Think I posted this to the wrong group first time around; let's see if anyone here can shed any light on this subject...
I'm just reading up on the subject of power factor correction in electrical supply lines and how as the PF heads down towards zero (as the phase difference between i and v gets more pronounced) more and more power is returned to the generator and how this can be fixed by adding capacitance or inductance in parallel with the load.
This struck me as incredibly similar to the situation with RF transmission lines where an antenna - if not at its resonant frequency - can be 'seen' as inductive or capacitive and when such is the case, power is reflected back to the transmitter, but this can be fixed by adding compensatory capacitance or inductance in parallel with the tx line.
Would I be right to infer from the above that these two phenomena are in fact one and the same? I just wonder if it's correct to do that, because we never seem to hear the terms 'power factor' applied to RF tx lines, nor 'reflections' applied to power lines. Can some kind soul clarify?
No. On the utility side, it would be necessary to have standing waves on the power lines in order to even use the term "reflected power" in the same sentence. Unlike RF, the impedance of the utility generator, transmission line, and load are NOT matched. If they were, you would dissipate the same amount of power in the generator as you would in the load, causing a rather unpleasant explosion at the generation facility. Also, in order to see standing waves on the power the frequency must be much higher than 60 Hz. Standing waves appear at
1/2 wave intervals. At 60 Hz, one wavelength is 5,000 km. Driving distance from Smog Angeles to New Yuck is about 4500 km so you're not going to see any reflections even with a cross country power line. This incidentally is why 60 Hz was selected as the AC line frequency. Higher would have used less iron and cost less to generate, but that would run the risk of standing waves on the power lines.
--
Jeff Liebermann jeffl@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558
FWIW, standing waves are not a necessary aspect of reflected power. All that matters is you have a defined system impedance.
Mismatch occurs just as well at DC as at RF -- it's just not very useful, becase no matter how long your transmission line is, it's always got a phase angle of zero...
It is useful, however, to keep in mind that this method / concept (of working with transmission lines) is completely frequency independent.
The main difference is, nothing is matched by resistance (which would cause losses!), the transmission line impedances are all over the place (different line spacings, cable sizes, etc.), the source and load power vary substantially (and without causing much change in the supply voltage), and the transmission line lengths are very short (under 1/8 wave).
AFAIK, mains distribution networks are designed as lumped element equivalent circuits. In the low frequency region, they exhibit equivalent parallel capacitance and series inductance. The capacitance is tuned out with reactors (large inductors), while the series inductance I think tends to not be as significant (which actually implies the lines' Zo is below whatever the operating |V| / |I| is on the line: the equivalent is capacitive if Zo < Rs and inductive if Zo > Rs), but contributes to droop (since the line inductance and reactor inductance act as an inductive voltage divider).
At substations, selectable tapped transformers are used to regulate amplitude, so that domestic power is largely stable against load variation. (Which will double the dV on the primary line side, but I suppose that doesn't matter.)
At the end user, the total transmission line length isn't very long, so it's not obvious what the characteristic impedance might be. A small SWR would be expected for any load, simply because there's not enough line to have a standing wave on. As a result, for example: the voltage doesn't go crazy when unloaded. Which is another way of saying: power that doesn't get used at the outlet, gets reflected back to the generator, where it arrives in phase, thus pushing power back into the propshaft, reducing the mechanical load.
(If this lesson has done its job, you should now be unable to unsee the AA cell sitting loose on the table, generating tens of watts continuously, only to be immediately recharged by that power reflecting off its open circuit terminals. ;-) )
** Because the AC supply voltage is fixed, doing this never makes any difference to operation of the load itself. The benefit of PF correction is that it reduces the current flowing in the supply cable and generator. The same system can then supply more customers with power and make more money.
** The AC power system is not a transmission line and loads are not matched to it. Resistive loads are he ideal in terms of efficient use of the installed hardware.
Forward and reverse power have opposite signs of current for the same voltage.
Short range power transmission systems don't really have a well-defined characteristic impedance, so resistive mismatches aren't really relevant--with a wavelength of 5000 km, even a whole city is essentially a lumped-element load. Maximum power transfer is also irrelevant--you don't want to match all the available power from the pole pig into your desk lamp, for instance. ;)
Reactive mismatches and harmonics cause the RMS current to be higher than P_load/Vrms, which causes excess loss, and that's what you care about in power distribution.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
You've fallen foul of Joerg's favourite fallacy: transmitters do _not_ have 50 ohm outputs. The antenna needs to be matched to the line, but the transmitter doesn't. Otherwise you could never have a transmitter efficiency as high as 50%.
Existence or nonexistence of standing waves doesn't depend on the length of the line. A fractional cycle of standing wave is still a standing wave.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
Just noticed this paragraph from the Wikipedia page on AC Power:
"Conventionally, capacitors are considered to generate reactive power and inductors to consume it. If a capacitor and an inductor are placed in parallel, then the currents flowing through the inductor and the capacitor tend to cancel rather than add. This is the fundamental mechanism for controlling the power factor in electric power transmission; capacitors (or inductors) are inserted in a circuit to partially compensate reactive power 'consumed' by the load."
I've never seen this explanation anywhere else and it doesn't read at all right to me. Anyone care to critique it?
A better explanation is reactive loads such as capacitors and inductors cause the flow of current to fall out of sync eith each other. This causes more current to flow that is actually being consumed as a load.
Large extra currents running between a load and the power generator causes plain old resisitive losses, which can't be billed to anybody, plus it will cause distribution equipment like transformers to overload quicker, or have to be swapped for larger units capable of the required VA load. Notice the use of Volt Amperes, not watts.
So, surprise, the power company likes to cancel out these reactive loads to increase efficiency and stability of the power grid. More customers are running motors and transfomers than capacitor banks so the power company will add power factor correction capacitors near the load. This is also done inside facilities or switching power supplies these days.
And, the load's dominant inductance is compensated by capacitor banks scattered all over the network, and generally set up for remote switching as the load varies. Fluorescent lighting and motors are badly inductive (lagging loads) and outweigh the occasional capacitive (leading) loads.
Well, not in the sense of a 50 or 75 Ohm transmission line from a radio transmitter to a transmitting antenna. but, long distribution lines DO exhibit transmission line behavior. Since transformers are generally wide bandwidth devices, faults and lightning strikes can inject much higher frequencies onto the lines and these can propagate and reflect through the network.
Well, a purely resistive load draws power from the network with minimal loss, as the current is pure real, with no imaginary current. Any reactive load (poor power factor) draws additional imaginary current, that the utility doesn't charge residential customers for. (They DO charge industrial customers heavily for bad power factor.) So, resistive loads are ideal for the sake of efficiency.
Reactive power has a sign, usually called leading or lagging. leading reactive power means the current leads the phase of the voltage. A capacitor DRAWS leading current, it does not "GENERATE" it. Inductors draw lagging power, the current lags the phase of the voltage.
So, your quote is not totally "wrong" but it is quite badly WORDED.
I went round and round with this one and never really came to an adequate conclusion. At least some of the following assertions is wrong as many of these observations on the topic are contradictory.
If the xmitter output were NOT 50 ohms, then a transmitter with zero output impedance will produce the same output voltage with any resistive load. Unless there's some kind of feedback mechanism operating, such as an ALC (automatic level control) zero output impedance means a perfect voltage source. Want twice the power output? No problem, just terminate the transmitter with half the load resistance. Since the transmitter output is allegedly not matched, and there's no coax cable involved to produce standing waves, then there's no need for the load to be matched to anything.
With no ALC and no load on the transmitter output, the transmitter will produce some output voltage. Ignoring reactances and considering only resistances (which are the only things that dissipate power), the transmitter output impedance could be any resistance. With no output current flowing, the output voltage is at the maximum voltage. If I attach a 50 ohm dummy load directly to the transmitter output, with no coax cable in between to confuse things with standing waves, I should see a 2:1 voltage divider between the 50 ohm transmitter output resistance and the 50 ohm load resistance. That results in half the voltage or 1/4th the power output. However, that doesn't agree with conventional wisdom, which suggests that half the power is dissipated in the load and the other half in the source.
I also find it odd that I've been designing power amps for years with impedance matching transformers or microstrip lines between the low output impedances of the RF output stage, and 50 ohms, just so I can attach a 50 ohm in and out RF lowpass filter. If the output impedance of the amp were something other than 50 ohms, the filter input would be mis-terminated and the LP filter ripple would be awful.
There's obviously something wrong in this mess, but I don't see it. I'll see if some measurements with a real transmitter show anything useful.
I beg to differ. When the 60 Hz wavelength is 5,000 km, there will be standing waves at half wave intervals or 2,500 km. If you have a very short transmission line, you'll never see the standing waves. I'll admit that they're present, but not in any form that could be easily measured.
Bah Humbug. (T'is the season).
--
Jeff Liebermann jeffl@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558
ALC is irrelevant. Output impedance is instantaneous, not time-averaged.
I'd be interested in the measurements--especially calorimetric ones.
You don't have to go as far as a node or antinode to have a standing wave. A standing wave is
V = A sin (kx + omega t + phi1) + B sin (-kx + omega t + phi2)
which is well defined at every point for any choice of A, B, phi1, and phi2.
With knobs on. ;)
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
That's the core of the issue here, I'd certainly agree on re-reading the passage. The energy is cycling back and forth between an inductive or capacitive load which isn't able to absorb any of it (if it's in quadrature) and the generator. The wiki text states that reactive power is "consumed" which to my mind is nonsense since reactive power doesn't get used-up or "consumed" in the same way as active/real power does. It would have been better if they said "temporarily stored in the form of a magnetic field in the reactive load" instead IMV.
I only mentioned ALC to avoid having someone suggest that my voltage source and low output impedance simplified model of an RF output stage was being output limited by a functional ALC. Just ignore ALC, turn if off, or run power levels well below the ALC threshold.
I'm not averaging impedances, although I'm tempted to try it with a square wave, where the output impedance is quite different during transitions and during constant output voltage. Pretend I didn't mention this (because I'm suppose to be fixing machines and doing year end billing, not ranting on Usenet).
Sigh. I have most of what is needed to build an RF calorimeter. Or, I could burn $300 on one that works: But, why would I need a calorimeter? It makes sense if I want to measure power independently of modulation characteristics, but for this exercise, it's strictly RF carrier with no modulation. A voltsguessser and known load resistance should suffice.
All I'm implying is that if I measure the voltage along the transmission line (even if it's 5,000 km long), and I see nodes and antinodes characteristic of standing waves, there has to be some VSWR in the system. Conversely, if there's VSWR, there will be standing waves with nodes and anti-nodes along the transmission line. However, if the wavelength is rather long compared to the transmission line length, they're negligible and I can safely pretend they don't exist.
I decoded that to mean you believe either my rant or my dismal holiday spirit to be rather extreme. Both are probably true, but please consider that someone has to play E. Scrooge or there would be no story. It's no fun being a caricature that is NOT extreme.
Incidentally, I have a pen and ink drawing of me as a caricature of Ebinezer Scrooge on my wall at home. If I ever need to act dour or depressing, I have only to look up at the drawing.
--
Jeff Liebermann jeffl@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558
Assuming it's long enough. Transmission line effects become important at line lengths well below lambda/4, however. Short coaxial stubs look like capacitors if they're open at the far end, and inductors if they're shorted.
It's sort of a UKism for "Yo' momma". ;)
The guy I shave with thinks I'm ugly enough already. ;)
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
RF amplifiers are designed with matching networks to convert whatever the device output impedance is to the line impedance. The device will have both a reactive (usually capacitive) and resistive component which depends on signal level and bias. The strategy is to cancel the reactive component (usually with an inductor or transmission line segment) and complete with a real to real match - either with a filter section or a transformer. If one measures S22 with a network analyzer the amplifier will look like a resistive load and will absorb reflected power (though not always happily - wrong phase -> unstable or cooked).
Measurement of the output impedance of a power amplifier is not done with a network analyzer. The load-pull method uses resistive terminations of several mismatched values presented to the PA output through a phase shifter (different lengths of cable for frequencies where that's practical). The forward and reflected wave amplitudes can be run through a bit of arithmetic to extract the PA output impedance.
A tech at a former employer did that. I vaguely recall that the damage to the network analyzer was in the thousands of dollars. Network analyzers are a very bad way to measure output impedance.
True, but it fails to address the original contention by Phil Hobbs: "You've fallen foul of Joerg's favourite fallacy: transmitters do _not_ have 50 ohm outputs. The antenna needs to be matched to the line, but the transmitter doesn't. Otherwise you could never have a transmitter efficiency as high as 50%."
Note that we are talking about transmitter efficiency, which generally means that something is converting power supply DC into heat instead of RF. The reactive components of the various impedances involved do not dissipate any power and are assumed not to radiate RF. Only the resistive (real) components contribute to heating.
I think life will be simpler if we assume that the reactive components are conjugate matched and are therefore eliminated from the efficiency calculation, leaving only the real resistances. While the reactive components will need to be resurrected in order to build a working system, I believe they can be temporarily ignored for an efficiency discussion.
--
Jeff Liebermann jeffl@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558
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