How Electricity Works? By Electrons or by Fields?

[Radio1950]

Hi Jeremy

thank you for your interesting posts.

You are correct.

I am wrong.

[Radio Wrangler]

There are two trip-ups to understanding this. The first one is how you describe the current of the reflected wave - either as a positive current means in the direction from source to load end, so reflected current and forwards current are opposite in sign from the reflection at an open load and therefore cancel. This gives zero net current into a load which can't take any, and is therefore correct. In this view you would say that an open circuit load is being hit simultaneously by a current of +I amps and a current wave of -I amps, leaving net zero amps at the load. You define both forwards wave and reverse wave in the same direction, the reverse wave is just differentiated by the sign of the current.

(Notice that voltage is transverse to the conductors of the line and is independent of how you view direction. Current is longitudinal and measured in phase with the voltage, it's sign defines current flow in terms of your definition of energy flow. Forwards and reverse energ flow on the line need either a reversal of the current sign or a reversal of the definition of direction of current in the line. You are free to view it either way.)

So as an alternative, you can define the current of the reflected wave in the opposite direction to that of the incident wave. In this view you would say that +I amps is hitting the load, and +I amps is leaving it in the reverse direction. This allows you to view the wave folding at an open load and continuing backwards.

The second confusion that hits people trying to understand transmission lines comes down to their modes. There are two simple modes. A simple line, whether balanced or unbalanced can shift energy is two modes... the differential mode, where the line voltage is considered to be transverse between the two conductors, and common mode, where we think of the voltage common to both conductors, against some external reference, like earth. When constructing equations for voltages and currents in the conductors of lines, you need to be careful to define them so they work for the mode you intend.

The two views thing goes further. If we have a line with a mismatched load and we draw equations for the voltage and current at all places and all times on the line, we can resolve them into EITHER two waves, one flowing to the load, and one flowing back from the load. OR we can resolve them into a wave representing the power going into the load, plus a 'standing wave' which is not moving, and has maxima and minima sitting at fixed locations. BOTH views are valid. Measuring the differences in peak-peak voltage by probing along a slotted waveguide or coax line will show you the Voltage Standing Wave Ratio, VSWR - long an easily measured indication of the accuracy of the match of a load. Zero reflection from a perfect wave gives zero amplitude returned wave, gives zero amplitude standing wave, so the measured voltage is the same along the line, so 1:1 VSWR. If you view the goings-on in the line with a directional power meter, it will resolve the voltage and current at the point you insert it, and tell you the powers of the forward and reflected waves but it needs to know the impedance of the line to do this. Similarly if you use a network analyser, it has a directional coupler hidden inside it for measuring S11 or S22, so the network analyser is wedded to the forwards and reflected wave viewpoint.

There's a lot here, but it all comes down to the definitions of how we view things.

Is that car going past you on the other side of the motorway actually moving in your direction, just with negative velocity

David

[regenfreak]

the differential mode, where the line voltage is considered to be transverse between the two conductors, and common mode, where we think of the voltage common to both conductors, against some external reference, like earth. When constructing equations for voltages and currents in the conductors of lines, you need to be careful to define them so they work for the mode you intend.

This has got me thinking..

The impedance of a line can depend on the relative voltage and not always referenced to the earth.

If I have two parallel common-mode (even) transmission lines with voltage vectors V1 and V2 relative to the earth potential, current I1 and I2, since the polarity and direction of travel are the same, the common-mode impedance of line 1 is

Zcommon = (V1 + V2)/I1

Similarly in differential mode (odd),

Zdifferential = (V1-V2)/I1= differential impedance.

As the spacing between the two differential lines is reduced, the electromagnetic field lines between the conductors are compressed until it creates a virtual ground reference plate VGRP. Heavy stuff, I will go, make a cup of tea and stop thinking about this.

[G0HZU_JMR]

Here's a simple analysis for the transmission line with an open at the far end. If you could 'see' the magnetic field (viewed from the source) as the positive pulse went through a short piece of wire into the transmission line you would see the magnetic field with a clockwise 'curl' as this obeys the right hand rule as the current direction is into the transmission line.

When the pulse reappears after its journey to the far end and back the magnetic field will be seen again but this time the curl will be anti-clockwise. This is because the current direction has flipped. At the far end of the transmission line where there is an open, the current can't flow so the magnetic field collapses and all the energy ends up in the E field and so you see twice the voltage at the open end when the pulse arrives there.

Here's an analysis for the case with a short at the far end: If you could 'see' the magnetic field (viewed from the source) as the positive pulse went through a short piece of wire into the transmission line you would again see the magnetic field with a clockwise 'curl' as this obeys the right hand rule as the current direction is into the transmission line.

When the pulse reappears after its journey to the far end and back the magnetic field will be seen again and the curl will still be clockwise because the current direction hasn't flipped even though the pulse is returning back to the source.

At the far end where there is a short, there can be no voltage so the electric field collapses and all the energy ends up in the magnetic field and you briefly see twice the current through the shorting link at the far end of the cable when the pulse arrives there.

[kalee20]

Originally Posted by G0HZU_JMR

Here's an analysis for the case with a short at the far end... you briefly see twice the current through the shorting link at the far end of the cable when the pulse arrives there.

I can't see that somehow.

When the pulse arrives at the far end, a reverse voltage is reflected and a zeroness of voltage starts travelling towards the source. But the current can't momentarily double - if it did, it would have to be double in the section of feeder immediately preceding the short.

There would also have to be an equation by which the duration of the brief double-current pulse could be calculated, and I can't derive one!

just with negative velocity

No, it has a positive scalar velocity (I don't think you can have a negative scalar value), chuck in a vector and then you know which way is is going. Grin.

[G0HZU_JMR]

There would also have to be an equation by which the duration of the brief double-current pulse could be calculated, and I can't derive one!

You don't need one. The duration of the doubled current is the same as the duration of the pulse itself.

[G0HZU_JMR]

Just to clarify, if you were to set up an experiment to send a 10ns wide pulse of current (10mA pulse) into the transmission line at the source then there would be a 10ns wide pulse of current seen through the shorting link wire at the end of the cable a short time later. This pulse would have double the 10mA current of the pulse that was fed in at the source. So a 20mA pulse of current 10ns wide would have been observed through the shorting link at the far end of the cable at the conclusion of the experiment.

[Radio Wrangler]

I was always taught that velocity is a vector quantity, speed is a scalar, but there's a lot of loose language on the loose.

Davi

[regenfreak]

(Notice that voltage is transverse to the conductors of the line and is independent of how you view direction. Current is longitudinal and measured in phase with the voltage, it's sign defines current flow in terms of your definition of energy flow.

I have to re-read this a few times to get my head around the idea that voltage vector and current vector are in phase. Current does not only flow along the line but also through the distributed capacitors to ground in the lumped element model. I can only understand the concept if it can be represented in a drawing or animation like this:

https://youtu.be/ozeYaikI11g

[G0HZU_JMR]

When the pulse arrives at the far end, a reverse voltage is reflected and a zeroness of voltage starts travelling towards the source. But the current can't momentarily double - if it did, it would have to be double in the section of feeder immediately preceding the short.

Are you happy that the voltage 'doubles' and the current falls to zero at the end of the transmission line in the case of the transmission line with an open at the end? In the case of the shorted transmission line it is the current that doubles and the voltage that falls to zero when the pulse reaches the short at the end of the cable.

If this still seems too crazy then consider that the role of the transmission line is to provide a means to efficiently transfer energy from a power source to an intended load that is some distance away. You can consider the source as a Thevenin Source or a Norton source and the transmission line can be thought of as a way to effectively move that source direct to the load.

If you have a (10ns pulsed) 1V Thevenin source with a 50R series resistor and feed this to a 50R transmission line you would end up with 1/(50+50) = a 0.01A (10mA) pulse travelling down the 50R transmission line. If the line is assumed to be lossless then you can model the far end as a Thevenin source or you could opt for a Norton source at both ends which would be 50R in parallel with 20mA.

It's easy to see that if you connected a short to the far end of the 50R cable the Norton Source would dump all 20mA into the short circuit. So 10mA goes into the transmission line at the source but 20mA actually comes out of the far end into the short circuit.

The 10mA 10ns pulse travels into the line and reflects back from the short as a 10mA pulse but with the same direction of current flow as the forward pulse. During the few ns where the pulse is colliding with the short circuit at the far end both of these currents sum together as they occupy the same final part of the transmission line. Therefore, for the whole 10ns of the pulse duration they will both be summing current into the short circuit at the far end of the cable as they slide past each other in the transmission line. So you see 20mA through the shorting link for 10ns.

[regenfreak]

It starts to get confusing when someone thinks about standing waves along a transmission line, where voltage and current are 90 degrees out of phase at short and open ends.

On the other hand, the instantaneous voltage and current are in phase for a pulse of travelling signal at any point along the line.

[kalee20]

OK thanks Jeremy - I'll ponder on that!

I'm wondering how it works with a constant-current drive to the cable, say 10mA imposed continuously. After approx. 1nsec per foot length the signal will appear at the other end, if it's shorted you say it will be 20mA, and continuously as it's an ongoing pulse. We'd then have the situation where 10mA is poured into the source end, and 20mA appears at the remote end, which is paradoxic!

I may resolve this myself - I just wanted to get some sort of reply done.

[regenfreak]

Are you happy that the voltage 'doubles' and the current falls to zero at the end of the transmission line in the case of the transmission line with an open at the end? In the case of the shorted transmission line it is the current that doubles and the voltage that falls to zero when the pulse reaches the short at the end of the cable.

(Notice that voltage is transverse to the conductors of the line and is independent of how you view direction. Current is longitudinal and measured in phase with the voltage,

Current and voltage are said to be in phase if their maxima and minima occur at the same time. This certainly is not the case at the open and short ends. I wonder if you meant voltage is the amplitude of the pulse(always positive), and current has either positive or negative sign? I don't know what is the meaning of "measured in phase with voltage" ?At the open end, the distributed capacitance is charged to twice the voltage when current is zero. At the short end, the distributed capacitance is shorted with all charge Q dumped to create twice the current at zero voltage.

[kalee20]

Originally Posted by regenfreak

It starts to get confusing when someone thinks about standing waves along a transmission line, where voltage and current are 90 degrees out of phase at short and open ends.

That also seems wrong.

At an open end, current is zero at all times so whatever the voltage is, the phase between it and zero current is undefined.

It's vice-versa with a shorted end, of course. Voltage is zero then.

So instead of open-circuiting or short-circuiting, we'll put a resistor of 1 megohm or 1 milliohm, way above or below characteristic impedance for the cable. Now the current or voltage is very small, but it's not quite zero, so a phase relationship can be defined.

Since the 1 megohm or 1 milliohm are resistors, current and voltage will be in-phase, not 90 deg out, no matter what the effect of the cable is. And that phase relationship will barely alter as you travel back along the cable by 1mm, 10cm, 1m... as long as they're small fractions of the wavelength.

[G0HZU_JMR]

In the case of the 10mA pulse travelling down the 50R transmission line, try replacing the short at the far end with a 50R resistor. You should see 10mA pulsed through the resistor at the far end. The power in the load would be 5mW and a 50R load is the optimal load in terms of power transfer as I'm sure we will all agree. Nothing is reflected with a 50R load.

Then reduce the 50R to 16.667R. Would you expect the current delivered to the 16.667R resistor be more than 10mA or less than 10mA or the same?

To answer this you should consider that the far end of the 50R transmission line will also have a source (surge) impedance of 50R. If you are happy with this then everything below should make sense. The load VSWR for 16.667 is 50/16.667 = 3:1 and the reflection coefficient is 0.5. Theory books tell us the
power delivered to the load with a 3:1 VSWR will be 75% of 5mW and so the power delivered to the load should be 3.75mW.

0.00375W/16.667R = I^2 so I = 15mA for 3.75mW. This is more than the 10mA going in at the source end of the transmission line.

If you believe what I'm saying there should also be a 5mA pulse going back towards the source because the 10mA FWD pulse summed with the reflected pulse creates the 15mA pulse seen at the 16.667R load. So the reflected pulse will be 15mA - 10mA = 5mA or 0.005A.

If you work out the power absorbed back in the source then it should be 0.005A*0.005A* 50R = 1.25mW.

So the power delivered to the load in this case is 3.75mW and 1.25mW is reflected to the source and these sum to the 5mW that would have been dissipated in a perfect 50R load with no reflected power. So despite the 10mA pulse going into the transmission line and a 15mA pulse coming out of it into the load there was no free lunch in terms of free energy anywhere. It still adds up to 5mW.

[regenfreak]

That also seems wrong.

At an open end, current is zero at all times so whatever the voltage is, the phase between it and zero current is undefined.

It's vice-versa with a shorted end, of course. Voltage is zero then.

So instead of open-circuiting or short-circuiting, we'll put a resistor of 1 megohm or 1 milliohm, way above or below characteristic impedance for the cable. Now the current or voltage is very small, but it's not quite zero, so a phase relationship can be defined.

Since the 1 megohm or 1 milliohm are resistors, current and voltage will be in-phase, not 90 deg out, no matter what the effect of the cable is. And that phase relationship will barely alter as you travel back along the cable by 1mm, 10cm, 1m... as long as they're small fractions of the wavelength.

This is magic. The transmission line behaves with reactive impedance everywhere until it becomes purely resistive at the open or 1M resistor termination.

[G0HZU_JMR]

If my above post #116 seems to be very complicated then simply remove the transmission line and connect a 16.667R load to the 1V pulsed 50R Thevenin source.

The current delivered to the load will be 1V/(50R + 16.667R) = 15mA and this agrees with the above analysis.

The power dissipated in the 16.667R load will be 0.015A*0.015A* 16.667R = 3.75mW and this agrees with my post above.

[G0HZU_JMR]

I'm wondering how it works with a constant-current drive to the cable, say 10mA imposed continuously.

I wouldn't recommend using a genuine 'constant current source' as this won't be matched to the transmission line Zo.

However, if you mean use a Thevenin source of 1V with a Thevenin resistance of 50R and connect it to a 50R transmission line and then send a 'really' long 10mA pulse down the cable then I guess it's a case of how long you want to run the test for before you accept the 20mA current (in the shorting wire at the far end) will be there for the same duration as the duration of the 'really' long pulse.

[kalee20]

No, it's a true constant-current source I was postulating.

Squirting 10mA into my local end of say a 20 foot length of 50 ohm coax, shorted at the far end, I'd expect to see 500mV at my local end (10mA x 50 ohms) immediately after switching on.

The 500mV / 10mA wavefront would travel along the coax and after 20nsec would reach the far end. With the gross mismatch, the 500mV would be reflected with reverse polarity and travel back to the local end, so the voltage would snap from 500mV to zero, 40nsec after switch-on.

I now see what you're getting at - if my local end was fed its 10mA by a 50 ohm source, the zero voltage would demand 20mA from the source and this would exactly match your claimed 20mA into the remote S/C. No electrons created or destroyed in the process, and reflections would stop.

But my local end isn't - it's 10mA constant so the zero voltage will again be reflected, the source-end voltage will drop to zero and this will travel down the line - the third journey. I've not yet worked out what happens at the remote end, but it's clear that with an ideal line, ideal current source, and ideal S/C the reflections will persist indefinitely.