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In Reply to: Re: You must mean milliohms posted by Mahatma Kane Jeeves on February 18, 2005 at 06:06:41:
Thevenin's theorum and Norton's Theorm are a duality. Thevenin says a source can be replaced for mathematical anylitical purposes with an ideal voltage source having zero internal impedence in series with an internal complex impedence (resistance plus reactance). Norton's theorem says it can be replaced with an ideal current source having infinite internal impedence in parallel also with an internal complex impedence. When people talk about the output impedence of audio amplifiers, they invariably are talking about the series impedence of the source referred to in Thevenins theorem. The two theorems are interconvertable because they are two different ways of looking at exactly the same thing although in the amplifier you specified, it is more convenient to look at Norton's theorem to analyze it. It is an unusual and possibly unique exception because of the method of controlling the output. In theory for example, if the load were to be disconnected, the output voltage would go to infinity trying to continue to drive current through an infinite load. Any amplifier (or any other circuit for that matter) having an internal (series)impedence (Thevenins equivalent) which is ten times the load impedence, cannot have an efficiency of delivering power to the load of more than 10%, that is 90% or more of the electrical power delivered by the power supply will be converted to heat internally. The example points out an intresting thing about so called "power" amplifiers. They don't actually amplify power, they amplify voltage and their final stage is capable of delivering a lot of current (and voltage) to a loudspeaker so it is a powerful signal. In the case of your amplifier, it actually amplifies current, not voltage and the output stage can also deliver both adequate current and voltage to power a loudspeaker system. In a sense it is a tradeoff between the advantages of voltage versus current amplification. Because the load is reactive, neither can produce a transfer of power to the load independent of frequency and as pointed out in the original technical paper, special consideration has to be given to the design of the loudspeaker drivers and crossover networks to exploit the advantage current controlled amplifiers have to offer. They need series rather than parallel networks. Most loudspeakers on the market fail to qualify by this criteria very badly.
Follow Ups:
"Any amplifier (or any other circuit for that matter) having an internal (series)impedence (Thevenins equivalent) which is ten times the load impedence, cannot have an efficiency of delivering power to the load of more than 10%, that is 90% or more of the electrical power delivered by the power supply will be converted to heat internally."This, of course, runs contrary to experimental fact.
The problem is that you're trying to extend the model too far. A Thevenin/Norton equivalent will allow you to predict how the source and the load will interact, but is not useful for determining efficiency of the source. Again, think of the Norton model which suggests a HIGH efficiency. And just as falsely. Do the experiment, which will convince you much more rapidly than I can.
You are correct in that using a current source amp will force you to rethink crossovers- topologies are NOT the same. And in the real world, CS amps are most useful for single-driver full-range systems. I had my first experience with them listening to Nelson Pass's Lowther-based Kleinhorn. Marvellous. I'm somewhat poorer than Nelson, so a Fostex FE166 is more my speed.
It feels soooooooooooo good when you stop hitting your head against a brick wall.
Then you can find another wall to hit your head on.My own results with a modified ST-70, EL34 output tubes, Curcio regulator, SYclotron driver stage. *Results*, not speculation:
With feedback, 8 ohm load, source Z = 0.5 ohm. Vout = 22V before clipping (about 30 watts). Pd in output stage = 55W.
No feedback, source Z = 150 ohm. Vout = 22V before clipping (about 30 watts). Pd in output stage = 55W.
You don't even need to learn AC circuits. DC will be enough to show you your mistake in this thread. Little more than Ohm's law.
The area where the curve flattens out and runs more or less horizontal is your 'constant current range' with an effective impedance of hundreds to many thousands of ohms, yet the current passing through the device can exceed several hundred milliamperes, even at low junction voltages. Such devices can make perfectly good and efficient voltage amplifiers with enough feedback.That's what *I'm* talking about.
You don't even need to learn AC circuits. DC will be enough to show you your mistake in this thread. Little more than Ohm's law.
No, it's not little more than Ohm's Law.
You seem to keep looking at output impedance as if it were some sort of literal resistance dissipating power. Well, that would be the case if you took a voltage source amplifier and just slapped a resistor on its output to give it a high output impedance, but that's not how to look at a proper current source amplifier.
Just consider a simple single transistor current source. Its output is from the transistor's collector. The impedance looking into the collector can be very high. Upwards of meghoms.
If we look at this as "little more than Ohm's Law" then a current source couldn't deliver but a tiny amount of current without having to dissipate tremendous amounts of power.
If the collector impedance were just 10k ohms, and we wanted to deliver 1 amp of current, that'd 10,000 watts in the "little more than Ohm's Law" way of looking at it.
But the high collector impedance isn't like a high value resistor being driven from a voltage source. Instead, the power dissipated by the transistor is essentially the voltage across it times the current flowing through it, not the square of the current flowing through it times the collector impedance.
Bottom line, a proper current source amplifier (i.e. not a voltage source with a resistor tacked onto its output) can be just as efficient as a voltage source amplifier.
se
It's a nice little essay worth writing for pedagogical purposes. I might do it if I can get more spare time.In brief: If we represent the amp as a Norton equivalent, then by an equally incorrect use of Norton's theorem, we determine that efficiency of my amp is 95%. Which one is an incorrect theorem, Norton or Thevenin? Clearly neither. Thevenin and Norton equivalents are just that: equivalents, not reality. They are useful and rigorous as equivalents when you use them for their derived and intended purpose- to predict the effects of changing a load on the applied voltages and currents.
My ST-70 agrees, since it has not melted down its output tubes in a vain attempt to dissipate 300 watts. As an aside, dynamic error in source impedance is reflected as good old fashioned distortion with CS drive. The ST-70 in question had unexceptional distortion figures (i.e., on the order of 1% at 10 watts out, 1kHz) when loaded with an 8 ohm resistor.
My preamp test circuit agrees, too (http://www.diyaudio.com/forums/showthread.php?s=&threadid=48502&highlight=). The source impedance of the current source is 2M and it's a 10mA CCS. Somehow, I don't think it's dissipating 200W, which is what the Thevenin equivalent would predict.
"You seem to keep looking at output impedance as if it were some sort of literal resistance dissipating power"Why do you think power transistors get hot when current flows through them? Hint, I squared times R. What is R? the junction resistance. The transistor doesn't know or care if thaey are in a current source or a voltage source amplifier. They work the same way in either kind of circuit and generate internal heat for the same reason. It's the same reason the plate on a vacuum tube glows red if it's overdriven to high plate current. I squared R. Don't believe it? Put a fresh 9 volt battery in your pocket when it is full of change. The battery gets hot because of the current flowing through its internal resistance. I squared R. (Don't try it with an automobile battery unless you want your wife to collect on your life insurance policy. Although it's only 3 volts more, R is much much lower and I will be enormous.)
"Just consider a simple single transistor current source. Its output is from the transistor's collector. The impedance looking into the collector can be very high. Upwards of meghoms."
Not when the junctions are forward biased. That's when the transistor is delivering power and at that time it is very low.
Before you take a refresher course in AC electricity, try one in basic DC electricity first. Perhaps you will be luckty and sit next to Mahatma.
"a proper current source amplifier (i.e. not a voltage source with a resistor tacked onto its output) can be just as efficient as a voltage source amplifier."
I never disagreed with or challenged that statement.
Why do you think power transistors get hot when current flows through them? Hint, I squared times R. What is R? the junction resistance.
Um, so? What's that to do with the impedance looking into the collector?
"Just consider a simple single transistor current source. Its output is from the transistor's collector. The impedance looking into the collector can be very high. Upwards of meghoms."Not when the junctions are forward biased. That's when the transistor is delivering power and at that time it is very low.
Um, during normal operation, the base-emitter junction is forward biased and the base-collector junction is reverse biased. That's why you have such a high impedance looking into the collector and a low impedance looking into the emitter. When looking into the collector, you're looking into a reverse biased diode. Not a forward biased diode.
Before you take a refresher course in AC electricity, try one in basic DC electricity first.
That's the problem. Basic DC electritity seems to be all you know anything about. Time to read a few more chapters and move up to solid state electronics.
Perhaps you will be luckty and sit next to Mahatma.
Cool! Mahatma's got a Ph.D. in physics. I'll just copy off his papers.
"a proper current source amplifier (i.e. not a voltage source with a resistor tacked onto its output) can be just as efficient as a voltage source amplifier."I never disagreed with or challenged that statement.
Did you not say that an amplifier with a 150 ohm output impedance driving an 8 ohm load could never be more than 10% efficient?
An amplifier which has a series output impedence (Thevenin's equivalent) of 150 ohms and drives an 8ohm load cannot be more than 10 percent efficient, I don't care what class the output devices are biased at. That's not an opinion, that's a fact.
Are those not your words?
se
Any circuit whether a transistor amplifier, a battery, a transformer which has an internal series impedence more than ten times its load cannot be more than ten percent efficient delivering power to the load.Take my advice, don't quit your day job.
Any circuit whether a transistor amplifier, a battery, a transformer which has an internal series impedence more than ten times its load cannot be more than ten percent efficient delivering power to the load.
Well then you just keep going on believing that and the Tooth Fairy will leave a quarter under your pillow.
se
Understand the difference between a model and reality. The resistor in a Thevenin source is not a real resistor, it's a model for how a circuit will act when hooked to a load. You need to explain why the Norton model gives you a different answer for efficiency.If you had been one of my electronics students, I would have flunked you- not for being wrong (that's OK, I'm wrong 10 times an hour), but for being wrong and not bothering to do an experiment to find out why.
"Understand the difference between a model and reality"A real professor of electrical engineering doesn't challenge a model, he explains it and its limitations. So far, I have not encountered a case where Thevenin's theorem is invalid and so far you haven't explained why you think it failed.
Either your statement that your amplifier's output impedence is 180 ohms is wrong because it is not the series impedence or it varies and decreases during its operating cycle as it delivers current (the most probable source of your error) or your statement that the amplifier is more than ten percent efficient cannot be true because most of the voltage drop is internal and not across the load. Just look at the output of any battery as it discharges and the truth of Thevenins theorem becomes obvious.
"If you had been one of my electronics students, I would have flunked you..."
Where do you teach, Whats-a-matta "U"? In the unlikely event that the gods saw fit to put me in a classroom and you were the electrical engineering professor, you would never get the chance to flunk me because after the first lecture, I would have demand my tuition money back.
Leave it to me to argue with a guy who has a one of a kind esoteric off the wall amplifier and uses it as an example to confuse some poor shnook who posted a question on an internet board. A real professor of electrical engineering would never do that because it doesn't teach anyone anything, it is merely designed to confuse. I ought to have my head examined for wasting my time.
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