Calculate Coaxial Cable Loss & Input Impedance at 1 MHz

[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.

 

[Baluncore]

The instruction to pilots regarding crash landings, is that they must look at the gap between the trees, rather than the trees that are in the front of their mind. It is well known that, where they focus on the obstacle, they tend to collide with it.

I want to navigate a safe course between Scylla and Charybdis. I do not doubt the MPTT.
There are often engineering solutions that avoid the close encounter and come through it economically.

The fascination with engineering a collision in order to prove the MPTT exists is quite unnecessary.
It completely distracts attention away from the methods available to avoid poor efficiency.

 

[Baluncore]

AlephZero and Averagesupernova;
I quite agree that an amplifier with a 10 ohm output resistance is going to be inefficient for obvious reasons.
It must be matched to the load in order to transfer maximum power with 50% efficiency.

But why have a 10 ohm Rs when you can have very much less ?

The fact is that amplifier class is critical to overall efficiency because it determines amplifier output impedance.
With class A the slope at the middle of the load-line must be matched to the load. That cripples the efficiency from the start.

With class D, only the ends of the load line are connected by the timing of the switching selected. That permits very high efficiency because it transfers the low output resistance point, from the output of the power supply, to the output of the amplifier.

The fundamental component that enables efficiency with class C or D is the inductor.
The PWM of the inductor regulates the power flow from the low impedance power supply.

So prohibition of inductors leads directly to prohibition of amplifier classes C and D.
That guarantees a class A collision with the MPTT at 50% efficiency.

 

[sophiecentaur]

The instruction to pilots regarding crash landings, is that they must look at the gap between the trees, rather than the trees that are in the front of their mind. It is well known that, where they focus on the obstacle, they tend to collide with it.

I want to navigate a safe course between Scylla and Charybdis. I do not doubt the MPTT.
There are often engineering solutions that avoid the close encounter and come through it economically.

The fascination with engineering a collision in order to prove the MPTT exists is quite unnecessary.
It completely distracts attention away from the methods available to avoid poor efficiency.

More stories and still no Maths. All you need to do is to show that an amplifier that is 90% efficient when feeding into a 50Ω load on the end of a 50Ω feeder (so the transmitter has a source resistance of 6Ω), when operating to its spec., can, itself, be made to present a 50Ω termination for signals reflected from the load. The problem can't be reduced further so it should be easy for you to show that you are correct.

Is that simple bit of analysis not allowed in your world? Whatever you remember - or think you remember having been told or having measured in the past, the above has to be proved for your story to hold. Of course, it cannot.

Steer between the trees of nonsense and aim at the reliable landing strip of Mathematics. It will not fail you.

 

[Baluncore]

There is no reflected energy.
QED.

 

[Baluncore]

All you need to do is to show that an amplifier that is 90% efficient when feeding into a 50Ω load on the end of a 50Ω feeder (so the transmitter has a source resistance of 6Ω), when operating to its spec., can, itself, be made to present a 50Ω termination for signals reflected from the load. The problem can't be reduced further so it should be easy for you to show that you are correct.

Is that simple bit of analysis not allowed in your world? Whatever you remember - or think you remember having been told or having measured in the past, the above has to be proved for your story to hold. Of course, it cannot.

If you want maximum power you should use a transformer to match the amplifier's Rs of 6 ohm to the 50 ohm line. Since the transformer is broadband and reciprocal it will correctly match the line to the amplifier's Rs for all reflected or received energy.

If the transmitter was operated near others in the same band, I would avoid a broadband transformer and use a hybrid cross connected transformer pair since the presence of the received signal from a nearby transmitter will cross modulate in the output stage of the transmitter and produce all sorts of out of band spurious signals. It is best to reflect the other signal or absorb it in a loaded hybrid than to allow it access to the final stage.I am not an imbecile.

 

[sophiecentaur]

You really don't get this, do you?
If you use a transformer to match the 6 Ω to a 50 Ω load, the transmitter will not be 90% efficient - will it? So it will not be operating as it was designed (to work into 50Ω). It will see 6Ω and not 50Ω. It will possibly melt, in fact because, instead of only 10kW dissipation in its anode, it will be dissipating around 100kW.
What have all your specifics got to do with the basic principles? We are operating a single transmitter under ideal conditions. The style of transformer is totally irrelevant. It has just two ports and is a reciprocal network. It is not magic. If you cannot show the analysis to demonstrate how, in your world, the MPT does not apply then it does apply. I thought Engineering was a discipline. Professional Engineers can justify their designs and practice with calculations. Where are yours?

 

[Baluncore]

You really don't get this, do you?

Oh yes I do.

All you need to do is to show that an amplifier that is 90% efficient when feeding into a 50Ω load on the end of a 50Ω feeder (so the transmitter has a source resistance of 6Ω), when operating to its spec., can, itself, be made to present a 50Ω termination for signals reflected from the load. The problem can't be reduced further so it should be easy for you to show that you are correct.

It is absolutely clear that there can be no reflected energy in this situation.
It does not take mathematics to show that 50 ohm = 50 ohm = perfect match.

Was this a trick question, or are you so buried in your worship of theorems and mathematics that you don't understand the real world ?

Now if you want to be even more malevolent and mismatch the load to reflect some energy back down the line, which would be a very poor engineering practice, then I can protect my valuable transmitter, while still maintaining the 90% efficiency, by using a hybrid cross connected transformer pair with a 1:1 ratio. If this was a microwave installation then instead of a hybrid I would use a circulator. That would divert all the reflected energy into a dummy load. It will also prevent cross-modulation problems from received signals because the reverse traveling energy will not reach the output stage.

 

[Baluncore]

The style of transformer is totally irrelevant. It has just two ports and is a reciprocal network. It is not magic.

A hybrid cross connected transformer pair has four ports and is the low frequency equivalent of a circulator. A hybrid will work from the audio frequencies in a telephone all the way up to VHF. A hybrid is not the most efficient way to handle reflected energy but it is reliable and quite independent of line length variation, which really helps in switched networks and multicouplers.

 

[sophiecentaur]

It is absolutely clear that there can be no reflected energy in this situation.
It does not take mathematics to show that 50 ohm = 50 ohm = perfect match.

So how do you claim 90% efficiency is possible for the transmitter?
You are trying to shift the goal posts now. Originally you were implying that everything was matched in practice (both ways) and yet a class C transmitter can still operate at high efficiency. What is your position now?
Why do you keep bringing up irrelevant things like hybrids and circulators? More smoke and mirrors rather than a short bit of Maths. HF transmitters use neither. Also, real HF antenna arrays do not present a perfect match over their band. VSWRs of worse than 2 are not uncommon. You will be aware that one HF array may need to operate at several different frequencies in one day. They cannot be tiffled every 15 minutes.
BTW, are there any Class C microwave amplifiers? Why do you need to introduce waveguides into this? More smoke and mirrors.

Can you just concentrate on the basic question here. Can you show that you can perfectly 'match' with a load and still get 90% efficiency? If reactances are necessary to prove your point then show how they work (quantitatively). None of the Maths involved need be too difficult and no magic is allowed. Also, no 'helpful' information about operating in the presence of other signals is needed. If there are some practicalities are involved, then include them in your calculations. (All this stuff is calculable, you realize) Failing that, give a good reference about standard transmitter practice with evidence of what you claim. The Marconi HF Transmitter spec that I read, recently, made no mention of output impedance - just the range of load impedances it works into.
That's all there is to it.

 

[Baluncore]

Was this a trick question, or are you so buried in your worship of theorems and mathematics that you don't understand the real world ?

You have not answered my question.
You are an behaving like an incessant bully.

 

[sophiecentaur]

You have not answered my question.
You are an behaving like an incessant bully.

I am only trying to insist that you justify your original groundless statement. If you just said you don't know or you are not sure, the I would give up. Someone could be reading your misinformation and believe it. Should we allow that to happen?
PF standards on veracity are high. You need to back up statements or withdraw them.

 

[ZapperZ]

Closed, pending moderation.

Zz.