# how to reduce the amplification of second order notch filter

I build a second order notch filter.it is centered at 50hz. When I gave a sine wave signal with a sweep from 20hz to 80hz.The signal's input amplitude is 1v and the signal's output amplitude is like below.

20hz--2.2v 25hz--2.1v 30hz--2.0v 35hz--1.9v 40hz--1.8v 45hz--1.0v 48hz--550mv 49hz--430mv 50hz--320mv 52hz--550mv 60hz--600mv 70hz--1v 80hz--1.5v 100hz--2v

c2=c3=100nf c1=200nf r1=r2 r1=r2=35.3ohm r3=17.1ohm variable resistance 100k(50:50)

But i want to know why the output is 2v for 100hz and 30hz??. But my input is 1v any help on this would be greatly appreciated.

• Do you mean 35.3 Ohm or 35.3kohm etc. for the resistor values? Have you measured the signal generator output voltage? If it is calibrated for 50 ohm and you feed into a high impedance load you will get about twice the output voltage. – Kevin White May 15 at 0:56
• The resistor values are in k ohm..I gave the changes in the output value with frequency above the figure .I can't understand your explanation – soba May 15 at 1:15
• DId you measure the output voltage of your signal generator? – Kevin White May 15 at 1:15
• Yes. I gave the values in the post . – soba May 15 at 1:22
• No, I mean did you actually measure the output directly from the generator or just set it for 1V and assume it gave out 1V? – Kevin White May 15 at 1:24

The original poster must have an error in the measurements or the schematic does not represent what is built.

Here is a simulation of the OP's circuit. Notice that the gain is close to unity at frequencies significantly different from the notch frequency.

A likely problem is that the input voltage is not what the OP thinks it is. It is easy to mistakenly get twice the intended output from many function generators when feeding into a high impedance load. I have done it myself and seen others make the same error.

Most function generators have 50 ohms output impedance and assume that the output is loaded with 50 ohm.

Any display of voltage on the instrument will take that into account and allow for the fact that the voltage at the load will only 50% of the actual voltage generated internally and so display 1V for example when it would actually deliver 2V into a high impedance load.

Some generators have a setting for the load impedance when they will then take this into account and display the correct voltage. I have experience with Keysight generators that do this.

In this question I expect that the signal generator is actually giving out 2V because it has been set for 1V but is feeding a high-impedance, not 50 ohms.

Here is the schematic of the circuit I simulated:

• This is not the reason why the output is >2V. He added R was missing the k which does not load down the signal. – Sunnyskyguy EE75 May 15 at 2:05
• @SunnyskyguyEE75 - I don't understand your comment. Initially he had the incorrect value of the resistors by a factor of 1000. The signal generator is not being loaded down and if he is going by the setting rather than measuring the input voltage he will be out by a factor of 2.The gain at DC is unity because the feedback is capacitor coupled - the gain will asymptotically get to this as the frequency is reduced. His output voltage at 20Hz is ~2V. I would just like the OP to confirm the input voltage. Your diagrams have the amplifier set to a gain of 1.5 or 2 so you will have a higher gain. – Kevin White May 15 at 19:44
• I now agree with you, measurement error. – Sunnyskyguy EE75 May 16 at 2:29