# Active low pass filter

1. Mar 28, 2012

### kimandrew20

1. The problem statement, all variables and given/known data

Design this circuit to give a cut off (-3dB) frequency of 1kHz, a low frequency gain of 20, and an input impedance equal to 22kΩ

2. Relevant equations

3. The attempt at a solution

So, I started off with Vout/Vin = [-R2/R1]/[1 + sCR2. I also know that the dc gain will be -R2/R1. I also know that frequency = 1/[2∏R2C]

Im a little lost as to where to start!

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Mar 28, 2012

### Staff: Mentor

Hi kimandrew20! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

What is the input impedance of the op-amp filter you show? Start with that.

Are you going to use just a single stage?

Last edited by a moderator: May 5, 2017
3. Mar 28, 2012

### kimandrew20

Is the input impedance just the value that the problem has specified? The 22kΩ value?

Yes, just a single stage.

4. Mar 28, 2012

### kimandrew20

I tried it again and I got a value for the capacitance and the resistors. Does this look correct?

R1 will equal the input impedence (The two inputs, V- and V+ are at approximately the same voltage, R1 is at ground voltage, therefore R1 = input impedence = 22 kΩ

Gain (A) = (Rf/ R1) → Rf/22k = 20
Therefore, Rf = 440 kΩ.

f = 1/ [2∏ * R * C] therefore
C = 1/ [2∏ * R * f] → 1/[2∏ * (22*1000) * (1 * 1000))] = 7.234 x 10-9 F

5. Mar 28, 2012

### rude man

You got the right Rf but the wrong C. Reconsider the R you used to compute C ...