I want to design a notch filter for eliminating 100Hz waveform. I am using the below schematic. I tried the circuit with bandwidth 300Hz but this will affect the frequency of 20Hz to 200Hz signal also. As well I am getting the -13dB at 100Hz( Vout(peak-peak) = 200mV and Vin(peak-peak) = 1V ) but I want the output signal as low as 10mV.

For reducing the effect of the filter on other frequency I decrease filter bandwidth(4Hz) by using below calculator and then change the R4 and R5 resistor value(mention in the schematic). But it is now giving the worst result. Now it is giving the output of -2.5dB(750mV(peak-peak)) at 100Hz. Or I can say there is negligible effect at 100hZ or other frequency.

Need to ask how to improve this circuit for deeper notch and approx (peak-peak) 10mV(or less) signal at the output for 100Hz frequency.
Thank you in advance.enter image description here

  • 1
    \$\begingroup\$ A bandwidth of 4Hz sets the -3dB point - you need a very narrow notch to get the 100Hz signal lower, so set the bandwidth to something like 0.4 Hz . \$\endgroup\$ Oct 10, 2017 at 9:42
  • \$\begingroup\$ ok but as per this calculator, the feedback reach at 99 at 4Hz BW can't go below this.Can it is possible to take a BW below 1Hz ?? \$\endgroup\$
    – user152036
    Oct 10, 2017 at 9:50
  • 1
    \$\begingroup\$ Have you considered how steep this filter needs to be? You want something like 30 or 40 dB, over a tiny fraction of an octave. This will need a higher order filter, if it is even possible in an analog circuit. \$\endgroup\$
    – JRE
    Oct 10, 2017 at 10:11
  • \$\begingroup\$ No, I have not considered even if it is possible or not. But I just want to use the maximum limit that can be reached by this second order low pass filter. Or I can make a fourth order low pass filter too!!! By the way, I just want to know why it is not giving the appropriate result at 4Hz BW even worse results than 300Hz BW. \$\endgroup\$
    – user152036
    Oct 10, 2017 at 11:02
  • 1
    \$\begingroup\$ It gives poor results for probably several reasons. First is that to actually hit the cutoff frequency and bandwidth you want you will need precision parts. A few percent can kill things. The other is that you probably (well) outside what a second order filter can deliver, and calculator just didn't warn you. \$\endgroup\$
    – JRE
    Oct 10, 2017 at 11:14

3 Answers 3


A Design Spec for any Active Filter must include many variables.

  • Gain PassBand ( I used 3dB )
  • PassBand Ripple
  • PassBand BW ( not defined here ) and relates with PassBand ripple or gain error
  • Attenuation BandStop ( 1/20 = -26 dB needed at 100Hz)
  • BW Bandstop ( not defined here which relates to 100 Hz signal f error. (I used 4Hz)

Some variations in spec desciptions may exist, but must define all important limits needed.

My specs

Pass Band gain: 1.5 ( 3.522 dB ) PB Ripple max: 1dB
center frequency : 100 Hz
Corner frequency attenuation 0.5 dB
Stages : 2 4th order
Input DC offset; minimal

enter image description here

This exceeds the requirements by a wide margin. and can be further optimized by offseting each notch e.g. +/-2 Hz if a wider/deeper f bandstop depth is needed.

Filter.exe Tools exist (free) at ti.com to design anything in minutes.

The secure https :// goo.gl/cHNTYx leads to my simulator above so you can change parts in real time. Java must be enabled .

  • \$\begingroup\$ Not sure why you put that link in such funny way. HERE it is as clickable. \$\endgroup\$
    – not2qubit
    Mar 26, 2019 at 8:29
  • \$\begingroup\$ Short url’s are rejected in answers but ok in comments \$\endgroup\$ Mar 26, 2019 at 8:31
  • \$\begingroup\$ Aha, ok. Another question, what kind of filter is that? (Usually people first look at twin-T filters for this, because of the circuit simplicity, but over-sensitive component values.) \$\endgroup\$
    – not2qubit
    Mar 26, 2019 at 8:33
  • 1
    \$\begingroup\$ Dual active twin T with less sensitive components and tolerances given \$\endgroup\$ Mar 26, 2019 at 8:35
  • \$\begingroup\$ What did you use to determine those component values for a given frequency? \$\endgroup\$
    – not2qubit
    Mar 26, 2019 at 8:44

To attain the type of notch performance that you are talking about you probably want to be looking at switched capacitor type filters. In these the accuracy of the notch frequency is controlled by an external clock frequency and thus can be very stable. With the type of analogue filter that you show the simulation results are all fine and good at the specified component values but will be unobtainable in real life circuits due to component variations and temperature/supply voltage effects. Of course you could add trimpots and adjustable capacitors to tweak each device you build but that adds expense. Trimming T networks can also be problematic where there are two resistors that are supposed to be exactly the same value.

This APP note from Maxim will help you get introduced to switched capacitor filters. In the past I have used this type of component to filter power line frequency effects (50/100Hz or 60/120Hz) from low level signals attained from sensors in a medical application. At the time there were single chip components that performed this function with few added components and I recall some devices even had a 50/60 selector pin on the package. Hopefully this information will enable you to do a search of modern devices suitable to your application.

  • \$\begingroup\$ Ok I will see the switched capacitor filter also. Regarding the modern chips that you are mentioning that can be directly used but I want to build it from scratch with basic electronics components like transistors and Op-amps which are the building block of every chip and IC's. I want to ask one more thing which is out of this topic actually. How this "Right Tick" sign comes up automatically to an answer. Is it really mean the answer is correct(Do not mind my ques it just I am curious about it ;-)) \$\endgroup\$
    – user152036
    Oct 10, 2017 at 11:19
  • \$\begingroup\$ The tick is a button you may press if you found the answer solves your problem. \$\endgroup\$
    – Janka
    Oct 10, 2017 at 11:33
  • \$\begingroup\$ If you refer to the check mark that can appear under the answer vote total - that is for you, the original question asker, to select which answer you want to select as best. If you get multiple answers you can see this pop-up for each one and then select the one you like. \$\endgroup\$ Oct 10, 2017 at 11:34
  • \$\begingroup\$ Alright, I am getting this check mark for only Michael answer, not for all answers. So that means it is already tracked best answer by stack exchange? So I have only one option to press that tick button. \$\endgroup\$
    – user152036
    Oct 10, 2017 at 11:41
  • \$\begingroup\$ Anyways, Michael, your answer is actually helpful for me. Thank you :-) and will come back here again after doing more experiment on this design. \$\endgroup\$
    – user152036
    Oct 10, 2017 at 11:48

An effective active filter needs, beyond the design, an opamp with

1) adequate DC gain to implement the accuracy (precision signal-energy cancellation)

2) adequate Unity Gain Bandwidth, to provide adequate gain at the AC region where you want some precision signal-energy cancellaion

3) low Rout (low Zout) of the opamp, to provide the needed energy to cancel the incoming energy

The designer needs to incorporate 1 & 2 & 3 into the design method.


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