0
\$\begingroup\$

I'm trying to understand general concepts involved in impedance matching and I have question about the conditions which should be satisfied by the matching network. Assume we have in most general setting the classical problem to match impedance between a source S with impedance \$Z_S\$ and load L with impedance \$Z_L\$:

enter image description here

The most general strategy is to put a impedance matching network between source and load (which is a circuit that can be relatively simple like L- or T-network, or a transformator, but even much more complicated in dependence of the actual problem) and the task of the practical impedance matching is then to adapt the parameters of the components in the matching network such that the impedances of source, load and matching network satisfy certain conditions.

Now I read that the conditions on the matching network that is required to be satisfied by involved impedances depend on two auxilary impedances: the input impedance \$Z_{in}\$ and the output impedance \$Z_{out}\$.

These two arise as follows. We can replace the network

enter image description here

Firstly (I) by an equivalent circuit

enter image description here

where \$Z_{in}\$ "summarizes" the impedance of the matching network & load replaced in one common impedance \$Z_{in}\$. In case of L-matching network here Andy aka provided an explicit calculation of \$Z_{in}\$

and secondly (II) we can also replace it by

enter image description here

the generator & the impedance \$Z_{out}\$ replace the source (generator & \$Z_{S}\$) and the matching network.

Note that here the calculation of \$Z_{out}\$ is a bit more subtile because here we have take into account the generator \$V_S\$. But the derivation of \$Z_{out}\$ from given \$Z_{S}\$ and the matching network is not the subject of this question. If one have fixed a matching box there is an arsenal of techniques available to do it, eg by transforming the source & matching network by a sequence of Norton and Theverin transformations until it has the desired shape in the picture above.

My concrete Question is: Assume we can calculate \$Z_{in}\$ and \$Z_{out}\$ as functions of the parameters of the components sitting in the matching network and our goal is to find the optimal parameters for these components in order to archieve an impedance match via maximum power transfer.

Which are in most general setting the conditions on \$Z_{S}\$, \$Z_{L}\$, \$Z_{in}\$ and \$Z_{out}\$ which should be satisfied in order to obtain the impedance match?

Proposal: should always simultaneously satisfied both - (I) \$Z_{S}^*= Z_{in}\$ and (II) \$Z_{out}= Z_{L}^*\$ - or does in suffice only to archieve (I) \$Z_{S}^*= Z_{in}\$?

Why I'm asking this: several times I found in the net tutorials about impedance matching methods that the authors only focus on \$Z_{in}\$ and waive the discussion about \$Z_{out}\$ completely.

\$\endgroup\$
1
  • \$\begingroup\$ @MathKeepsMeBusy: yes, or even more simpler: assume I fix for all next considerations a certain frequency and try to reach maximize power transfer only just for it. the question is about which explicit comditions should be them tried to be satisfied when designing the matching network. I proposed (I) and (II) but I'm not sure if these two are exactly which we should try to archieve. it looks logical to me but I nowhere found a source which exactly states that these (I) and (II) are exactly the conditions which we should try to satisfy. \$\endgroup\$ Dec 9, 2021 at 3:28

3 Answers 3

1
\$\begingroup\$

If you do matching at one port the other port will also get matched. Consider a simple L impedance transformer:

schematic

simulate this circuit – Schematic created using CircuitLab

It is easy to show that around resonance frequency: $$R_{in} \approx \frac{R_P}{\omega^2C^2R_P^2} = \frac{R_P}{Q^2}$$

Now, suppose you have a matched source at input, as shown below:

schematic

simulate this circuit

The output impedance becomes \$R_{out} = Q^2R_{in} = R_P\$, which is matched to load impedance. So you need to match just once either at output or input. This can be proved in general for any matching network.
There could be other ways to prove it but this is what I can think of. Consider the matching network as shown below which is transforming source impedance \$Z_s\$ to load impedance \$Z_l\$. Let's say that the input is matched for maximum power so \$Z_{in}= Z_s^*\$

schematic

simulate this circuit

Now, let's switch and apply the voltage at the other port of the matching network as shown below:

schematic

simulate this circuit

Since impedance looking into the input port is still \$Z_s^*\$. This means that maximum power is transferred from the matching network into the source impedance. Since the matching network itself is lossless, this implies that this power is same as that transferred to the output impedance of the matching network, \$Z_{out}\$.
Thus, the only way that the power transferred to the source can be maximum is if power transferred to the load \$Z_{out}\$ is maximum. That is if, \$Z_{out} = Z_l^*\$.

\$\endgroup\$
8
  • \$\begingroup\$ nice, so the point you want to make clear is that in case of \$Z_S\$ and \$Z_L\$ resistive (i.e. \$Z_S=R_S\$ and \$Z_L=R_L\$ it happens that it indeed suffice to archieve only the first condition (I) \$ Z_{in} =Z_{S}^*\$ which in resistive case is equivalent to \$ Z_{in} =R_S\$. and then the second condition I suggeted \$ Z_{out}= Z_{L}^*\$ is then automatically also fullfuled? \$\endgroup\$ Dec 9, 2021 at 12:29
  • \$\begingroup\$ if yes, does this holds only for \$Z_S\$ and \$Z_L\$ resistive or for arbitrary \$Z_S\$ and \$Z_L\$ as well? \$\endgroup\$ Dec 9, 2021 at 12:31
  • \$\begingroup\$ Yes that is correct... This holds for arbitrary Zs and Zl as long as the matching network is lossless. Obviously, nothing is lossless in reality but losses are pretty low for a well defined matching network. And this would be true for all practical purposes. \$\endgroup\$
    – sarthak
    Dec 9, 2021 at 14:11
  • \$\begingroup\$ is there an easy way to check this statement that if our network satisfies \$Z_{in} =Z_{S}^*\$, then \$Z_{out} =Z_{L}^* \$ must be true as consequence and vice versa? already in order to calculate \$Z_{out}\$ alone one has to calculate Thévenin equivalent circuit of the whole stuff left to the \$Z_{L}\$ and that's tortuous in general. do you know a more elegant way to verify that the condition \$Z_{in} =Z_{S}^*\$ suffice to garantee that \$Z_{out} =Z_{L}^* \$ must be also hold? \$\endgroup\$ Dec 17, 2021 at 0:20
  • \$\begingroup\$ could you lose a couple of words on why in your conderations the input and output impedances are related by \$R_{out}=Q^2 R_{in}\$ to each other? can this argument be generalized to general not pure resistive \$Z_{out},Z_{in}\$? \$\endgroup\$ Dec 19, 2021 at 13:53
1
\$\begingroup\$

For Reflectionless Matching in both directions, all ports are matched Z(f) both in and out. But may be constrained over a limited range of frequencies. . Although a Rule of thumb for Return Loss is -15 dB minimum. So the reflection echo must be lost by at least 15 dB. The down side of course is 50% of the power is lost at the source.

Have you heard of conjugate matching? That is for Maximum Power Transfer by Conjugal vists ;) .. strike that

by CONJUGATE IMPEDANCE MATCHING

It can be mathematically explained as:

If \$Z_S=R_S+jX_S ~and~ Z_L=R_L+jX_L,\$ then the condition for max power txfr or MPT is by conjugate matching: \$R_S=R_L ~and~ X_S= - X_L\$. .

nitty gritty http://bwrcs.eecs.berkeley.edu/Classes/icdesign/ee142_f10/Lectures/Lect8_2up.pdf

Although you would never considering try to get maximum power transfer from the grid or a battery as this also implies with Matched Impedances you get a 50% drop in voltage and equal power dissipated in source and load at 50% efficiency, but it would be technically maximum power transfer, just not efficient.

Thus rated power is usually defined by some tolerance like 10% voltage or 85'C power KVA output rating. Load regulation is thus a ratio of these impedances , as a voltage divider R(source / source + load)*100%) = worst case Load regulation error Typically DC power supplies are 2% range or so and you can estimate source impedance from this ratio . The same is true for battery load regulation (effective series resistance in source of power to load to load resistance ratio.) Although not an example of impedance matching, when you compute Vout / I max to compute load to match that rated power with Voltage Regulation error, these all use the same variables.

Although for AC grid control of power factor (p.f.), it requires conjugate . impedance balancing of just X(f) for a small range of frequencies. Thus large tank capacitance is added to connjugate match the inductance from industrial motors

Another application of Matched Impedances. Solar Cells for DC power.

But what if you have a solar powered current source? Yes MPT also uses Matched Imepdances. The voltage usally starts around 82% of Voc then drops to about 70% of Voc for the useful range.

\$\endgroup\$
1
\$\begingroup\$

My concrete Question is: Assume we can calculate Zin and Zout as functions of the parameters of the components sitting in the matching network and our goal is to find the optimal parameters for these components in order to achieve an impedance match via maximum power transfer..... Which are in most general setting the conditions on ZS, ZL, Zin and Zout which should be satisfied in order to obtain the impedance match?

Quite simply this (where impedances are purely resistive): -

$$Z_S = Z_{IN} \hspace{1cm}\text{ and }\hspace{1cm} Z_{OUT} = Z_{L}$$

If you want to complicate things by a assuming source and load are not purely resistive, then make them resistive by cancelling out their capacitive reactance with series inductive reactance (or vice versa) at the operating frequency required. This can also be done using a parallel opposing reactance to make a parallel tuned circuit.

It's a thoroughly tried and tested technique used by many RF engineers who need to match complex antenna impedances to complex (and different) chip impedances.

the generator & the impedance Zout replace the source (generator & ZS) and the matching network.

Not quite true; the original source voltage generator does not necessarily become the same source voltage shown in your diagram here: -

enter image description here

\$V_S\$ will be different depending on how you decide to implement the matching components. You appear to be assuming that matching components will be reactive components but, many methods use resistive attenuators or transformers.

or does in suffice only to achieve (I) \$Z_S=Z_{IN}\$?

For a matching network like an L-pad (example): -

enter image description here

The formulas are: -

  • \$L = R_{IN}\cdot R_L\cdot C\$
  • \$C = \dfrac{1}{\omega\cdot R_L}\sqrt{\dfrac{R_L}{R_{IN}-1}}\$

As you should be able to see, both \$R_{IN}\$ and \$R_L\$ are numerically involved in the solution of the values of the two reactive components (L and C).

Hence, if the power into the input of the impedance matching network is (say) 1 watt, then that 1 watt is passed to an input impedance that is wholly resistive at the right frequency (\$\omega\$). Because that power cannot be dissipated in either L or C, it follows that it can only be dissipated in the output loading resistor. So, 1 watt (in this example) is dissipated in the 300 Ω resistor.

From this, there can be no other meaning other than the output impedance of the LC network is 300 Ω. I also showed you this simulation previously: -

enter image description here

Simulation of the impedance looking back into the matching network's output: -

enter image description here enter image description here

Above images from this answer.

It demonstrates that the impedance looking into capacitor C1 is 300 Ω (when the source is set to be 50 Ω). If you are insisting on a mathematical proof, then please be clear about this.

Note that here the calculation of \$Z_{OUT}\$ is a bit more subtle because here we have take into account the generator \$V_S\$

No you don't, you can assume the generator is 0 volts and hence a short to ground. Subtle isn't maybe the right word. Using the circuit above: -

enter image description here

Summary

  • I've given a wordy explanation of what the output impedance is
  • I've given a simulation (that agrees)
  • I've given a mathematical proof (that also agrees)
  • In another of your questions I've shown that a series to parallel conversion also produces the correct value.
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.