L Impedance matching network problem

In summary, the L impedance matching network is used to match the impedance of the source and load. The calculation shows that there is no reflection between the source and the rest of the components because of the matching, and that there is also no reflection between the load and the rest of the components because of the matching. Power is fully transferred to the load from the source, and there is no local standing wave between the start point of the reactive components and the load.
  • #1
goodphy
216
8
Hello.

The attached image is L impedance matching network, use for impedance matching between source and load impedance.

The calculation shows that in this circuit, load impedance Rp sees whole impedance (equivalent impedance) of the source impedance and reactive components as matched to itself and the source impedance also see matched equivalent impedance for other impedance components.

Thus I have a conclusion:

1. There is no reflection between source impedance and rest of the components because of matching.
2. There is also no reflection between load impedance and rest of the components because of matching.
3. Power is fully transferred to the load from source.

Could you tell me my interpretation is right? Actually I've wonder if there is local standing wave between start point of the reactive components and the load...
 

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  • #2
goodphy said:
Hello.

The attached image is L impedance matching network, use for impedance matching between source and load impedance.

The calculation shows that in this circuit, load impedance Rp sees whole impedance (equivalent impedance) of the source impedance and reactive components as matched to itself and the source impedance also see matched equivalent impedance for other impedance components.

Thus I have a conclusion:

1. There is no reflection between source impedance and rest of the components because of matching.
2. There is also no reflection between load impedance and rest of the components because of matching.
3. Power is fully transferred to the load from source.

Could you tell me my interpretation is right? Actually I've wonder if there is local standing wave between start point of the reactive components and the load...
goodphy said:
Hello.

The attached image is L impedance matching network, use for impedance matching between source and load impedance.

The calculation shows that in this circuit, load impedance Rp sees whole impedance (equivalent impedance) of the source impedance and reactive components as matched to itself and the source impedance also see matched equivalent impedance for other impedance components.

Thus I have a conclusion:

1. There is no reflection between source impedance and rest of the components because of matching.
2. There is also no reflection between load impedance and rest of the components because of matching.
3. Power is fully transferred to the load from source.

Could you tell me my interpretation is right? Actually I've wonder if there is local standing wave between start point of the reactive components and the load...
You do not give numbers, so you need to check that Rs and Rp are the correct way round. The resistor across the shunt reactance needs to be expressed as a parallel resistance, and that in series with the series reactive element needs to be expressed as a series resistance. The parallel one must be the greater. (Any reactances in source or load need to be "added" to the network).
Your conclusions look basically correct. In Item 3, there is always some loss in the L-network, mainly arising from the losses in the inductor. Regarding standing waves, this is basically correct, although strictly speaking the network is a lumped element network rather than a transmission line network. The network stores reactive energy in the same way as do standing waves on a transmission line. The network could perhaps be viewed as a lumped element line.
 
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  • #3
@ goodphy. You are generally correct, but making an assumption that your signal is a continuous pure sinewave without any signal bandwidth.
For wider band signals you should use several stages of L network with intermediate impedances being the geometric mean of the end impedances. By staggering the resonant frequencies of elements you can then correct gain and phase shift over the band.
goodphy said:
Actually I've wonder if there is local standing wave between start point of the reactive components and the load...
There is a circulating current between the L and C that could be seen as the equivalent of a standing wave on an equivalent transmission line. But one L and one C does not a transmission line make. It is better to consider the L network as a flywheel.
 
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  • #4
tech99 said:
You do not give numbers, so you need to check that Rs and Rp are the correct way round. The resistor across the shunt reactance needs to be expressed as a parallel resistance, and that in series with the series reactive element needs to be expressed as a series resistance. The parallel one must be the greater. (Any reactances in source or load need to be "added" to the network).
Your conclusions look basically correct. In Item 3, there is always some loss in the L-network, mainly arising from the losses in the inductor. Regarding standing waves, this is basically correct, although strictly speaking the network is a lumped element network rather than a transmission line network. The network stores reactive energy in the same way as do standing waves on a transmission line. The network could perhaps be viewed as a lumped element line.

Oh I thanks. I really appreciate you!

The last question is somewhat involved to what happens on the extent over impedance matching is done on the circuit. For example, it is known that quart-wave transformer is used for impedance matching between transmission line and the load. This is nothing but additional λ/4 length of small transmission line of certain characteristic impedance determined by impedances of both original transmission line and load. Anyway, the textbook shows graphic that there is standing wave only over this transformer. How is it possible? The presence of standing wave there means there is reflected wave from the load and that wave is confined only over the transformer?

Do you have any idea of this?
 
  • #5
Baluncore said:
@ goodphy. You are generally correct, but making an assumption that your signal is a continuous pure sinewave without any signal bandwidth.
For wider band signals you should use several stages of L network with intermediate impedances being the geometric mean of the end impedances. By staggering the resonant frequencies of elements you can then correct gain and phase shift over the band.

There is a circulating current between the L and C that could be seen as the equivalent of a standing wave on an equivalent transmission line. But one L and one C does not a transmission line make. It is better to consider the L network as a flywheel.

Oh..thanks! And...do you mean that there is actually power loss (not energy) mechanism involving energy storing and exchange between L and C although no reflection occurs at both ends of these matching network?
 
  • #6
goodphy said:
And...do you mean that there is actually power loss (not energy) mechanism involving energy storing and exchange between L and C although no reflection occurs at both ends of these matching network?
For a forward traveling wave there is a partial reflection from both ends of the L network, but they cancel on the input line because the step size from the input line to the network and from the network to the output line impedance is carefully controlled. The same is true for the reverse wave.
When a 1/4 wave TL transformer is used, it will have an impedance that is the geometric mean of the terminal impedances. ZGM = √(Z1 * Z2)
 
  • #7
Baluncore said:
For a forward traveling wave there is a partial reflection from both ends of the L network, but they cancel on the input line because the step size from the input line to the network and from the network to the output line impedance is carefully controlled. The same is true for the reverse wave.
When a 1/4 wave TL transformer is used, it will have an impedance that is the geometric mean of the terminal impedances. ZGM = √(Z1 * Z2)

Thanks!
But unfortunately I need little bit more assistance. Could you tell me more about "they cancel on the input line because the step size from the input line to the network and from the network to the output line impedance is carefully controlled."? What does input and output line and most importantly, step size mean here? It is very nice if you specify them in a detailed way.
 
  • #8
goodphy said:
What does input and output line and most importantly, step size mean here?
Think of Rs as being a transmission line with impedance Rs. Think of Rp as being a transmission line with impedance Rp. A 1/4 wave line to match those two lines would have an impedance of Rq = √(Rs * Rp).
The step impedance mismatch from Rs to Rq would be the same ratio as that from Rq to Rp. Rq/Rs = Rp/Rq.
A forward wave entering Rq from Rs will have a partial reflection at the mismatch. 1/4 wave period later the forward wave will be again partially reflected passing from Rq to Rp. That second reflection will travel back to the Rq Rp junction where it will be 1/2 wavelength late and so the two reflections will have opposite signs and so cancel.
 
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  • #9
Baluncore said:
Think of Rs as being a transmission line with impedance Rs. Think of Rp as being a transmission line with impedance Rp. A 1/4 wave line to match those two lines would have an impedance of Rq = √(Rs * Rp).
The step impedance mismatch from Rs to Rq would be the same ratio as that from Rq to Rp. Rq/Rs = Rp/Rq.
A forward wave entering Rq from Rs will have a partial reflection at the mismatch. 1/4 wave period later the forward wave will be again partially reflected passing from Rq to Rp. That second reflection will travel back to the Rq Rp junction where it will be 1/2 wavelength late and so the two reflections will have opposite signs and so cancel.

Oh...Thanks! Thus eventually local standing wave is possible as 1st reflected wave is canceled out with the 2nd reflected wave which is 1/2 wavelength delayed! I'm still wandered how not only phase (out-of-phase) but amplitude of both reflected waves are matched such that they are exactly canceled out.
 
  • #10
goodphy said:
I'm still wandered how not only phase (out-of-phase) but amplitude of both reflected waves are matched such that they are exactly canceled out.
You must also consider the reflections of the reflections, of the reflections.

This may help; https://en.wikipedia.org/wiki/Quarter-wave_impedance_transformer
 

Related to L Impedance matching network problem

1. What is an impedance matching network?

An impedance matching network is a circuit that is used to match the impedance of a source and a load in order to maximize power transfer. It is commonly used in electronic devices to ensure that the output from the source is efficiently delivered to the load without any loss of power.

2. How does an impedance matching network work?

An impedance matching network works by adjusting the impedance of the source or the load to match the impedance of the other component. This is achieved by using various components such as resistors, capacitors, and inductors in a specific configuration to achieve the desired impedance.

3. Why is impedance matching important?

Impedance matching is important because it ensures that maximum power is transferred from the source to the load without any loss. If the impedance of the source and load do not match, it can result in reflections and loss of power, which can affect the performance of the electronic device.

4. What are the consequences of not using an impedance matching network?

Not using an impedance matching network can result in poor performance, loss of power, and potentially damage to the electronic components. It can also cause signal distortion and reduce the efficiency of the device.

5. How do you design an impedance matching network?

The design of an impedance matching network depends on the specific source and load impedances and the desired frequency range. It involves using various calculations and simulations to determine the appropriate components and configuration for the network. It also requires knowledge of circuit theory and experience in designing electronic circuits.

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