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Circuit to "zoom in" on mV fluctuations of a DC signal?

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Circuit to “zoom in” on mV fluctuations of a DC signal?

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2

$begingroup$

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

share|improve this question

edited 37 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.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  41 mins ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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

share|improve this question

edited 37 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.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  41 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

share|improve this question

edited 37 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.

$endgroup$

share|improve this question

edited 37 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.

share|improve this question

edited 37 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

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

3 Answers
3

active

oldest

votes

3

$begingroup$

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:

$$Ftext{(Hz)} = frac{1}{2 pi R C}$$

share|improve this answer

edited 27 mins ago

Dave Tweed♦

125k10155269

answered 58 mins ago

evildemonicevildemonic

2,643922

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  37 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
  35 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, I am still trying to figure out how to insert the nice looking equations others use here.
  $endgroup$
  – evildemonic
  31 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♦
  26 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
  24 mins ago
  

  
  
  
  

 | 
show 2 more comments

1

$begingroup$

Use a coupling capacitor prior to the amplifier. The DC signal will be blocked but the fluctuations will pass through.

share|improve this answer

answered 57 mins ago

Charles HCharles H

511

$endgroup$

add a comment | 

0

$begingroup$

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.

share|improve this answer

answered 32 mins ago

MattMatt

31016

$endgroup$

add a comment | 

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3

$begingroup$

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:

$$Ftext{(Hz)} = frac{1}{2 pi R C}$$

share|improve this answer

edited 27 mins ago

Dave Tweed♦

125k10155269

answered 58 mins ago

evildemonicevildemonic

2,643922

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  37 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
  35 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, I am still trying to figure out how to insert the nice looking equations others use here.
  $endgroup$
  – evildemonic
  31 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♦
  26 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
  24 mins ago
  

  
  
  
  

 | 
show 2 more comments

3

$begingroup$

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:

$$Ftext{(Hz)} = frac{1}{2 pi R C}$$

share|improve this answer

edited 27 mins ago

Dave Tweed♦

125k10155269

answered 58 mins ago

evildemonicevildemonic

2,643922

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  37 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
  35 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, I am still trying to figure out how to insert the nice looking equations others use here.
  $endgroup$
  – evildemonic
  31 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♦
  26 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
  24 mins ago
  

  
  
  
  

 | 
show 2 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:

$$Ftext{(Hz)} = frac{1}{2 pi R C}$$

share|improve this answer

edited 27 mins ago

Dave Tweed♦

125k10155269

answered 58 mins ago

evildemonicevildemonic

2,643922

$endgroup$

share|improve this answer

edited 27 mins ago

Dave Tweed♦

125k10155269

answered 58 mins ago

evildemonicevildemonic

2,643922

edited 27 mins ago

Dave Tweed♦

125k10155269

edited 27 mins ago

Dave Tweed♦

125k10155269

answered 58 mins ago

evildemonicevildemonic

2,643922

answered 58 mins ago

evildemonicevildemonic

2,643922

1

$begingroup$

Use a coupling capacitor prior to the amplifier. The DC signal will be blocked but the fluctuations will pass through.

share|improve this answer

answered 57 mins ago

Charles HCharles H

511

$endgroup$

add a comment | 

1

$begingroup$

Use a coupling capacitor prior to the amplifier. The DC signal will be blocked but the fluctuations will pass through.

share|improve this answer

answered 57 mins 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.

share|improve this answer

answered 57 mins ago

Charles HCharles H

511

$endgroup$

share|improve this answer

answered 57 mins ago

Charles HCharles H

511

answered 57 mins ago

Charles HCharles H

511

answered 57 mins ago

Charles HCharles H

511

0

$begingroup$

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.

share|improve this answer

answered 32 mins ago

MattMatt

31016

$endgroup$

add a comment | 

0

$begingroup$

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.

share|improve this answer

answered 32 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.

share|improve this answer

answered 32 mins ago

MattMatt

31016

$endgroup$

share|improve this answer

answered 32 mins ago

MattMatt

31016

answered 32 mins ago

MattMatt

31016

answered 32 mins ago

MattMatt

31016