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I'm failing to understand something. Impedance matching by complex conjugate can be defined as this: "For two impedances to be complex conjugates, their resistances must be equal, and their reactances must be equal in magnitude but of opposite signs. " But, I have seen many Smith chart examples, where it looks like people add only reactive elements (while starting out with different resistive impedances in the source and load), and somehow achieve the complex conjugate. I do not understand why. It seems contradictory to the definition: " their resistances must be equal, and their reactances must be equal in magnitude but of opposite signs". In the video screenshot below for example, the resistive impedances in the source and load are different, but somehow a complex conjugate is achieved anyway.

I might be misunderstanding something fundamental. I usually tend to be able to understand things. I asked two other questions on a related topic, but, they all derive from this fundamental issue that I cannot understand, so it is better that I ask about it directly.

An easy example of how reactive elements can change resistive impedance, could be good. The simplest possible, with as few parts as possible.

enter image description here

Edit: I've now had the effect demonstrated to myself by simulating Andy's example in Falstad. And that is what I was looking for. The effect is similar to a boost converter or other voltage increasers. It wastes a lot of current to achieve the voltage boost, just like a boost converter.

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    \$\begingroup\$ Yes reactive components can change the resistive component. Designing VHF and UHF Class C transmitters, we regularly transformed the transistor output impedance ( 1 or 2 ohms ) using shunt capacitors and series inductors to 50 ohms impedance. \$\endgroup\$
    – Marla
    May 5, 2023 at 23:32
  • \$\begingroup\$ I mean to ask more: how? I know they can, I can easily follow the lines in the Smith chart. As for how, I see some contradictions. It seems to have to do with shunts, parallel reactive elements. But, during impedance matching, the reactance should be neutralized and the elements at resonance. Then, any shunt should act like a short circuit. This not just decreases the resistance of the load, it removes it completely. To not remove it, there has to be some reactance in shunt. Then, maybe the resonance is when source+load is combined, but not locally in the load. Overall, I am missing something. \$\endgroup\$
    – BipedalJoe
    May 5, 2023 at 23:41
  • \$\begingroup\$ We were making wideband transmitters, resonance was undesirable. \$\endgroup\$
    – Marla
    May 6, 2023 at 0:00
  • \$\begingroup\$ It's simply arithmetic on the complex plane. You have two operations available to you: parallel (Zeff = Z1 Z2 / (Z1 + Z2)) and series (Zeff = Z1 + Z2). All else can be derived from this. If you aren't understanding impedances and matching, you really aren't understanding complex arithmetic. Admittedly it's one of the worst named sets, but therein seems to lie your confusion. \$\endgroup\$ May 6, 2023 at 0:20
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    \$\begingroup\$ @BipedalJoe I agree with your statement about not using math but, on this particular occasion it's not possible; it's fundamentally a math thing. \$\endgroup\$
    – Andy aka
    May 6, 2023 at 10:13

2 Answers 2

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I have seen many Smith chart examples, where it looks like people add only reactive elements (while starting out with different resistive impedances in the source and load), and somehow achieve the complex conjugate.

When matching different resistive impedances, the use of complex conjugates is not as obvious as when matching same-value resistive impedances. The approach taken uses reactive components in a different way. Nevertheless, complex conjugate matching does take place.

An easy example of how reactive elements can change resistive impedance, could be good. The simplest possible, with as few parts as possible.

It comes down to circuit examples like this (called L-pads): -

enter image description here

The above image (taken from my basic website calculator) uses an inductor and parallel capacitor to match a 50 Ω source to a 300 Ω load at at a particular operating frequency.

The math is on the link above but, taking a different route, just consider this well-known fact; a parallel capacitor and resistor (\$C_{PAR}, R_{PAR}\$) can be converted to a series capacitor and resistor (\$C_{SER}, R_{SER}\$) with an identical complex impedance at a particular frequency.

And, we choose \$C_{PAR}\$ so that the resulting equivalent \$R_{SER}\$ matches the value of \$R_{SOURCE}\$. Thus we achieve a complex conjugate situation but via a slightly more deviating route.

I might be misunderstanding something fundamental.

Parallel values (\$C_{PAR}, R_{PAR}\$) \$\rightarrow\$ series values (\$C_{SER}, R_{SER}\$) at a particular frequency: -

enter image description here

The image above (edited by me) was taken from this calculator. Basically I started with \$C_{PAR}\$ of 118.627 pF in parallel with 300 Ω resistive and, after pressing the button I got an equivalent series resistance of 50 Ω.

In short, at 10 MHz, a parallel capacitor of 118.627 pF and 300 Ω resistor is equivalent to a series circuit of 50 Ω and 142.353 pF. And, if you explore this a tad more, you will find that the equivalent series capacitance of 142.353 pF (-j111.8) is an equal magnitude reactance to the 1.779406 μH (+j111.8) inductor shown in the calculator below (from my basic website): -

enter image description here.

Hence, complex conjugates are used although it's a little tricky to see on first view. The title question (that I initially forgot to answer) is this: -

Can reactive components affect resistive impedance of circuit

Absolutely they can and hopefully, that should now be clear.

I asked two other questions on a related topic, but, they all derive from this fundamental issue that I cannot understand, so it is better that I ask about it directly.

Yes, please revisit them and note that on the one I answered I have tried to make my answer clearer in lieu of this question.


Why a simple L-pad provides 100% energy efficient impedance matching

The example above makes a 300 Ω load to look like a 50Ω load. So, if 1 volt RMS is applied to the input, 20 mA RMS is drawn i.e. it appears to be a 50 Ω load. That's a power in of 20 mW.

Because it's a 100% efficient process, the power to the 300 Ω load is also 20 mW and, by simple math, we know that the load voltage is 2.44949 volts.

  • Using ohm's law, the current is 8.165 mA.
  • So, the input current is 20 mA and the output current is 8.165 mA.
  • That's a current ratio of 2.44949:1

The voltage step-up ratio of 1:2.44949 might be usefully verified using a simulator: -

enter image description here

So, although the current in the load is reduced by 2.44949, the voltage increases by 2.44949 hence, the power remains the same i.e. a 100% energy efficient process. While I'm at it it's worth showing the input impedance in terms of reactance magnitude and phase angle: -

enter image description here

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  • \$\begingroup\$ power of 5 mW on stades.co.uk/Impedance%20TX/Zmatch%20Lpad%20LP.html, is 25% of 1 V through 50 Ohm (and 100% of 0.5 V through 50 Ohm), is maximum power that can be transferred theoretically around 25% then? and any non-matches will be less than 25%? any conceptualization around that 25% figure in literature or is it a dumb way to think about it? \$\endgroup\$
    – BipedalJoe
    May 6, 2023 at 19:06
  • \$\begingroup\$ A power of 5 mW in a 50 ohm resistor is 0.5 volts. For the rest of what you said I'm not understanding what you mean or how it connects to the website. The number "25" does not appear on that webpage @BipedalJoe \$\endgroup\$
    – Andy aka
    May 6, 2023 at 19:31
  • \$\begingroup\$ well you don't have to. \$\endgroup\$
    – BipedalJoe
    May 6, 2023 at 19:50
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    \$\begingroup\$ @Andyaka - If you have concerns with another answer, you are allowed to politely & constructively raise those concerns in comments on that answer. You have done that, which is fine. It is not allowed to turn one answer into arguments against another answer. We've seen it before, it ended badly (and with suspensions for site disruption IIRC). Therefore I have rolled-back your edit. In this case, if you are unable, with polite comments (or now chat, as comments have become excessive) to reach a meeting of minds on another answer, please disengage. The OP can't be forced to change their answer. \$\endgroup\$
    – SamGibson
    May 8, 2023 at 15:59
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    \$\begingroup\$ @Andyaka - Re: "is it OK if I reinstate the amendment [...]" I can't say yes or no until you do it, to see exact wording etc. Doing that would be a risk. || Clarification of an answer is allowed & encouraged, but an answer must be an answer to the question, not a disagreement with another answer. The more you phrase any edit of your answer to better address the question, rather than responding to another answer, the less likely it is to appear as a disagreement with another answer or as "meta". It would be less risky to either disengage or to reply politely in the chat instead. Thanks. \$\endgroup\$
    – SamGibson
    May 8, 2023 at 16:34
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The short answer is: yes, but, the terminology might be misleading. Andy gave a great example of a simple circuit where reactive components "change the real component of the impedance". But, what the reactive components were really doing, was act like a "voltage increaser", much like a boost converter. At the cost of drawing a lot more current (6x more in Andy's example), they could pump up the voltage ahead of the load, so that the power P = V^2/R was increased. A new person learning about the topic might be looking more for a resistance decrease than voltage increase (even if they are the same from one point of view, it might not be the easiest point of view to take for the person asking the question) and that is why I mention that for anyone else who had the same question I had.

The concept of "maximum power transmission" can be explained in a simple way. The source has a resistance that you cannot change (although you can remove the reactance. ) This means it will act as a voltage divider together with the load resistance. And, the power in the load happens to peak at 50% voltage divider. If more than half is lost at source, there is less voltage left at load. If less than half is lost at load, there is higher series resistance between source and load.

Since 50% voltage divider gives highest power in load, the resistance/impedance of the load should ideally be changed to the same value as the source (so, either increased, or decreased, unless it is exact same to start with. )

The reactance of the source can conveniently be removed, by the opposite phase in the load. This reduces source impedance, and increases power in load by decreasing the series resistance of source and load (but, from that increased power, we still have to divide the source voltage by half with a load resistance matched to the source, but we start off from a better position. )

And, then back to the question: can reactive components affect resistive impedance of circuit. They can for example, as in Andy's example, act as a "voltage increaser", to make the 300 Ohm load "behave" as if it was a 50 Ohm resistor. But, it only "behaves" that way, because the voltage was increased before it by 2.45x. I think the terminology used is a bit misleading, and that it can make it harder for beginners to learn the topic. At least I feel like if I had it explained to myself in the way I explained it here, I would have gotten it quicker.

Credit to Andy's example on his website: http://www.stades.co.uk/Impedance%20TX/Zmatch%20Lpad%20LP.html.

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    \$\begingroup\$ There is no power loss in the circuit example I showed. \$\endgroup\$
    – Andy aka
    May 6, 2023 at 22:47
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    \$\begingroup\$ You said "At the cost of wasting a lot more current" and, there is no current wasted at all; all the energy taken from the output of the 50 ohm source is transferred to the higher resistance load. Nothing is wasted. The analogy with a boost converter is not great; it's much closer to a step up transformer that is optimized at one frequency. And no, absolutely no power increase at all. I don't know how you could possibly think that power can be increased. All my comments are on your first paragraph because I gave up reading your answer after that due to the anomalies. \$\endgroup\$
    – Andy aka
    May 7, 2023 at 9:31
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    \$\begingroup\$ And, I don't know why you are making such a significantly flawed commentary of my answer in your answer. Then you criticize my answer by saying it only behaves that way because the circuit increases the output voltage; this shows a lack of comprehension of what energy efficient impedance converters do; there is NO OPTION other than to increase the voltage when stepping up a lower impedance to a higher one. \$\endgroup\$
    – Andy aka
    May 7, 2023 at 9:36
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    \$\begingroup\$ You are criticizing my answer nevertheless and therefore, I am entitled to point out that your criticism is baseless. \$\endgroup\$
    – Andy aka
    May 7, 2023 at 14:57
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    \$\begingroup\$ You said this: At the cost of wasting a lot more current (6x more in Andy's example), they could pump up the voltage ahead of the load, so that the power P = V^2/R was increased. <-- and that is an utter falsehood. This is the sort of thing that I strongly object to about your answer. The method I show is 100% energy efficient. \$\endgroup\$
    – Andy aka
    May 8, 2023 at 12:07

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