If you assign "positive" to "generate" and "negative" to "consume", it works. The wording isn't strictly correct but it does make some sense, without going into any unreal math. Don't try to read too much into such explanations.
It's not a fallacy if he actually built RF amplifiers using triodes. ;-)
Output impedance of an active system can be arbitrary; audio power amps are traditionally ~0 ohms, RF generators 50 (or 75, or 600, or..) ohms, transconductance amps (most RF amps, pretty much any active device besides a triode) ~infty ohms.
A friend of mine designed and built an audio amp with continuously variable output resistance (i.e., varying the NFB source between 100% voltage feedback and 100% current feedback). You get interesting frequency response curves (usually horribly lumpy and resonant ;-) ) when trying that with different speaker designs.
Tim
-- Seven Transistor Labs Electrical Engineering Consultation Website: http://seventransistorlabs.com
Wrong. Unless the capacitor has a lousy dissipation factor, or the inductor has some series resistance in the windings, neither the capacitor or inductor is going to generate or consume anything. Reactances do not dissipate power, only resistances do that.
Yep, in a perfect world. If the capacitor has a dissipation factor problem, and the inductor has some series resistance, the currents are not going to be equal. What's important is the phase of the currents (relative to the voltage across the line). The only time the reactive phase currents cancel is at resonance (which is the definition of resonance).
In my mis-spent youth, I worked for my father in his lingerie factory keeping about 200 active sewing machines alive (and pushing a broom). Southern Cal Edison had some nasty penalties if we let the power factor get out of hand. SCE helped considerably by putting PF correction caps on the poles. The rest was up to us. The problem was that the load (mostly sewing machines and fluorescent lights) were not constant. We had to monitor the power factor and switch in banks of capacitors as necessary (roughly weekly). Switching caps under load was risky, so we would yell for everyone to take a break, add or remove caps, and continue. Zero crossing switching was a dream at the time. No problems, except when a sand filled 300A(?) fuse exploded about 2ft from my head for reasons unknown. For monitoring power factor, we had a magic meter:
Wrong. Even if the power factor were totally screwed up, a typical inductive load, such as motors, xformers, electromagnets, etc will draw the same amount of power as when the power factor is a perfect
1.0. Your electric bill will not change because the common electric power meter measures real power (watts), not apparent power (volt-amps). Even if you have a totally inductive or capacitive load, with a power factor = 0, you will still get the same bill in the mail.
I guess I should mention there is one big difference between correcting for VSWR with RF, and correcting for PF with AC power. With RF, you're trying to conjugate match both the reactances as well as match the real part of the impedance. With AC power, the PF correction tries to produce a purely resistive load (thus adjusting the phase between the voltage and current to zero). Because the source, line, and loads are intentionally mismatched in AC power, there is no attempt to match the load to the line or to the source.
-- 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
Correct. This is the case with most conventional (CV) sources. Batteries, AC mains, audio power amplifiers...
^ ^ ^ ^ ^ ^ ^ ^
Because you weren't drawing 1x power output before, you were drawing 0x power before. Current was zero dot nil, remember? ;-)
If the transmitter has an internal resistance characteristic, then you always get maximum power output when it's matched. This might be undesirable for other reasons, though.
Audio power amplifier: suppose Ro = 30mohm. Okay, we attach a 30mohm... hunk of wire, across the output terminals. Let's turn this up slowly. Oh man, getting awfully hot just from a few mV input... *click*, oh, the amp's protection circuit tripped. Seems it was drawing too much current.
This is where you meet a disconnect, between ideal linear system theory and real nonlinear systems. The real system does indeed exhibit considerably increased power output (and thus power gain) into such a load, but only as long as it is able to deliver the voltage and current. As soon as one or the other starts banging its head against the wall, the theory falls apart.
Likewise, if we take a normal everyday 2N3904, and bias it with an inductive load in parallel with a load resistance, we have an AC coupled amplifier with a very high output resistance. It looks like a Norton source, where the only resistance is due to Early effect. The load resistance is dominant, and power output and gain continually increase as load resistance is *increased*. (An RF amplifier can't do this forever, because sooner or later, Ccb feedback turns it into an oscillator. Neutralization helps though.)
This is usually the case with RF amplifiers (which range from pentodes to MOSFETs), but made complicated by other intrinsic device properties, and package parasitics. The s22 parameters of RF MOSFETs in the 500MHz+ range tend to be complex. Below (at least for the well-behaved flat leaded packages -- not like, shitty TO-220s and stuff), they tend to look capacitive. Indeed, you can read the Cdss and Req (actually coming from 'channel length modulation constant' lambda) off a J/MOS FET's s22 or y22 plot.
As a somewhat aside:
A switching supply might be intended for a given load (say, 160V peak and
2A, for a ~100W forward conveter, perhaps?), but the [average, equivalent] output resistance is very small, perhaps a few ohms (the Rds(on), or equivalent, of the switches).But at frequencies on par with the rise time, it is often valuable to match to that impedance: not so much for power matching purposes, but for minimizing the impact of parasitics. Consider: if the impedance is very low (large Cdss, small Lstray, in the switching loop), then peak currents will be higher than the load currents, and the devices experience additional strain. If the impedance is very high (large Lstray, small Cdss), the voltage overshoot will be greater than the supply voltage.
If sqrt(Lstray / Cdss) = Vsupply / Iload, then both are matched to the supply and load requirements, and the parasitics are, well.. equally bothersome, rather than being more of one or the other (V/I).
The rule of thumb "eliminate stray inductance" has its origins in attempting to reduce _the time constant sqrt(L*C)_ below the switching speed. That was fine, back in the days of ~200ns bipolar, and fat, high capacitance DMOS. With ~1ns GaN devices being introduced these days, this is now completely impossible!
So, even in extremely nonlinear places, like switching converters, matching still pays off. It also gives you some flavor for what resistors you should need in your snubber, and what kind of equivalent loads they should present. Especially if you want to, say, use a regenerative snubber to save power!
This implies a couple of possibilities:
There are standard filter designs out there for ~infty source impedance (filter starts with parallel element) or ~0 source (filter starts with series element). Or other matches. If you can model your RF output transistor as a capacitive Norton source (i.e., CCS || Req || Cdss), then the best filter is one which terminates into R_L on the load end (R_L being the maximum power load for the transistor, usually Vdd / Ipk, or a few times Rds(on), or..), and which terminates into Req (~infty) on the source end, and which has the first capacitor equal to Cdss (so you don't need to add any extra, and since you can't remove any less!). The cutoff frequency or bandwidth will be on the order of 1 / (2*pi*R_L*Cdss) -- independent of what the band center is at (whether it's lowpass/baseband or bandpass).
Riddle: suppose you have a class D power amplifier, but you can't afford the space or expense for a proper lowpass filter. But you have EMI problems over 5MHz say. Can you filter just that? Sure: use a filter designed for ~0 ohm source, and whatever the load is (8 ohms?). The load might not be 8 ohms at that frequency, of course, and you may need to add a stabilizing R+C across the output to provide it. But anyway, this is how you'd filter a low impedance square wave, while dodging harmonics (i.e., cutoff frequency > switching frequency -- just to "take the edge off").
Tim
-- Seven Transistor Labs Electrical Engineering Consultation Website: http://seventransistorlabs.com
Well, if you include the PFC components in the source, you'd get a conjugate. Well, sort of. Because a voltage source is a voltage source, and you can't tell there's a capacitor in parallel with one just by watching the terminals. But the amount of reactance added for PFC is certainly complementary, all the same.
Anyway, if you tuned the reactance out of both the transmitter and antenna sides, the transmission line will see the least possible Irms, and VSWR =
1.00, which is what the power company wants. They just gauge it by Irms, because that's what makes lines heat up and sag.Tim
-- Seven Transistor Labs Electrical Engineering Consultation Website: http://seventransistorlabs.com
The keyword in the sentence is "reactive".
No, it's not wrong. The fundamental mechanism for controlling the power factor is capacitors or inductors and again, the keyword is "reactive". Nothing was said about billed electricity. Even there, while consumers aren't charged for reactive power, industrial customers are.
Nope. Efficiency is not controlled by the device type (triode, tetrode, pentode, etctode) but by the operating class of the amplifier based on conduction angle: Class Max theoretical efficiency (approx) A 50% B 78.5% C 99%
However, Joerg's comment has nothing to do with the class of operation. It's based on the simple observation that a perfect voltage source (the RF output xsistor), in series with 50 ohms (the output impedance), driving a dummy load of 50 ohms, will dissipate equal power in the series output resistance as in the dummy load. Since all the RF output power is in the dummy load, the overall efficiency is 50%.
Yep. As I mumbled elsewhere in this thread, I have to match the power amps fairly low output impedance to the input impedance of the low pass filter found in most radios necessary to keep the FCC happy. The input impedance of this filter is usually 50 ohms, which would look rather ugly if the output impedance of the PA was not matched to this
50 ohms.-- 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
Nope. The key word is "generate", which on my planet means to convert some form of kinetic energy into electrical energy. I have never seen a capacitor do that[1].
Methinks we're both saying the same thing. I guess equating billed electricity from the power meter, with consumed power (watts) was a bit obscure. So, that there's no misunderstanding, you can adjust the PF all you want, it's not going to reduce your electric bill or measured real power.
[1] Yes, I know about electret microphones and piezoelectric dielectrics on MLCC caps. Such dielectric materials are not common in big PF correcting caps or in RF amplifiers.-- 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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I intend to do that one of these days. There has been some discussion among audiophilia about why some like the tube sound better and this has got to be part of it.
Just how tube amps clip cannot be the whole story because they don't clip t hem. Just frequency response can't be it because some of them essentially h ave flat response as good as solid state amps. Can't be distortion because they are quite low, even in single ended amps. (like using 6BQ5s, they are getting popular among the tube guys)
And this is not confined to what some here call "audiophools", they still s ell guitar amps with tube output. Some are all tube, some are solid state d riven tubes and actually I have seen some with a tube preamp and solid stat e output. Usually those are MOSFET.
What's more, no matter the output I have seen them use a mixture of voltag e and current feedback. About a month ago I had one intermittent and couldn 't find the problem anywhere expected until I found the sense resistor. Cur rent feedback in solid state amps, I would presume to make it sound like a tube amp. In some that feature is switchable.
Why ? I am at a loss here to understand why they would give a shit. Bothers the transformers or something ? Doesn't seem to make sense. Of course that is a deficiency on my part so someone will probably be along to call me an idiot, but if they explain it satisfactorily that would be OK.
No, that wasn't what I was winking about. Triodes have anomalously low output resistance, so that Ro ~= R_L. At least for a class AB2-ish amplifier. (You tend to use higher loads, about 3*Rp, for class A, and lower loads for C.)
A triode is something like a JFET with the gate strapped to the drain with resistors, you know like the variable resistor circuit. Except it's an internal phenomenon governed by electric fields, not external resistor dividers.
Like I said, the Zo of an active system can be anything. Most of our op-amp based systems are voltage source type, while most of our RF systems are current source type. The Zo of the amp is only a characteristic of Vo / Io, it has nothing to do with the efficiency of the amplifier (yet another discrepancy from linear circuit theory!).
The joke is just that triodes happen to have a low Rp among amplifiers, and tend to use loads near Rp. So I'm right, but for the wrong reason.
And that Joerg has a history of using tubes for RF power, and would likely understand the joke too. :)
Tim
-- Seven Transistor Labs, LLC Electrical Engineering Consultation and Contract Design Website: http://seventransistorlabs.com
it woeks on immaginary power, not real power.
-- \_(?)_
that's a very narrow definition, dicionaries must be brief on your planet.
-- \_(?)_
Yes, I think it's just different names for essentially the same things.
In RF the lines are expressed as infinitesimal small combinations of R, L and C, which form the transmission line along which the wave travels. The coefficients found in the EM equations lead to the term 'reflection'.
In Power Generation and Distribution normally the network elements are lumped together and different equations are used and consequentially the term 'reflection' is not found but 'power factor' is used.
But I think basically it's the same thing.
joe
Because a low power factor means a high RMS current, therefore higher dissipation in the power lines.
joe
Power companies consider the 'apparent power', that is the voltage you measure with your multimeter over the load X the current you measure with another multimeter (or the same if you want to economize) through the load.
The current can be split in two components: one that is in phase with the voltage over the load and one that is in counter-phase with that voltage. The sum of the two is the total current, which is always greater than (or equal to, if the load is perfectly resistive) the in-phase current.
Mind that in the complex plane the in- and out-phase currents are perpendicular.
The power which is associated with the in-phase current is called the 'active power', as this is the 'real' power consumed by the device, and which has to be generated in the generator station using real fuel.
The out-phase current gives the 'reactive power' if multiplied with the voltage. This current is useless as seen from the utility's perspective, and even consumes real fuel as it leads to resistive power losses along the lines and transformers. Therefore you have to pay for it.
The 'apparent power' is the root of (active squared + reactive squared).
So, the capacitor, with it's I = C (du/dt) causes an out-phase current, hence 'generates' reactive power.
joe
The electrical generator generates 'active' power or real power. The capacitor 'generates' (or gives rise to) a flow of reactive power, which you also can compare with the elsewhere mentioned 'reflected power' analogues to RF-technology. This reflected or reactive power has a current associated with it which gives rise to increased resistive losses in cables, transformers and generators and therefore needs a higher investment in equipment (over-rating) and also leads to more fuel consumption, as the result of the reactive power is a real power loss. Only some percentage, but nevertheless, power companies will bill you, if you're an industrial consumer and they think your current is too much out of phase.
joe
Actually I think it does. :)
As well as you can draw up balances of 'consumed' and 'generated' active power, so you can with reactive power. Power transmitting and distributing companies do this all the time.
An induction motor 'consumes' reactive power which has to be either generated by some capacitor(s) between it and the generator, or the generator has to get a higher exciter current in order to have it generated there. (Solar power plants would have their power electronic converters--I think the Aglo-Saxons call it 'DC/AD-inverters'--loaded with an excess current.) If that would be ignored, voltage stability problems would immediately occur.
joe
As this 30mOhm has been obtained by a negative feedback loop, why not adapt the loop such that the Ro = 8 Ohm for instance? Then how would your argument go?
joe
Highly efficient transmitters aren't running Class A, so the instantaneous output impedance of the final isn't constant within a cycle.
It's low and mostly resistive during conduction, and almost purely reactive otherwise.
Cheers
Phil Hobbs
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