Single supply active filter design with gain

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gn.times

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Hi,

I want to design an active Sallen-Key 2nd order low pass filter with single supply opamp. I am using OP213 from AD with +5V and 0V as the supply voltage.

Cutt-off frequency: 2kHz
Gain: 10 (V/V)

I designed the below circuit using AD filter design tool and connected 2.5V for Vref from a voltage divider. But I did not get the output.



I tried using the below circuit from TI with Unity gain and it worked fine. I don't know whether biasing is required for the feedback resisstor network if I want to increase the gain beyond 1.



Could you please add the feedback resistor network so that I can have a gain of more than 1?
 

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  • AD_LP_Ckt.bmp
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  • Single supply LPF_TI.bmp
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That circuit won't work. What you call "REF" needs to be a low impedance, NOT a voltage divider. What is your input voltage?
 

The (+) input of the opamp on your first schematic does not have an important "ref" voltage. Its DC voltage is whatever is the input source voltage which might be 0V.

You need VERY accurate RC parts values to get a gain as high as 10 from a Sallen-Key filter.
 

The (+) input of the opamp on your first schematic does not have an important "ref" voltage. Its DC voltage is whatever is the input source voltage which might be 0V.

That's a perfectly valid topology-the only issue is the common (ref).
 

That circuit won't work. What you call "REF" needs to be a low impedance, NOT a voltage divider. What is your input voltage?

Can I use a 2.5V voltage regulator IC (such as LM336-2.5) for 'REF' supply?
My input signal voltage range will be between 1V to 2V

- - - Updated - - -


In the 1st schematic, the capacitor C2A is connected between (+) terminal of opamp and "REF" where I will connect a 2.5V voltage regulator o/p

The assumed gain is 10 (max) but it will be less in practical
 

1) If your input signal is 1-2volts, you don't need any offset, since the opamp common mode range is 0 to 4 volts.
2) If your input is 1 volt and your gain is 10, you are not going to get 10 volts out with a 5 volt supply.
 

The impedance of the signal source must be much less than the value of R1A. Its DC voltage must be +2.5V/10. You have nothing to bias at +2.5V.
If this is an AC filter then why have a DC gain of 10? Connect R3A in series with a capacitor to ground then the DC gain is 1 and the signal source can be biased at +2.5V.
 

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