How do microstrips actually represent lumped components?
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I'm having trouble understand intutively how microstrips repersent lumped element components.
All we are told is that, high impedance sections on a microstrip line repersent inductors and low-impedance repersent capacitors. How though? How does this flat piece of conductive material with a dielectric below it actually repersent a capacitor/inductor.
rf
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up vote
4
down vote
favorite
I'm having trouble understand intutively how microstrips repersent lumped element components.
All we are told is that, high impedance sections on a microstrip line repersent inductors and low-impedance repersent capacitors. How though? How does this flat piece of conductive material with a dielectric below it actually repersent a capacitor/inductor.
rf
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I'm having trouble understand intutively how microstrips repersent lumped element components.
All we are told is that, high impedance sections on a microstrip line repersent inductors and low-impedance repersent capacitors. How though? How does this flat piece of conductive material with a dielectric below it actually repersent a capacitor/inductor.
rf
I'm having trouble understand intutively how microstrips repersent lumped element components.
All we are told is that, high impedance sections on a microstrip line repersent inductors and low-impedance repersent capacitors. How though? How does this flat piece of conductive material with a dielectric below it actually repersent a capacitor/inductor.
rf
rf
edited 2 hours ago
Seth
1,542515
1,542515
asked 5 hours ago
AlfroJang80
47429
47429
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4 Answers
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up vote
2
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How do microstrips actually repersent lumped components?
It is actually the other way round!
A microstrip is a distributed component. It has a length and that length has (continous) capacitance and inductance all over its length/width/height
This is difficult to do calculations on as the number of elements (inducators, capacitors, resistors) is basically infinite.
To simplify things a lumped components model can be used, which simplifies the infinite number of components to a finite number.
add a comment |
up vote
2
down vote
We can picture all transmissions lines as having some inductance and capacitance per unit length.
If we consider a microstrip line; the lower impedance it is, the wider it will be. This results in greater capacitance because, as you said, there are two pieces of conductor with a dielectric between them which is exactly the structure of a capacitor. So a lower impedance means a larger capacitor area and therefore larger capacitance.
For higher impedance lines, the capacitance is negligible compared to the inductance so we can model it as a lumped inductor in some circumstances. I don't know off the top of my head what impact the dimensions of microstrip have on the inductance.
Many microwave texts, such as https://www.amazon.com/Microwave-Engineering-4th-David-Pozar-ebook/dp/B008ACCCHO, will discuss this more formally and actually show the equivalence but I think the intuition is sufficient here.
add a comment |
up vote
2
down vote
A microstrip is a form of transmission line in that the conductor has series inductance and there is capacitance to ground. The ratio of inductance to capacitance determine the characteristic impedance. If you terminate a transmission line with a resistor equal to its impedance, the input impedance remains constant and resistive as the length is changes. If the strip is narrow, its inductance will dominate and the input will look inductive. Conversely a broad strip will add more capacitance than inductance.
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up vote
2
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A micro strip line is the physical implementation of a transmission line. Suppose you have a load and a transmission line.
The impedance at distance $l$ is
$$Z(l) = Z_0 frac{1 + Gamma e^{-2gamma l}}{1 - Gamma e^{-2 gamma l}},$$
where $Gamma$ is the reflection coefficient and $gamma$ is the propagation constant.
If you assume that the transmission line is lossless and by setting the load to either open or close you get a purely imaginary impedance. Since the impedance of an inductor and a capacitor is
$$X_mathrm{L} = j omega L,$$
$$X_mathrm{C} = frac{1}{j omega C},$$
the transmission line impedance looks like a capacitor or an inductor at some specific frequency. For example
$$Z(l_0) = j x = j omega_0 L.$$
It is important to note that this simple implementation of a capacitor or inductor is only valid in the neighborhood of said specific frequency. There are methods to improve the bandwidth, but they are outside the scope of this answer.
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
How do microstrips actually repersent lumped components?
It is actually the other way round!
A microstrip is a distributed component. It has a length and that length has (continous) capacitance and inductance all over its length/width/height
This is difficult to do calculations on as the number of elements (inducators, capacitors, resistors) is basically infinite.
To simplify things a lumped components model can be used, which simplifies the infinite number of components to a finite number.
add a comment |
up vote
2
down vote
How do microstrips actually repersent lumped components?
It is actually the other way round!
A microstrip is a distributed component. It has a length and that length has (continous) capacitance and inductance all over its length/width/height
This is difficult to do calculations on as the number of elements (inducators, capacitors, resistors) is basically infinite.
To simplify things a lumped components model can be used, which simplifies the infinite number of components to a finite number.
add a comment |
up vote
2
down vote
up vote
2
down vote
How do microstrips actually repersent lumped components?
It is actually the other way round!
A microstrip is a distributed component. It has a length and that length has (continous) capacitance and inductance all over its length/width/height
This is difficult to do calculations on as the number of elements (inducators, capacitors, resistors) is basically infinite.
To simplify things a lumped components model can be used, which simplifies the infinite number of components to a finite number.
How do microstrips actually repersent lumped components?
It is actually the other way round!
A microstrip is a distributed component. It has a length and that length has (continous) capacitance and inductance all over its length/width/height
This is difficult to do calculations on as the number of elements (inducators, capacitors, resistors) is basically infinite.
To simplify things a lumped components model can be used, which simplifies the infinite number of components to a finite number.
answered 5 hours ago
Bimpelrekkie
46.4k240103
46.4k240103
add a comment |
add a comment |
up vote
2
down vote
We can picture all transmissions lines as having some inductance and capacitance per unit length.
If we consider a microstrip line; the lower impedance it is, the wider it will be. This results in greater capacitance because, as you said, there are two pieces of conductor with a dielectric between them which is exactly the structure of a capacitor. So a lower impedance means a larger capacitor area and therefore larger capacitance.
For higher impedance lines, the capacitance is negligible compared to the inductance so we can model it as a lumped inductor in some circumstances. I don't know off the top of my head what impact the dimensions of microstrip have on the inductance.
Many microwave texts, such as https://www.amazon.com/Microwave-Engineering-4th-David-Pozar-ebook/dp/B008ACCCHO, will discuss this more formally and actually show the equivalence but I think the intuition is sufficient here.
add a comment |
up vote
2
down vote
We can picture all transmissions lines as having some inductance and capacitance per unit length.
If we consider a microstrip line; the lower impedance it is, the wider it will be. This results in greater capacitance because, as you said, there are two pieces of conductor with a dielectric between them which is exactly the structure of a capacitor. So a lower impedance means a larger capacitor area and therefore larger capacitance.
For higher impedance lines, the capacitance is negligible compared to the inductance so we can model it as a lumped inductor in some circumstances. I don't know off the top of my head what impact the dimensions of microstrip have on the inductance.
Many microwave texts, such as https://www.amazon.com/Microwave-Engineering-4th-David-Pozar-ebook/dp/B008ACCCHO, will discuss this more formally and actually show the equivalence but I think the intuition is sufficient here.
add a comment |
up vote
2
down vote
up vote
2
down vote
We can picture all transmissions lines as having some inductance and capacitance per unit length.
If we consider a microstrip line; the lower impedance it is, the wider it will be. This results in greater capacitance because, as you said, there are two pieces of conductor with a dielectric between them which is exactly the structure of a capacitor. So a lower impedance means a larger capacitor area and therefore larger capacitance.
For higher impedance lines, the capacitance is negligible compared to the inductance so we can model it as a lumped inductor in some circumstances. I don't know off the top of my head what impact the dimensions of microstrip have on the inductance.
Many microwave texts, such as https://www.amazon.com/Microwave-Engineering-4th-David-Pozar-ebook/dp/B008ACCCHO, will discuss this more formally and actually show the equivalence but I think the intuition is sufficient here.
We can picture all transmissions lines as having some inductance and capacitance per unit length.
If we consider a microstrip line; the lower impedance it is, the wider it will be. This results in greater capacitance because, as you said, there are two pieces of conductor with a dielectric between them which is exactly the structure of a capacitor. So a lower impedance means a larger capacitor area and therefore larger capacitance.
For higher impedance lines, the capacitance is negligible compared to the inductance so we can model it as a lumped inductor in some circumstances. I don't know off the top of my head what impact the dimensions of microstrip have on the inductance.
Many microwave texts, such as https://www.amazon.com/Microwave-Engineering-4th-David-Pozar-ebook/dp/B008ACCCHO, will discuss this more formally and actually show the equivalence but I think the intuition is sufficient here.
answered 5 hours ago
jramsay42
426126
426126
add a comment |
add a comment |
up vote
2
down vote
A microstrip is a form of transmission line in that the conductor has series inductance and there is capacitance to ground. The ratio of inductance to capacitance determine the characteristic impedance. If you terminate a transmission line with a resistor equal to its impedance, the input impedance remains constant and resistive as the length is changes. If the strip is narrow, its inductance will dominate and the input will look inductive. Conversely a broad strip will add more capacitance than inductance.
add a comment |
up vote
2
down vote
A microstrip is a form of transmission line in that the conductor has series inductance and there is capacitance to ground. The ratio of inductance to capacitance determine the characteristic impedance. If you terminate a transmission line with a resistor equal to its impedance, the input impedance remains constant and resistive as the length is changes. If the strip is narrow, its inductance will dominate and the input will look inductive. Conversely a broad strip will add more capacitance than inductance.
add a comment |
up vote
2
down vote
up vote
2
down vote
A microstrip is a form of transmission line in that the conductor has series inductance and there is capacitance to ground. The ratio of inductance to capacitance determine the characteristic impedance. If you terminate a transmission line with a resistor equal to its impedance, the input impedance remains constant and resistive as the length is changes. If the strip is narrow, its inductance will dominate and the input will look inductive. Conversely a broad strip will add more capacitance than inductance.
A microstrip is a form of transmission line in that the conductor has series inductance and there is capacitance to ground. The ratio of inductance to capacitance determine the characteristic impedance. If you terminate a transmission line with a resistor equal to its impedance, the input impedance remains constant and resistive as the length is changes. If the strip is narrow, its inductance will dominate and the input will look inductive. Conversely a broad strip will add more capacitance than inductance.
answered 5 hours ago
Steve Hubbard
94717
94717
add a comment |
add a comment |
up vote
2
down vote
A micro strip line is the physical implementation of a transmission line. Suppose you have a load and a transmission line.
The impedance at distance $l$ is
$$Z(l) = Z_0 frac{1 + Gamma e^{-2gamma l}}{1 - Gamma e^{-2 gamma l}},$$
where $Gamma$ is the reflection coefficient and $gamma$ is the propagation constant.
If you assume that the transmission line is lossless and by setting the load to either open or close you get a purely imaginary impedance. Since the impedance of an inductor and a capacitor is
$$X_mathrm{L} = j omega L,$$
$$X_mathrm{C} = frac{1}{j omega C},$$
the transmission line impedance looks like a capacitor or an inductor at some specific frequency. For example
$$Z(l_0) = j x = j omega_0 L.$$
It is important to note that this simple implementation of a capacitor or inductor is only valid in the neighborhood of said specific frequency. There are methods to improve the bandwidth, but they are outside the scope of this answer.
add a comment |
up vote
2
down vote
A micro strip line is the physical implementation of a transmission line. Suppose you have a load and a transmission line.
The impedance at distance $l$ is
$$Z(l) = Z_0 frac{1 + Gamma e^{-2gamma l}}{1 - Gamma e^{-2 gamma l}},$$
where $Gamma$ is the reflection coefficient and $gamma$ is the propagation constant.
If you assume that the transmission line is lossless and by setting the load to either open or close you get a purely imaginary impedance. Since the impedance of an inductor and a capacitor is
$$X_mathrm{L} = j omega L,$$
$$X_mathrm{C} = frac{1}{j omega C},$$
the transmission line impedance looks like a capacitor or an inductor at some specific frequency. For example
$$Z(l_0) = j x = j omega_0 L.$$
It is important to note that this simple implementation of a capacitor or inductor is only valid in the neighborhood of said specific frequency. There are methods to improve the bandwidth, but they are outside the scope of this answer.
add a comment |
up vote
2
down vote
up vote
2
down vote
A micro strip line is the physical implementation of a transmission line. Suppose you have a load and a transmission line.
The impedance at distance $l$ is
$$Z(l) = Z_0 frac{1 + Gamma e^{-2gamma l}}{1 - Gamma e^{-2 gamma l}},$$
where $Gamma$ is the reflection coefficient and $gamma$ is the propagation constant.
If you assume that the transmission line is lossless and by setting the load to either open or close you get a purely imaginary impedance. Since the impedance of an inductor and a capacitor is
$$X_mathrm{L} = j omega L,$$
$$X_mathrm{C} = frac{1}{j omega C},$$
the transmission line impedance looks like a capacitor or an inductor at some specific frequency. For example
$$Z(l_0) = j x = j omega_0 L.$$
It is important to note that this simple implementation of a capacitor or inductor is only valid in the neighborhood of said specific frequency. There are methods to improve the bandwidth, but they are outside the scope of this answer.
A micro strip line is the physical implementation of a transmission line. Suppose you have a load and a transmission line.
The impedance at distance $l$ is
$$Z(l) = Z_0 frac{1 + Gamma e^{-2gamma l}}{1 - Gamma e^{-2 gamma l}},$$
where $Gamma$ is the reflection coefficient and $gamma$ is the propagation constant.
If you assume that the transmission line is lossless and by setting the load to either open or close you get a purely imaginary impedance. Since the impedance of an inductor and a capacitor is
$$X_mathrm{L} = j omega L,$$
$$X_mathrm{C} = frac{1}{j omega C},$$
the transmission line impedance looks like a capacitor or an inductor at some specific frequency. For example
$$Z(l_0) = j x = j omega_0 L.$$
It is important to note that this simple implementation of a capacitor or inductor is only valid in the neighborhood of said specific frequency. There are methods to improve the bandwidth, but they are outside the scope of this answer.
edited 5 hours ago
answered 5 hours ago
user110971
3,1791717
3,1791717
add a comment |
add a comment |
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