Impedance matching theory.

[jimmc101]

Originally Posted by jimmc101

If the amplifier is producing maximum possible power into the optimum load resistance then it will be on the verge of both voltage and current limiting (clipping).
Jim

Sorry there is some ambiguity in my previous post (#97), I think this removes it. But I could be wrong.

If an amplifier designed for maximum efficiency is producing maximum possible power into the optimum load resistance then it will be on the verge of both voltage and current limiting (clipping).

Jim

[ukcol]

What a great thread this is, 100 not out and still going strong!

[kalee20]

Originally Posted by anthonys radios

Sorry if I has missed this point in my quick look, but I don't think anyone has mentioned distortion, or has actually made any reference to distortion

Well, yes we have, very much so! When a waveform starts to be clipped, this is the onset of distortion.

Up to the point that an ideal valve swings smoothly down to zero current, it's distortionless. But at zero current, it stops, and then you get clipping (distortion). The same happens on the other half-cycles for voltage - the anode swings down smoothly towards zero, but stops there. And again you get distortion.

Look at jimmc101's posts, and my posts, in this thread. The whole idea of distortion is fundamental to the idea of optimum load for a Class A audio amplifier. Less than the optimum, and distortion sets in at a lower power level. More than the optimum, and again it sets in at a lower power level.

[anthonys radios]

Yes Yes of course, I was a bit tired after a 10 hour drive in traffic that should have been 6 hours, and forgot to engage my brain before letting my fingers loose on the keyboad.
However, Having stood in the corner for a few hours, I am ready to come out to play !

[anthonys radios]

If an amplifier designed for maximum efficiency is producing maximum possible power into the optimum load resistance then it will be on the verge of both voltage and current limiting (clipping).

Can I suggest a typical amplifier would be designed for maximum efficiency anyway, and therefore could produce it's maximum power output relative to it's rated distortion figure, and may not be on the verge of clipping at that point at all, since it is still operating within specification.

But producing the maximum possible power into the load needs some elaboration.

If the rated output of the amplifier was not exceeded, and therefore the rated distortion level also not exceeded, then clipping would not occur.
If on the other hand the rated output was exceeded by over driving (excessive input), then the amplifier may well produce even greater output (up the the absolute maximum possible) with exceesive distortion occuring,
(and therefore clipping)

Some cheaper or poorly designed amplifiers may well do as you say, being rated beyond there true ability.

[kalee20]

Originally Posted by anthonys radios

Can I suggest a typical amplifier would be designed for maximum efficiency anyway

Definitely not!

Designing an amplifier, once you know the requirements, generally starts with defining the circuit topology. If you were designing for maximum efficiency, you would probably go for a switched, PWM type thing.

But if you were making a MW/LW AM radio, this might be ruled out because it could interfere with incoming transmissions.

Again, even after setting out the basic topology, there could be variations and constraints. Ultra-linears, for instance, can be more efficient than plain pentodes or beam tetrodes (because the screen current contributes to the output, whereas otherwise it's lost). But, the UL may be ruled out owing to the output transformer being more expensive.

Single-ended triodes get more efficient the higher the HT voltage. But many triode amplifiers are designed at lower voltages than the valve's maximum ratings, either due to cost, or to make the output transformer turns ratio less extreme (with its associated frequency response penalties).

[anthonys radios]

Designing an amplifier, once you know the requirements, generally starts with defining the circuit topology. If you were designing for maximum efficiency, you would probably go for a switched, PWM type thing.

Please! we are talking Valve Amplifiers ! and certainly not lw/mw radios except possible the audio amplifier, and more specifically the output stage.
Very sorry but I cannot equate you post with my previous one Im afraid

[kalee20]

Well, I suppose in spirit, the thread is about valve amplifiers, but nowhere have we said explicitly that it is. In particular, the optimum load for a pentode or beam tetrode in a single-ended Class A output stage (Rl = 0.9 x Va / Ia) is the same principle as a single-ended Class A output transistor (Rl = 0.95 x Vc / Ic), such as you'd once see in car radio output amplifiers. The reason for the 0.9 and 0.95 factors is because a 'rule of thumb' for a pentode is that it 'bottoms' at about 10% of the HT supply, and a transistor bottoms at 5% of the supply. An ideal device, would of course work down to zero.

For a single-ended Class A stage, operating into its optimum load, winding up the drive once clipping has set in, does increase the output power (as you say) and distortion sets in rapidly. Assuming that the optimum load is used, and symmetric clipping occurs, the power increases, the waveform gets steeper, and the output becomes more and more a square wave. And as for this, the power is just Vpk x Ipk, and for the sine-wave it's Vpk x Ipk / 2, you could screw up a 10W class A amplifier into giving a power approaching 20W.

It's interesting to note that at zero output, the supply power (constant 20W in an ideal world) is dissipated entirely within the output device. At full-power, sinewave output (threshold of clipping), the supply power 20W divides with 10W dissipated and 10W getting to the load, and with square-wave output (again at the threshold of clipping), the 20W of supply power all gets to the load. The efficiency of a Class A amplifier is waveform dependent!

[anthonys radios]

Hi Kalee20, Implicit in your posting 108 is that for a Class A operating at the maximum efficiency possible (50%), maximum power will be delivered to the load, when the source impedance equals the load impeadance.

i.e from your statement for sinewave output.
"10W dissipated and 10W getting to the load,

In the practical case, the best compromise between power output and distortion is to match the load to the optimum value, as per the data supplied by the valve manufacturer ?

So, refering to my earlier postings, are we agreeing in posting 108 after all ?

[kalee20]

Hi Anthony,

Well, the maximum efficiency of a Class A amplifier operating with sinewave drive is 50%, but not for the reason you say. It's 50% because of the √2 relationship betwen peak and RMS of the sinewave, and this factor for the current and for the voltage multiplies out to give 2.

If you drive your 20W-input Class A amplifier with a triangular wave, at the point of clipping, you will have a maximum power of 6.667W with the optimum load. A higher load resistance - and after adjusting the input so that you're on the threshold of clipping, the power will be less. Lower load resistance - the same. The maximum efficiency is only 33%and can never be greater.

On the other hand, for a constant voltage generator with (real) internal resistance connected to the load, you get maximum power at the load when the load resistance equals the source resistance - and efficiency then is 50%, whether the source provides DC, sinewave, square wave, sawtooth, or white noise!

[anthonys radios]

HI Kalee20,
Best if you re read my posting!
It refered to a Class A stage already running at the maximum efficiency possible, 50%, (your words), maximum power, will be transfered to the load, when the source impedance, matches the load impedance. [I]
I put it to you, that you cannot disagree with this, since your own assertion, regarding 10 watts in load, with 10 watts dissipated in the valve confirms this !

[jimmc101]

Originally Posted by anthonys radios

If an amplifier designed for maximum efficiency is producing maximum possible power into the optimum load resistance then it will be on the verge of both voltage and current limiting (clipping).

Can I suggest a typical amplifier would be designed for maximum efficiency anyway, and therefore could produce it's maximum power output relative to it's rated distortion figure, and may not be on the verge of clipping at that point at all, since it is still operating within specification.

But producing the maximum possible power into the load needs some elaboration.

If the rated output of the amplifier was not exceeded, and therefore the rated distortion level also not exceeded, then clipping would not occur.
If on the other hand the rated output was exceeded by over driving (excessive input), then the amplifier may well produce even greater output (up the the absolute maximum possible) with exceesive distortion occuring,
(and therefore clipping)

Some cheaper or poorly designed amplifiers may well do as you say, being rated beyond there true ability.

Please look at the 6V6 data sheet... (post#48)
Taking the 250v / 45mA bias point.
Load resistance is given as 5000ohm and maximum power 4.5W.
The knee of the Ia/Va curve is approx 35V at 90mA

RMS current = sqrt(P/R) = sqrt(4.5/5000) = 30mA so peak current = 42.5mA
RMS voltage = sqrt(PR) = sqrt(4.5*5000) = 150V so peak voltage = 212V

This means that the valve at rated output is swinging to within 2.5mA of cutoff and within 3V of saturation.

I'd call that on the verge of both current and voltage clipping. Wouldn't you?

If you think I've chosen a special case, try 180V and 315V bias points.
I think you'll find 315V result is into clipping (12% distortion)!

Jim

[G8HQP Dave]

After all that has been said in this thread, it still seems that there is confusion between source impedance and optimum load impedance.

If you vary the load impedance either side of the source impedance of an unlimited power source then you will find that the peak power transfer occurs when they (conjugate) match. This is the maximum power transfer theorem.

If you vary the load impedance either side of the optimum load impedance for a real power amplifier then you will find that the peak power transfer may be somewhere near the optimum load impedance, but almost certainly not exactly at it. This is NOT the maximum power transfer theorem at work, but instead a case of hitting soft or hard limits in current or voltage. In addition, the distortion will vary. The optimum load impedance is the one which gives the desired compromise between power and distortion. If you change your compromise criteria then you will get a different optimum load impedance. This demonstrates that this is not a fundamental circuit theorem, but merely an example of the compromises inherent in any real engineering. The 50% maximum efficiency of a perfect Class A amplifier when driven by a sine wave has absolutely nothing whatsoever to do with the maximum power transfer theorem! It is merely a numerical coincidence.

[MichaelR]

well said

Mike

[PJL]

Agree with that, we had already concluded the effective output impedance was the plate resistance at the bias condition which is much higher than the recommended anode load.

I thought we had also agreed that the maximum power transfer theorem could not be applied here as it is a non-linear system. The valve is an active device and as my earlier post and others that have followed suggest including discussions of class D output, the actual power loss has no bearing on the apparant output impedance.

Kalee20's observation that dissipated power in a class A output valve is reduced by the output power is one I hadn't considered before. So the way to increase the life of your output valve is to wind the volume up - could have used that knowledge in my youth.

Finally, I am not convinced the transmission line rules relate at all to the maximum power transfer theorem? This is surely a completely different phenomenum which I have never managed to get my head around.

[anthonys radios]

Originally Posted by G8HQP Dave

After all that has been said in this thread, it still seems that there is confusion between source impedance and optimum load impedance.

If you vary the load impedance either side of the source impedance of an unlimited power source then you will find that the peak power transfer occurs when they (conjugate) match. This is the maximum power transfer theorem.

If you vary the load impedance either side of the optimum load impedance for a real power amplifier then you will find that the peak power transfer may be somewhere near the optimum load impedance, but almost certainly not exactly at it. This is NOT the maximum power transfer theorem at work, but instead a case of hitting soft or hard limits in current or voltage. In addition, the distortion will vary. The optimum load impedance is the one which gives the desired compromise between power and distortion. If you change your compromise criteria then you will get a different optimum load impedance. This demonstrates that this is not a fundamental circuit theorem, but merely an example of the compromises inherent in any real engineering. The 50% maximum efficiency of a perfect Class A amplifier when driven by a sine wave has absolutely nothing whatsoever to do with the maximum power transfer theorem! It is merely a numerical coincidence.

Well, what can anyone say but three cheers

There seems to have been be a degree of unnecessary complication in the thread !
As a schoolboys we made audio amplifiers, that worked, we studied valve makers data, bought suitable output valves and transformers and away we went.
20 or 30 amps down the line we became apprentices, and learnt more about amps (and a lot else), but my view is, in the practical world, we dont need to revert to first principles, every time we embark on a project, be it amps or anything else.
We all did our theory studies and interpreted things in different ways of course. We know where to lay our hands on information, sometimes that has long since slipped our memories, but it's still nice if we can learn something new as often as possible)
In the practical world, we may be lucky if our class A amp achieves 25% efficiency, ie converting DC power to usefull AC but we automatically make choices, without, I say again, reverting to first principles, unless we hit a sticky patch,

[MichaelR]

There seems to have been be a degree of unnecessary complication in the thread !
As a schoolboys we made audio amplifiers, that worked, we studied valve makers data, bought suitable output valves and transformers and away we went.
20 or 30 amps down the line we became apprentices, and learnt more about amps (and a lot else), but my view is, in the practical world, we dont need to revert to first principles, every time we embark on a project, be it amps or anything else.
We all did our theory studies and interpreted things in different ways of course. We know where to lay our hands on information, sometimes that has long since slipped our memories, but it's still nice if we can learn something new as often as possible)
In the practical world, we may be lucky if our class A amp achieves 25% efficiency, ie converting DC power to usefull AC but we automatically make choices, without, I say again, reverting to first principles, unless we hit a sticky patch,[/quote]

Facts are not Understanding !!

In the real world where things go wrong understanding the Science really helps.

I think this thread has brought out ways of looking at the theorem I would never have thought of . My understanding of this has been transformed for which I am very grateful to the contributors.

Mike

[G8HQP Dave]

Kalee20's observation that dissipated power in a class A output valve is reduced by the output power is one I hadn't considered before. So the way to increase the life of your output valve is to wind the volume up - could have used that knowledge in my youth.

First sentence true. Second sentence not necessarily true. I'm sure your neighbours will be pleased about this!

It all depends on what is killing off the valve. If simply heat, causing outgassing which then leads to excess grid current and cathode damage due to ion bombardment then turning up the volume might help. If it is simply the cathode emission layer wearing out then turning up the volume will wear it out slightly more quickly. This is because a perfect Class A output will keep the same average current (so same wear rate) with volume, but a real Class A output will have an increase in average current due to the second-order distortion of the valve.

[G8HQP Dave]

In the real world where things go wrong understanding the Science really helps.

Hear, hear! As someone originally trained in theoretical physics who then went into engineering, I keep coming across examples where engineers (whether academic, technician or amateur) get in a mess because they don't understand the basic science (or maths) of what they are doing as well as they think they do. Far too often they learn an approximation or theorem which is only true under certain circumstances, then they try to apply it to different circumstances. When it doesn't work they either don't notice, or complain that theory is no good in the real world. The CORRECT theory, properly applied, is always useful! (Dave gets off soap-box.)

[MichaelR]

Originally Posted by PJL

Finally, I am not convinced the transmission line rules relate at all to the maximum power transfer theorem? This is surely a completely different phenomenum which I have never managed to get my head around.

On the basis that maximum power is transferred with match or conjugate match it does apply surely. Power propagating in a waveguide , coax etc will be reflected at every discontinuity ( point where the impedance is not matched). If you use an adapter to connect to a coax cable and it is not a perfect match there will be power reflected back into the source and the power transmitted into the cable will be that much smaller. If you have an open or a short at a point the reflected power is 100% and no power is transmitted beyond that point.Vector network analysers can be used as essentialy frequency domain reflectometers to identify discontinuities and plot reflections against distance , very useful for cable testing.

Mike