I have made low pass filter both on Breadboard and in Proteus.Both of them have different problems.
I have kept voltage gain at "3".
In breadboard,voice output in the earphone is lower than input.In input I have a microphone When I blow air by mouth, It give me rough voice hearing by earphone.
In case of simulation in Proteus,When I applied above 333 Hz freq, output in speaker is still beep.
I cant understanding what is this happening..
Please help me…I am beginner.
Electrical – Low pass filter not working correctly
breadboardcircuit-designlow pass
Related Solutions
If it's a DC signal, then, yes, there should be no change in the measured signal. If, however, you have other signals, such as, for example, a 1 kHz signal, this would be significantly attenuated. If you want a bit more of the math here, try to think of your signal in the Fourier domain. The Fourier transform of a signal breaks a signal into a number of sinusoids at many different frequencies that, when added together, recreate your signal. The zero-th term corresponds to your DC signal. Higher terms represent AC components of the signal, starting at the fundamental frequency, a sinusoidal signal with period equal to your measurement duration. These higher terms will also be present in the output signal, until the frequency of the term is higher than the cutoff frequency of your low-pass filter. Higher frequencies are attenuated.
As for "mean value", you need to be careful of your word choice. The mean value of a signal is its average value. More commonly encountered is the root mean square (RMS) value. So, for a signal V(t): $$V_{RMS}= \sqrt{{1\over{T}}\int_0^TV(t)^2dt}$$ This is a bit different from your zero-th term DC signal.
A 2nd order low-pass butterworth filter as you have shown will have a similar frequency response as shown below in trace 2: -
So, if 500 Hz is re-normalized to coincide with "1" along the x-axis, you will see that at 5000 Hz, the attenuation is 40 dB down and below 500 Hz it's fairly flat all the way to DC (but with some gain dictated by R3 and R4 in your circuit).
In the specification it says "Signals need to be attenuated by at least 80dB relative to difference frequency of 400Hz" but I'm not sure what this means?
It certainly doesn't apply to a 2nd order filter as you have drawn in your question. It might mean that at 900 Hz (500 Hz plus 400 Hz) the attenuation needs to be at least 80 dB and, if that is the case then a fifth order butterworth filter would be required.
It could even imply that what you want is a pass band filter with 80 dB (or greater) attenuation 400 Hz either side of the centre at 500 Hz.
But, at the end of the day, it's down to you understanding the full specification you are given so, maybe publish a little bit more of it.
Best Answer
1) You are using a single supply. Analyze the circuit for positive inputs. This requires negative outputs. Where will the - voltage come from?
2) You do not specify your op amp. Most op amps are completely incapable of driving 8 ohms, which the usual nominal speaker impedance. You need a power stage for the output - subject, of course, to fixing part 1).
3) Using 500 ohms for feedback and 150 ohms input is too low. For instance, you need to check to see if your AC source is happy driving 150 ohms. A good function generator will drive 50 ohms, but many sources aren't happy doing that. Use 10k and 30k. This will also let you use a smaller capacitor.