Reflected Power

Reactive power (which is the power coming back to the generator) causes the power companies headaches for 3 main reasons.

Firstly, the increase in current demands higher ratings for all the current-carrying components in the distribution network; cables, transformers, fusing, switchboxes etc.

Secondly, it creates problems in regulating the outgoing line voltage and getting the compensation wrong can result in brownouts and in extreme cases blackouts.

Thirdly, increased transmission losses, since they are dependent on the current rather than the power and more current = more losses.

Paradoxically, power companies actually desire a certain (predictable) amount of reactive power coming back at them, but that's a bit outside the scope of this thread.

Then the amplifier (if it's designed for maximum power into 8 ohms) will have a normal amount of gain / power sensitivity, and will reach clipping/limiting at the rated power level.

Other loads will deliver less power, and this will be true under both conditions: in the linear range (where the power transfer theorem applies), and in clipping (where either V or I runs out first).

In the linear range, you'll have a nice smooth parabola of P_o vs. R_L, but in clipping, it'll be a somewhat straighter curve, or more lumpy anyway, due to harmonic distortion. (If the amplifier clips in a simple fashion, with no phase reversal or recovery effects, then the worst case clipping waveform is a square wave, which delivers twice the power of a sine wave having the same peak voltage or current amplitude.)

Kind of more like how you might naively expect a "linear" amplifier to behave (i.e., "linear with exceptions", so to speak), where there's a sweet spot in either condition.

So the fact that Zo can be designed arbitrarily, just has the effect of spreading the max power curves for linear and nonlinear operation.

Tim

--
Seven Transistor Labs 
Electrical Engineering Consultation 
Website: http://seventransistorlabs.com

Yup, more of the same miswording, by somebody who really didn't understand what they were writing about.

Jon

Here's the simple answer:

- Say a light bulb draws 120 watts at a line voltage of 120 volts. Great, it draws 1amp. Power is amps times volts so 1 amp time 120 volts is 120 watts. Perfect. Resisitive loads behave this way, the same as if they were running off DC.

- things with coils do NOT behave this way. Take for example a real induction motor I have. It's rated 1/3 HP and at 120 volts has a nameplate current rating of 7.3 amps. At first glance it sounds like the motor will consume 120V x 7.3A for 876 watts. If it can output 1/3HP (or 249 watts of work) somewhere else, like running a pump. The motor is using 876-249 or

627 watts too much power and must be giving that extra off as heat.

Not true at all though. My motor is not a 627 watt heater when running a pump.

While it may be drawing 7.3 amps, it's actually giving some of that back during each half cycle which is why it's really only drawing a couple hundred extra watts over the 1/3HP and not really using 876 watts, but

876 VA. So yes, motors, or transformers and capacitors actually store energy and return it back to the power company. This back and forth and worthless activity can be explained in another way.

So you run a restaurant and expect 50 pounds of potatos delivered each day.

If a perfect world you delivery company drops off one 50 pound bag each day. Easy, like a pure resistive load.

If they had a lower power factor, they might deliver you two 50 pound bags, you keep one, and reject or send the other back. It's alot more work being done to get the net delivery of just one bag per day. You don't have to do anymore work, the delivery guy does and the truck fills up faster and gets more wear and tear on it.

Like I said before, inductive and capacitive are in sense inefficient like a stupid delivery service. There's lots of extra back and forth activity with nothing to show for in the end.

[...]

This is indeed more or less the way a transmitter final is designed. The remaining step is to design a matching network that makes the

50 Ohms of the cabling and antenna look like that load resistance. The matching network usually has the topology of a low-pass filter. Of course, the final device has to be up to the task, i.e., it must withstand the voltages and currents involved.

It depends. You are working with power amplifiers in the several watts or higher range? Then the output impedance is pretty low and the filter may be designed to have the desired response for a low or zero source impedance. The filter would still have a 50 Ohm input impedance in the pass band.

People are often confused by filter impedances. One should distinguish between the source and load impedances that the filter was designed for and the impedances one would see looking into the input or output of that filter. Those are entirely different things.

Jeroen Belleman

On 01/12/15 04:14, Jeff Liebermann wrote: [...]

I haven't seen anyone mention that PF correction is often also done to suppress harmonic currents.

Jeroen Belleman

I don't think so. Actually the process of PFC can give rise to unwanted harmonics which then require additional filtering.

Sigh. So many words for essentially the same thing. So far we have real, imaginary, reactive, effective, apparent, and now active power. I'm sure I probably missed a few other names. When I see such a surplus of mostly redundant names, I invariably find that the original terms were either inadequately explained, or generally misunderstood. On my planet, we like to call a spade a spade, and not invent fanciful names that only serve to confuse. Those who fail to comply are dealt with appropriately, in this case, boiled in transformer oil.

The analogy doesn't quite work. In power transmission, adjusting the power factor so that the alleged reactive power is zero does not match the load to the transmission line or source. In RF transmission, adjusting the load impedance (with an antenna tuner) for minimum VSWR and best match only matches the load to the transmission line impedance. The load can still have reactive components, as long as the vector sum is equal to the transmission line characteristic impedance.

If there were reflected power in a power transmission system, as one might expect from a highly inductive load presented by fluorescent ballasts, motors, and xformers, one would also expect to see standing waves on the transmission line in the form of voltage nodes and anti-nodes at half wave intervals. With the wavelength of 5,000 km at

60 Hz, it is unlikely that you'll see any such voltage nulls and peaks. The power transmission system is certainly not impedance matched between the source, transmission lines, and load. If it were, the generator would dissipate as much power as the load which would burn up the generator. So, the generator runs at the lowest possible impedance, the transmission lines at the lowest possible inductance (to reduce transmission line losses), and the load impedance is whatever happens to be turned on and running. Synchronous generators, running in parallel and in phase reduce the source impedance dramatically. Therefore, adjusting the power factor to "match" something doesn't work, because the load is a moving target, the transmission lines are almost random, and the source depends however the grid is configured at the moment.

SCE (Southern Calif Edison) was not very nice about this. Instead of singling out the single customer that was not bothering to fix their power factor, they would send a giant bill to everyone in the general area, and let them fight among each other. I must admit that it worked, but did create some rather ill will among neighboring factories.

--
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

Finally, some one gets it!

Phase distortion caused by reactive loads not properly loaded.

Jamie

In the past, the typical range was 5 watts to several thousand watts. Mostly marine HF and VHF radios, with a smattering of land mobile, datacomm, and NMR drivers. As I recall, the typical bipolar RF transistor will have an output impedance of 50 ohms at about the 250 mw level. As the power output goes higher additional bipolar devices are connected in parallel inside the power transistor, thus lowering the output impedance.

We never did that. Most power amps had a 50 or 75 ohm output impedance as provided by either an ugly RF transformer made from ferrite beads, brass tubing, and PTFE wire, or a microstrip impedance matching section on a PCB (depending on frequency). Smaller power amps, usually VHF and up, had room for the low pass filter on the PA board, so those were impedance matched to the RF power xsistor without going through 50 ohms. The problem with doing that is that at low impedances, the inductors tend to grow large in order to handle the higher currents.

I once did an HF PA->LPF->antenna combination, that was designed to eliminate the usual antenna tuner. The antenna was electrically small, so the antenna impedance was low. With a low PA output impedance, and a low antenna impedance, all that was needed was a low impedance HF filter and it might be possible to eliminate the usual antenna tuner. Ignoring the huge size and cost of 10 ohm coax cable, I built a prototype, which sorta worked. The problem was that the very high currents now going though the coupling caps would unsolder the caps, and the high currents would saturate the toroid cores. Project was shelved when I dragged in a sample of what 10 ohm coax cable might look like.

Unless I missed something here, the input and source impedances are the same, and the output and load impedances are also the same. They have to be matched or the filter ripple and losses would not be optimum. It is possible to make the filter asymmetrical and have different input and output impedances, but I didn't do that too often (except when they were on the same PCB).

--
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

It's an explanation without the use of imaginary numbers. The language works for what it was designed to do.

Only if you're not billed for VARs *AND* if you neglect heating of the wires in your house (after the meter) caused by those VARs. Again, many customers are charged for VARs.

OK, we can neglect that. ;-)

I understand what's being said even if I don't agree with the way it's being said. I don't expect Ma and Pa to understand imaginary power, either (I've had that discussion with non-technical people - not worth it).

If they're sending reactive power down the line, they must be generating it, no? ;-)

...and transformers and everything else in the system. Their loss is related to current. Reactive power means more current.

No, phase distortion is cause by nonlinear loads. Reactive loads can only cause phase displacement.

** You are stubbornly making stuff up, to suit you crazy heading. Reactive and "reflected" are not equivalent.

Seems you are having way too much fun with this troll to let a simple facts interfere.

.... Phil

There are some explanations for power factor correction that can be understood by the GUM (great unwashed masses). For example, I like th horse and the barge analogy near the bottom of:

Nope. It's "imaginary power" which is mostly in one's imagination.

--
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

Yes.

joe

Then why don't they just charge for volt*amps ?

They really are allowed to do whatever the f*ck they want.

[...]

A filter can be designed to have the desired response for a zero source impedance and a 50 Ohm load impedance, say, or for a 50 Ohm source and load. The component values and possibly also the schematic will be different. Of course, looking into the source end of a properly terminated filter for zero source impedance, in the pass-band one sees the impedance of the termination.

That was my point: A filter designed for a given port impedance does not necessarily show that impedance at that port.

The same is true for a power final. It may be designed to drive a 50 Ohm load, but that does not imply its source impedance is 50 Ohms.

Jeroen Belleman