Yeah that makes sense now - reactance decreases with increased frequency (1.59MΩ @ 1kHz, 159KΩ @ 10kHz and 15k9Ω @ 100kHz). So, with a larger impedance at the input, higher frequencies take the path to ground, right? If that's correct, does that mean that I should make C2 a smaller value as with a 1M2Ω impedance, I might be losing some higher frequencies of interest, e.g. c.1.3kHz - 15kHz? Or have a got that wrong..!

Thanks.

 

C1 and R3 are making a high-pass filter at the same time (cutting off lower frequencies). I dont' know exactly how they will work together but somebody has (hopefully) put some work into this circuit. You can similuate the frequency response and try to make it flat over the audio range (or what ever shape you want over the audio range) in your favorite version of SPICE or TINA, ...

 

C1 and R3 are making a high-pass filter at the same time (cutting off lower frequencies). I dont' know exactly how they will work together but somebody has (hopefully) put some work into this circuit. You can similuate the frequency response and try to make it flat over the audio range (or what ever shape you want over the audio range) in your favorite version of SPICE or TINA, ...

Which schematic are we talking about?

 

See post #19.

 

C1 and R3 are making a high-pass filter at the same time (cutting off lower frequencies). I dont' know exactly how they will work together but somebody has (hopefully) put some work into this circuit. You can similuate the frequency response and try to make it flat over the audio range (or what ever shape you want over the audio range) in your favorite version of SPICE or TINA, ...

C1*R3 make a high pass filter with the corner at 1.3Hz.

 

Yes the circuit I was referring to is the one in post #19, but the corner frequency is not just C1 and R3, it is C1 in series with the equivalent capacitance of the sensor (a PVDF piezo sensor) which has a value of 1.38nF (see notes on schematic Preamp-5-violin.pdf).

I have calculated the corner frequency by considering the sensor equivalent capacitance (Cs) in series with C1:

C1*Cs / C1+Cs = Ctotal = (100nF * 1.38nF) / (100nF + 1.38nF) = 1.36nF

I have then calculated the corner frequency (fc) as:

fc= 1/2*pi*Ctotal*R3 = 1/2*pi*1.36nF*1M2 = 97.5Hz

In my earlier post, I was referring to the capacitor C2=100pF which goes from input signal to ground and is not part of the HP filter network at the input to the circuit (or at least I don't think it is...). I was wondering if this value should be changed as GopherT said that it is there to filter out unwanted high frequencies. My concern was that at its present value, it may be filtering desired frequencies to ground.

Hope that makes sense...

Thanks

 

I thin that you shoudl remove C2 and C10 capacitors form the circuit.
C2 will form a AC voltage divider with piezo sensor equivalent capacitance.
Also piezo sensor will have some resistance, which in combination with C2 will form low-pass filter. C10 also will form with a C3 capacitor AC voltage divider. So any power supply noise will be only attenuates by 0.5 and next the noise will be amplified by the circuit.