How does an equivalent impedance consume a real power?

[goodphy]

Hello.

Please looks at the attached image first.

The left image is LCC impedance matching network (one of T-type impedance matching networks) and the right is the corresponding equivalent circuit when C_T and C_L are adjusted such that an equivalent impedance of the sub-circuit surrounded by dashed-line becomes R_S.

You don't need to take care of what Z_L really is, that was not important for my question I would like to ask you here. In the left circuit, there is only one resistance, R_L, located in Z_L so that's the only point that a real power (active power) is consumed. But in the right, Z_eq itself is a purely resistive. In order to make sense in terms of power consumption, a real power at Z_eq, I²Z_eq = I²R_S is same to I₁²R_L, I think. But..R_L can be a lower than R_S and I₁ is lower than I. As a result, two consumed powers are not equal.

How can I solve this apparent contradiction? I think reactive powers for reactive components have some role to match these two power. For example, reactive components receive power over a quarter cycle and release it to the R_L for another quarter cycle, not to the source. But I don't know how such an energy flow occurs.

Could you please help to clarify this?

 

[sophiecentaur]

That diagram may be more complicated than necessary to resolve this dilemma.
You can think in terms of a simple, ideal transformer., which avoids the need to do complex calculations and the operative word here is 'transformation'. The load RL is presented with a voltage V2 (the secondary volts). The secondary current will be V2/ RL. and the power delivered will be (V2)^2/RL. The power that flows into the transformer primary will be V1I1, which must be the same as the power dissipated in the load RL. So the resistance presented to the source is V1/I1, which will be RL T² where T is the turns ratio.
The voltage across the load is not the same as the voltage across the primary winding and, if you want to impedance match, the source resistance Rs will not be the same as RL
The different matching network you use in your diagram will still behave in the same way, with RL being presented as a different value to the source. It's what you would expect. If you are bothered about the "Energy Flow" then you can consider the energy that is not dissipated in the load as being stored within the reactive components. There is stored energy but no Power dissipation. (Perhaps that last sentence may be enough explanation)

 

[BvU]

You don't need to take care of what ZL really is, that was not important for my question I would like to ask you here. In the left circuit, there is only one resistance, RL, located in ZL so that's the only point that a real power (active power) is consumed. But in the right, Zeq itself is a purely resistive. In order to make sense in terms of power consumption, a real power at Zeq, I2Zeq = I2RS is same to I12RL, I think. But..RL can be a lower than RS and I1 is lower than I. As a result, two consumed powers are not equal.

How can I solve this apparent contradiction?

You can solve this by not jumping to conclusions as you appear to do. Check the proper expressions including the phases. For complex numbers the > and < aren't directly applicable or useful if you forget about the phases.

 

[goodphy]

That diagram may be more complicated than necessary to resolve this dilemma.
You can think in terms of a simple, ideal transformer., which avoids the need to do complex calculations and the operative word here is 'transformation'. The load RL is presented with a voltage V2 (the secondary volts). The secondary current will be V2/ RL. and the power delivered will be (V2)^2/RL. The power that flows into the transformer primary will be V1I1, which must be the same as the power dissipated in the load RL. So the resistance presented to the source is V1/I1, which will be RL T² where T is the turns ratio.
The voltage across the load is not the same as the voltage across the primary winding and, if you want to impedance match, the source resistance Rs will not be the same as RL
The different matching network you use in your diagram will still behave in the same way, with RL being presented as a different value to the source. It's what you would expect. If you are bothered about the "Energy Flow" then you can consider the energy that is not dissipated in the load as being stored within the reactive components. There is stored energy but no Power dissipation. (Perhaps that last sentence may be enough explanation)

Hello. Thanks for giving a quick answer:)

I'm sorry but I'm afraid I couldn't get your point yet. I know how the transformer works and it seems that you're equating any complex network with a load Z_L = R_L + iX_L to some ideal transformer with a load R_L in its secondary side. I don't know how this correspondence is made.

You use an expression of power as voltage squared divided by an impedance while I use it as current squared multiplied by an impedance, they are both equal. With sticking on my power expression, I have to say these: the power consumed by the sub-circuit surrounded by the dashed-line in left diagram must be P₁ = I²R_S , according to the right equivalent diagram and it is an active power as the load R_S is purely resistive (the active power is the power consumed by a resistance, not reactance). Obviously, the active power of P₁ must be consumed by the sub-circuit somehow. I think P₁ should be consumed by R_L in the left diagram as it is the only resistance in the circuit, so I₁²R_L = I²R_S where I₁ is the current going through R_L. If R_L is lower than R_S , then I₁ is higher than I. However..how this is possible? I is split into I₁ and I₂ so I₁ is lower than I.

I'm sorry not to understand your comment, but I need to get a little more help to follow your sense. What did I miss?

 

[goodphy]

You can solve this by not jumping to conclusions as you appear to do. Check the proper expressions including the phases. For complex numbers the > and < aren't directly applicable or useful if you forget about the phases.

Hello.

I'm sorry but phase? Do you want me to write phase of ..impedance? I'm talking about general case so a specific phase is not relevant. And what do you mean > and < ?

 

[sophiecentaur]

I know how the transformer works

I think that statement should read "I am familiar with what a basic ideal transformer does". A transformer that you would recognise as such is only one of many ways to achieve its function of Impedance Transformation.
A transformer can be regarded as a set of impedances, connected in such a way as to change volts. current and resistances. Yes, there is Mutual Impedance involved in a common or garden transformer but, only when the transformer is 'ideal' (where the various capacitances and self inductances play a minor part in its operation) can we assume it's the only element involved. Then we can make do with the simple 'transformer equations' that we start off with. My example of a familiar thing like a transformer was one which you (I mean you and not "one") felt you 'understand'. But many other arrangements of impedances can produce the same effect as a transformer with an iron core and two windings. In all of them, there is EM energy stored, temporarily, in reactances and that energy can influence the power being dissipated in the resistive elements. That's the philosophy behind reactive circuits and how they can change the apparent resistance values.
It's a pain in the butt to do the full analysis of such circuits but you could choose appropriate reactances such that the V and I referred to the point where the circuit emerges through that box such that V/I = R_S. You could actually produce any resistance you wanted.
Is this the first complex circuit that you have considered? If so, it would be better to start a bit further back in the learning process and do some easier examples.

Do you want me to write phase of ..impedance?

If you plot X and R on an Argand diagram, the angle between them and the resultant Z can be referred to as Phase.
A "specific phase" is relevant when it's referred to the input signal.
I think you need to go back to basics for this.

 

[jim hardy]

In order to make sense in terms of power consumption, a real power at Zeq, I2Zeq = I2RS is same to I12RL, I think. But..RL can be a lower than RS and I1 is lower than I. As a result, two consumed powers are not equal.

maybe i misunderstand the question... but here's how i would answer the question as i understand it..

Of course RL can be lower than RS and the two consumed(dissipated?) powers therefore unequal. But then you have mismatched impedances, and the only downside to that is you have not optimal transfer of power.

Observe that with matched impedance your efficiency is 50% because half the power gets dissipated inside the source.
I assure you that's unacceptable for a multimegawatt central station generator. We want to sell ALL the power not just half of it.
But it's fine for a low power RF signal line, and even desirable because it suppresses reflections.

old jim

 

[jim hardy]

How can I solve this apparent contradiction? I think reactive powers for reactive components have some role to match these two power. For example, reactive components receive power over a quarter cycle and release it to the RL for another quarter cycle, not to the source. But I don't know how such an energy flow occurs.

I well remember struggling with these concepts.

Yes reactive components have some role. You can deduce it intuitively.

Reactance opposes flow of current, so minimizing reactance will maximize current flow. When reactances of load and source have opposite sign and equal magnitude they'll add to zero and that's pretty well minimized, eh ?
Therefore for optimum power transfer you want Zload to be complex conjugate of Zsource.

That logic is easy enough to remember ?

 

[goodphy]

maybe i misunderstand the question... but here's how i would answer the question as i understand it..

Of course, RL can be lower than RS and the two consumed(dissipated?) powers therefore unequal. But then you have mismatched impedances, and the only downside to that is you have not optimal transfer of power.

Observe that with matched impedance your efficiency is 50% because half the power gets dissipated inside the source.
I assure you that's unacceptable for a multimegawatt central station generator. We want to sell ALL the power not just half of it.
But it's fine for a low power RF signal line, and even desirable because it suppresses reflections.

old jim

Hello. Thanks for giving comments.

I think I need to think this problem more carefully, before asking further and further.

However, in order to start right reasoning, I need to confirm one thing: In impedance matching by adjusting C_T and C_L so that Z_eq = R_S, a power dissipated in Z_eq is P₁ = I²R_S, which is a pure active power (non-zero net power transfer). As Z_eq is an equivalent impedance of the sub-circuit (presented by the dashed-line in my figure) and the power on Z_eq is the active power, P₁ must be consumed by the sub-circuit without going back to the power source (unlikely to the active power, a reactive power goes back to the source after being received by a reactive circuit or component so net power transfer is zero). In the sub-circuits, R_L is the only one resistive component so whole amount of power P₁ is dissipated on R_L, eventually.

Could you tell me whether these statements I made are correct or not ? If these are correct, I'm on a right track. If not, there is something big I missed so I need to check up what I was totally wrong with.

 

[jim hardy]

Could you tell me whether these statements I made are correct or not ?

I don't know. Still trying to parse them.
Also your power flows right to left,, unconventional but i finally figured it out.

In impedance matching by adjusting CT and CL so that Zeq = RS, a power dissipated in Zeq is P1 = I2RS, which is a pure active power (non-zero net power transfer).

hmm latex doesn't quote well ?
Regardless of that, yes that'll be the power transferred when impedances are matched.
I'm not sure what your term "pure active" means, though. Are you suggesting the currents I and I1 are in phase ? I don't see that as a condition , yet.

As Zeq is an equivalent impedance of the sub-circuit (presented by the dashed-line in my figure) and the power on Zeq is the active power, P1 must be consumed by the sub-circuit without going back to the power source

I think i agree with that one, because in order for Zeq to match a resistive source impedance Zeq must appear resistive also. Complex conjugate of R +j0 is R-j0, and that's resistive.

(unlikely to the active power, a reactive power goes back to the source after being received by a reactive circuit or component so net power transfer is zero).

did you mean "(unlikely to the REactive power,.." ?
Yes that's what reactance does, shuttles energy back and forth between components that store it.

In the sub-circuits, RL is the only resistive component so whole amount of power P1 is dissipated on RL, eventually.

Sounds right to me. P1 = I1²R_L and R_L is the only circuit element that dissipates power.
Current through R_L ≠ current through R_S though; as Sophie observed your matching circuit behaves like a transformer.

We're speaking of course about average not instantaneous power. That's what you meant when you said "net power" , Correct ?

If these are correct, I'm on a right track. If not, there is something big I missed so I need to check up what I was totally wrong with.

I think you're on the right track. I hope above was some help. Sophie and the radio guys here are better with the maths than i am.

A word on style:

When writing technical stuff i try to break it up into single stepwise thoughts and place one thought per sentence, starting each sentence on a fresh line..
It requires a lot of editing because I'm prone to run-on sentences..
But the payoff is i can read it a week later and probably figure out what i was talking about.
I think also it makes it easier for others to follow my thoughts.
Try it, you might like it. After all, we do our maths that way don't we ?

old jim

 

[goodphy]

I'm not sure what your term "pure active" means, though. Are you suggesting the currents I and I1 are in phase ? I don't see that as a condition , yet.

Please never mind. "Pure active" just means that the power on Z_eq has no reactive power, it only has an active power.

did you mean "(unlikely to the REactive power,.." ?

The original sentence I made was right. I was trying to emphasize a difference between an active power and a reactive power.

A word on style:

When writing technical stuff i try to break it up into single stepwise thoughts and place one thought per sentence, starting each sentence on a fresh line..
It requires a lot of editing because I'm prone to run-on sentences..
But the payoff is i can read it a week later and probably figure out what i was talking about.
I think also it makes it easier for others to follow my thoughts.
Try it, you might like it. After all, we do our maths that way don't we ?

old jim

Thanks for giving me some advice. I'll consider your suggestion of writing style when I have a new post.

All right. It seems my original reasoning is right, but I still have to figure out how I₁ can be larger than I. Instanteously, I₁ can be smaller than I but on average, it must be higher than I in order to match up powers, P_1,net = <I²R_S> and P_2,net = <I₁²R_L>. Maybe I₁ can be larger in other instantaneous time when currents from reactive components reinforce it.

I appreciate your kind help. You made me felt more clear view.

 

[jim hardy]

but I still have to figure out how I1 can be larger than I. Instanteously, I1 can be smaller than I but on average, it must be higher than I in order to match up powers, P1,net = <I2RS> and P2,net = <I12RL>

Hmmm i was thinking the other way, your matching network acts as a step up transformer not step down .

Maybe I1 can be larger in other instantaneous time when currents from reactive components reinforce it.

or maybe you'd need a different circuit topology.
Wikipedia suggests that step-down uses reactance in parallel not series with the source

L-section

Basic schematic for matching R1 to R2 with an L pad. R1 > R2, however, either R1 or R2 may be the source and the other the load. One of X1 or X2 must be an inductor and the other must be a capacitor.

L networks for narrowband matching a source or load impedance Z to a transmission line with characteristic impedance Z0. X and B may each be either positive (inductor) or negative (capacitor). If Z/Z0 is inside the 1+jx circle on the Smith chart (i.e. if Re(Z/Z0)>1), network (a) can be used; otherwise network (b) can be used.[3]
A simple electrical impedance-matching network requires one capacitor and one inductor. In the figure to the right, R1 > R2, however, either R1 or R2 may be the source and the other the load. One of X1 or X2 must be an inductor and the other must be a capacitor. One reactance is in parallel with the source (or load), and the other is in series with the load (or source). If a reactance is in parallel with the source, the effective network matches from high to low impedance.

The amateur radio buffs here are versed in practical side of impedance matching.http://www.hamuniverse.com/kb1lqc41balun.html

 

[sophiecentaur]

that's unacceptable for a multimegawatt central station generator. We want to sell ALL the power not just half of it.

Yes. That point was made to me the first time the maximum power theorem was introduced.
But the problem of reflections is solved with a good match into the load, at the other end of the feeder line. There are then, no reflected signals hitting the source (=transmitter). This applies to high powered RF transmitters. When appropriate, they are made with high efficiency / very non-linear output stages and they, again, cannot afford to be only 50% efficient. So impedance matching is only vital at the antenna end of the transmission line. For a high power valve transmitter, the output stage needs to be 'tuned' to present the right impedance for the efficient operation of the valve rather than to match the Z₀ of the line
Audio amplifiers seldom have an 8Ω output impedance - often, they are 'Voltage sources" with an output impedance of less than 1Ω. You can expect a (basically) emitter follower style of output stage.
There are many occasions when matching at the source end is important - particularly with measuring equipment, where a standing wave can be a huge embarrassment.