# Filter arrangement

Discussion in 'Homework Help' started by ali8tor, Apr 15, 2013.

1. ### ali8tor Thread Starter Member

May 29, 2012
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0
hi i need help understanding the filter arrangement of this circuit

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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C2 and R1 form a High pass filter.
Fc = 0.16/RC = 0.034Hz
Next C1 and R7 form a low pass filter Fc = 0.16/RC = 10.6MHz

Next we have a op-amp non-inverting amp. The C3 and C4 and R2 and R4 form a bandpass type response filter.
With maximum gain equal to

Av = 1 + (C3*R3)/(R2*C3 + R4*C4) = 2.4 [V/V]

At frequency equal to

$Fo = \frac{1}{2 *\pi \sqrt{C3*C4*R2*R4}} = 4.9KHz$

And at the end we have another High pass filter (C6, R6).
Fc = 0.16/RC = 0.16Hz And that is the end of this simplified analysis.

Is it those information you were looking for?

Last edited: Apr 15, 2013
screen1988 likes this.
3. ### screen1988 Member

Mar 7, 2013
310
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Hi Jony,
I am curious to know why C1 and R7 are separated.
Why don't we put C1 in parallel with R8?
And could you explain the role of R8 here?

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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You can place C1 in parallel with R8 if you like. This change don't have any effect on the circuit operation.
Because C2>>C1 we can treat C2 as a short circuit.
So C1 time constant is equal to:
t = C1 * Rth and Fc = 1/(2 * pi *t)

Rth = R7||R8||R1 ≈ R7

R8 provide a path to ground for C2 left plate.
So when we disconnect input source (signal). We don't leave input (C2) floating. So we use R8 as a pull-down resistor.

R6 at the output is also pull-down resistor.

Last edited: Apr 29, 2013
screen1988 likes this.

Mar 7, 2013
310
3