Johnson-Nyquist noise for a lossy inductor?How Johnson measured Johnson-Nyquist noise?Thermal noise (Johnson noise)Thermal Johnson-Nyquist Noise GeneratorReduction of DC motor noiseJohnson noise with amplifier circuitIs it sensible to always use larger diameter conductors for carrying smaller signals?Old Inductor, replace with newer current-production?What would an inductor be doing here, and can I replace it with a jumper?Johnson–Nyquist noise contribution Antenna noise temperatures?Ferrite vs. iron powder toroid for buck converters?
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Johnson-Nyquist noise for a lossy inductor?
How Johnson measured Johnson-Nyquist noise?Thermal noise (Johnson noise)Thermal Johnson-Nyquist Noise GeneratorReduction of DC motor noiseJohnson noise with amplifier circuitIs it sensible to always use larger diameter conductors for carrying smaller signals?Old Inductor, replace with newer current-production?What would an inductor be doing here, and can I replace it with a jumper?Johnson–Nyquist noise contribution Antenna noise temperatures?Ferrite vs. iron powder toroid for buck converters?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
In searching the internet for thermal noise on an inductor and I'm not finding much on the basics for a real lossy inductor. I'm interested in thermal noise for a fixed inductor in a surface mount package. Here is a diagram TDK provides for their surface mount inductors that shows a loss model for the inductor..
At point 2 in the above diagram, is the Johnson-Nyquist noise the following?
$overlinev^2_n = 4k_BTR1 + 4k_BTR2$
I'm working on building my first LNA for a hobby project and wanting a basic understanding of noise added by inductors. I know there are other sources of noise for a LNA. This question is about the noise contribution by the inductors in particular.
noise inductor noise-spectral-density
asked May 19 at 17:15
acker9acker9
1098
$endgroup$
add a comment |
$begingroup$
In searching the internet for thermal noise on an inductor and I'm not finding much on the basics for a real lossy inductor. I'm interested in thermal noise for a fixed inductor in a surface mount package. Here is a diagram TDK provides for their surface mount inductors that shows a loss model for the inductor..
At point 2 in the above diagram, is the Johnson-Nyquist noise the following?
$overlinev^2_n = 4k_BTR1 + 4k_BTR2$
I'm working on building my first LNA for a hobby project and wanting a basic understanding of noise added by inductors. I know there are other sources of noise for a LNA. This question is about the noise contribution by the inductors in particular.
noise inductor noise-spectral-density
asked May 19 at 17:15
acker9acker9
1098
$endgroup$
1
$begingroup$
Where's the bandwidth factor?
$endgroup$
– jonk
May 19 at 17:43
1
$begingroup$
Is this inductor near some sort of shield? that will cause losses. Do you need to include shield resistance (and skin depth) in your design?
$endgroup$
– analogsystemsrf
May 19 at 21:40
add a comment |
$begingroup$
In searching the internet for thermal noise on an inductor and I'm not finding much on the basics for a real lossy inductor. I'm interested in thermal noise for a fixed inductor in a surface mount package. Here is a diagram TDK provides for their surface mount inductors that shows a loss model for the inductor..
At point 2 in the above diagram, is the Johnson-Nyquist noise the following?
$overlinev^2_n = 4k_BTR1 + 4k_BTR2$
I'm working on building my first LNA for a hobby project and wanting a basic understanding of noise added by inductors. I know there are other sources of noise for a LNA. This question is about the noise contribution by the inductors in particular.
noise inductor noise-spectral-density
asked May 19 at 17:15
acker9acker9
1098
$endgroup$
In searching the internet for thermal noise on an inductor and I'm not finding much on the basics for a real lossy inductor. I'm interested in thermal noise for a fixed inductor in a surface mount package. Here is a diagram TDK provides for their surface mount inductors that shows a loss model for the inductor..
At point 2 in the above diagram, is the Johnson-Nyquist noise the following?
$overlinev^2_n = 4k_BTR1 + 4k_BTR2$
I'm working on building my first LNA for a hobby project and wanting a basic understanding of noise added by inductors. I know there are other sources of noise for a LNA. This question is about the noise contribution by the inductors in particular.
noise inductor noise-spectral-density
noise inductor noise-spectral-density
asked May 19 at 17:15
acker9acker9
1098
asked May 19 at 17:15
acker9acker9
1098
asked May 19 at 17:15
acker9acker9
1098
asked May 19 at 17:15
acker9acker9
1098
asked May 19 at 17:15
acker9acker9
1098
1098
1
$begingroup$
Where's the bandwidth factor?
$endgroup$
– jonk
May 19 at 17:43
1
$begingroup$
Is this inductor near some sort of shield? that will cause losses. Do you need to include shield resistance (and skin depth) in your design?
$endgroup$
– analogsystemsrf
May 19 at 21:40
add a comment |
1
$begingroup$
Where's the bandwidth factor?
$endgroup$
– jonk
May 19 at 17:43
1
$begingroup$
Is this inductor near some sort of shield? that will cause losses. Do you need to include shield resistance (and skin depth) in your design?
$endgroup$
– analogsystemsrf
May 19 at 21:40
1
1
$begingroup$
Where's the bandwidth factor?
$endgroup$
– jonk
May 19 at 17:43
$begingroup$
Where's the bandwidth factor?
$endgroup$
– jonk
May 19 at 17:43
1
1
$begingroup$
Is this inductor near some sort of shield? that will cause losses. Do you need to include shield resistance (and skin depth) in your design?
$endgroup$
– analogsystemsrf
May 19 at 21:40
$begingroup$
Is this inductor near some sort of shield? that will cause losses. Do you need to include shield resistance (and skin depth) in your design?
$endgroup$
– analogsystemsrf
May 19 at 21:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I want to edit @Bimplerekkie's answer, but cannot.
From a noise perspective, the coil will act like the equivalent circuit schematic. This means that at very low frequencies R1 will be shorted out by the equivalent coil -- I would say that at very high frequencies it would be shorted out by the equivalent cap, but I suspect the model breaks down above resonance.
At any given frequency, it's safe to calculate the effective resistance of the circuit -- i.e., reduce it to an inductor (or cap, above resonance) either in parallel or series with an appropriate resistance. The noise characteristics of that resistor at that frequency and around it will be valid.
This sort of thing will work for just about any passive circuit that's all at a constant temperature -- if you burrow deeply enough into it you'll find out that it's a consequence of the laws of thermodynamics.
And note:
All of this equivalent resistance circuit stuff doesn't necessarily apply to active circuits. The "equivalent resistance is equivalent noise resistance" rule works for a circuit whose elements are in thermodynamic equilibrium. It's a stretch, but by definition an amplifying element that has current flowing through it is not at thermodynamic equilibrium, because energy is flowing from the voltage source through the amplifying element; this is why you can build LNAs that have noises temperatures below ambient, and even ones (through clever use of transformer feedback) that have controllable input or output impedances.
answered May 19 at 21:00
TimWescottTimWescott
8,7611720
$endgroup$
add a comment |
$begingroup$
You're correct in thinking that the noise can be evaluated like that. The ideal L1 and C1 in the model are noiseless, all the noise originates from R1 and R2. For an inductor these resistors will have quite a low value so their noise contribution is small compared to other elements in a typical LNA circuit.
I am quite sure your formula
$overlinev^2_n = 4k_BTR1 + 4k_BTR2$
isn's correct as it doesn't show the frequency dependency caused by L1 and C1 and also noise cannot be added like that as the sources are uncorrelated.
In general experienced LNA designers do not worry about the noise of the inductors as they already know it can be neglected. That is not to say it is not a good exercise to do once.
Usually in a properly designed LNA the active element (NPN, NMOS etc.) will be the most dominant noise contributor.
answered May 19 at 17:36
BimpelrekkieBimpelrekkie
54.2k251122
$endgroup$
$begingroup$
When you say there's a frequency dependency, this is because R1 and R2 themselves change as the frequency changes? Or is it because of a bandwidth factor that needs to be considered, as @jonk suggests?
$endgroup$
– acker9
May 19 at 18:18
$begingroup$
It is both! To properly specify noise, I either use spot noise, that's the noise (voltage, current or power) at a certain frequency, for example at 1 kHz. Or integrated noise and that's the noise (again voltage, current or power) in a frequency band, for example 1 kHz to 1 MHz.
$endgroup$
– Bimpelrekkie
May 19 at 19:03
$begingroup$
The parallel resistor would be high in value relative to the reactance of the inductor, not low. It would act as a current noise source in parallel with the inductor.
$endgroup$
– Kevin White
May 19 at 19:07
$begingroup$
@KevinWhite That was what I thought too. If higher inductor values have a larger Rp, it seems like the noise of inductors goes up as the inductance goes up for a given frequency, correct? Which, it seems, the lower the frequency for a NMOS-based LNA, the greater the inductance needed, and the greater the noise from said inductors.
$endgroup$
– acker9
May 19 at 20:49
$begingroup$
True, although I've never encountered a situation where the noise from the inductor was significant compared to other components in a circuit.
$endgroup$
– Kevin White
May 19 at 21:21
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I want to edit @Bimplerekkie's answer, but cannot.
From a noise perspective, the coil will act like the equivalent circuit schematic. This means that at very low frequencies R1 will be shorted out by the equivalent coil -- I would say that at very high frequencies it would be shorted out by the equivalent cap, but I suspect the model breaks down above resonance.
At any given frequency, it's safe to calculate the effective resistance of the circuit -- i.e., reduce it to an inductor (or cap, above resonance) either in parallel or series with an appropriate resistance. The noise characteristics of that resistor at that frequency and around it will be valid.
This sort of thing will work for just about any passive circuit that's all at a constant temperature -- if you burrow deeply enough into it you'll find out that it's a consequence of the laws of thermodynamics.
And note:
All of this equivalent resistance circuit stuff doesn't necessarily apply to active circuits. The "equivalent resistance is equivalent noise resistance" rule works for a circuit whose elements are in thermodynamic equilibrium. It's a stretch, but by definition an amplifying element that has current flowing through it is not at thermodynamic equilibrium, because energy is flowing from the voltage source through the amplifying element; this is why you can build LNAs that have noises temperatures below ambient, and even ones (through clever use of transformer feedback) that have controllable input or output impedances.
answered May 19 at 21:00
TimWescottTimWescott
8,7611720
$endgroup$
add a comment |
$begingroup$
I want to edit @Bimplerekkie's answer, but cannot.
From a noise perspective, the coil will act like the equivalent circuit schematic. This means that at very low frequencies R1 will be shorted out by the equivalent coil -- I would say that at very high frequencies it would be shorted out by the equivalent cap, but I suspect the model breaks down above resonance.
At any given frequency, it's safe to calculate the effective resistance of the circuit -- i.e., reduce it to an inductor (or cap, above resonance) either in parallel or series with an appropriate resistance. The noise characteristics of that resistor at that frequency and around it will be valid.
This sort of thing will work for just about any passive circuit that's all at a constant temperature -- if you burrow deeply enough into it you'll find out that it's a consequence of the laws of thermodynamics.
And note:
All of this equivalent resistance circuit stuff doesn't necessarily apply to active circuits. The "equivalent resistance is equivalent noise resistance" rule works for a circuit whose elements are in thermodynamic equilibrium. It's a stretch, but by definition an amplifying element that has current flowing through it is not at thermodynamic equilibrium, because energy is flowing from the voltage source through the amplifying element; this is why you can build LNAs that have noises temperatures below ambient, and even ones (through clever use of transformer feedback) that have controllable input or output impedances.
answered May 19 at 21:00
TimWescottTimWescott
8,7611720
$endgroup$
add a comment |
$begingroup$
I want to edit @Bimplerekkie's answer, but cannot.
From a noise perspective, the coil will act like the equivalent circuit schematic. This means that at very low frequencies R1 will be shorted out by the equivalent coil -- I would say that at very high frequencies it would be shorted out by the equivalent cap, but I suspect the model breaks down above resonance.
At any given frequency, it's safe to calculate the effective resistance of the circuit -- i.e., reduce it to an inductor (or cap, above resonance) either in parallel or series with an appropriate resistance. The noise characteristics of that resistor at that frequency and around it will be valid.
This sort of thing will work for just about any passive circuit that's all at a constant temperature -- if you burrow deeply enough into it you'll find out that it's a consequence of the laws of thermodynamics.
And note:
All of this equivalent resistance circuit stuff doesn't necessarily apply to active circuits. The "equivalent resistance is equivalent noise resistance" rule works for a circuit whose elements are in thermodynamic equilibrium. It's a stretch, but by definition an amplifying element that has current flowing through it is not at thermodynamic equilibrium, because energy is flowing from the voltage source through the amplifying element; this is why you can build LNAs that have noises temperatures below ambient, and even ones (through clever use of transformer feedback) that have controllable input or output impedances.
answered May 19 at 21:00
TimWescottTimWescott
8,7611720
$endgroup$
I want to edit @Bimplerekkie's answer, but cannot.
From a noise perspective, the coil will act like the equivalent circuit schematic. This means that at very low frequencies R1 will be shorted out by the equivalent coil -- I would say that at very high frequencies it would be shorted out by the equivalent cap, but I suspect the model breaks down above resonance.
At any given frequency, it's safe to calculate the effective resistance of the circuit -- i.e., reduce it to an inductor (or cap, above resonance) either in parallel or series with an appropriate resistance. The noise characteristics of that resistor at that frequency and around it will be valid.
This sort of thing will work for just about any passive circuit that's all at a constant temperature -- if you burrow deeply enough into it you'll find out that it's a consequence of the laws of thermodynamics.
And note:
All of this equivalent resistance circuit stuff doesn't necessarily apply to active circuits. The "equivalent resistance is equivalent noise resistance" rule works for a circuit whose elements are in thermodynamic equilibrium. It's a stretch, but by definition an amplifying element that has current flowing through it is not at thermodynamic equilibrium, because energy is flowing from the voltage source through the amplifying element; this is why you can build LNAs that have noises temperatures below ambient, and even ones (through clever use of transformer feedback) that have controllable input or output impedances.
answered May 19 at 21:00
TimWescottTimWescott
8,7611720
answered May 19 at 21:00
TimWescottTimWescott
8,7611720
answered May 19 at 21:00
TimWescottTimWescott
8,7611720
answered May 19 at 21:00
TimWescottTimWescott
8,7611720
8,7611720
add a comment |
add a comment |
$begingroup$
You're correct in thinking that the noise can be evaluated like that. The ideal L1 and C1 in the model are noiseless, all the noise originates from R1 and R2. For an inductor these resistors will have quite a low value so their noise contribution is small compared to other elements in a typical LNA circuit.
I am quite sure your formula
$overlinev^2_n = 4k_BTR1 + 4k_BTR2$
isn's correct as it doesn't show the frequency dependency caused by L1 and C1 and also noise cannot be added like that as the sources are uncorrelated.
In general experienced LNA designers do not worry about the noise of the inductors as they already know it can be neglected. That is not to say it is not a good exercise to do once.
Usually in a properly designed LNA the active element (NPN, NMOS etc.) will be the most dominant noise contributor.
answered May 19 at 17:36
BimpelrekkieBimpelrekkie
54.2k251122
$endgroup$
$begingroup$
When you say there's a frequency dependency, this is because R1 and R2 themselves change as the frequency changes? Or is it because of a bandwidth factor that needs to be considered, as @jonk suggests?
$endgroup$
– acker9
May 19 at 18:18
$begingroup$
It is both! To properly specify noise, I either use spot noise, that's the noise (voltage, current or power) at a certain frequency, for example at 1 kHz. Or integrated noise and that's the noise (again voltage, current or power) in a frequency band, for example 1 kHz to 1 MHz.
$endgroup$
– Bimpelrekkie
May 19 at 19:03
$begingroup$
The parallel resistor would be high in value relative to the reactance of the inductor, not low. It would act as a current noise source in parallel with the inductor.
$endgroup$
– Kevin White
May 19 at 19:07
$begingroup$
@KevinWhite That was what I thought too. If higher inductor values have a larger Rp, it seems like the noise of inductors goes up as the inductance goes up for a given frequency, correct? Which, it seems, the lower the frequency for a NMOS-based LNA, the greater the inductance needed, and the greater the noise from said inductors.
$endgroup$
– acker9
May 19 at 20:49
$begingroup$
True, although I've never encountered a situation where the noise from the inductor was significant compared to other components in a circuit.
$endgroup$
– Kevin White
May 19 at 21:21
add a comment |
$begingroup$
You're correct in thinking that the noise can be evaluated like that. The ideal L1 and C1 in the model are noiseless, all the noise originates from R1 and R2. For an inductor these resistors will have quite a low value so their noise contribution is small compared to other elements in a typical LNA circuit.
I am quite sure your formula
$overlinev^2_n = 4k_BTR1 + 4k_BTR2$
isn's correct as it doesn't show the frequency dependency caused by L1 and C1 and also noise cannot be added like that as the sources are uncorrelated.
In general experienced LNA designers do not worry about the noise of the inductors as they already know it can be neglected. That is not to say it is not a good exercise to do once.
Usually in a properly designed LNA the active element (NPN, NMOS etc.) will be the most dominant noise contributor.
answered May 19 at 17:36
BimpelrekkieBimpelrekkie
54.2k251122
$endgroup$
$begingroup$
When you say there's a frequency dependency, this is because R1 and R2 themselves change as the frequency changes? Or is it because of a bandwidth factor that needs to be considered, as @jonk suggests?
$endgroup$
– acker9
May 19 at 18:18
$begingroup$
It is both! To properly specify noise, I either use spot noise, that's the noise (voltage, current or power) at a certain frequency, for example at 1 kHz. Or integrated noise and that's the noise (again voltage, current or power) in a frequency band, for example 1 kHz to 1 MHz.
$endgroup$
– Bimpelrekkie
May 19 at 19:03
$begingroup$
The parallel resistor would be high in value relative to the reactance of the inductor, not low. It would act as a current noise source in parallel with the inductor.
$endgroup$
– Kevin White
May 19 at 19:07
$begingroup$
@KevinWhite That was what I thought too. If higher inductor values have a larger Rp, it seems like the noise of inductors goes up as the inductance goes up for a given frequency, correct? Which, it seems, the lower the frequency for a NMOS-based LNA, the greater the inductance needed, and the greater the noise from said inductors.
$endgroup$
– acker9
May 19 at 20:49
$begingroup$
True, although I've never encountered a situation where the noise from the inductor was significant compared to other components in a circuit.
$endgroup$
– Kevin White
May 19 at 21:21
add a comment |
$begingroup$
You're correct in thinking that the noise can be evaluated like that. The ideal L1 and C1 in the model are noiseless, all the noise originates from R1 and R2. For an inductor these resistors will have quite a low value so their noise contribution is small compared to other elements in a typical LNA circuit.
I am quite sure your formula
$overlinev^2_n = 4k_BTR1 + 4k_BTR2$
isn's correct as it doesn't show the frequency dependency caused by L1 and C1 and also noise cannot be added like that as the sources are uncorrelated.
In general experienced LNA designers do not worry about the noise of the inductors as they already know it can be neglected. That is not to say it is not a good exercise to do once.
Usually in a properly designed LNA the active element (NPN, NMOS etc.) will be the most dominant noise contributor.
answered May 19 at 17:36
BimpelrekkieBimpelrekkie
54.2k251122
$endgroup$
You're correct in thinking that the noise can be evaluated like that. The ideal L1 and C1 in the model are noiseless, all the noise originates from R1 and R2. For an inductor these resistors will have quite a low value so their noise contribution is small compared to other elements in a typical LNA circuit.
I am quite sure your formula
$overlinev^2_n = 4k_BTR1 + 4k_BTR2$
isn's correct as it doesn't show the frequency dependency caused by L1 and C1 and also noise cannot be added like that as the sources are uncorrelated.
In general experienced LNA designers do not worry about the noise of the inductors as they already know it can be neglected. That is not to say it is not a good exercise to do once.
Usually in a properly designed LNA the active element (NPN, NMOS etc.) will be the most dominant noise contributor.
answered May 19 at 17:36
BimpelrekkieBimpelrekkie
54.2k251122
answered May 19 at 17:36
BimpelrekkieBimpelrekkie
54.2k251122
answered May 19 at 17:36
BimpelrekkieBimpelrekkie
54.2k251122
answered May 19 at 17:36
BimpelrekkieBimpelrekkie
54.2k251122
54.2k251122
$begingroup$
When you say there's a frequency dependency, this is because R1 and R2 themselves change as the frequency changes? Or is it because of a bandwidth factor that needs to be considered, as @jonk suggests?
$endgroup$
– acker9
May 19 at 18:18
$begingroup$
It is both! To properly specify noise, I either use spot noise, that's the noise (voltage, current or power) at a certain frequency, for example at 1 kHz. Or integrated noise and that's the noise (again voltage, current or power) in a frequency band, for example 1 kHz to 1 MHz.
$endgroup$
– Bimpelrekkie
May 19 at 19:03
$begingroup$
The parallel resistor would be high in value relative to the reactance of the inductor, not low. It would act as a current noise source in parallel with the inductor.
$endgroup$
– Kevin White
May 19 at 19:07
$begingroup$
@KevinWhite That was what I thought too. If higher inductor values have a larger Rp, it seems like the noise of inductors goes up as the inductance goes up for a given frequency, correct? Which, it seems, the lower the frequency for a NMOS-based LNA, the greater the inductance needed, and the greater the noise from said inductors.
$endgroup$
– acker9
May 19 at 20:49
$begingroup$
True, although I've never encountered a situation where the noise from the inductor was significant compared to other components in a circuit.
$endgroup$
– Kevin White
May 19 at 21:21
add a comment |
$begingroup$
When you say there's a frequency dependency, this is because R1 and R2 themselves change as the frequency changes? Or is it because of a bandwidth factor that needs to be considered, as @jonk suggests?
$endgroup$
– acker9
May 19 at 18:18
$begingroup$
It is both! To properly specify noise, I either use spot noise, that's the noise (voltage, current or power) at a certain frequency, for example at 1 kHz. Or integrated noise and that's the noise (again voltage, current or power) in a frequency band, for example 1 kHz to 1 MHz.
$endgroup$
– Bimpelrekkie
May 19 at 19:03
$begingroup$
The parallel resistor would be high in value relative to the reactance of the inductor, not low. It would act as a current noise source in parallel with the inductor.
$endgroup$
– Kevin White
May 19 at 19:07
$begingroup$
@KevinWhite That was what I thought too. If higher inductor values have a larger Rp, it seems like the noise of inductors goes up as the inductance goes up for a given frequency, correct? Which, it seems, the lower the frequency for a NMOS-based LNA, the greater the inductance needed, and the greater the noise from said inductors.
$endgroup$
– acker9
May 19 at 20:49
$begingroup$
True, although I've never encountered a situation where the noise from the inductor was significant compared to other components in a circuit.
$endgroup$
– Kevin White
May 19 at 21:21
$begingroup$
When you say there's a frequency dependency, this is because R1 and R2 themselves change as the frequency changes? Or is it because of a bandwidth factor that needs to be considered, as @jonk suggests?
$endgroup$
– acker9
May 19 at 18:18
$begingroup$
When you say there's a frequency dependency, this is because R1 and R2 themselves change as the frequency changes? Or is it because of a bandwidth factor that needs to be considered, as @jonk suggests?
$endgroup$
– acker9
May 19 at 18:18
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It is both! To properly specify noise, I either use spot noise, that's the noise (voltage, current or power) at a certain frequency, for example at 1 kHz. Or integrated noise and that's the noise (again voltage, current or power) in a frequency band, for example 1 kHz to 1 MHz.
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– Bimpelrekkie
May 19 at 19:03
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It is both! To properly specify noise, I either use spot noise, that's the noise (voltage, current or power) at a certain frequency, for example at 1 kHz. Or integrated noise and that's the noise (again voltage, current or power) in a frequency band, for example 1 kHz to 1 MHz.
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– Bimpelrekkie
May 19 at 19:03
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The parallel resistor would be high in value relative to the reactance of the inductor, not low. It would act as a current noise source in parallel with the inductor.
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– Kevin White
May 19 at 19:07
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The parallel resistor would be high in value relative to the reactance of the inductor, not low. It would act as a current noise source in parallel with the inductor.
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– Kevin White
May 19 at 19:07
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@KevinWhite That was what I thought too. If higher inductor values have a larger Rp, it seems like the noise of inductors goes up as the inductance goes up for a given frequency, correct? Which, it seems, the lower the frequency for a NMOS-based LNA, the greater the inductance needed, and the greater the noise from said inductors.
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– acker9
May 19 at 20:49
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@KevinWhite That was what I thought too. If higher inductor values have a larger Rp, it seems like the noise of inductors goes up as the inductance goes up for a given frequency, correct? Which, it seems, the lower the frequency for a NMOS-based LNA, the greater the inductance needed, and the greater the noise from said inductors.
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– acker9
May 19 at 20:49
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True, although I've never encountered a situation where the noise from the inductor was significant compared to other components in a circuit.
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– Kevin White
May 19 at 21:21
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True, although I've never encountered a situation where the noise from the inductor was significant compared to other components in a circuit.
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– Kevin White
May 19 at 21:21
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Where's the bandwidth factor?
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– jonk
May 19 at 17:43
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Is this inductor near some sort of shield? that will cause losses. Do you need to include shield resistance (and skin depth) in your design?
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– analogsystemsrf
May 19 at 21:40