Why 50 Ω termination results in less noise than 1 MΩ termination on the scope reading?Accurately measure relative intensity noise (RIN) using RF spectral analyzer?Ground noise measured on scope50Hz hum on oscilloscope probe only when using clipWhy is a lower Noise Figure considered better than a higher Noise Figure?A question about digital scope input and common mode noiseWhy am I calculating different sampling rate than the scope shows for a given memory depth?Practices for using a scope, scope probe and termination resistor for a proper measurement setupIsolating A Noisy Component In A Circuit
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Why 50 Ω termination results in less noise than 1 MΩ termination on the scope reading?
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Why 50 Ω termination results in less noise than 1 MΩ termination on the scope reading?
Accurately measure relative intensity noise (RIN) using RF spectral analyzer?Ground noise measured on scope50Hz hum on oscilloscope probe only when using clipWhy is a lower Noise Figure considered better than a higher Noise Figure?A question about digital scope input and common mode noiseWhy am I calculating different sampling rate than the scope shows for a given memory depth?Practices for using a scope, scope probe and termination resistor for a proper measurement setupIsolating A Noisy Component In A Circuit
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Regarding the following section from the Keysight document "Making Your Best Power Integrity Measurements":
It says that using 50 Ω termination we will see less noise on the scope comparing to 1 MΩ.
Can this be explained by modeling what meant here as an electrical circuit? I'm trying to understand why lower resistance causes less noise on the scope screen.
noise oscilloscope
edited 8 hours ago
SamGibson
12.4k4 gold badges18 silver badges44 bronze badges
asked 8 hours ago
panic attackpanic attack
4013 silver badges12 bronze badges
$endgroup$
add a comment |
$begingroup$
Regarding the following section from the Keysight document "Making Your Best Power Integrity Measurements":
It says that using 50 Ω termination we will see less noise on the scope comparing to 1 MΩ.
Can this be explained by modeling what meant here as an electrical circuit? I'm trying to understand why lower resistance causes less noise on the scope screen.
noise oscilloscope
edited 8 hours ago
SamGibson
12.4k4 gold badges18 silver badges44 bronze badges
asked 8 hours ago
panic attackpanic attack
4013 silver badges12 bronze badges
$endgroup$
$begingroup$
e2e.ti.com/blogs_/b/powerhouse/archive/2016/07/27/… also talks about a 50 Ohms while doing a accurate mesurement of noise
$endgroup$
– Umar
5 hours ago
add a comment |
$begingroup$
Regarding the following section from the Keysight document "Making Your Best Power Integrity Measurements":
It says that using 50 Ω termination we will see less noise on the scope comparing to 1 MΩ.
Can this be explained by modeling what meant here as an electrical circuit? I'm trying to understand why lower resistance causes less noise on the scope screen.
noise oscilloscope
edited 8 hours ago
SamGibson
12.4k4 gold badges18 silver badges44 bronze badges
asked 8 hours ago
panic attackpanic attack
4013 silver badges12 bronze badges
$endgroup$
Regarding the following section from the Keysight document "Making Your Best Power Integrity Measurements":
It says that using 50 Ω termination we will see less noise on the scope comparing to 1 MΩ.
Can this be explained by modeling what meant here as an electrical circuit? I'm trying to understand why lower resistance causes less noise on the scope screen.
noise oscilloscope
noise oscilloscope
edited 8 hours ago
SamGibson
12.4k4 gold badges18 silver badges44 bronze badges
asked 8 hours ago
panic attackpanic attack
4013 silver badges12 bronze badges
edited 8 hours ago
SamGibson
12.4k4 gold badges18 silver badges44 bronze badges
asked 8 hours ago
panic attackpanic attack
4013 silver badges12 bronze badges
edited 8 hours ago
SamGibson
12.4k4 gold badges18 silver badges44 bronze badges
edited 8 hours ago
SamGibson
12.4k4 gold badges18 silver badges44 bronze badges
edited 8 hours ago
SamGibson
12.4k4 gold badges18 silver badges44 bronze badges
12.4k4 gold badges18 silver badges44 bronze badges
asked 8 hours ago
panic attackpanic attack
4013 silver badges12 bronze badges
asked 8 hours ago
panic attackpanic attack
4013 silver badges12 bronze badges
asked 8 hours ago
panic attackpanic attack
4013 silver badges12 bronze badges
4013 silver badges12 bronze badges
$begingroup$
e2e.ti.com/blogs_/b/powerhouse/archive/2016/07/27/… also talks about a 50 Ohms while doing a accurate mesurement of noise
$endgroup$
– Umar
5 hours ago
add a comment |
$begingroup$
e2e.ti.com/blogs_/b/powerhouse/archive/2016/07/27/… also talks about a 50 Ohms while doing a accurate mesurement of noise
$endgroup$
– Umar
5 hours ago
$begingroup$
e2e.ti.com/blogs_/b/powerhouse/archive/2016/07/27/… also talks about a 50 Ohms while doing a accurate mesurement of noise
$endgroup$
– Umar
5 hours ago
$begingroup$
e2e.ti.com/blogs_/b/powerhouse/archive/2016/07/27/… also talks about a 50 Ohms while doing a accurate mesurement of noise
$endgroup$
– Umar
5 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Because Johnson-Nyquist noise. Resistors generate thermal noise, with a noise power that's proportional to the absolute temperature. A higher-valued resistor will then generate more noise voltage, with the voltage proportional to $sqrtR$.
Setting your O-scope up as a $1mathrmMOmega$ instrument gets you the noise from that $1mathrmMOmega$ resistor; if you're measuring some super low-impedance node like you'd find in a power supply, you gain absolutely no accuracy from the high impedance. So you measure at $50Omega$, and get less noise.
answered 6 hours ago
TimWescottTimWescott
13.4k1 gold badge11 silver badges26 bronze badges
$endgroup$
$begingroup$
Interesting, so it is due to the Johnson noise. But if you in practice measure a power supply’s noise on scope do you really terminate with 50 Ohm resistance?
$endgroup$
– panic attack
6 hours ago
$begingroup$
Normal scope probes have a built in 1MΩ resistance (I'm pretty sure some or all is actually built into the cable). So using a normal cable puts that in series with the source. Fancy scopes let you switch in a 50Ω terminator so you can plug a 50Ω cable into it, and if you're using a cheaper scope you can get an in-line 50Ω terminator.
$endgroup$
– TimWescott
6 hours ago
$begingroup$
To the OP: There are two main considerations when using 50 Ohm mode. First, will the resulting current fry the oscilloscope input? You don't want to measure 120V line voltage in 50 Ohm mode, obviously. Second, will the thing you are measuring tolerate a 50 Ohm load? Measuring a signal with a 50 Ohm termination is equivalnet to placing a 50 Ohm resistor directly across the thing you are measuring. Sometimes that is OK and sometimes it is not OK because the 50 Ohms will totally change the very voltage you are trying to measure.
$endgroup$
– mkeith
3 hours ago
add a comment |
$begingroup$
Because it takes a lot more induced noise current to produce the same noise voltage across 50 Ohms compared to 1 megaOhm. The tradeoff is that it is more difficult to drive.
A brick and a piece of paper: Which one is more resistant against disturbances and undesired movement in a breeze? Which is easier to move when you actually want to move it? You can't have it both ways. Same idea.
answered 8 hours ago
DKNguyenDKNguyen
6,6231 gold badge7 silver badges28 bronze badges
$endgroup$
$begingroup$
Thanks but I dont get the analogy. I need a simple circuit model with maybe a noise source if needed.
$endgroup$
– panic attack
8 hours ago
$begingroup$
@panicattack To the system, there is no difference between desired and undesired change. Something that is difficult to be disturbed by noise will also be difficult to be disturbed by you. Likewise, something easy for you to manipulate will also be easy for noise to manipulate.
$endgroup$
– DKNguyen
8 hours ago
$begingroup$
@panicattack From V=IR, if I is constant then a larger R gives a larger V.
$endgroup$
– Andrew Morton
7 hours ago
$begingroup$
@AndrewMorton Why do you model the noise as constant current source?
$endgroup$
– panic attack
6 hours ago
$begingroup$
@panicattack I am no expert on the matter, but the material containing the electrons will be at the same temperature, so the thermal noise generated by the electrons is the same, and the flow of electrons is current. Bear in mind that I could be completely wrong about the mechanism.
$endgroup$
– Andrew Morton
6 hours ago
|
show 1 more comment
$begingroup$
The first thing to consider is the Johnson noise of the resistor which cannot be eliminated. The higher the resistance the greater the noise. Reducing bandwidth will also reduce Johnson noise. So if your scope has bandwidth settings, and if you don't need high bandwidth for your signal, you can get cleaner results using the reduced bandwidth modes.
The second thing to consider is noise which couples in to the oscilloscope, particularly if it couples through the probe wiring arrangement by way of a magnetic field. The time-varying magnetic field will induce a current in the probe. The termination resistance inside the oscilloscope will convert that current to a voltage. If the termination resistor is 50 Ohms, that will lead to a much smaller voltage than if it is 1M Ohm.
In general, lower impedance termination is more resistant to noise. This is a very important concept when you encounter situations where noise immunity is required. Usually any noise coupling path will have some series resistance or fundamental power limiting just by its nature. So the lower your termination resistance, the lower the voltage due to noise coupling. Sometimes a 20 pF capacitor on a digital input can make the difference between a flaky and totally unreliable piece of junk and a rock solid product.
Often if I need to put the oscilloscope on a shunt resistor, I will use the 50 Ohm termination feature of the oscilloscope. This greatly reduces noise, and since the shunt resistance is much less than 50 Ohms (for the shunts I deal with) there is no worry of excessive current flowing into the oscilloscope, even if the shunt current may be high.
This image was formally assigned to the public domain by its creator (not me). Retrieved here: https://upload.wikimedia.org/wikipedia/commons/f/f6/JohnsonNoiseEquivalentCircuits.svg)
answered 3 hours ago
mkeithmkeith
11.9k1 gold badge11 silver badges34 bronze badges
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Because Johnson-Nyquist noise. Resistors generate thermal noise, with a noise power that's proportional to the absolute temperature. A higher-valued resistor will then generate more noise voltage, with the voltage proportional to $sqrtR$.
Setting your O-scope up as a $1mathrmMOmega$ instrument gets you the noise from that $1mathrmMOmega$ resistor; if you're measuring some super low-impedance node like you'd find in a power supply, you gain absolutely no accuracy from the high impedance. So you measure at $50Omega$, and get less noise.
answered 6 hours ago
TimWescottTimWescott
13.4k1 gold badge11 silver badges26 bronze badges
$endgroup$
$begingroup$
Interesting, so it is due to the Johnson noise. But if you in practice measure a power supply’s noise on scope do you really terminate with 50 Ohm resistance?
$endgroup$
– panic attack
6 hours ago
$begingroup$
Normal scope probes have a built in 1MΩ resistance (I'm pretty sure some or all is actually built into the cable). So using a normal cable puts that in series with the source. Fancy scopes let you switch in a 50Ω terminator so you can plug a 50Ω cable into it, and if you're using a cheaper scope you can get an in-line 50Ω terminator.
$endgroup$
– TimWescott
6 hours ago
$begingroup$
To the OP: There are two main considerations when using 50 Ohm mode. First, will the resulting current fry the oscilloscope input? You don't want to measure 120V line voltage in 50 Ohm mode, obviously. Second, will the thing you are measuring tolerate a 50 Ohm load? Measuring a signal with a 50 Ohm termination is equivalnet to placing a 50 Ohm resistor directly across the thing you are measuring. Sometimes that is OK and sometimes it is not OK because the 50 Ohms will totally change the very voltage you are trying to measure.
$endgroup$
– mkeith
3 hours ago
add a comment |
$begingroup$
Because Johnson-Nyquist noise. Resistors generate thermal noise, with a noise power that's proportional to the absolute temperature. A higher-valued resistor will then generate more noise voltage, with the voltage proportional to $sqrtR$.
Setting your O-scope up as a $1mathrmMOmega$ instrument gets you the noise from that $1mathrmMOmega$ resistor; if you're measuring some super low-impedance node like you'd find in a power supply, you gain absolutely no accuracy from the high impedance. So you measure at $50Omega$, and get less noise.
answered 6 hours ago
TimWescottTimWescott
13.4k1 gold badge11 silver badges26 bronze badges
$endgroup$
$begingroup$
Interesting, so it is due to the Johnson noise. But if you in practice measure a power supply’s noise on scope do you really terminate with 50 Ohm resistance?
$endgroup$
– panic attack
6 hours ago
$begingroup$
Normal scope probes have a built in 1MΩ resistance (I'm pretty sure some or all is actually built into the cable). So using a normal cable puts that in series with the source. Fancy scopes let you switch in a 50Ω terminator so you can plug a 50Ω cable into it, and if you're using a cheaper scope you can get an in-line 50Ω terminator.
$endgroup$
– TimWescott
6 hours ago
$begingroup$
To the OP: There are two main considerations when using 50 Ohm mode. First, will the resulting current fry the oscilloscope input? You don't want to measure 120V line voltage in 50 Ohm mode, obviously. Second, will the thing you are measuring tolerate a 50 Ohm load? Measuring a signal with a 50 Ohm termination is equivalnet to placing a 50 Ohm resistor directly across the thing you are measuring. Sometimes that is OK and sometimes it is not OK because the 50 Ohms will totally change the very voltage you are trying to measure.
$endgroup$
– mkeith
3 hours ago
add a comment |
$begingroup$
Because Johnson-Nyquist noise. Resistors generate thermal noise, with a noise power that's proportional to the absolute temperature. A higher-valued resistor will then generate more noise voltage, with the voltage proportional to $sqrtR$.
Setting your O-scope up as a $1mathrmMOmega$ instrument gets you the noise from that $1mathrmMOmega$ resistor; if you're measuring some super low-impedance node like you'd find in a power supply, you gain absolutely no accuracy from the high impedance. So you measure at $50Omega$, and get less noise.
answered 6 hours ago
TimWescottTimWescott
13.4k1 gold badge11 silver badges26 bronze badges
$endgroup$
Because Johnson-Nyquist noise. Resistors generate thermal noise, with a noise power that's proportional to the absolute temperature. A higher-valued resistor will then generate more noise voltage, with the voltage proportional to $sqrtR$.
Setting your O-scope up as a $1mathrmMOmega$ instrument gets you the noise from that $1mathrmMOmega$ resistor; if you're measuring some super low-impedance node like you'd find in a power supply, you gain absolutely no accuracy from the high impedance. So you measure at $50Omega$, and get less noise.
answered 6 hours ago
TimWescottTimWescott
13.4k1 gold badge11 silver badges26 bronze badges
answered 6 hours ago
TimWescottTimWescott
13.4k1 gold badge11 silver badges26 bronze badges
answered 6 hours ago
TimWescottTimWescott
13.4k1 gold badge11 silver badges26 bronze badges
answered 6 hours ago
TimWescottTimWescott
13.4k1 gold badge11 silver badges26 bronze badges
13.4k1 gold badge11 silver badges26 bronze badges
$begingroup$
Interesting, so it is due to the Johnson noise. But if you in practice measure a power supply’s noise on scope do you really terminate with 50 Ohm resistance?
$endgroup$
– panic attack
6 hours ago
$begingroup$
Normal scope probes have a built in 1MΩ resistance (I'm pretty sure some or all is actually built into the cable). So using a normal cable puts that in series with the source. Fancy scopes let you switch in a 50Ω terminator so you can plug a 50Ω cable into it, and if you're using a cheaper scope you can get an in-line 50Ω terminator.
$endgroup$
– TimWescott
6 hours ago
$begingroup$
To the OP: There are two main considerations when using 50 Ohm mode. First, will the resulting current fry the oscilloscope input? You don't want to measure 120V line voltage in 50 Ohm mode, obviously. Second, will the thing you are measuring tolerate a 50 Ohm load? Measuring a signal with a 50 Ohm termination is equivalnet to placing a 50 Ohm resistor directly across the thing you are measuring. Sometimes that is OK and sometimes it is not OK because the 50 Ohms will totally change the very voltage you are trying to measure.
$endgroup$
– mkeith
3 hours ago
add a comment |
$begingroup$
Interesting, so it is due to the Johnson noise. But if you in practice measure a power supply’s noise on scope do you really terminate with 50 Ohm resistance?
$endgroup$
– panic attack
6 hours ago
$begingroup$
Normal scope probes have a built in 1MΩ resistance (I'm pretty sure some or all is actually built into the cable). So using a normal cable puts that in series with the source. Fancy scopes let you switch in a 50Ω terminator so you can plug a 50Ω cable into it, and if you're using a cheaper scope you can get an in-line 50Ω terminator.
$endgroup$
– TimWescott
6 hours ago
$begingroup$
To the OP: There are two main considerations when using 50 Ohm mode. First, will the resulting current fry the oscilloscope input? You don't want to measure 120V line voltage in 50 Ohm mode, obviously. Second, will the thing you are measuring tolerate a 50 Ohm load? Measuring a signal with a 50 Ohm termination is equivalnet to placing a 50 Ohm resistor directly across the thing you are measuring. Sometimes that is OK and sometimes it is not OK because the 50 Ohms will totally change the very voltage you are trying to measure.
$endgroup$
– mkeith
3 hours ago
$begingroup$
Interesting, so it is due to the Johnson noise. But if you in practice measure a power supply’s noise on scope do you really terminate with 50 Ohm resistance?
$endgroup$
– panic attack
6 hours ago
$begingroup$
Interesting, so it is due to the Johnson noise. But if you in practice measure a power supply’s noise on scope do you really terminate with 50 Ohm resistance?
$endgroup$
– panic attack
6 hours ago
$begingroup$
Normal scope probes have a built in 1MΩ resistance (I'm pretty sure some or all is actually built into the cable). So using a normal cable puts that in series with the source. Fancy scopes let you switch in a 50Ω terminator so you can plug a 50Ω cable into it, and if you're using a cheaper scope you can get an in-line 50Ω terminator.
$endgroup$
– TimWescott
6 hours ago
$begingroup$
Normal scope probes have a built in 1MΩ resistance (I'm pretty sure some or all is actually built into the cable). So using a normal cable puts that in series with the source. Fancy scopes let you switch in a 50Ω terminator so you can plug a 50Ω cable into it, and if you're using a cheaper scope you can get an in-line 50Ω terminator.
$endgroup$
– TimWescott
6 hours ago
$begingroup$
To the OP: There are two main considerations when using 50 Ohm mode. First, will the resulting current fry the oscilloscope input? You don't want to measure 120V line voltage in 50 Ohm mode, obviously. Second, will the thing you are measuring tolerate a 50 Ohm load? Measuring a signal with a 50 Ohm termination is equivalnet to placing a 50 Ohm resistor directly across the thing you are measuring. Sometimes that is OK and sometimes it is not OK because the 50 Ohms will totally change the very voltage you are trying to measure.
$endgroup$
– mkeith
3 hours ago
$begingroup$
To the OP: There are two main considerations when using 50 Ohm mode. First, will the resulting current fry the oscilloscope input? You don't want to measure 120V line voltage in 50 Ohm mode, obviously. Second, will the thing you are measuring tolerate a 50 Ohm load? Measuring a signal with a 50 Ohm termination is equivalnet to placing a 50 Ohm resistor directly across the thing you are measuring. Sometimes that is OK and sometimes it is not OK because the 50 Ohms will totally change the very voltage you are trying to measure.
$endgroup$
– mkeith
3 hours ago
add a comment |
$begingroup$
Because it takes a lot more induced noise current to produce the same noise voltage across 50 Ohms compared to 1 megaOhm. The tradeoff is that it is more difficult to drive.
A brick and a piece of paper: Which one is more resistant against disturbances and undesired movement in a breeze? Which is easier to move when you actually want to move it? You can't have it both ways. Same idea.
answered 8 hours ago
DKNguyenDKNguyen
6,6231 gold badge7 silver badges28 bronze badges
$endgroup$
$begingroup$
Thanks but I dont get the analogy. I need a simple circuit model with maybe a noise source if needed.
$endgroup$
– panic attack
8 hours ago
$begingroup$
@panicattack To the system, there is no difference between desired and undesired change. Something that is difficult to be disturbed by noise will also be difficult to be disturbed by you. Likewise, something easy for you to manipulate will also be easy for noise to manipulate.
$endgroup$
– DKNguyen
8 hours ago
$begingroup$
@panicattack From V=IR, if I is constant then a larger R gives a larger V.
$endgroup$
– Andrew Morton
7 hours ago
$begingroup$
@AndrewMorton Why do you model the noise as constant current source?
$endgroup$
– panic attack
6 hours ago
$begingroup$
@panicattack I am no expert on the matter, but the material containing the electrons will be at the same temperature, so the thermal noise generated by the electrons is the same, and the flow of electrons is current. Bear in mind that I could be completely wrong about the mechanism.
$endgroup$
– Andrew Morton
6 hours ago
|
show 1 more comment
$begingroup$
Because it takes a lot more induced noise current to produce the same noise voltage across 50 Ohms compared to 1 megaOhm. The tradeoff is that it is more difficult to drive.
A brick and a piece of paper: Which one is more resistant against disturbances and undesired movement in a breeze? Which is easier to move when you actually want to move it? You can't have it both ways. Same idea.
answered 8 hours ago
DKNguyenDKNguyen
6,6231 gold badge7 silver badges28 bronze badges
$endgroup$
$begingroup$
Thanks but I dont get the analogy. I need a simple circuit model with maybe a noise source if needed.
$endgroup$
– panic attack
8 hours ago
$begingroup$
@panicattack To the system, there is no difference between desired and undesired change. Something that is difficult to be disturbed by noise will also be difficult to be disturbed by you. Likewise, something easy for you to manipulate will also be easy for noise to manipulate.
$endgroup$
– DKNguyen
8 hours ago
$begingroup$
@panicattack From V=IR, if I is constant then a larger R gives a larger V.
$endgroup$
– Andrew Morton
7 hours ago
$begingroup$
@AndrewMorton Why do you model the noise as constant current source?
$endgroup$
– panic attack
6 hours ago
$begingroup$
@panicattack I am no expert on the matter, but the material containing the electrons will be at the same temperature, so the thermal noise generated by the electrons is the same, and the flow of electrons is current. Bear in mind that I could be completely wrong about the mechanism.
$endgroup$
– Andrew Morton
6 hours ago
|
show 1 more comment
$begingroup$
Because it takes a lot more induced noise current to produce the same noise voltage across 50 Ohms compared to 1 megaOhm. The tradeoff is that it is more difficult to drive.
A brick and a piece of paper: Which one is more resistant against disturbances and undesired movement in a breeze? Which is easier to move when you actually want to move it? You can't have it both ways. Same idea.
answered 8 hours ago
DKNguyenDKNguyen
6,6231 gold badge7 silver badges28 bronze badges
$endgroup$
Because it takes a lot more induced noise current to produce the same noise voltage across 50 Ohms compared to 1 megaOhm. The tradeoff is that it is more difficult to drive.
A brick and a piece of paper: Which one is more resistant against disturbances and undesired movement in a breeze? Which is easier to move when you actually want to move it? You can't have it both ways. Same idea.
answered 8 hours ago
DKNguyenDKNguyen
6,6231 gold badge7 silver badges28 bronze badges
edited 8 hours ago
answered 8 hours ago
DKNguyenDKNguyen
6,6231 gold badge7 silver badges28 bronze badges
answered 8 hours ago
DKNguyenDKNguyen
6,6231 gold badge7 silver badges28 bronze badges
answered 8 hours ago
DKNguyenDKNguyen
6,6231 gold badge7 silver badges28 bronze badges
6,6231 gold badge7 silver badges28 bronze badges
$begingroup$
Thanks but I dont get the analogy. I need a simple circuit model with maybe a noise source if needed.
$endgroup$
– panic attack
8 hours ago
$begingroup$
@panicattack To the system, there is no difference between desired and undesired change. Something that is difficult to be disturbed by noise will also be difficult to be disturbed by you. Likewise, something easy for you to manipulate will also be easy for noise to manipulate.
$endgroup$
– DKNguyen
8 hours ago
$begingroup$
@panicattack From V=IR, if I is constant then a larger R gives a larger V.
$endgroup$
– Andrew Morton
7 hours ago
$begingroup$
@AndrewMorton Why do you model the noise as constant current source?
$endgroup$
– panic attack
6 hours ago
$begingroup$
@panicattack I am no expert on the matter, but the material containing the electrons will be at the same temperature, so the thermal noise generated by the electrons is the same, and the flow of electrons is current. Bear in mind that I could be completely wrong about the mechanism.
$endgroup$
– Andrew Morton
6 hours ago
|
show 1 more comment
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Thanks but I dont get the analogy. I need a simple circuit model with maybe a noise source if needed.
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– panic attack
8 hours ago
$begingroup$
@panicattack To the system, there is no difference between desired and undesired change. Something that is difficult to be disturbed by noise will also be difficult to be disturbed by you. Likewise, something easy for you to manipulate will also be easy for noise to manipulate.
$endgroup$
– DKNguyen
8 hours ago
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@panicattack From V=IR, if I is constant then a larger R gives a larger V.
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– Andrew Morton
7 hours ago
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@AndrewMorton Why do you model the noise as constant current source?
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– panic attack
6 hours ago
$begingroup$
@panicattack I am no expert on the matter, but the material containing the electrons will be at the same temperature, so the thermal noise generated by the electrons is the same, and the flow of electrons is current. Bear in mind that I could be completely wrong about the mechanism.
$endgroup$
– Andrew Morton
6 hours ago
$begingroup$
Thanks but I dont get the analogy. I need a simple circuit model with maybe a noise source if needed.
$endgroup$
– panic attack
8 hours ago
$begingroup$
Thanks but I dont get the analogy. I need a simple circuit model with maybe a noise source if needed.
$endgroup$
– panic attack
8 hours ago
$begingroup$
@panicattack To the system, there is no difference between desired and undesired change. Something that is difficult to be disturbed by noise will also be difficult to be disturbed by you. Likewise, something easy for you to manipulate will also be easy for noise to manipulate.
$endgroup$
– DKNguyen
8 hours ago
$begingroup$
@panicattack To the system, there is no difference between desired and undesired change. Something that is difficult to be disturbed by noise will also be difficult to be disturbed by you. Likewise, something easy for you to manipulate will also be easy for noise to manipulate.
$endgroup$
– DKNguyen
8 hours ago
$begingroup$
@panicattack From V=IR, if I is constant then a larger R gives a larger V.
$endgroup$
– Andrew Morton
7 hours ago
$begingroup$
@panicattack From V=IR, if I is constant then a larger R gives a larger V.
$endgroup$
– Andrew Morton
7 hours ago
$begingroup$
@AndrewMorton Why do you model the noise as constant current source?
$endgroup$
– panic attack
6 hours ago
$begingroup$
@AndrewMorton Why do you model the noise as constant current source?
$endgroup$
– panic attack
6 hours ago
$begingroup$
@panicattack I am no expert on the matter, but the material containing the electrons will be at the same temperature, so the thermal noise generated by the electrons is the same, and the flow of electrons is current. Bear in mind that I could be completely wrong about the mechanism.
$endgroup$
– Andrew Morton
6 hours ago
$begingroup$
@panicattack I am no expert on the matter, but the material containing the electrons will be at the same temperature, so the thermal noise generated by the electrons is the same, and the flow of electrons is current. Bear in mind that I could be completely wrong about the mechanism.
$endgroup$
– Andrew Morton
6 hours ago
|
show 1 more comment
$begingroup$
The first thing to consider is the Johnson noise of the resistor which cannot be eliminated. The higher the resistance the greater the noise. Reducing bandwidth will also reduce Johnson noise. So if your scope has bandwidth settings, and if you don't need high bandwidth for your signal, you can get cleaner results using the reduced bandwidth modes.
The second thing to consider is noise which couples in to the oscilloscope, particularly if it couples through the probe wiring arrangement by way of a magnetic field. The time-varying magnetic field will induce a current in the probe. The termination resistance inside the oscilloscope will convert that current to a voltage. If the termination resistor is 50 Ohms, that will lead to a much smaller voltage than if it is 1M Ohm.
In general, lower impedance termination is more resistant to noise. This is a very important concept when you encounter situations where noise immunity is required. Usually any noise coupling path will have some series resistance or fundamental power limiting just by its nature. So the lower your termination resistance, the lower the voltage due to noise coupling. Sometimes a 20 pF capacitor on a digital input can make the difference between a flaky and totally unreliable piece of junk and a rock solid product.
Often if I need to put the oscilloscope on a shunt resistor, I will use the 50 Ohm termination feature of the oscilloscope. This greatly reduces noise, and since the shunt resistance is much less than 50 Ohms (for the shunts I deal with) there is no worry of excessive current flowing into the oscilloscope, even if the shunt current may be high.
This image was formally assigned to the public domain by its creator (not me). Retrieved here: https://upload.wikimedia.org/wikipedia/commons/f/f6/JohnsonNoiseEquivalentCircuits.svg)
answered 3 hours ago
mkeithmkeith
11.9k1 gold badge11 silver badges34 bronze badges
$endgroup$
add a comment |
$begingroup$
The first thing to consider is the Johnson noise of the resistor which cannot be eliminated. The higher the resistance the greater the noise. Reducing bandwidth will also reduce Johnson noise. So if your scope has bandwidth settings, and if you don't need high bandwidth for your signal, you can get cleaner results using the reduced bandwidth modes.
The second thing to consider is noise which couples in to the oscilloscope, particularly if it couples through the probe wiring arrangement by way of a magnetic field. The time-varying magnetic field will induce a current in the probe. The termination resistance inside the oscilloscope will convert that current to a voltage. If the termination resistor is 50 Ohms, that will lead to a much smaller voltage than if it is 1M Ohm.
In general, lower impedance termination is more resistant to noise. This is a very important concept when you encounter situations where noise immunity is required. Usually any noise coupling path will have some series resistance or fundamental power limiting just by its nature. So the lower your termination resistance, the lower the voltage due to noise coupling. Sometimes a 20 pF capacitor on a digital input can make the difference between a flaky and totally unreliable piece of junk and a rock solid product.
Often if I need to put the oscilloscope on a shunt resistor, I will use the 50 Ohm termination feature of the oscilloscope. This greatly reduces noise, and since the shunt resistance is much less than 50 Ohms (for the shunts I deal with) there is no worry of excessive current flowing into the oscilloscope, even if the shunt current may be high.
This image was formally assigned to the public domain by its creator (not me). Retrieved here: https://upload.wikimedia.org/wikipedia/commons/f/f6/JohnsonNoiseEquivalentCircuits.svg)
answered 3 hours ago
mkeithmkeith
11.9k1 gold badge11 silver badges34 bronze badges
$endgroup$
add a comment |
$begingroup$
The first thing to consider is the Johnson noise of the resistor which cannot be eliminated. The higher the resistance the greater the noise. Reducing bandwidth will also reduce Johnson noise. So if your scope has bandwidth settings, and if you don't need high bandwidth for your signal, you can get cleaner results using the reduced bandwidth modes.
The second thing to consider is noise which couples in to the oscilloscope, particularly if it couples through the probe wiring arrangement by way of a magnetic field. The time-varying magnetic field will induce a current in the probe. The termination resistance inside the oscilloscope will convert that current to a voltage. If the termination resistor is 50 Ohms, that will lead to a much smaller voltage than if it is 1M Ohm.
In general, lower impedance termination is more resistant to noise. This is a very important concept when you encounter situations where noise immunity is required. Usually any noise coupling path will have some series resistance or fundamental power limiting just by its nature. So the lower your termination resistance, the lower the voltage due to noise coupling. Sometimes a 20 pF capacitor on a digital input can make the difference between a flaky and totally unreliable piece of junk and a rock solid product.
Often if I need to put the oscilloscope on a shunt resistor, I will use the 50 Ohm termination feature of the oscilloscope. This greatly reduces noise, and since the shunt resistance is much less than 50 Ohms (for the shunts I deal with) there is no worry of excessive current flowing into the oscilloscope, even if the shunt current may be high.
This image was formally assigned to the public domain by its creator (not me). Retrieved here: https://upload.wikimedia.org/wikipedia/commons/f/f6/JohnsonNoiseEquivalentCircuits.svg)
answered 3 hours ago
mkeithmkeith
11.9k1 gold badge11 silver badges34 bronze badges
$endgroup$
The first thing to consider is the Johnson noise of the resistor which cannot be eliminated. The higher the resistance the greater the noise. Reducing bandwidth will also reduce Johnson noise. So if your scope has bandwidth settings, and if you don't need high bandwidth for your signal, you can get cleaner results using the reduced bandwidth modes.
The second thing to consider is noise which couples in to the oscilloscope, particularly if it couples through the probe wiring arrangement by way of a magnetic field. The time-varying magnetic field will induce a current in the probe. The termination resistance inside the oscilloscope will convert that current to a voltage. If the termination resistor is 50 Ohms, that will lead to a much smaller voltage than if it is 1M Ohm.
In general, lower impedance termination is more resistant to noise. This is a very important concept when you encounter situations where noise immunity is required. Usually any noise coupling path will have some series resistance or fundamental power limiting just by its nature. So the lower your termination resistance, the lower the voltage due to noise coupling. Sometimes a 20 pF capacitor on a digital input can make the difference between a flaky and totally unreliable piece of junk and a rock solid product.
Often if I need to put the oscilloscope on a shunt resistor, I will use the 50 Ohm termination feature of the oscilloscope. This greatly reduces noise, and since the shunt resistance is much less than 50 Ohms (for the shunts I deal with) there is no worry of excessive current flowing into the oscilloscope, even if the shunt current may be high.
This image was formally assigned to the public domain by its creator (not me). Retrieved here: https://upload.wikimedia.org/wikipedia/commons/f/f6/JohnsonNoiseEquivalentCircuits.svg)
answered 3 hours ago
mkeithmkeith
11.9k1 gold badge11 silver badges34 bronze badges
answered 3 hours ago
mkeithmkeith
11.9k1 gold badge11 silver badges34 bronze badges
answered 3 hours ago
mkeithmkeith
11.9k1 gold badge11 silver badges34 bronze badges
answered 3 hours ago
mkeithmkeith
11.9k1 gold badge11 silver badges34 bronze badges
11.9k1 gold badge11 silver badges34 bronze badges
add a comment |
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$begingroup$
e2e.ti.com/blogs_/b/powerhouse/archive/2016/07/27/… also talks about a 50 Ohms while doing a accurate mesurement of noise
$endgroup$
– Umar
5 hours ago