Calculate Coaxial Cable Loss & Input Impedance at 1 MHz

[Averagesupernova]

Thank you sophie. I will take a little more time later too look at the links you have provided. I skimmed through them very quickly.

 

[the_emi_guy]

Here are a few links on the subject of matching. Read what they are actually saying about the significance of transmitter / load matching and efficiency. They all agree with what I have been saying.
http://www.jaycar.co.uk/images_uploaded/impmatch.pdf
http://users.tpg.com.au/users/ldbutler/OutputLoadZ.htm
http://urgentcomm.com/test-amp-measurement-mag/maximum-power-transfer

Thank you sophie. I will take a little more time later too look at the links you have provided. I skimmed through them very quickly.

Agreed, I would like to digest these but it will take me some time. There does seem to be a divergence of opinion on this interesting topic (or maybe I am just not up to speed).

In the meanwhile here is what I have done:

I used Motorola AN267 which was created to help impedance match RF power amplifiers to transmission line. I designed two cases:

Case1: 1 ohm source impedance to 50 ohm line.
Case2: 5+5j source impedance (10 ohms parallel with 10j capacitive reactance) to 50 ohm line.

The results are attached (clipped from Mathcad).

Images 1,2,3 are case1.
Images 4,5,6 are case2.

Note that in both cases the network presents the desired impedance to the output of the amp while simultaneously presenting 50 ohms to any signal traveling in the reverse direction from far end reflection.

I would say that this is a "first order" method of matching. More elaborate schemes are used that take into account non-linearities and load pull characterization, but it seems that the basic scheme provides backmatch.

 

[sophiecentaur]

@the_emi_guy
The calculations are probably right; you can clearly drive the software correctly and your network will present the 50 Ohm load to the transmitter as a 1 Ohm load and vice versa. That would be a match for maximum power. It would result in an equal amount of power being dissipated in the 1 Ohm source resistance. No problem for a low power amplifier and it would be a well behaved piece of equipment - like a good Test Signal Generator, for instance but, with 50% efficiency, a high power transmitter would produce loads of embarrassing heat in its output stage and cost a lot of money to run. That was the point I have been making. I have never claimed that you can't match power amplifiers to deliver maximum power but I know that high power transmitters are built to be efficient (90% or so) and so they cannot be matched in that way - else, where do they get the 40% lost power back from and why do they not melt into the ground? Would you comment on that issue please and not on the nuts and bolts of matching networks?

 

[the_emi_guy]

@the_emi_guy
Would you comment on that issue please and not on the nuts and bolts of matching networks?

I hear you sophie, I understand your point and I need to think about this some more...

 

[sophiecentaur]

OK
No problem.
I think this is just to do with the directions we are coming from. I have no problem with your last post at all.

 

[the_emi_guy]

Here is what I am trying to get my arms around:

Assuming that the impedance that the amp want to see hanging on its output for optimum, high efficiency, performance is higher than its impedance as seen from outside looking in (these were assumed to be the same in my previous analysis).

For example an RF amp with 1ohm output impedance specified to drive 1Kohm load (not maximum power transfer but really efficient).

We need a matching network to make 50 ohm cable look like 1K load. Is it not possible to design this network to simultaneously provide 50 ohm backmatch?

Maybe not, that is what I am hung up on.

 

[Averagesupernova]

Here is what I am trying to get my arms around:

Assuming that the impedance that the amp want to see hanging on its output for optimum, high efficiency, performance is higher than its impedance as seen from outside looking in (these were assumed to be the same in my previous analysis).

For example an RF amp with 1ohm output impedance specified to drive 1Kohm load (not maximum power transfer but really efficient).

We need a matching network to make 50 ohm cable look like 1K load. Is it not possible to design this network to simultaneously provide 50 ohm backmatch?

Maybe not, that is what I am hung up on.

Any time we match with a transformer, Q-section, pi network, T-network, L-network, etc. we will match in both directions. If something is spec'd to drive a 1000 ohm load and you have a 50 ohm antenna system you will most certainly need to transform the impedance.

 

[sophiecentaur]

Here is what I am trying to get my arms around:

Assuming that the impedance that the amp want to see hanging on its output for optimum, high efficiency, performance is higher than its impedance as seen from outside looking in (these were assumed to be the same in my previous analysis).

For example an RF amp with 1ohm output impedance specified to drive 1Kohm load (not maximum power transfer but really efficient).

We need a matching network to make 50 ohm cable look like 1K load. Is it not possible to design this network to simultaneously provide 50 ohm backmatch?

Maybe not, that is what I am hung up on.

If you think of any matching network as a transformer and if you ignore reactances then the ratio looking one way will be the inverse of the ratio looking the other way. If you transform the 50Ω feeder to look like 1kΩ from the transmitter's point of view (that's 1:√20 turns ratio) then the 1Ω output resistance of the transmitter will look like 1/20Ω (20:1 times 1), from the feeder side of the transformer. Afaics, you are stuck with that and it means that designers have to take that sort of thing into account and make allowance for reflected power. But, as long as the antenna match is good, there shouldn't be any signals arriving back at the Tx.
For UHF frequencies and above, it is possible to include an isolator or circulator to avoid the problems associated with mismatch at the transmitter. But then you need to accept a 1dB (iirc) or more extra loss.

 

[Baluncore]

My analysis shows that there are two quite distinct situations involved here.

Firstly; For the RF signal path through multiple stages to beyond the output transmission line, I believe that my assertion of “impedance matching is required for optimum efficiency” is correct for any RF signal path.

Secondly; for regional AC power distribution grids and local DC power supply rails, including 12V car batteries, I believe that sophicentaur's assertion of “zero impedance source is required for optimum efficiency” is correct.


RF signal paths.

Firstly, we should not dismiss the reactive coupling systems that transform the complex impedance between RF stages as being irrelevant. They are vitally important to the RF energy economy. It is the reactive components in switching power supplies and class C or D amplifiers that make those devices so highly efficient.

The second point I want to make regards a fundamental principle of RF design, one that has now been in use successfully for almost a century. When considering the design of a two stage RF amplifier, the load line of the first stage output is known and the input impedance of the second stage is known. A network that will efficiently match those quite different complex impedances is required in order to fully utilise the capabilities of the active elements in both stages. Likewise, we know the load line of the final output stage and we must match that to the transmission line impedance. I agree that without careful matching it will still work, but it will be wasteful of RF energy and equipment resources. The argument that the output impedance of signal path modules must be as low as possible for optimum efficiency is clearly false.

I therefore make the substantial claim that the RF impedances throughout the system need to be matched. Failure to efficiently match at RF represents an underutilisation of the components available and a reduction in the maximum RF energy that can be passed to the transmission line.

The characteristic impedance of a linear transmission line does not effect the energy it will dissipate as heat. It simply sets the relative phase and magnitudes of the voltage and current propagating in its two independent directions. Likewise, in free space, the ratio of the E to M field strengths represents an impedance of approximately 120*Pi ohms. That does not in any way represent a dissipation of power. An amplifier with a load line having a negative Gm represents power gain, not dissipation.


Power Distribution Systems.

A power supply can be derived from the regional AC distribution grid which has an extremely low impedance. The impedance of a power supply should not be matched to the grid, for obvious reasons. So let's buy a 90% efficient switching power supply that generates a DC supply voltage with an effectively zero resistance or impedance at the maximum current we specify, and pay for. We all should recognise that a switching power supply is a class D amplifier and as such it can have very high efficiency. But all this purchase has done is extended the “zero impedance source needed for efficiency” situation from the AC distribution grid to an internal DC supply distribution rail, and yes, as expected, it has done it efficiently without any need to waste 50% of the energy in the process.

Wherever modules share a common energy source, or distribute power to many users or stages with variable requirements, efficiency is gained by having a low supply resistance. Apart from thick wires and plenty of parallel capacitance, the low output impedance is often achieved by providing the source with some form of voltage control feedback. Two examples are the control of field current of an alternator, or the error voltage comparator that adjusts the duty cycle of a switching voltage regulator.


Summary.

The boundary between these two distinct regions occurs near to or within the RF amplifiers or modules. Just where the regions approach is actually determined by the operating class of the amplifiers employed. Close examination of the transition within an amplifier module reveals that for optimal efficiency the two regions are separated by a reactive network. That reactive isolation prevents the low impedance power distribution supply from short circuiting the signal path. It also prevents the RF signal from influencing other modules through the power supply rails.

Class A amplifiers will always be inefficient, Class B will be better while classes C and D can be very efficient. By mismatching the power distribution supplies, while matching the RF signal path, it is possible, (by using class C or D amplifiers), to make equipment that operates at significantly better than 50% efficiency.

To ascribe to the RF signal path the “zero impedance = high efficiency” concepts applicable to power distribution systems is a mistake.

 

[sophiecentaur]

You clearly have lots of experience and knowledge about transmitter practice but the notion that the Maximum Power theorem somehow only works under certain circumstances is a step too far. It involves Conjugate matching, of course, Introducing reactance is a red herring. Resistive components are the only ones that dissipate Power.
"Underuse of" components is not as heinous a crime as 'overusing' Power. Good Engineering aims at minimising the appropriate losses. Electrical supply costs soon outweigh the cost of a powerful broadcast transmitter - the electricity bills would make your eyes water. No one would ever chuck away 500kW of RF power into the cooling water.

Likewise, in free space, the ratio of the E to M field strengths represents an impedance of approximately 120*Pi ohms. That does not in any way represent a dissipation of power.

If you look at the Impedance, looking into a transmitting antenna, you will come across a resistive component (if not, then it is not radiating any power). That resistive component is referred to as Radiation Resistance. The Characteristic impedance of a 50Ω line is the ratio of Reactances, as is the impedance of free space, effectively. But, in both cases, the Energy transfer is due to where the energy goes and not to the characteristic impedance.

 

[Averagesupernova]

I have a question for you baluncore. If zero output impedance is not applying to RF transmitters, then why are we not dissipating the same amount of power in the class C amplifier that drives the antenna system that we are radiating in the antenna? You said it yourself in a previous post.

A 1 kW RF transmitter driving a matched line does not generate >= 1 kW internally.
Class C and D amplifiers can operate at efficiencies better than 80% into matched lines.

I call you on this by saying: The reason it does not generate >= 1kW internall is because it has a Zout below 50 ohms which is exactly what sophie has been saying. The load at the end of the line may be 50 ohms and the line may be 50 ohms and we can call that matched but that does not mean the Zout of the transmitter is 50 ohms. It may well be spec'd to drive a 50 ohm load, but that is irrelevant.
...and then you go on to say:

The transmitter output stage has an impedance transformer called a tank circuit. To cancel the inductive component of the wound transformer it has a parallel capacitor, adjustable for different frequencies.

The load line or V:I ratio of the amplifier's active element is transformed to the impedance of the transmission line by the tank.

A reflected signal returning from the output t'line is reverse transformed by the tank from a V:I ratio of 50 ohm in the t'line to the V:I ratio of the amplifier's active element, and so is not reflected.

The reflected signal is irrelevant since the assumption is that there is no reflected signal in this discussion. In the real world there will always be some reflected power but it is low enough to be disregarded with respect to the subject being discussed.
-
As for the tank circuit behaving as an impedance transformer. I suppose one can look at it in this manner. However, you cannot have it both ways. If the tank is acting as a transformer and matching the low impedance line to the transistor/tube, then you have to treat it as an actual transformer in that it is only changing the V:I ratio and not the power. So, by your logic the active element would once again be dissipating the same amount of power that the antenna is radiating and you clearly said this is not the case:

Class C and D amplifiers can operate at efficiencies better than 80% into matched lines.

 

[Baluncore]

As this subject is fraught by the many devils in the details, any general statement in a non-mathematical language must be expected to fail under some interpretation. So, full speed ahead and damn the torpedoes.

Unlike a linear supply regulator, a switching power supply can drop the voltage without the need to dissipate energy. It does it by employing a series inductor that acts as a reactive current limiter. The duty cycle of the switch, in conjunction with another switch or diode, effectively pulse-width-modulates the average value of the fixed inductive reactance at the switching frequency. That regulates the average current and so can be used to regulate the voltage efficiently.

The load line of an active device can also be modulated by oscillating between two points on the line rather than operating on the linear middle part of the line. That is a difference between the class A,B and class C,D amplifiers. The tank circuit will be different for different class amplifiers. That is because the effective output impedance of class A and B will appear resistive while the output impedance of classes C and D will be a very low resistance in series with a reactive network that functions as a flywheel.

Class A is inefficient because it operates in the linear mode. It forms a simple model of an externally matched amplifier in which a significant proportion of the energy is wasted in the output of the active device. Class D is efficient because the switching is between a point with high voltage with very low current and a point of high current with very low voltage, neither of which dissipates high power in the output of the active device. The output impedance is effectively the ratio of the pulsed current average to the pulsed average voltage. The lack of in-phase voltage and current is consistent with the use of a reactive element to limit the output power.

There is another way of viewing the design of a linear RF power amplifier. We can consider it to be two amplifiers in one. One is a voltage amplifier, the other is a current amplifier. The one active element performs both those tasks. The operating conditions of the active element can be chosen to maximise the output power, W_out, which is the product of V_out and I_out. The selection of active element operating conditions decides the input impedance. Matching between stages then becomes a case of juggling the voltage and current gain to maximise power output for hardware investment and/or energy expenditure. It should be clear that the ratio of V to I is the terminal impedance, and that it does not matter what V to I ratio is used so long as output power is optimised.

It should also be clear that when considering power output, neither voltage nor current is more important than the other. As such, arguing for a minimum ratio of voltage to current is unrealistic. What we actually need for economy is a minimum real loss resistance in series with the output. The output impedance Zo is then the ratio of output voltage to output current, which is quite different to the lossy real Rs component.

This demonstrates that a fixed voltage power distribution system should have a minimum series loss resistance. The output impedance as an output voltage to output current ratio is then unimportant.

On the other hand, the transfer of power between amplifier stages needs to transfer all available signal V and I for maximum power. That requires the V to I ratio, the impedance, be matched between sources and their loads for efficient power transfer. I agree that Rs << Zo to minimise the waste of the hard-won RF signal energy.

 

[AlephZero]

As this subject is fraught by the many devils in the details, any general statement in a non-mathematical language must be expected to fail under some interpretation.

Indeed - so why don't you back up your assertions with some math? (Give some references, if you don't want to duplicate standard theory)

Otherwise, you are imposing on the rest of the PF community to pick through your non-mathematical assertions and figure out which of them are correct and/or relevant.

For example, IMO much of your last post is confused through mixing different definitions of "efficiency", but I'm not interested in spending an hour of my time (literally) unpicking it in detail.

 

[sophiecentaur]

@Baluncore
You are totally missing the wood for the trees.
1. You can never ignore the maximum power theorem. It applies everywhere - with or without the detail.
2. Reactive components dissipate no power.
3. A low efficiency transmitter can be probably be well matched from the feeder as well as into the feeder
4. High efficiency transmitters (amplifiers) must have very low (or very high, aamof) source resistances. The way they are tuned to the feeder will (usually) assume the antenna is matched to the feeder and will be done to achieve efficiency. If the 1Ohm source is matched to the 50 Ohm feeder then they cannot be working at 90% efficiency because they will be seeing a 1 Ohm Load.
5. Introducing switched mode PSUs into your argument can only go against your ideas because they are designed to have a low internal resistance as possible (i.e. they don't get hot).


You are introducing so many irrelevant factors to back up the ideas you got from somewhere about matching. The only way you can prove this for yourself, and get it right, is to do the actual calculations with some real values. The analysis will be very hard. At any particular frequency, you can reduce all your networks to a simple emf and one source Z so you don't need to over-complicate things to find the ultimate truth in this.

I have very little argument with you except when you claim that matching both ways is possible to achieve with high efficiency. That just has to be wrong on basic, almost philosophical, grounds.

 

[Averagesupernova]

I just wish my questions would be answered.

 

[Baluncore]

I just wish my questions would be answered.

In post #46 you asked only one question.

I have a question for you baluncore. If zero output impedance is not applying to RF transmitters, then why are we not dissipating the same amount of power in the class C amplifier that drives the antenna system that we are radiating in the antenna?

My reply was;

The load line of an active device can also be modulated by oscillating between two points on the line rather than operating on the linear middle part of the line. That is a difference between the class A,B and class C,D amplifiers. The tank circuit will be different for different class amplifiers. That is because the effective output impedance of class A and B will appear resistive while the output impedance of classes C and D will be a very low resistance in series with a reactive network that functions as a flywheel.

Class A is inefficient because it operates in the linear mode. It forms a simple model of an externally matched amplifier in which a significant proportion of the energy is wasted in the output of the active device. Class D is efficient because the switching is between a point with high voltage with very low current and a point of high current with very low voltage, neither of which dissipates high power in the output of the active device. The output impedance is effectively the ratio of the pulsed current average to the pulsed average voltage. The lack of in-phase voltage and current is consistent with the use of a reactive element to limit the output power.

 

[Baluncore]

You can never ignore the maximum power theorem. It applies everywhere - with or without the detail.

I do not ignore it. I simply do not misapply it to concepts such as matching characteristic impedance.
The MPTT applies to the real dissipative internal resistance of the output of power distribution systems, NOT to the complex ratio of the voltage and current waves propagating toward the end of a lossless transmission line.

The problem arises because Impedance has two meanings that share the same dimensions and unit, the ohm.
Firstly, the complex vector sum; Impedance = Resistance + Reactance.
And secondly, the complex ratio;
Impedance = Voltage / Current, for example, the characteristic impedance of a transmission line, or;
Impedance = Electric / Magnetic field strengths, for example the intrinsic impedance of free space.

The MPTT indicates, that to be efficient, the internal series resistance, Rs, of a line must be much less than the characteristic impedance, Zo. I have absolutely no argument with that.

I have very little argument with you except when you claim that matching both ways is possible to achieve with high efficiency. That just has to be wrong on basic, almost philosophical, grounds.

It depends on the situation you are referring to. A fixed voltage power source, with unlimited current capability should never be matched to any load in any direction. I am talking about “matching both ways” along a signal path that has a limited RF energy available. The example I used was between two stages of an RF amplifier.

I see no reason why “efficient matching both ways” must be philosophically impossible. Where two ports are coupled by a lossless transformer that has an appropriate ratio, and any reactive mismatch is neutralised with a conjugate reactance, then there can be no real loss, hence there must be high efficiency.

Perfectly matched interfaces are invisible from both sides. We live in a world composed of mismatched interfaces.
It is all the more beautiful because we can see that many of the mismatches are EM wavelength dependent.

 

[sophiecentaur]

I see no reason why “efficient matching both ways” must be philosophically impossible. Where two ports are coupled by a lossless transformer that has an appropriate ratio, and any reactive mismatch is neutralised with a conjugate reactance, then there can be no real loss, hence there must be high efficiency.

That statement includes nothing about the source impedance relative to the load impedance, which is what the whole of the MPT is about - so we can ignore it. Obviously, a lossless transformer is 100% efficient but it can't impose efficiency onto anything else.

If you see no reason why not then you must be able to give an example where it actually happens, involving some numbers.

It strikes me that you were either mis-taught or mis understood this particular thing, a long time ago and the misconception has stuck in your mind. You are treating it as an article of faith and you keep trying to justify it by digging further and further into irrelevant complexities of non-linearity and Impedance transformation rather than aiming at the sort of simplification which is the essence of good theory.

This nonsensical distinction between the two 'sorts of' Impedance doesn't help in any way. If you were really across this stuff you would see there is no such distinction. 'Purple passages' do not cut any ice in Science and Engineering proofs.

Give us a credible reference to all this. That's the acid test, as usual.

 

[Baluncore]

This nonsensical distinction between the two 'sorts of' Impedance doesn't help in any way. If you were really across this stuff you would see there is no such distinction.

I'm sorry that you are blind to the distinction.

 

[sophiecentaur]

I'm sorry that you are blind to the distinction.

Haha
I am sorry you are blind to the equivalence.

Still waiting for a reference, btw.

 

[Baluncore]

This thread is quite confused. A reference to precisely what. Life, the universe and everything ?

 

[sophiecentaur]

A reference to the meaning / relevance / possibility of matching for high efficiency and perfect termination, that you claim. A reference that provides an exception to the MPT, even.

 

[Baluncore]

A reference to the meaning / relevance / possibility of matching for high efficiency and perfect termination , that you claim.

Matching what ? for high efficiency. Matching at one frequency or broadband ? One way or reciprocal matching ?

 

[sophiecentaur]

I think you are being disingenuous here. I thought we were well aware of the difference of opinion.

The root of the disagreement is that you seem to claim that a transmitter can be operated 1.) at high efficiency 2.) see the load as the same as the transmitter internal resistance 3.) present a perfect termination when looking back from the load.
The three are not mutually compatible. A reciprocal match can only be achieved with 50% efficiency (as the MPT tells us)
The MPT can be derived in three lines andI don't think is in doubt.

If the above is not what you are claiming and I have misunderstood your posts then we have no argument.

 

[Baluncore]

I think you are being disingenuous here. I thought we were well aware of the difference of opinion.

It has been quite clear to me that there has been a misunderstanding for some time. That is why I have been trying to isolate a simple situation where we can either clarify the misunderstanding or resolve the difference. You call it “disingenuous”, I call it reductionist.

The root of the disagreement is that you seem to claim that a transmitter can be operated
1.) at high efficiency
2.) see the load as the same as the transmitter internal resistance
3.) present a perfect termination when looking back from the load.

I do claim points 1 and 3. I don't understand precisely what you mean by your point 2.
By “transmitter internal resistance”, do you mean the “load line” of the active device ?

To “match” something actually requires the definition of two things, the output port of a source and the input port to a load.
The hypothetical matching network is placed between those two defined ports.
What are those ports in your example ?

 

[sophiecentaur]

2.) is what is required for maximum power transfer.

 

[sophiecentaur]

We could make this simpler if you replace matching network, feeder and antenna with a resistance R(L). You can replace the transmitter by an emf and a resistance R(T). We can assume you have eliminated all reactances. For maximum power transfer, R(L) = R(T). That will involve 50% Power loss. Take it from there.

 

[Averagesupernova]

Baluncore it seems to me you continue to throw in non-relevant trivia into this discussion which makes it more complicated than necessary. Let's go back to the three points sophie listed:

The root of the disagreement is that you seem to claim that a transmitter can be operated 1.) at high efficiency 2.) see the load as the same as the transmitter internal resistance 3.) present a perfect termination when looking back from the load.
The three are not mutually compatible. A reciprocal match can only be achieved with 50% efficiency (as the MPT tells us)
The MPT can be derived in three lines andI don't think is in doubt.

You claim to understand points 1 and 3. Point 2 you didn't. I don't understand why you wouldn't since point 3 cannot exist without point 2. You throw matching networks into the mix which is not necessary for the sake of the discussion. But, giving you that freedom still does not support your view. Take a transmitter with a Zout of 10 ohms for instance. It can be a class C amplifier so it has a tuned output. It is non-reactive at the frequency of operation. Your view is that a matching network is absolutely required to hook this thing to a 50 ohm load. Ok, so we will do this with a 5:1 impedance matching transformer between the transmitter and the load. Now this is NOT a 5:1 voltage ratio. This means that as the transmitter looks into the matching transformer it sees a 10 ohm load. The V:I ratio before the transformer is one fifth of what it is after the transformer. We will plug some numbers in now. Suppose the voltage at the load is 10 VAC. Math tells us that the antenna receives 2 watts. We will assume for the sake of discussion a lossless transformer. 2 watts in and 2 watts out. This means that the transmitter has to be outputting 4.47 VAC since we know we have a 10 ohm load when looking into the matching transformer and (4.47 squared) / 10 = 2 watts. The transmitter is sourcing a current of .447 amps. The .447 amps is through a real resistance in the Zout of the transmitter. Since (.447 squared) * 10 ohms Zout = 2 watts. 2 watts dissipated in the Zout of the transmitter. No matter how you do it, if you have a reciprocal match you will lose the same power in the transmitter that you are dissipating in the load.

 

[Baluncore]

We could make this simpler if you replace matching network, feeder and antenna with a resistance R(L). You can replace the transmitter by an emf and a resistance R(T). We can assume you have eliminated all reactances. For maximum power transfer, R(L) = R(T). That will involve 50% Power loss. Take it from there.

Now I see what you mean.
By eliminating all reactance you can deny the existence of all switch mode power conversion.
That eliminates all the efficient class C and D amplifiers, along with all resonant converters.
I am left to use class A and B amplifiers which are inefficient because of the MPTT.

The reference I offer you is; Hamlet Act 1, scene 5. Shakespeare, W.
“There are more things in heaven and earth, Horatio, Than are dreamt of in your philosophy”.

Switching power supplies along with class C and D amplifiers, provide a way to avoid the MPTT by using a switched reactance.

 

[Baluncore]

No matter how you do it, if you have a reciprocal match you will lose the same power in the transmitter that you are dissipating in the load.

I disagree.
By taking the analytic path that you have, you are setting out to encounter the MPTT when it could be avoided.

 

[sophiecentaur]

Now I see what you mean.
By eliminating all reactance you can deny the existence of all switch mode power conversion.
That eliminates all the efficient class C and D amplifiers, along with all resonant converters.
I am left to use class A and B amplifiers which are inefficient because of the MPTT.

The reference I offer you is; Hamlet Act 1, scene 5. Shakespeare, W.
“There are more things in heaven and earth, Horatio, Than are dreamt of in your philosophy”.

Switching power supplies along with class C and D amplifiers, provide a way to avoid the MPTT by using a switched reactance.

If I use a mains transformer to provide a 12V AC supply from the 240V mains, I am making use of reactive components to achieve an impedance transformation from the , say 10Ω load on the 12V secondary to the 4kΩ that the 240 V mains supply sees. I could do the equivalent by transforming a 240V DC supply to a 12V DC supply, with a SMPSU. Are you trying to say there is an essential difference in the efficiency of the process (making the PSU and Transformer as expensive and ideal as necessary)? In the above setups, the efficiency is high because the source resistance is a small fraction of the load resistance. A class C transmitter (perhaps, even more, a class D amplifier) is virtually the same as a SMPSU - it just does without the rectifier on the output - and exactly the same considerations apply.
The nuts and bolts (reactances / oscillators / amplifiers / switches have nothing to do with the basic principles involved but you keep introducing irrelevant mechanisms into your argument. Earlier on, you claimed to be reductionist. Well - just reduce your argument as far as you can and explain how the MPT suddenly ceases to apply. How many reactances are needed and where would you say they need to be placed? Could you just put in a few and reduce the consequences of the MPT by ' just a bit'.

You would need to provide some Maths to show how that is supposed to work.
Else you could provide a single reference which shows the measured output impedance of a class C transmitter and the load it is designed to feed. Just what evidence have you?

What has a quote from Hamlet got to do with this?

 

[sophiecentaur]

I disagree.
By taking the analytic path that you have, you are setting out to encounter the MPTT when it could be avoided.

Along with the second law of thermodynamics and the effects of relativity, no doubt.
Come off it. You can't ignore fundamentals just by using flowery language and unsupported assertions.

 

[Averagesupernova]

Baluncore, it doesn't matter if you have or have not eliminated a certain class of amplifier. If the Zout of an amplifier is 10 ohms, then you will have lost power in that 10 ohms as my math shows. Now, let's see your math.

 

[Baluncore]

As you suggest sophiecentaur, I have a hypothetical SMPS.
It operates from a very low output impedance AC supply.
It provides 12 volt DC at a very low output impedance.
I load it with a 12 ohm resistor that draws 1 amp from the supply.
The resistor dissipates 12 W.
There is much less power dissipated in the SMPS.

We do have a very low impedance power supply available. To insist on throwing away that huge advantage by effectively inserting a series resistor as the RF system would be very poor engineering.

We could benefit by using a switched class D amplifier to generate RF power with very low loss in exactly the same way that a SMPS operates. It could generate 12 volt RMS RF into a 12 ohm resistive load, a current of 1 amp RMS will flow. That makes 12 watt of efficiently generated RF.

I can replace the 12 ohm load resistor with a lossless impedance matching network into a transmission line, that is then matched to the antenna. The antenna matching unit improves economy by preventing reflected energy traveling on the line twice.

 

[AlephZero]

Baluncore, it doesn't matter if you have or have not eliminated a certain class of amplifier. If the Zout of an amplifier is 10 ohms, then you will have lost power in that 10 ohms as my math shows. Now, let's see your math.

Agreed. The whole diversion about different classes of amplifiers is just that, a diversion - whether it is deliberate obfuscation, or just lack of clear thinking.

The fact that different types of amplifier and/or power supply designs have different quiescent power consumptions has nothing to do with the OP's question. The matching issues are exactly the same, whether you have an ultra-linear class A amp providing 1μW of power to a sensitive experimental device, or a high power broadcast radio transmitter putting out hundreds of kW.