I don't see the relevance. Loudspeakers are not resistors. They are not even purely electric/electronic devices. As such their impedance contains terms which are not normally included in the statement that they are 4ohm, 8ohm or whatever. Strictly these terms reflect back into and modify the electrical component of impedance they possess and should be included in any network analysis that includes them.

Another way of viewing this is to say that modelling a loudspeaker as an equivalent resistor will get you the wrong answer. You have to use the correct model, just as with ac analysis of transistors you need the appropriate model.

I did publish the appropriate equations back along I will try to find the thread.

The second paragraph on this page starts with this sentence:
http://www.allaboutcircuits.com/vol_1/chpt_10/12.html

This is essentially what is aimed for in stereo system design, where speaker “impedance” is matched to amplifier “impedance” for maximum sound power output.

The stereo system design is not a relevant example of the real-world application of the MPTT. for these reasons:
1. "Loudspeakers are not resistors. They are not even purely electric/electronic devices."
2. "As such their impedance contains terms which are not normally included in the statement that they are 4ohm, 8ohm or whatever."
3. "Another way of viewing this is to say that modelling a loudspeaker as an equivalent resistor will get you the wrong answer."
4. My complaint: It is not customary to drive a 4-Ohm speaker with a 4-Ohm output impedance amplifier; the driver is 0.004 to 0.004 Ohm to achieve a 100 to 1000 damping factor according to the audio experts.

Therefore, the reference to a stereo system / speaker needs to be removed. It is an inappropriate example of the application of the MPTT.

 

But an amplifier cannot be modelled as Bill has done as a fixed voltage supply in series with a fixed impedance (resistance).

This causes the fixed supply to be divided between the load and the series resistance i.e a variable voltage across the load, depending upon the load, leading to the conditions for the max power tfer theorem to apply.

In fact an amplifier tries to supply a fixed voltage to the load, regardless of its value. This has nothing to do with damping. Nor does the idea that the amplifier somehow 'damps out' the back emf of the inductive component in the speaker impedance. The ratio = damping factor would still be the same if both the speaker and amp impedances were pure resistance and so there was no back emf.

 

But an amplifier cannot be modelled as Bill has done as a fixed voltage supply in series with a fixed impedance (resistance).

All practical voltage sources have an internal non-zero resistance. Small dry cells have a relatively large internal resistance. A 12V Lead-Acid (starting, lighting, ignition) SLI battery has a relatively low internal resistance. The audio experts say that the amplifier is designed to look like it has 0.004 to 0.04 Ohms internal resistance (for use with a 4 Ohm speaker). And I would expect the speaker to present a 4 Ohm impedance at 1000 Hz. (I think the speaker Z is specified at that frequency.

This causes the fixed supply to be divided between the load and the series resistance i.e a variable voltage across the load, depending upon the load, leading to the conditions for the max power tfer theorem to apply.

Taking Rg (amplifier internal esistance)= 0.04 Ohm and Rspeaker = 4 Ohm:
{4/(4+0.04)}Vg= 0.99Vg across the speaker, 0.1Vg across the amplifier internal resistance. For the maximum power transfer theorem to apply, the voltages need to divide 50/50 across equal amplifier and load resistances. Not the case for a high-fi amplifier.

In fact an amplifier tries to supply a fixed voltage to the load, regardless of its value. This has nothing to do with damping.

Rg = 0.04 Rspkr=4 Vspkr= {4/(4+0.04)}Vg= 0.990Vg
Rg = 0.04 Rspkr=8 Vspkr= {8/(8+0.04)}Vg= 0.995Vg
What you say seems to be true for the above two lines where the Rg internal amplifier impedance is low (0.04 Ohm). The speaker voltage doesn't change that much in changing it form 4 to 8 Ohms.

However, as soon as I change Rg internal amplifier resistance to MPTT required 4 Ohm = speaker load = 4 Ohm, it is no longer constant at about 0.99Vg, but drops to 0.5Vg and changes to 0.67Vg with changing the speaker from 4 to 8 Ohm as above -- not constant.
Rg = 4 Rspkr=4 Vspkr= {4/(4+4)}Vg= 0.5Vg
Rg = 4 Rspkr=8Vspkr= {8/(8+4)}Vg= 0.67Vg

Nor does the idea that the amplifier somehow 'damps out' the back emf of the inductive component in the speaker impedance. The ratio = damping factor would still be the same if both the speaker and amp impedances were pure resistance and so there was no back emf.

I don't think it is the ringing of the inductive component, but the mechanical ringing produced by the moving speaker cone which could even continue after excitation is removed. This is analogous to the old analytic balances which had a magnetic damper to stop the beam holding the pans from swinging. This damper was a loop of aluminum sheet flattened to fit between the poles of a magnet. Its movement through the poles damped out the swinging of the balance beam. To damp out a moving cone in an isolated speaker, I would short the terminals to damp in a similar manner. (In the day of the D'Arsonval meter, they were shipped with the movement terminals shorted to damp movement.) Once, the speaker is connected to an amplifier, I can no longer short the terminals. The best I can do is to make the internal impedance of the driving amplifier as low as practical to provide damping.

A high-fi stereo amplifier is not an example of an application of the maximim power transfer theorem. I'm sure a person could design an amplifier with an internal impedance of 4 Ohms to match a 4 Ohm speaker to meet the requirement of the MPTT. However, the audio experts will not let us call it hi-fi.

 

I would remove the reference to the electric vehicle, as in this context you don't have matched loads, but what is usually a PWM controller feeding a electric motor.

Stating the obvious (which doesn't need to be in the article) one of the reasons both have to be matched in a RF transmitter application is so the excess power can be absorbed. Reflections on an RF line are disruptive in several ways.

An audio amp with low ohmage (call it zero) will absorb the EMF generated by a vibrating speaker, and would absorb the energy. Any resistance at all would allow the speaker to free wheel. It goes back to power supply phylosophy vs. matched impedances.

Thank you Dennis, insolation, learned a new word!

 

Well I donno about you guys, But all I deal is with Amps and speakers, and the theory works for me.
RF ? well I'm not so into it.

Rifaa

 

I (think) I have understood Ray's difficulty. And I now see that my comments, although true, were a red herring.

Sorry.

I am not a teacher so I don't naturally see someone else's difficulty for something I understand my way.

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.

This means that the source voltage is fixed as is the source resistance (thevenins theorem). We vary the load resistance and find that when it is equal to the given source resistance max power is transferred.

Herein lies Ray's dificulty. He is trying to apply the theorem to fix the load resistance and study the effect of varying the source resistance.

Of course he gets the strange answers.

 

I (think) I have understood Ray's difficulty. And I now see that my comments, although true, were a red herring.

Sorry.

I am not a teacher so I don't naturally see someone else's difficulty for something I understand my way.

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.

This means that the source voltage is fixed as is the source resistance (thevenins theorem). We vary the load resistance and find that when it is equal to the given source resistance max power is transferred.

Herein lies Ray's dificulty. He is trying to apply the theorem to fix the load resistance and study the effect of varying the source resistance.

Of course he gets the strange answers.

In fact, the theorem even works in NON-linear networks, such as Class-C R.f. amplifiers. (You just have to include a duty cycle factor, which turns out to be a simple coefficient).

Again, I think we need to clearly differentiate between the DEFINITION of a conjugate match and the DESIRABILITY of one.

A perfect example of this is a classic A.M. broadcast transmitter. We had an old Gates VP-50 which was capable of cranking out 100KW without even asking "mother may I". Since we were licensed at 50KW, we had to run it at MUCH less than optimum matching...which we did by "deloading" (intentionally mismatching) the tuning network. A perfect conjugate match would have been patently ILLEGAL among other things. However, if our goal was cranking out every possible watt, we would have adjusted our output network for a conjugate match. The definition remains unaltered.


Hope this helps.

Eric

 

For better or worse, I have updated the master copy.

Thanks for the input, Ray, and others.

Dennis Crunkilton