Grabbing the ends of a transmission line

[RedX]

If you have a two-wire transmission line that is open-ended at the load, and connected to an alternating voltage generator at the source, then the voltage between the open ends can be as high as twice the voltage of the generator. However, if the two ends are shorted, then the voltage would be zero at the ends. So if you were to grab both ends at the same time, would you be shocked? That is, would the voltage through you be the open-circuit voltage, or the short-circuit voltage?

 

[Studiot]

What are you studying?

 

[RedX]

What are you studying?

Transmission lines.

There's an experiment done at 41 minutes, 30 seconds into this video:

http://ocw.mit.edu/courses/physics/8-03-physics-iii-vibrations-and-waves-fall-2004/video-lectures/lecture-16/

that attaches a light bulb to two wires coming out of an AC generator. As the light bulb slides along the wires, there are voltage nodes and the light bulb gets dark at those locations. However, the location of the nodes depends on whether the ends of the two wires are left open, or are shorted. In particular, if the ends are shorted, there should be a node at the end of the wires. If they are not shorted, the node is a quarter wavelength (of the AC frequency) from the end of the line.

So that just seems strange to me, so I wonder what happens if you touched the two ends of the line: would it be a voltage node or not?

 

[vk6kro]

It depends on the length of the transmission line.

If it is a quarter wavelength long, you do get effects like you mention.

However, if the input did look like a short circuit, it is hard to imagine how you could get any power into the line.

Assuming you managed it somehow, then the voltage at the other end could get very high, not just double.

Suppose there was a 1000 ohm resistor across the open end of the line (possibly your fingers) and it was 50 ohm line, then the input would look like (50 ohms * 50 ohms / 1000 ohms) or 2.5 ohms.

If you put 1 watt into the line, that would be 1.58 volts (1.58^2 / 2.5 ohms = 1 watt) at the input.

But at the 1000 ohm resistor, the voltage would be 31 volts because the power would be the same, (31^2 / 1000 = 1 watt), but the impedance is higher so the voltage has to be higher.

So, you wouldn't have to have a very high input voltage to get a big voltage out.

Even 31 volts RMS would be enough for you to feel.

To answer your question, you would possibly get a large voltage across your fingers, depending on the length of the transmission line and the input power.

 

[Studiot]

OK, somehow I got the impression you were talking about power transmission lines, not lecher lines.

I have posted links to these several times recently.

http://www.google.co.uk/#sclient=ps...gc.r_pw.&fp=6aca5766706a5eaa&biw=1024&bih=585

go well

 

[RedX]

It depends on the length of the transmission line.

If it is a quarter wavelength long, you do get effects like you mention.

However, if the input did look like a short circuit, it is hard to imagine how you could get any power into the line.

Assuming you managed it somehow, then the voltage at the other end could get very high, not just double.

Suppose there was a 1000 ohm resistor across the open end of the line (possibly your fingers) and it was 50 ohm line, then the input would look like (50 ohms * 50 ohms / 1000 ohms) or 2.5 ohms.

If you put 1 watt into the line, that would be 1.58 volts (1.58^2 / 2.5 ohms = 1 watt) at the input.

But at the 1000 ohm resistor, the voltage would be 31 volts because the power would be the same, (31^2 / 1000 = 1 watt), but the impedance is higher so the voltage has to be higher.

So, you wouldn't have to have a very high input voltage to get a big voltage out.

Even 31 volts RMS would be enough for you to feel.

To answer your question, you would possibly get a large voltage across your fingers, depending on the length of the transmission line and the input power.

A 1000 ohm resistor and a 50 ohm line is a huge mismatch of impedance, so I'm not sure you can say the power from the generator is equal to to the power burned by the resistor, since a lot of that power would be reflected.

Although for a quarter-wavelength line you can't get any power from a short circuit, for a different length such as a half-wavelength line you could. In fact, theoretically, if would be infinite power if you short a half-wavelength line.

 

[vk6kro]

There is a reflection of the voltage back to the source and this is what sets up the standing wave on the line.

As a result of this, there is a low voltage at the sending end and a high voltage at the 1000 ohm resistor end and an approximately sinusoidal quarter wave of voltage in between.

Apart from transmission line losses, which we can assume are negligible because the line is very short, the majority of the power will be dissipated in the 1000 ohm resistor. It has a high voltage across it, so it can't do anything else but dissipate the power.

This is a quarter wave transformer and they are routinely used for antenna matching because they are extremely efficient.

Although for a quarter-wavelength line you can't get any power from a short circuit, for a different length such as a half-wavelength line you could. In fact, theoretically, if would be infinite power if you short a half-wavelength line.

If you shorted the end of a half wave line, it would look like a short circuit to the voltage generator, too.

In fact, theoretically, if would be infinite power if you short a half-wavelength line.
Would you like to explain this comment? Where is this infinite power going to come from?

 

[sophiecentaur]

A 1000 ohm resistor and a 50 ohm line is a huge mismatch of impedance, so I'm not sure you can say the power from the generator is equal to to the power burned by the resistor, since a lot of that power would be reflected.

The power 'from' the transmitter would have to be the power dissipated in the load because that would be the only resistive element in the circuit and, thus, the only place that power could go. However, there would be an indeterminate amount of power dissipated in the output stage of the transmitter, depending on its actual impedance and the design specifics. IF the system were designed for maximum efficiency feeding a 50Ω load then there would be very little power getting into the 1kΩ load and the transmitter could easily be 'using' about the same amount of power from its supply as before. The volts appearing at the transmitter output could be around twice the normal operating value- which may be a problem.

 

[RedX]

In fact, theoretically, if would be infinite power if you short a half-wavelength line.
Would you like to explain this comment? Where is this infinite power going to come from?

I have a formula for the impedance that a generator connected to a line and load sees:

$$Z=Z_0 \frac{Z_l \cos(kl)+iZ_0\sin(kl)}{Z_0\cos(kl)+iZ_l \sin(kl)}$$
where Z_o is the impedance of the line (characteristic impedance) and Z_l is the impedance of the load.

So at half wave-length this would reduce to: $$Z=Z_l$$ so the power delivered by the transmitter for a short would be: $$P=.5|V|^2/Z=.5|V|^2/Z_l =\infty$$

Similarly for quarter-wavelength I get $$P=.5|V|^2/Z=.5|V|^2/(Z_0^2/Z_l)=0$$ so the power delivered by the transmitter for a short would be zero.

 

[RedX]

The power 'from' the transmitter would have to be the power dissipated in the load because that would be the only resistive element in the circuit and, thus, the only place that power could go. However, there would be an indeterminate amount of power dissipated in the output stage of the transmitter, depending on its actual impedance and the design specifics. IF the system were designed for maximum efficiency feeding a 50Ω load then there would be very little power getting into the 1kΩ load and the transmitter could easily be 'using' about the same amount of power from its supply as before. The volts appearing at the transmitter output could be around twice the normal operating value- which may be a problem.

Are you talking about the internal impedance of the generator? I wasn't aware you had to match that to the transmission line. So the internal impedance of the generator should match the 50Ω impedance of the line, ideally? If it matches, then the voltage at output would be around twice the generator's value? If it doesn't match, do you get multiple reflections from the interface between source/line and line/load that can make the value of the voltage at the end of the line greater than twice the value of the generator voltage? For that matter, with multiple reflections at the generator/line end, can the output voltage at the interface between generator and line exceed the nominal voltage output of the generator?

 

[sophiecentaur]

Are you talking about the internal impedance of the generator? I wasn't aware you had to match that to the transmission line.

No. I wasn't saying that. I was saying that the impedance of the generator would determine how much power was 'wasted'. I think it's easy to get hold of the wrong end of the stick when talking about what should be happening with the transmitter. The antenna /load end is easy - you just need to produce as good a match as possible so that no power is reflected. This helps with efficiency, of course, but also it stops reflections which produce standing waves (over voltage problems).* Also, because the transmitter cannot be relied on to be a good match, these reflected signals can get back up the feeder and lead to, either ghosting or frequency response distortions in the transmitted signal.
* With a really bad match at each end, there is no limit to the maximum value of voltage that could appear at either end of the line.

The transmitter situation is much harder because what do you want? Ideally, you would have a voltage source (source resistance zero) so that there would be no resistive losses in the transmitter. This is the ideal for an AC Power Generator, of course, and you have a chance of making one with thick enough wire to achieve nearly this. With a transmitter, however, you are usually dealing with an active device (valve, transistor) which is controlling the current from a power supply to produce an amplified version of some input signal. You are often just trying to arrange it so that you get as much power as you can into your 50Ω feeder with a given output device which could have a ghastly impedance in its own right and also needs to operate with certain DC conditions on it. Transmitter engineers have to be pretty smart about this - not my field - but it depends upon the technology and frequency involved. If you are just 'line driving' where power is not an issue, you would make sure your source was 50Ω so that reflections would not be an issue and also so that any filters in the way could be relied on to work as intended.

 

[sophiecentaur]

However, if the input did look like a short circuit, it is hard to imagine how you could get any power into the line.

The input could be a Voltage Source- zero resistance - which could be ideal, in some cases. You'd achieve a power input of V²/50W into a matched load and the volts at the transmitter would be V, whatever you hung on it.