I am a beginner in filter design. This is the bandpass filter and its central frequency is 2kHz. Would you please give me some comments on this design?
Need comments on low noise band-pass filter design
analogfilter
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Question: Is it possible that the required corner frequency (3 dB) is 1 kHz and the stopband frequency (with a certain required damping specification) is given with 5 kHz ? Please note that it is NOT possible to "filter out" a frequency of 2.5 kHz. You only can provide a certain degree of damping - unless you design a filter with a passband zero at 2.5 kHz. However, this has no Butterworth response.
The filter order necesary to meet your requirements is determined by the stop band attenuation, which is not given up to now.
As an example: A Butterworth low pass of 4th order can be realized as a series connection of two active lowpass stages with the same 3dB cut-off frequency (1 kHz) but with two different Qp values (Qp: pole Q).
Yes, we only want to eliminate 1kHz
A notch filter can be used for this: -
This is a 10 kHz design and note figure 6 - how it responds as you get between 9 kHz and 11 kHz with the input signal.
This article describes the circuit in more detail and provides examples of other notch filters across the audio range. The article is by TI and is entitled: -
An audio circuit collection, Part 2
Regarding the op-amps, because you are using a lock-in amplifier it's not critical BUT just in case go for op-amps that are below 10 nV/\$\sqrt{Hz}\$ specified on noise. Devices that spring to mind are OP1177, AD8605, ADA4528 but there are plenty with lower noise.
Best Answer
1st stage is a 1.9995 kHz bandpass filter using a standard multiple-feedback topology. It has a Q of about 20 and a gain of about 40 dB.
2nd stage is a 2.009 kHz BP filter with a Q of about 25 and a gain of about 40 dB.
Total gain of first stage is 40 dB which makes a 10nV 2 kHz signal into 1 microvolt.
Bandwidth of 1st stage is \$\dfrac{f_C}{Q}\$ = 100Hz and for the purposes of noise calculation you can assume the bandwidth is 1.6x greater at 160 Hz.
\$E_{NOISE}\$ from ADA4004 is 1.8nV /\$\sqrt{Hz}\$. This means noise in a 160 Hz bandwidth is 22.8 nV
This is bigger than your signal (10nV) therefore this isn't going to be a great design.
Even if you took into account the Q of the 2nd stage, the bandwidth would only be halved i.e. a half power point would become a quarter-power point. Noise voltage would be about 16 nV and still a significantly bigger value than your signal. This is made worse by the 2nd stage being 9.5 Hz different to the 1st stage.