Impedance matching theory.

[Ray Cooper]

Originally Posted by GMB

...having said that I calculate that the example of 42KW in to get 25KW into 50 ohms could have a simple output impedance of 34 ohms - so not really that low at all...

No, no, no... that 60% efficiency figure quoted was largely the conversion efficiency (DC to RF) of the device used - a klystron, actually - and has nothing to do with power lost in any source impedance.

[studiot]

I see discussion of the max power transfer theorem is popular these days.

Whilst there are many first class engineers in this forum there also appear to be a few blind spots, particularly about this particular theorem or its application.

The Maximum Power Theorem is a linear network theorem that applies to all networks with linear components or networks that we regard as having linear components.

Thus it applies to our (hopefully) linear audio amplifier and to mains power supplies.

It only applies to electric machinery (including loudspeakers) if we use a linear model for that machinery.

It does not apply to nonlinear components eg thermistors.

It does not apply to any network where there is interaction between the source and load (eg feedback) other than across the two terminals of connection.

This last condition is important as you have to be able to split the network into two sub networks connected by only two terminals. One sub network must contain all the sources and the other must contain all the loads.

Let us call these sub networks the source and the load. The theorem states that maximum power is transferred from the source subnetwork to the load subnetwork when the impedances seen looking from the source sub network into the other is the complex conjugate of the impedance looking into the source subnetwork from the load.

Translating this mouthful into the example with a battery and resistance it states that the max power is transferred when the load resistance equals the source resistance.
Not the other way round.

We should also remember that the source voltage and impedances are Thevenin equivalents for the whole source network so may well not be the same as the actual circuit values.

[Ray Cooper]

Just a couple of extra pebbles to toss into the pool...

One definition of Maximum Power Transfer theorem that seems to be about (admittedly, it sidesteps the issue of conjugate impedances) runs as follows:-

"The Maximum Power Transfer Theorem states that the maximum amount of power will be dissipated by a load resistance if it is equal to the Thevenin or Norton resistance of the network supplying power."

Which is strictly provable, beyond doubt. But note that it defines the problem back-to-front from the direction that we normally use in practice: it assumes that the source impedance is known and fixed, and attempts to find the value of load impedance that satisfies the MPT criterion.

But in the real world, the problem is often different: we know what the load impedance is, and the source impedance is often less well-defined or controllable. Using this criterion, we find a different solution for the MPT requirement, one where the source impedance has to be zero for maximum power in the load. The sums aren't difficult to do: I shall shame you into doing them for yourselves...

And if that's too much trouble, Googling 'Maximum Power Transfer Theorem' produces a host of solutions, but note that they are all derived from the condition where source impedance is known, and load impedance has to be determined...

A couple of quite contentious entries are as follows:-

http://www.analog-rf.com/match.shtml

http://faraday.ee.emu.edu.tr/EENG224...20Transfer.pdf

... both of which discuss the Godlike status that the MPT theorem has acquired, and the actual limitations to its use in practice. Enjoy.

[studiot]

But this is not necessarily true unless the other conditions I mentioned are also satisfied.

In fact my old electrical engineering textbook goes further. It also requires all the components to be bidirectional.

It is easy to cook up examples where the resistances are equal but minimum power is transferred, or the max occurs at a resistance ratio of other than unity, simply by choosing an example that violates one of my posted conditions.

[MichaelR]

As a non practicing professional engineer but if you like a student wanting more understanding to develop my knowledge , I hope the following is a foundation for correct use of the theorem.

Only Applicable to linear networks.
When reactive component involved we must have conjugate match.

Studiot I understand and accept your more detailed reply.

Whatever use we think of the theorem I just need to clarify when it is applicable to use.

My last post concerning reflected power from a matched aerial was in response to a statement that I still cannot understand. maybe I am reading the post too literally

Mike

[studiot]

True understanding comes when one can explain why

We do not use impedance matching for mains supply or HIFI amps/speakers, although the theorem applies to these situations. Consequently we do not achieve max power transfer in these situations.

The charging of a rechargeable battery violates one of my conditions and the theorem is not applicable. Minumum (perhaps zero) power is transferred at unity impedance ratio.

[jimmc101]

Originally Posted by MichaelR

Excuse my ignorance but I do not understand this statement.

Assuming a perfectly matched antenna surely there is no reflected power back to the source.

My limited understanding of antennas is that it may aborb the power but how well it raditaes that power is entirely dependent upon the antenna. You make the statement that half the power is radiated and half is relected back to source with a matched antenna.
Mike

Mike, I think your confusion is because Dave is talking about an aerial used to receive a signal.

Its output voltage behaves as if the aerial were a generator in series with an internal resistance (equal to the characteristic impedance of the line if it is matched).
As has been said before when matched, equal power is dissipated in this resistance and the load (in this case the receiver).
If the antenna is lossless then this energy cannot be dissipated as heat, the only way to 'get rid' of it is to radiate it back into the aether.

Jim

[MichaelR]

Eureka!!! many thanks Jimmc... i need to learn to practice my reading as well

Mike

[jimmc101]

In several previous posts it has been said that for an amplifier connected to a load equal to it's output impedance; the power dissipated in the amplifier is equal to that in the load. This is not always true!!!

Consider an operational amplifier with negative feedback applied:
If the feedback is derived from the output voltage (shunt feedback) the output impedance is lowered.
Conversely if the feedback is derived from the output current (series feedback) the output impedance is raised.

Combine the two and you can set the output impedance without any dissipative components.

For a fuller explanation see http://sound.westhost.com/project56.htm

A similar idea using feedback can be applied to defining the input impedance of an amplifier. This in much more often used as it enables the input impedance of an amplifier to be matched to the source an still have noise figure of less than 3dB.

Jim

[Skywave]

Originally Posted by GMB

. . . . I calculate that in the example of 42KW in to get 25KW out into 50 ohms could have a simple output impedance of 34 ohms - so not really that low at all.

Correct me if I'm wrong - and I'm sure you will - but I assume that your thinking to arrive at that figure goes like this:

D.C. input = 42 kW; RF output = 25 kW. Therefore 'lost' power = 42 minus 25 = 17 kW. Let's assume that this 17 kW is dissipated in the output impedance, R0 of the transmitter. Then, since this output resistance is in series with the 50 - ohm load, RL, the current is same in each. Therefore 25 kW divided by 50-ohms = 17 kW divided by R0, which gives R0 = 34 ohms.

However, there is an error in the thinking here. The 17 kW 'lost' is not dissipated in the output resistance, R0. This 'lost' power is d.c. - not RF - and becomes heat dissipated inside the transmitter, not in R0.

Al. / Skywave.

PS This Post written whilst Posts #81 (and on) were being submitted.

[Skywave]

Originally Posted by jimmc101

Mike, I think your confusion is because Dave is talking about an aerial used to receive a signal.

Its output voltage behaves as if the aerial were a generator in series with an internal resistance (equal to the characteristic impedance of the line if it is matched).
As has been said before when matched, equal power is dissipated in this resistance and the load (in this case the receiver).
If the antenna is lossless then this energy cannot be dissipated as heat, the only way to 'get rid' of it is to radiate it back into the aether.

Jim

Basically, I'd go along with that from Jim. My understanding - which may be wrong - goes thus:

An E-M wave impinging on a conductor will induce a voltage in it. Since it is a conductor, a current will flow. This current is not infinite; the conductor under the specified conditions (angle of strike of wave, etc.) will exhibit an impedance. When we try to extract power from the aerial, the current that flows in the aerial's external load is in series with the aerial's impedance. If the two impedances are the same, only half of the available power is produced in the load. The remaining half of the power has to go somewhere: it is simply re-radiated.

A very similar situation exists with a director on a Yagi-type aerial, but here there is no external load - so all of the power is re-radiated - and by choosing the appropriate length of the director and its spacing from the 'active element', where the power is finally extracted, the current in the active element is thus increased.

Al. / Skywave.

[GMB]

T

he 17 kW 'lost' is not dissipated in the output resistance, R0. This 'lost' power is d.c. - not RF - and becomes heat dissipated inside the transmitter, not in R0.

My little calculation wasn't meant to be taken too seriously. For all I knew it had a 17KW heater! I was just making the point that the information given was insufficient to deduce that it had a very low output impedance.

However, it feels misleading to me to make a distinction between heat dissipated inside the transmitter and R0. There is no separate R0, it IS the transmitter. Losses in the operation of the output stage manifest themselves as the source impedance of the output. OK, pure wasted DC won't do that, but other aspects of its operation will do.

But I suppose it just comes down to how you like to think about this stuff.

[G8HQP Dave]

Combine the two and you can set the output impedance without any dissipative components.

Interesting point! I'm still thinking about it. A real amplifier would probably dissipate more than the supposed half lost in the output impedance anyway.

What if you had a perfect lossless amplifier (i.e. it converted its power supply into signal with 100% efficiency, like an ideal SMPS), with both current and voltage feedback? Its output impedance looks like a resistance, so can you drive a signal back into this resistance and get dissipation? If not, in what sense is it a resistance? If so, where actually does the dissipation take place?

My gut feeling is that dissipation would take place, if necessary by the perfect amplifier working backwards and returning power to its supply. An imperfect (i.e. real) amplifier would just dissipate the power inside itself somewhere, probably in the active devices. So I think that this 'manufactured (by feedback) output resistance' behaves just like a real resistance. Anone care to comment? I know this is getting a bit theoretical, but this is how we test the edges of an idea.

[studiot]

In several previous posts it has been said that for an amplifier connected to a load equal to it's output impedance; the power dissipated in the amplifier is equal to that in the load. This is not always true!!!

But is the amplifier output impedance the Thevenin equivalent for the source network?

[Skywave]

Originally Posted by GMB

The 17 kW 'lost' is not dissipated in the output resistance, R0. This 'lost' power is d.c. - not RF - and becomes heat dissipated inside the transmitter, not in R0. My little calculation wasn't meant to be taken too seriously.

OK; so what was your calculation that produced your figure of 34 ohms (approx.) for the output impedance, R0, of the transmitter?

Al. / Skywave.

[MichaelR]

Originally Posted by G8HQP Dave

My gut feeling is that dissipation would take place, if necessary by the perfect amplifier working backwards and returning power to its supply. An imperfect (i.e. real) amplifier would just dissipate the power inside itself somewhere, probably in the active devices. So I think that this 'manufactured (by feedback) output resistance' behaves just like a real resistance. Anone care to comment? I know this is getting a bit theoretical, but this is how we test the edges of an idea.

There is no such thing as a perfect amplifier. With Opamps the open loop output impedance is not theoretically zero in reality it can be a few hundred ohms and then manufacturers specify subject to a fixed load for that quoted value.I think you are right power would be dissipated inside the amp.

Mike

[jimmc101]

Originally Posted by G8HQP Dave

What if you had a perfect lossless amplifier (i.e. it converted its power supply into signal with 100% efficiency, like an ideal SMPS), with both current and voltage feedback? Its output impedance looks like a resistance, so can you drive a signal back into this resistance and get dissipation? If not, in what sense is it a resistance? If so, where actually does the dissipation take place?

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).
Any increase in the load resistance will cause voltage limiting, any decrease current limiting.
Similarly any power fed back into the output will cause one or the other limit to be hit.
Under these circumstances the output resistance cannot be measured, it has no meaning.

Now consider what happens if the drive is reduced, we can now measure the output voltage and current for a range of load resistances without running into clipping.
From these readings we can calculate the output impedance.

Now if we have switchmode amplifier where the output is obtained by averaging (with a low pass filter) a variable Mark:Space ratio rectangular wave, it is in theory this is 100% efficient at all output levels from zero to maximum.
This is still true when the output resistance is modified by negative feedback, no losses have been introduced.
So where does the reverse power go?
It is my belief that the power drawn from the DC supply will reduce to compensate, an equal amount of power will simply not be generated in the first place.

Jim

[kalee20]

Originally Posted by jimmc101

It is my belief that the power drawn from the DC supply will reduce to compensate, an equal amount of power will simply not be generated in the first place.

Dead right.

In fact, you don't even need an amplifier - consider the DC case, with a switch-mode power supply with voltage and current feedback to give a drooping output characteristic.

This can easily be made to give a straight-line, resistive output. And when you load it so that the voltage collapses to half the open-circuit value, you'll get maximum power in the load, negligible dissipation in the SMPS, and the power drawn will be only exceed the load power by thge SMPS losses.

[jimmc101]

Can't get simpler than that, I wish I'd thought of explaining it that way!

Jim

[anthonys radios]

HI,
Just back from another hol ! and had a quick look at the posts.
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,
Matching the speaker to the output valve implies that distortion versus power output is considered together
This is where the valve makers "Optimum Load" is important, the valve maker having calulated the best (optimum) impedance that should be matched to the speaker, for an acceptable percentage distortion level.
So unfortunately, simply arriving at the "static" resistance value, based upon anode voltage and current, may not enable on it's own, a satisfactory impedance to be calculated in matching the speaker.

The output transformer supplies the necessary turns ratio to achieve the impedance ratio required to match the speaker to this optimum load resistance.

Of course this will only hold over the limited range of frequencies that the speaker impedance remains at the nominal quoted impedance, in practice it does not cause too much of a problem at other frequencies, especially if the optimum load impedance quoted by the manufacture is used as a basis.
The power output available from the valve is all that is available, and to convey this maximum power to the speaker, requires that the load be properly matched, therefore maximum power transfer implies correct matching
An incorrcectly matched speaker may therefore increase or decrease distortion, may increase or decrease power output, but if you want the best mix of distortion and power output, match the speaker to the optimum load.