Impedance matching theory.

[anthonys radios]

REF jimmc101 No, sorry I disagree.

Can you say if you disagree with part or all of that posting,?
If part, which parts do you agree with, and which you disagree with ?

[MichaelR]

Originally Posted by G8HQP Dave

For linear conditions (e.g. small signal) the theorem applies, but it may be irrelevant because usually one wants to obtain some output/distortion compromise at higher power (where 'impedance' varies over the AC cycle).

It gets harder for RF output stages, as the 'impedance' varies even more (e.g. Class C - even when smoothed by the flywheel effect of the output tank circuit). Then another spanner gets thrown into the works, as sometimes we want maximum efficiency rather than maximum power. For some designs, such as Class E, these two maxima occur at different tunings but not everyone realises this.

I think the maximum power transfer theorem should always be taught with a health warning: in a given situation this may be true and helpful, true but misleading, or not applicable!

Dave you are right , for me the most important information that has come out of this whole thread is the detailed understanding rather than general answers and JIMMC has focused on the requirement for linearity is essential for max power transfer to be relevant.If you remember you kindly focused on a similar point with respect to heater power in another recent thread whereby my statement then was only applicable to linear conditions.

If we take the simple example shown in this thread with a battery and internal impedance and the measured volta and amps. Perfect it is linear. However If you inreased the current being drawn so that you are overdriving the battery certainly with some chemestries you will get a non linear behaviour occuring.

great thread

Mike

[jimmc101]

Sorry I should have been more specific.

Maximum power transfer takes place if the source impedance matches the load.

As we are talking about power amplifiers I disagree.
If you only want to talk about amplifiers that meet the criteria below then I will agree.

For a small signal device operating in its linear region where there is no need to consider any limits imposed by the device bias point the maximum power transfer theorem applies.

Anode impedance is related/changes according to DC conditions.
Specifically Anode impedance varies with changes in anode voltage in practice.
Distortion can increases as signal amplitude increase.

Agree, but I'm not sure of the relevance here.

Valves in general are not linear devices.

Agree, this is the reason why the maximum power transfer theorem does not apply.

It is not practical to match the Anode Impedance of a Pentode output valve to the loudspeaker.

Agree but there is no reason why we would want to, we want to choose a load so that the valve saturates on the negative swing at the same power level as it cuts off on the positive swing.

Valve makers suggest the Optimum Anode Resistance that should be used for impedance matching calculations, being the best compromise between power output and distortion.

Maybe, if you mean that it is chosen so that, at maximum power, clipping occurs equally on the positive and negative half cycles.

The reason I disagreed in general terms is that I think that the maximum power transfer theorem has no place in the design of power amplifiers. We can always turn the volume control up until clipping occurs and this is what limits the output power.

I wonder if you may be using Anode Impedance to mean different things at different times, my definition is always the slope of the Ia/Va curve at the operating point being considered.

Jim

[kalee20]

There's ben much confusion in this thread about Maximum Power Transfer theorem, valve linearity, meaning of 'impedance', what is ra, etc.

But, matching pentodes is fairly straightforward, if you assume that you can give as much grid drive as the thing needs.

Consider a Class A, single-ended stage, with transformer load. Suppose the load impedance is pretty low.

Then, wind up the grid drive. Since the load impedance is lowish, the anode voltage swing won't be much, but as you increase drive, at some point the current will swing to zero on one half-cycle. And any more drive can't swing it less than zero, so the output will be clipped on these half-cycles (which are whan the anode is swinging positive). So this represents the limit of distortionless output.

Now, suppose you have instead, a highish load impedance. Again, winding up the drive, you'll get plenty of voltage swing (for very little current swing). And it won't be long before the anode voltage swings downward to the 'knee' of the characteristic. More drive to the valve can't make it swing lower, so once again clipping sets in. But, this is distortion on the negative voltage swings, ie the opposite half-cycle to the first case.

So, somewhere between these two extremes is a situation where the onset of clipping just starts on both half-cycles at the same drive level. With this load, the valve is giving maximum voltage swing AND maximum current swing, and any further drive will clip the output symmetrically.

If you assume that the valve is linear right down to zero current, and also the 'knee' is at zero voltage, this optimum load is exactly equal to Va / Ia. In practice, most valves are linearish down to zero current, but the 'knee' is at about 25V (for a 250V HT rail), giving a rule of thumb that the optimum load is 0.9Va / Ia.

[jimmc101]

Totally agree.

Jim

[anthonys radios]

OK, let me put it another way, you have been commisioned to refurbish an audio amplifier, single ended, 6V6 for example, it has an 8 ohm speaker, but the output transformer is missing, however a nearby transformer manufacturer will wind you a replacement transformer.
(PS the amplifier is completely unknown to you, or anyone else, so you are in the dark.)
What information would you give them regarding winding details ?

[kalee20]

Anthony,

This is what I would do.

The output transformer is missing - so connect the 6V6 anode to HT +ve via a milliammeter. Then switch on.

Let the thing warm up, and note the quiescent anode current (don't disconnect it, or all the current will go to the screen grid, which will overheat!). Suppose this is 45mA.

Also measure the HT voltage (say 272V), and the 6V6 cathode voltage (11V). Switch off.

OK, then the 6V6 is effectively running with 272 - 11V = 261V of HT.

Then the optimum load is going to be 0.9 x 261V / 45mA = 5.22kΩ.

If you have an 8Ω loudspeaker, the turns ratio is going to be √ (5220 / 8) = 25.5 : 1. But, really that's doing the transformer manufacturer's work for them! All you should need, is to tell them:

* To match 8Ω to 5.22kΩ

* 45mA DC current will be flowing (they need this so that they can design the transformer so it won't saturate)

* 3db cut-off frequency 40Hz (or whatever you want).

(The LF cutoff frequency is needed so that they can ensure there's sufficient primary inductance. Otherwise, there'll be bass cut. If you want your frequency response to extend down to 10Hz for instance, you'll end up with a much bigger transformer - but that's a new topic!)

[jimmc101]

Assuming that all other components are know to be good.

Connect ammeter between anode and supply measure anode current Ia and anode voltage Va
Calculate R load = Va/Ia and reduce by 10% to allow for the knee as kalee20 has suggested.
Calculate maximum rms signal power P as (Va x Ia)/2 i.e. (peak voltage x peak current) /2
Calculate turns ratio T = SQRT(R load/8)

Transformer can now be specified

Turns ratio T
DC current handling Ia
AC power handling P
(Primary inductance as required for LF response)

Leave core size, wire gauge etc to the manufacturer.

That's it

If you want to check try some figures from the 6V6 datasheet attached.

Damn too slow again! At least our answers agreed.

Jim

[kalee20]

Yes.

Strictly, the power handling info is redundant - if the manufacturer knows what the application is, he can calculate the maximum power (because the current swing can never be greater than the 45mA).

(Strictly, if I was asked to design this transformer, I might also ask what is the HIGHEST frequency needed. If somebody wanted 20kHz, than I'd have to interleave the windings to reduce leakage inductance, which hurts HF response).

[PJL]

The power transfer theorem assumes the source comprises a fixed voltage and a series impedance which in this case with audio frequency I think we can agree is resistive.

The output voltage is given by Vs x Rl / (Rs + Rl)

In this case, the source is the output transformer secondary and the speaker is the load.

A valve is a current device so the AC output voltage is given by Ia/(Tr x (1/Ri + Tr^2/Rl)) where Tr is the turns ratio, Ia is the AC current, Ri is the plate resistance and Rl is the load. (the reflected load impedance is in parallel with Ri)

I am not a believer yet! This does not look anything like a voltage source with a series resistor (if only I could remember some maths I could probably prove it).

But that aside, if we take an EL84 and assume a 25:1 ideal transformer (8 ohm match), Ri is 40K, 40mA bias current and a small 1mA signal then.

Vs would be the open circuit voltage = Ia x Ri/Tr = 1mA * 40K = 40/25V
And the optimal load would occur when the voltage is 1/2 x Vs or when Ri = reflected impedance or Rl = Ri/Tr^2 = 64 ohms which doesn't seem to make a lot of sense!

[kalee20]

Originally Posted by PJL

The power transfer theorem assumes the source comprises a fixed voltage and a series impedance which in this case with audio frequency I think we can agree is resistive.

Actually, the Max Power Transfer Theorem just as validly asumes the source comprises a fixed current and a parallel impedance.

This is exactly equivalent to a source voltage and a series resistance.

The rest of your post is a bit confused, but I think I can follow your thoughts. You say, with your 25:1 transformer, the output voltage is 40/25V (= 1.6V), this is correct, and it's the open-circuit voltage.

Then you say, the optimum load would be 64 ohms, being the valve's ra (40k) divided by the square of the turns ratio (25). Again, you are actually correct, and you would get maximum power at this load!

But, if you assume an HT voltage of 250V, then the maximum peak voltage you could possibly get at the load is 250V/25 = 5V (3.5V RMS), which in 64 ohms is 0.19W. And then distortion sets in. With a lower load, sure, the small-signal power at fixed grid drive will be less, but you compensate for this by screwing up the volume control - and then you'll find distortion sets in at a greater output power. That's why the Max Power Theorem is just a red herring!

[anthonys radios]

R--i--g---h---t,
I agree with a good amount that these last posts say, especially with regards to the highest frequency needs, thats absolutely spot on , and of course the required inductance is most important for low frequency response also.
Actually, now I dont really see why the previous post were so involved, as since I have promoted/presented it from a different viewpoint, we seem to largely agree !
A slight difference for me would be that I would take a couple of miniutes and just looked up the valve makers optimum load, and calculated the turns ratio from there without the extra trouble of measuring the anode voltage/current, but their you are,

In the process I would be matching the optimum load (ohms) to the speaker (ohms) knowing the maximum power was available and the distortion levels accompanying this power had been experimented on by the valve maker, I call this process"The the easy way" , of course I would also call this process "matching the load to the source", just as you have! and as far as I am concerned, I would have satisfied criterior for maximum power transfer in the process .

Which is what I thought I was saying before, obviously not very well or you would have agreed with me !

Oh yes ref Ra, Dynamic resistance anode resistance surely measured by dVa/dVi
ie a small change in anode voltage devided by a small change in anode current, that way you get the dynamic anode resistance, rather than deviding anode volts by anode current as mentioned ?

Also indeed I would leave wire size and core air gaps etc to the manufacturer, but only if he was more than just a rewind shop.
I have had to write this in double quick, fast march time, sorry if there are two many errors.
Regards.

[jimmc101]

One last try then I'll promise I'll keep quiet.

Imagine you have an impedance meter, connect this in place of the speaker and the amplifier powered with no signal in.
(Assume the transformer is perfect)

Will it be 8 ohm [(R load) 5000 / 25^2) or 64 ohm [(ra) 40000 / 25^2]?

I am confident you will measure 64 ohm, so source and load impedances are not matched!

With regard to using the data sheet, it gives recommended loads for 180v/29mA, 250v/45mA, 315v/34mA.
How do you know which one to use if you haven't measured the operating conditions?
Remember you said '(PS the amplifier is completely unknown to you, or anyone else, so you are in the dark.)'


Jim

[kalee20]

Originally Posted by anthonys radios

A slight difference for me would be that I would take a couple of miniutes and just looked up the valve makers optimum load, and calculated the turns ratio from there without the extra trouble of measuring the anode voltage/current, but their you are

I thought you wanted to know how the optimum load figure came about though ??!

In any case, looking up the valve maker's figure is only OK if you operate it under the stated conditions. ISTR the 6V6 is quoted for a 250V HT supply and also a 315V supply. But what if you have a mains transformer/rectifier combination which gives 270V? Or, you want to run at a somewhat lower current to reduce heat output.

Under these conditions, you can use the ideas I've outlined to get the best performance out of the stage.

I can also see from your post how easy it is to misunderstand, when using terminology like 'matching the load to the source', in fact I may have misinterpreted this as per the Maximum Power Transfer theorem case (which is definitely not the optimum load).

The other things - you are right with ra (= dVa / dIa), though for pentodes and beam tetrodes it plays no part in calculating the optimum load. It's the quiescent DC values which are used, because the DC anode voltage defines the maximum voltage swing, and the DC anode current, because that defines the maximum current swing. And at the optimum load, the two limits come into play together, and nothing is wasted!

[kalee20]

Originally Posted by jimmc101

Imagine you have an impedance meter, connect this in place of the speaker and the amplifier powered with no signal in.
(Assume the transformer is perfect)

Will it be 8 ohm [(R load) 5000 / 25^2) or 64 ohm [(ra) 40000 / 25^2]?

I am confident you will measure 64 ohm, so source and load impedances are not matched!

Yes, you will measure 64 ohms. The source impedance is not equal to the intended load impedance. But that doesn't stop the 8 ohm load from being the optimum. I've realised that the term 'matched' can easily be misinterpreted!

[jimmc101]

I can't edit my last post, so I'll repost it with one line added to try and remove the ambiguity that kalee20 referred to.

One last try then I'll promise I'll keep quiet.

Imagine you have an impedance meter, connect this in place of the speaker and the amplifier powered with no signal in.
(Assume the transformer is perfect)

Will it be 8 ohm [(R load) 5000 / 25^2) or 64 ohm [(ra) 40000 / 25^2]?

I am confident you will measure 64 ohm, so source and load impedances are not matched!

So the conditions of the maximum power transfer theory are not met even though the valve is correctly matched for maximum power.

With regard to using the data sheet, it gives recommended loads for 180v/29mA, 250v/45mA, 315v/34mA.
How do you know which one to use if you haven't measured the operating conditions?
Remember you said '(PS the amplifier is completely unknown to you, or anyone else, so you are in the dark.)'


Jim

[G8HQP Dave]

As I said, it doesn't apply when there are voltage or current limits, which means almost all output stages! For almost all output stages the best match is set by the limits (modified somewhat by distortion considerations), not the source impedance.

Why do colleges/textbooks teach it? It is a neat result, and a good illustration of simple calculus. It does have applications, mainly in the input stages of small signal circuits (but there it gets mixed up with optimum noise matching which can be different from optimum power matching). So like most simple engineering formulas: apply with caution! (And understand its limitations - which the textbooks sometimes gloss over.)

[MichaelR]

Actually Dave, was maximum power transfer theorem more introduced with transmission line theory with the now obvious constraints

Mike

[GMB]

I think "maximum power" is a very misleading phase.

The situation of equal load and source impedance would be better called the "absolute maximum power transfer theory" since it addresses the specific problem of the absolute maximum power that can be delivered to a load given the source is lumbered with an impedance and you don't care how hot it gets.

The equal source/load impedance situation is also the point of lowest overall efficiency, in that half the power is lost in the source. When you look at it like that it doesn't look so clever.

A valve output stage bias is backed into a corner by such considerations as maximum anode dissipation. So now we are trying for the maximum power given the limitations, and hence this is different from the absolute maximum power that might have been possible if things were different.

[kalee20]

Originally Posted by jimmc101

So the conditions of the maximum power transfer theory are not met even though the valve is correctly matched for maximum power.

Yes. The valve is loaded for maximum power, given the limitations that it swings smoothly down to zero current and then stops, and smoothl down to zero(ish) voltage and then stops.

The conditions of the Max Power Transfer theorem are not met. But then who cares anyway??!