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Circuit to “zoom in” on mV fluctuations of a DC signal?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Increasing precision of a practical opamp circuit when the input signal is very small40kHz signal amplifier with ua741Amplifying a decaying signal to a signal of uniform amplitudeHelp comparator circuit for this PWM signal inverterCircuit design question - low pass filterVirtual Earth - Signal ConnectionA question about choosing, implementing and placing a strain-gauge amplifierCircuit for squaring (raise to power 2) signalHow can I use a comparator in a circuit?Quadrature Encoder Interface Circuit
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$begingroup$
I have a signal that is roughly 0.2V + noise fluctuations of order 0.1-2 mV. Ideally I want to amplify this signal such that the mV fluctuations become about 1V. In other words I want to amplify the signal by about 1000x.
However, if I flat out amplify the signal, the total signal becomes 200V + 1V fluctuations, which I can't reasonably read on some bench top DAQ (0-10V range).
Is there some combination of circuit elements that can take my input 0.2V + 1mV signal and spit out only the amplified fluctuations (i.e. 0V + 1V fluctuations)?
edit: I should say that these fluctuations are controlled by me physically squeezing a pressure gauge, so they aren't necessarily high frequency. Basically the signal rises to 0.202V when I squeeze, and 0.200V when I let go. I want to see that excess 0.002V blown up to 1V, but I may be squeezing and letting go slowly in general.
operational-amplifier amplifier circuit-design signal-processing
asked 1 hour ago
MartyMarty
112
New contributor
Marty is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I have a signal that is roughly 0.2V + noise fluctuations of order 0.1-2 mV. Ideally I want to amplify this signal such that the mV fluctuations become about 1V. In other words I want to amplify the signal by about 1000x.
However, if I flat out amplify the signal, the total signal becomes 200V + 1V fluctuations, which I can't reasonably read on some bench top DAQ (0-10V range).
Is there some combination of circuit elements that can take my input 0.2V + 1mV signal and spit out only the amplified fluctuations (i.e. 0V + 1V fluctuations)?
edit: I should say that these fluctuations are controlled by me physically squeezing a pressure gauge, so they aren't necessarily high frequency. Basically the signal rises to 0.202V when I squeeze, and 0.200V when I let go. I want to see that excess 0.002V blown up to 1V, but I may be squeezing and letting go slowly in general.
operational-amplifier amplifier circuit-design signal-processing
asked 1 hour ago
MartyMarty
112
New contributor
Marty is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Are you interested in the signal? Or the noise? I can't tell from the writing. I'd normally assume that you don't want the signal part. But I'd rather not assume. Instead, just ask.
$endgroup$
– jonk
45 mins ago
add a comment |
$begingroup$
I have a signal that is roughly 0.2V + noise fluctuations of order 0.1-2 mV. Ideally I want to amplify this signal such that the mV fluctuations become about 1V. In other words I want to amplify the signal by about 1000x.
However, if I flat out amplify the signal, the total signal becomes 200V + 1V fluctuations, which I can't reasonably read on some bench top DAQ (0-10V range).
Is there some combination of circuit elements that can take my input 0.2V + 1mV signal and spit out only the amplified fluctuations (i.e. 0V + 1V fluctuations)?
edit: I should say that these fluctuations are controlled by me physically squeezing a pressure gauge, so they aren't necessarily high frequency. Basically the signal rises to 0.202V when I squeeze, and 0.200V when I let go. I want to see that excess 0.002V blown up to 1V, but I may be squeezing and letting go slowly in general.
operational-amplifier amplifier circuit-design signal-processing
asked 1 hour ago
MartyMarty
112
New contributor
Marty is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have a signal that is roughly 0.2V + noise fluctuations of order 0.1-2 mV. Ideally I want to amplify this signal such that the mV fluctuations become about 1V. In other words I want to amplify the signal by about 1000x.
However, if I flat out amplify the signal, the total signal becomes 200V + 1V fluctuations, which I can't reasonably read on some bench top DAQ (0-10V range).
Is there some combination of circuit elements that can take my input 0.2V + 1mV signal and spit out only the amplified fluctuations (i.e. 0V + 1V fluctuations)?
edit: I should say that these fluctuations are controlled by me physically squeezing a pressure gauge, so they aren't necessarily high frequency. Basically the signal rises to 0.202V when I squeeze, and 0.200V when I let go. I want to see that excess 0.002V blown up to 1V, but I may be squeezing and letting go slowly in general.
operational-amplifier amplifier circuit-design signal-processing
operational-amplifier amplifier circuit-design signal-processing
asked 1 hour ago
MartyMarty
112
New contributor
Marty is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
MartyMarty
112
New contributor
Marty is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 42 mins ago
Marty
asked 1 hour ago
MartyMarty
112
New contributor
Marty is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
MartyMarty
112
asked 1 hour ago
MartyMarty
112
112
New contributor
Marty is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Marty is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Marty is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Are you interested in the signal? Or the noise? I can't tell from the writing. I'd normally assume that you don't want the signal part. But I'd rather not assume. Instead, just ask.
$endgroup$
– jonk
45 mins ago
add a comment |
$begingroup$
Are you interested in the signal? Or the noise? I can't tell from the writing. I'd normally assume that you don't want the signal part. But I'd rather not assume. Instead, just ask.
$endgroup$
– jonk
45 mins ago
$begingroup$
Are you interested in the signal? Or the noise? I can't tell from the writing. I'd normally assume that you don't want the signal part. But I'd rather not assume. Instead, just ask.
$endgroup$
– jonk
45 mins ago
$begingroup$
Are you interested in the signal? Or the noise? I can't tell from the writing. I'd normally assume that you don't want the signal part. But I'd rather not assume. Instead, just ask.
$endgroup$
– jonk
45 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Capacitors block DC and pass AC.
You can use a series capacitor into an opamp with whatever gain you need.
Even better might be a simple RC high-pass filter...One capacitor (series) and one resistor (to ground) in front of your amplifier.
Like this:
simulate this circuit – Schematic created using CircuitLab
R2 and R3 set your gain. C1 and R1 set your low frequency cut-off. The formula you use to find the cutoff is:
$$Ftext(Hz) = frac12 pi R C$$
edited 32 mins ago
Dave Tweed♦
125k10155269
answered 1 hour ago
evildemonicevildemonic
2,653922
$endgroup$
$begingroup$
Thank you for your answer! If you see my edit: will the capacitor block out the fluctuations if they aren't very fast (maybe a quick squeeze/release every 2 seconds)? i.e. a voltage difference when I squeeze a pressure gauge (squeezing vs not squeezing is only a ~1mV signal added to the 0.2V DC)
$endgroup$
– Marty
41 mins ago
$begingroup$
Yes, you will need to choose C1 and R1 based on the slowest change you wish to see. The formula you use to find the cutoff is: F(Hz) = 1 / (2 * pi * R * C)
$endgroup$
– evildemonic
40 mins ago
$begingroup$
Sorry, I am still trying to figure out how to insert the nice looking equations others use here.
$endgroup$
– evildemonic
36 mins ago
1
$begingroup$
It's called "MathJax". I've added your formula to your answer to show you how it's done. You can learn more by clicking on the help icon in the editor, select "Advanced Help" and scroll down to the section labeled "LaTeX", which also has a link to MathJax specifically. There's also this post on meta, which provides links to a number of quick references and other resources.
$endgroup$
– Dave Tweed♦
30 mins ago
1
$begingroup$
So if I wanted a gain of 1000 and a cutoff of 1 Hz, the following values might work? C1=100 uF, R1=1.5k ohm, R2=100k ohm, R3=100 ohm
$endgroup$
– Marty
29 mins ago
|
show 3 more comments
$begingroup$
Use a coupling capacitor prior to the amplifier. The DC signal will be blocked but the fluctuations will pass through.
answered 1 hour ago
Charles HCharles H
511
$endgroup$
add a comment |
$begingroup$
Digital designer here so I'm not certain, but...
The other answers assume high-frequency fluctuations. Instead you want to subtract the 0.2 V and amplify that. You can use a summing amplifier to subtract the offset, if you've got positive and negative supply voltages. I think you can also use an inverting configuration where the non-inverting input is at 0.2V instead of ground.
answered 37 mins ago
MattMatt
31016
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Capacitors block DC and pass AC.
You can use a series capacitor into an opamp with whatever gain you need.
Even better might be a simple RC high-pass filter...One capacitor (series) and one resistor (to ground) in front of your amplifier.
Like this:
simulate this circuit – Schematic created using CircuitLab
R2 and R3 set your gain. C1 and R1 set your low frequency cut-off. The formula you use to find the cutoff is:
$$Ftext(Hz) = frac12 pi R C$$
edited 32 mins ago
Dave Tweed♦
125k10155269
answered 1 hour ago
evildemonicevildemonic
2,653922
$endgroup$
$begingroup$
Thank you for your answer! If you see my edit: will the capacitor block out the fluctuations if they aren't very fast (maybe a quick squeeze/release every 2 seconds)? i.e. a voltage difference when I squeeze a pressure gauge (squeezing vs not squeezing is only a ~1mV signal added to the 0.2V DC)
$endgroup$
– Marty
41 mins ago
$begingroup$
Yes, you will need to choose C1 and R1 based on the slowest change you wish to see. The formula you use to find the cutoff is: F(Hz) = 1 / (2 * pi * R * C)
$endgroup$
– evildemonic
40 mins ago
$begingroup$
Sorry, I am still trying to figure out how to insert the nice looking equations others use here.
$endgroup$
– evildemonic
36 mins ago
1
$begingroup$
It's called "MathJax". I've added your formula to your answer to show you how it's done. You can learn more by clicking on the help icon in the editor, select "Advanced Help" and scroll down to the section labeled "LaTeX", which also has a link to MathJax specifically. There's also this post on meta, which provides links to a number of quick references and other resources.
$endgroup$
– Dave Tweed♦
30 mins ago
1
$begingroup$
So if I wanted a gain of 1000 and a cutoff of 1 Hz, the following values might work? C1=100 uF, R1=1.5k ohm, R2=100k ohm, R3=100 ohm
$endgroup$
– Marty
29 mins ago
|
show 3 more comments
$begingroup$
Capacitors block DC and pass AC.
You can use a series capacitor into an opamp with whatever gain you need.
Even better might be a simple RC high-pass filter...One capacitor (series) and one resistor (to ground) in front of your amplifier.
Like this:
simulate this circuit – Schematic created using CircuitLab
R2 and R3 set your gain. C1 and R1 set your low frequency cut-off. The formula you use to find the cutoff is:
$$Ftext(Hz) = frac12 pi R C$$
edited 32 mins ago
Dave Tweed♦
125k10155269
answered 1 hour ago
evildemonicevildemonic
2,653922
$endgroup$
$begingroup$
Thank you for your answer! If you see my edit: will the capacitor block out the fluctuations if they aren't very fast (maybe a quick squeeze/release every 2 seconds)? i.e. a voltage difference when I squeeze a pressure gauge (squeezing vs not squeezing is only a ~1mV signal added to the 0.2V DC)
$endgroup$
– Marty
41 mins ago
$begingroup$
Yes, you will need to choose C1 and R1 based on the slowest change you wish to see. The formula you use to find the cutoff is: F(Hz) = 1 / (2 * pi * R * C)
$endgroup$
– evildemonic
40 mins ago
$begingroup$
Sorry, I am still trying to figure out how to insert the nice looking equations others use here.
$endgroup$
– evildemonic
36 mins ago
1
$begingroup$
It's called "MathJax". I've added your formula to your answer to show you how it's done. You can learn more by clicking on the help icon in the editor, select "Advanced Help" and scroll down to the section labeled "LaTeX", which also has a link to MathJax specifically. There's also this post on meta, which provides links to a number of quick references and other resources.
$endgroup$
– Dave Tweed♦
30 mins ago
1
$begingroup$
So if I wanted a gain of 1000 and a cutoff of 1 Hz, the following values might work? C1=100 uF, R1=1.5k ohm, R2=100k ohm, R3=100 ohm
$endgroup$
– Marty
29 mins ago
|
show 3 more comments
$begingroup$
Capacitors block DC and pass AC.
You can use a series capacitor into an opamp with whatever gain you need.
Even better might be a simple RC high-pass filter...One capacitor (series) and one resistor (to ground) in front of your amplifier.
Like this:
simulate this circuit – Schematic created using CircuitLab
R2 and R3 set your gain. C1 and R1 set your low frequency cut-off. The formula you use to find the cutoff is:
$$Ftext(Hz) = frac12 pi R C$$
edited 32 mins ago
Dave Tweed♦
125k10155269
answered 1 hour ago
evildemonicevildemonic
2,653922
$endgroup$
Capacitors block DC and pass AC.
You can use a series capacitor into an opamp with whatever gain you need.
Even better might be a simple RC high-pass filter...One capacitor (series) and one resistor (to ground) in front of your amplifier.
Like this:
simulate this circuit – Schematic created using CircuitLab
R2 and R3 set your gain. C1 and R1 set your low frequency cut-off. The formula you use to find the cutoff is:
$$Ftext(Hz) = frac12 pi R C$$
edited 32 mins ago
Dave Tweed♦
125k10155269
answered 1 hour ago
evildemonicevildemonic
2,653922
edited 32 mins ago
Dave Tweed♦
125k10155269
edited 32 mins ago
Dave Tweed♦
125k10155269
edited 32 mins ago
Dave Tweed♦
125k10155269
125k10155269
answered 1 hour ago
evildemonicevildemonic
2,653922
answered 1 hour ago
evildemonicevildemonic
2,653922
answered 1 hour ago
evildemonicevildemonic
2,653922
2,653922
$begingroup$
Thank you for your answer! If you see my edit: will the capacitor block out the fluctuations if they aren't very fast (maybe a quick squeeze/release every 2 seconds)? i.e. a voltage difference when I squeeze a pressure gauge (squeezing vs not squeezing is only a ~1mV signal added to the 0.2V DC)
$endgroup$
– Marty
41 mins ago
$begingroup$
Yes, you will need to choose C1 and R1 based on the slowest change you wish to see. The formula you use to find the cutoff is: F(Hz) = 1 / (2 * pi * R * C)
$endgroup$
– evildemonic
40 mins ago
$begingroup$
Sorry, I am still trying to figure out how to insert the nice looking equations others use here.
$endgroup$
– evildemonic
36 mins ago
1
$begingroup$
It's called "MathJax". I've added your formula to your answer to show you how it's done. You can learn more by clicking on the help icon in the editor, select "Advanced Help" and scroll down to the section labeled "LaTeX", which also has a link to MathJax specifically. There's also this post on meta, which provides links to a number of quick references and other resources.
$endgroup$
– Dave Tweed♦
30 mins ago
1
$begingroup$
So if I wanted a gain of 1000 and a cutoff of 1 Hz, the following values might work? C1=100 uF, R1=1.5k ohm, R2=100k ohm, R3=100 ohm
$endgroup$
– Marty
29 mins ago
|
show 3 more comments
$begingroup$
Thank you for your answer! If you see my edit: will the capacitor block out the fluctuations if they aren't very fast (maybe a quick squeeze/release every 2 seconds)? i.e. a voltage difference when I squeeze a pressure gauge (squeezing vs not squeezing is only a ~1mV signal added to the 0.2V DC)
$endgroup$
– Marty
41 mins ago
$begingroup$
Yes, you will need to choose C1 and R1 based on the slowest change you wish to see. The formula you use to find the cutoff is: F(Hz) = 1 / (2 * pi * R * C)
$endgroup$
– evildemonic
40 mins ago
$begingroup$
Sorry, I am still trying to figure out how to insert the nice looking equations others use here.
$endgroup$
– evildemonic
36 mins ago
1
$begingroup$
It's called "MathJax". I've added your formula to your answer to show you how it's done. You can learn more by clicking on the help icon in the editor, select "Advanced Help" and scroll down to the section labeled "LaTeX", which also has a link to MathJax specifically. There's also this post on meta, which provides links to a number of quick references and other resources.
$endgroup$
– Dave Tweed♦
30 mins ago
1
$begingroup$
So if I wanted a gain of 1000 and a cutoff of 1 Hz, the following values might work? C1=100 uF, R1=1.5k ohm, R2=100k ohm, R3=100 ohm
$endgroup$
– Marty
29 mins ago
$begingroup$
Thank you for your answer! If you see my edit: will the capacitor block out the fluctuations if they aren't very fast (maybe a quick squeeze/release every 2 seconds)? i.e. a voltage difference when I squeeze a pressure gauge (squeezing vs not squeezing is only a ~1mV signal added to the 0.2V DC)
$endgroup$
– Marty
41 mins ago
$begingroup$
Thank you for your answer! If you see my edit: will the capacitor block out the fluctuations if they aren't very fast (maybe a quick squeeze/release every 2 seconds)? i.e. a voltage difference when I squeeze a pressure gauge (squeezing vs not squeezing is only a ~1mV signal added to the 0.2V DC)
$endgroup$
– Marty
41 mins ago
$begingroup$
Yes, you will need to choose C1 and R1 based on the slowest change you wish to see. The formula you use to find the cutoff is: F(Hz) = 1 / (2 * pi * R * C)
$endgroup$
– evildemonic
40 mins ago
$begingroup$
Yes, you will need to choose C1 and R1 based on the slowest change you wish to see. The formula you use to find the cutoff is: F(Hz) = 1 / (2 * pi * R * C)
$endgroup$
– evildemonic
40 mins ago
$begingroup$
Sorry, I am still trying to figure out how to insert the nice looking equations others use here.
$endgroup$
– evildemonic
36 mins ago
$begingroup$
Sorry, I am still trying to figure out how to insert the nice looking equations others use here.
$endgroup$
– evildemonic
36 mins ago
1
1
$begingroup$
It's called "MathJax". I've added your formula to your answer to show you how it's done. You can learn more by clicking on the help icon in the editor, select "Advanced Help" and scroll down to the section labeled "LaTeX", which also has a link to MathJax specifically. There's also this post on meta, which provides links to a number of quick references and other resources.
$endgroup$
– Dave Tweed♦
30 mins ago
$begingroup$
It's called "MathJax". I've added your formula to your answer to show you how it's done. You can learn more by clicking on the help icon in the editor, select "Advanced Help" and scroll down to the section labeled "LaTeX", which also has a link to MathJax specifically. There's also this post on meta, which provides links to a number of quick references and other resources.
$endgroup$
– Dave Tweed♦
30 mins ago
1
1
$begingroup$
So if I wanted a gain of 1000 and a cutoff of 1 Hz, the following values might work? C1=100 uF, R1=1.5k ohm, R2=100k ohm, R3=100 ohm
$endgroup$
– Marty
29 mins ago
$begingroup$
So if I wanted a gain of 1000 and a cutoff of 1 Hz, the following values might work? C1=100 uF, R1=1.5k ohm, R2=100k ohm, R3=100 ohm
$endgroup$
– Marty
29 mins ago
|
show 3 more comments
$begingroup$
Use a coupling capacitor prior to the amplifier. The DC signal will be blocked but the fluctuations will pass through.
answered 1 hour ago
Charles HCharles H
511
$endgroup$
add a comment |
$begingroup$
Use a coupling capacitor prior to the amplifier. The DC signal will be blocked but the fluctuations will pass through.
answered 1 hour ago
Charles HCharles H
511
$endgroup$
add a comment |
$begingroup$
Use a coupling capacitor prior to the amplifier. The DC signal will be blocked but the fluctuations will pass through.
answered 1 hour ago
Charles HCharles H
511
$endgroup$
Use a coupling capacitor prior to the amplifier. The DC signal will be blocked but the fluctuations will pass through.
answered 1 hour ago
Charles HCharles H
511
answered 1 hour ago
Charles HCharles H
511
answered 1 hour ago
Charles HCharles H
511
answered 1 hour ago
Charles HCharles H
511
511
add a comment |
add a comment |
$begingroup$
Digital designer here so I'm not certain, but...
The other answers assume high-frequency fluctuations. Instead you want to subtract the 0.2 V and amplify that. You can use a summing amplifier to subtract the offset, if you've got positive and negative supply voltages. I think you can also use an inverting configuration where the non-inverting input is at 0.2V instead of ground.
answered 37 mins ago
MattMatt
31016
$endgroup$
add a comment |
$begingroup$
Digital designer here so I'm not certain, but...
The other answers assume high-frequency fluctuations. Instead you want to subtract the 0.2 V and amplify that. You can use a summing amplifier to subtract the offset, if you've got positive and negative supply voltages. I think you can also use an inverting configuration where the non-inverting input is at 0.2V instead of ground.
answered 37 mins ago
MattMatt
31016
$endgroup$
add a comment |
$begingroup$
Digital designer here so I'm not certain, but...
The other answers assume high-frequency fluctuations. Instead you want to subtract the 0.2 V and amplify that. You can use a summing amplifier to subtract the offset, if you've got positive and negative supply voltages. I think you can also use an inverting configuration where the non-inverting input is at 0.2V instead of ground.
answered 37 mins ago
MattMatt
31016
$endgroup$
Digital designer here so I'm not certain, but...
The other answers assume high-frequency fluctuations. Instead you want to subtract the 0.2 V and amplify that. You can use a summing amplifier to subtract the offset, if you've got positive and negative supply voltages. I think you can also use an inverting configuration where the non-inverting input is at 0.2V instead of ground.
answered 37 mins ago
MattMatt
31016
answered 37 mins ago
MattMatt
31016
answered 37 mins ago
MattMatt
31016
answered 37 mins ago
MattMatt
31016
31016
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Marty is a new contributor. Be nice, and check out our Code of Conduct.
Marty is a new contributor. Be nice, and check out our Code of Conduct.
Marty is a new contributor. Be nice, and check out our Code of Conduct.
Marty is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Are you interested in the signal? Or the noise? I can't tell from the writing. I'd normally assume that you don't want the signal part. But I'd rather not assume. Instead, just ask.
$endgroup$
– jonk
45 mins ago