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How to add a low pass filter to this non-inverting amplifier circuit?

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2

$begingroup$

I've come up with this simple circuit as an amplifier before an 8bit ADC which receives audio signals. R2 is actually a variable resistor (pot.) of 100k. The circuit also biases the signal to VCC/2 and uses large resistors to set the input impedance is pretty high all over the working audio frequency range. This works well for my application, however I would like to add a single low pass filter stage to this amplifier so I can remove some high freq. content that I would not be able to sample. How to approach this without having to use another op-amp and without changing the characteristics of the circuit in the spectrum I would like to pass?

operational-amplifier filter low-pass non-inverting

share|improve this question

edited May 18 at 23:36

K H

2,430315

asked May 18 at 22:16

user733606user733606

108111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what is the amplitude of the high frequency content? 8 LSBs?
  $endgroup$
  – analogsystemsrf
  May 18 at 23:53
  

  
  
  
  

add a comment | 

2

$begingroup$

I've come up with this simple circuit as an amplifier before an 8bit ADC which receives audio signals. R2 is actually a variable resistor (pot.) of 100k. The circuit also biases the signal to VCC/2 and uses large resistors to set the input impedance is pretty high all over the working audio frequency range. This works well for my application, however I would like to add a single low pass filter stage to this amplifier so I can remove some high freq. content that I would not be able to sample. How to approach this without having to use another op-amp and without changing the characteristics of the circuit in the spectrum I would like to pass?

operational-amplifier filter low-pass non-inverting

share|improve this question

edited May 18 at 23:36

K H

2,430315

asked May 18 at 22:16

user733606user733606

108111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what is the amplitude of the high frequency content? 8 LSBs?
  $endgroup$
  – analogsystemsrf
  May 18 at 23:53
  

  
  
  
  

add a comment | 

$begingroup$

I've come up with this simple circuit as an amplifier before an 8bit ADC which receives audio signals. R2 is actually a variable resistor (pot.) of 100k. The circuit also biases the signal to VCC/2 and uses large resistors to set the input impedance is pretty high all over the working audio frequency range. This works well for my application, however I would like to add a single low pass filter stage to this amplifier so I can remove some high freq. content that I would not be able to sample. How to approach this without having to use another op-amp and without changing the characteristics of the circuit in the spectrum I would like to pass?

operational-amplifier filter low-pass non-inverting

share|improve this question

edited May 18 at 23:36

K H

2,430315

asked May 18 at 22:16

user733606user733606

108111

$endgroup$

share|improve this question

edited May 18 at 23:36

K H

2,430315

asked May 18 at 22:16

user733606user733606

108111

share|improve this question

edited May 18 at 23:36

K H

2,430315

asked May 18 at 22:16

user733606user733606

108111

edited May 18 at 23:36

K H

2,430315

edited May 18 at 23:36

K H

2,430315

asked May 18 at 22:16

user733606user733606

108111

asked May 18 at 22:16

user733606user733606

108111

3 Answers
3

active

oldest

votes

3

$begingroup$

The correct approach is to choose a LPF and sampling frequency such that the maximum signal at fs/2 is less than your ADC resolution. This means you need a brick wall filter at 3x your -3dB BW or 128 x faster sampling rate than your signal -3dB BW for a 20dB decade filter for an 8bit ADC... not 2x faster with a 1st order filter.

share|improve this answer

answered May 18 at 22:41

Sunnyskyguy EE75Sunnyskyguy EE75

75.1k229106

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  OK, so let's assume my sampling freq. is 10KHz, so Nyquist freq. is 5KHz . My input range is 5V so LSB is ~19.53mV. So you are saying I need to select a cut freq. so that I can have about -48 dB attenuation at 5KHz? And the filter order depends on how low (or high) I select the cut freq. Is that correct?
  $endgroup$
  – user733606
  May 18 at 23:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Correct so 6dB/octave or 8th order at 2.5kHz for less than telephone quality audio which uses log ADC with 8bits > >72 dB I think
  $endgroup$
  – Sunnyskyguy EE75
  May 18 at 23:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  There are telephony filter IC’s that will work. I believe they are 8th order 2k3Hz but don’t quote me
  $endgroup$
  – Sunnyskyguy EE75
  May 19 at 3:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have one of those switches capacitors filters around that are tuned using a clock signal. It's 8th order and have a very sharp cut-off point. Right now I have no LPF at the input and I've managed to get audio from Youtube and process it and it didn't sound that bad to be honest. I guess the only way to figure this out is by testing different Fcut points (if not going "brick-wall").
  $endgroup$
  – user733606
  May 19 at 13:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can analyze the spectrum of audio in and out on Audacity and try generated test tones and measure image effects and distortion.
  $endgroup$
  – Sunnyskyguy EE75
  May 20 at 17:41
  

  
  
  
  

add a comment | 

0

$begingroup$

Put about 1k in series with C1 and a cap in parallel with R3. 800 ohms and 0.01 uF will give about 20kHz. But this is only a first-order filter, not very useful, as SunnySKyGuy says.

Edit: this assumes that the driving impedance is low. If not and you know what it is, then just put an appropriate cap.

share|improve this answer

edited May 18 at 23:02

answered May 18 at 22:48

Mattman944Mattman944

92117

$endgroup$

add a comment | 

0

$begingroup$

Add a low pass filter to a non-inverting amplifier circuit.

Without giving it much thought you already have a high pass filter on the input, and the simplest way to apply a low pass is to use the output of the opamp.

This image is from this calculator which may help you pick your filter component values easily. The low pass filter is R2C2 in the image above and is buffered by the opamp reducing the interaction of gain and filter components.

share|improve this answer

edited May 18 at 23:49

answered May 18 at 23:00

Jack CreaseyJack Creasey

16.2k2824

$endgroup$

add a comment | 

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3

$begingroup$

The correct approach is to choose a LPF and sampling frequency such that the maximum signal at fs/2 is less than your ADC resolution. This means you need a brick wall filter at 3x your -3dB BW or 128 x faster sampling rate than your signal -3dB BW for a 20dB decade filter for an 8bit ADC... not 2x faster with a 1st order filter.

share|improve this answer

answered May 18 at 22:41

Sunnyskyguy EE75Sunnyskyguy EE75

75.1k229106

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  OK, so let's assume my sampling freq. is 10KHz, so Nyquist freq. is 5KHz . My input range is 5V so LSB is ~19.53mV. So you are saying I need to select a cut freq. so that I can have about -48 dB attenuation at 5KHz? And the filter order depends on how low (or high) I select the cut freq. Is that correct?
  $endgroup$
  – user733606
  May 18 at 23:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Correct so 6dB/octave or 8th order at 2.5kHz for less than telephone quality audio which uses log ADC with 8bits > >72 dB I think
  $endgroup$
  – Sunnyskyguy EE75
  May 18 at 23:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  There are telephony filter IC’s that will work. I believe they are 8th order 2k3Hz but don’t quote me
  $endgroup$
  – Sunnyskyguy EE75
  May 19 at 3:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have one of those switches capacitors filters around that are tuned using a clock signal. It's 8th order and have a very sharp cut-off point. Right now I have no LPF at the input and I've managed to get audio from Youtube and process it and it didn't sound that bad to be honest. I guess the only way to figure this out is by testing different Fcut points (if not going "brick-wall").
  $endgroup$
  – user733606
  May 19 at 13:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can analyze the spectrum of audio in and out on Audacity and try generated test tones and measure image effects and distortion.
  $endgroup$
  – Sunnyskyguy EE75
  May 20 at 17:41
  

  
  
  
  

add a comment | 

3

$begingroup$

The correct approach is to choose a LPF and sampling frequency such that the maximum signal at fs/2 is less than your ADC resolution. This means you need a brick wall filter at 3x your -3dB BW or 128 x faster sampling rate than your signal -3dB BW for a 20dB decade filter for an 8bit ADC... not 2x faster with a 1st order filter.

share|improve this answer

answered May 18 at 22:41

Sunnyskyguy EE75Sunnyskyguy EE75

75.1k229106

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  OK, so let's assume my sampling freq. is 10KHz, so Nyquist freq. is 5KHz . My input range is 5V so LSB is ~19.53mV. So you are saying I need to select a cut freq. so that I can have about -48 dB attenuation at 5KHz? And the filter order depends on how low (or high) I select the cut freq. Is that correct?
  $endgroup$
  – user733606
  May 18 at 23:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Correct so 6dB/octave or 8th order at 2.5kHz for less than telephone quality audio which uses log ADC with 8bits > >72 dB I think
  $endgroup$
  – Sunnyskyguy EE75
  May 18 at 23:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  There are telephony filter IC’s that will work. I believe they are 8th order 2k3Hz but don’t quote me
  $endgroup$
  – Sunnyskyguy EE75
  May 19 at 3:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have one of those switches capacitors filters around that are tuned using a clock signal. It's 8th order and have a very sharp cut-off point. Right now I have no LPF at the input and I've managed to get audio from Youtube and process it and it didn't sound that bad to be honest. I guess the only way to figure this out is by testing different Fcut points (if not going "brick-wall").
  $endgroup$
  – user733606
  May 19 at 13:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can analyze the spectrum of audio in and out on Audacity and try generated test tones and measure image effects and distortion.
  $endgroup$
  – Sunnyskyguy EE75
  May 20 at 17:41
  

  
  
  
  

add a comment | 

$begingroup$

The correct approach is to choose a LPF and sampling frequency such that the maximum signal at fs/2 is less than your ADC resolution. This means you need a brick wall filter at 3x your -3dB BW or 128 x faster sampling rate than your signal -3dB BW for a 20dB decade filter for an 8bit ADC... not 2x faster with a 1st order filter.

share|improve this answer

answered May 18 at 22:41

Sunnyskyguy EE75Sunnyskyguy EE75

75.1k229106

$endgroup$

share|improve this answer

answered May 18 at 22:41

Sunnyskyguy EE75Sunnyskyguy EE75

75.1k229106

answered May 18 at 22:41

Sunnyskyguy EE75Sunnyskyguy EE75

75.1k229106

answered May 18 at 22:41

Sunnyskyguy EE75Sunnyskyguy EE75

75.1k229106

0

$begingroup$

Put about 1k in series with C1 and a cap in parallel with R3. 800 ohms and 0.01 uF will give about 20kHz. But this is only a first-order filter, not very useful, as SunnySKyGuy says.

Edit: this assumes that the driving impedance is low. If not and you know what it is, then just put an appropriate cap.

share|improve this answer

edited May 18 at 23:02

answered May 18 at 22:48

Mattman944Mattman944

92117

$endgroup$

add a comment | 

0

$begingroup$

Put about 1k in series with C1 and a cap in parallel with R3. 800 ohms and 0.01 uF will give about 20kHz. But this is only a first-order filter, not very useful, as SunnySKyGuy says.

Edit: this assumes that the driving impedance is low. If not and you know what it is, then just put an appropriate cap.

share|improve this answer

edited May 18 at 23:02

answered May 18 at 22:48

Mattman944Mattman944

92117

$endgroup$

add a comment | 

$begingroup$

Put about 1k in series with C1 and a cap in parallel with R3. 800 ohms and 0.01 uF will give about 20kHz. But this is only a first-order filter, not very useful, as SunnySKyGuy says.

Edit: this assumes that the driving impedance is low. If not and you know what it is, then just put an appropriate cap.

share|improve this answer

edited May 18 at 23:02

answered May 18 at 22:48

Mattman944Mattman944

92117

$endgroup$

share|improve this answer

edited May 18 at 23:02

answered May 18 at 22:48

Mattman944Mattman944

92117

answered May 18 at 22:48

Mattman944Mattman944

92117

answered May 18 at 22:48

Mattman944Mattman944

92117

0

$begingroup$

Add a low pass filter to a non-inverting amplifier circuit.

Without giving it much thought you already have a high pass filter on the input, and the simplest way to apply a low pass is to use the output of the opamp.

This image is from this calculator which may help you pick your filter component values easily. The low pass filter is R2C2 in the image above and is buffered by the opamp reducing the interaction of gain and filter components.

share|improve this answer

edited May 18 at 23:49

answered May 18 at 23:00

Jack CreaseyJack Creasey

16.2k2824

$endgroup$

add a comment | 

0

$begingroup$

Add a low pass filter to a non-inverting amplifier circuit.

Without giving it much thought you already have a high pass filter on the input, and the simplest way to apply a low pass is to use the output of the opamp.

This image is from this calculator which may help you pick your filter component values easily. The low pass filter is R2C2 in the image above and is buffered by the opamp reducing the interaction of gain and filter components.

share|improve this answer

edited May 18 at 23:49

answered May 18 at 23:00

Jack CreaseyJack Creasey

16.2k2824

$endgroup$

add a comment | 

$begingroup$

Add a low pass filter to a non-inverting amplifier circuit.

Without giving it much thought you already have a high pass filter on the input, and the simplest way to apply a low pass is to use the output of the opamp.

This image is from this calculator which may help you pick your filter component values easily. The low pass filter is R2C2 in the image above and is buffered by the opamp reducing the interaction of gain and filter components.

share|improve this answer

edited May 18 at 23:49

answered May 18 at 23:00

Jack CreaseyJack Creasey

16.2k2824

$endgroup$

share|improve this answer

edited May 18 at 23:49

answered May 18 at 23:00

Jack CreaseyJack Creasey

16.2k2824

answered May 18 at 23:00

Jack CreaseyJack Creasey

16.2k2824

answered May 18 at 23:00

Jack CreaseyJack Creasey

16.2k2824