shavarna.jiit
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Thanks BradtheRadI have been experimenting with a simulation.
Going with your value for C1 and C2 as 22 nF...
To make the notch at 5 kHz...
R1 and R2 need to be 1450 ohms. (The webpage gives a slightly different result.)
I used a bipolar supply to make things easy.
Screenshot (Falstad's simulator). The scope trace shows an AC sweep from 4 kHz to 6 kHz.
I cannot explain why your layout acted as a narrow bandpass.
- - - Updated - - -
I just recalculated.
The webpage method of calculating the resistors results in almost exactly the same values which work in my simulation.
- - - Updated - - -
I see you wanted to get 5.6 kHz.
1300 ohm resistors will do.
I think the problem is the S11 response is very high impedance at resonance and the coupling caps are too high impedance in this frequency range, so a voltage source low impedance is required.
Thanks again.
Now I am an Embedded Guy who need some conceptual help with filter design. Can you tell me why did you replace Cin and Cout in my circuit with 2 resistances of 100 here?
Hello,
I am trying to design a Fliege Notch Filter at 5.6 KHz for design of neuronal signal acquisition system. I am using op-amp OPA2333 with single supply. However the output to the circuit amazed me because it acts like an narrow band pass filter instead of notch filter.
I really appreciate if someone could point out the error. Attached is the simulation Circuit and Output.
I have done this design on the basis of basic steps mentined in https://www.ti.com/lit/an/sloa093/sloa093.pdf Page 9
I would suggest shavarna.jiit sticks to TINA (or maybe LTspice). That Java applet is extremely slow and that is with a simple circuit. Also, you would want to be doing an AC analysis with log scales for filter design and evaluation.
Keith
Hi, the circuit as shown in the TI documentation for single supply does not work (as you have noticed).
Do the following:
*Remove V3 and ground the resistor R3 instead.
*Use a high-resistive voltage divider 2:1 (two equal resistors) directly at the input of the circuit to bias this input with 0.5*Vcc.
*Use a large coupling capacitor for the input signal.
This should work. More than that, I recommend the dimensioning as shown in my former post.
Good success.
Hi, the circuit as shown in the TI documentation for single supply does not work (as you have noticed).
Do the following:
*Remove V3 and ground the resistor R3 instead.
*Use a high-resistive voltage divider 2:1 (two equal resistors) directly at the input of the circuit to bias this input with 0.5*Vcc.
*Use a large coupling capacitor for the input signal.
This should work. More than that, I recommend the dimensioning as shown in my former post.
Good success.
I relocated the 0.5*Vcc level as you suggested. This has greatly improved performance.
Hi, the circuit as shown in the TI documentation for single supply does not work (as you have noticed).
Do the following:
*Remove V3 and ground the resistor R3 instead.
*Use a high-resistive voltage divider 2:1 (two equal resistors) directly at the input of the circuit to bias this input with 0.5*Vcc.
*Use a large coupling capacitor for the input signal.
This should work. More than that, I recommend the dimensioning as shown in my former post.
Good success.
Thanks a lot LvW. It worked perfectly. Could you explain the changes you made here. Please , its important for me to understand the concept.
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