do so and don't compound any ohm's law violation problems.

That is called a reflection and travels from load to source.

Clearly, if \$V_F\$ and \$I_F\$ were originally of a ratio that matched the load (\$Z_L\$) impedance, we wouldn't need to set up the algebra that figured out how to deal with the "extra" signals and, there would be no thoughts of violating ohm's law nor talk of reflections.

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When \$Z_L\$ is an open circuit, we have to reflect the full applied voltage back to the source (in order to also return the full current) hence,

• That reflected voltage superimposes itself on top of the incident voltage and,
• For a short period of time we get twice the voltage at the end of the t-line
• The action of doing that means there is no net current flowing into an open circuit
• This is the conservation of ohm's law when an open circuit is encountered
• I.e. there is a cancellation of currents flowing down the t-line to the load

What happens next depends on how the initial reflection is dealt with at the source end. If the source is a pure voltage (no series resistance), the initial reflection from the load gets fully reflected back to the load end and, things carry-on sloshing around indefinitely.

Answer 2

It's simple: the wave is reflected from the end, and for an open circuit the voltage of the reflected wave is the same as and in phase with the transmitted wave. So, the two waves together have twice the voltage.

Answer 3

I believe the problem is that you're thinking in terms of electrons. But unlike water pressure, which is caused by the density of water increasing, voltage is not caused by the density of electrons increasing. It's a measurement of the electromotive force that encourages electrons to move.

Thank goodness, or increasing/decreasing voltage would lead to changing isotopes of the conductor... I'm imagining mushroom clouds... and it's one reason why you can't think in terms of electrons, but the force trying to move electrons.

We can start our visualization of what is physically happening by considering a newton's cradle.

Each of those balls is an electron prepared to travel down a conductor — but at the end of the conductor is (for the sake of argument) a perfect terminus. Unlike the image above, the last electron doesn't move because there's nowhere for the electron to go — but the force is pushing it and conservation of energy requires the force to go somewhere. Where?

For every action there is an equal and opposite reaction. – Isaac Newton

You've likely never thought about what happens when you hit a brick wall with a baseball bat. The bat is moving forward with a potential energy, a force (if you will...) and when it hits the wall there is an equal and opposite reaction stopping the motion of the bat. In an ideal world, the energy at the point of impact momentarily doubles. Or it would if you looked at the absolute value of all the energies. (No metaphor is perfect!)

So the electromotive force identified by the word "voltage" hits the proverbial brick wall and must have somewhere to go. The electrons don't move because, physically, there isn't anywhere for them to move to. But the force doubles and reflects back up the conductor. We can visualize something like this when we watch a ditch fill with water. The gate at the end is closed and the initial wall of water hits it. What happens? The volume quickly increases and that increase (reflection) "travels" back along the ditch. (Of course, all this is happening at just under the speed of light, so it looks instantaneous, and water doesn't represent electrons but the electromotive force. Yup, no metaphor is perfect.)

So, it's not the electrons you should be thinking about... it's the push against the electrons. If you wish, think of it this way. A line of people ends at a closed door. The person furthest from the door pushes against the next person in line, and so forth, until the nearest person pushes against the door — and the door pushes back with an equal and opposite force.

We just happen to be able to see that with an oscilloscope. (That ought to show my age.)

Answer 4

1. The voltage at the start capacitor starts to fall because it is being driven down by the supply voltage and supply current.

2. The voltage at the end capacitor continues to rise because it is being driven up by the line inductance, and unlike the "properly terminated" sub-lengths of the line in the middle of the line, there is no where for the current to go: the line is cut off, current is coming into the end capacitance but no current is flowing on.