# Active Low Pass Filter Design: 1kHz, Gain 20, 22kΩ Input Impedance

• kimandrew20
In summary, the circuit should be designed to have a cut off frequency of 1kHz, a low frequency gain of 20, and an input impedance of 22kΩ. The input impedance will be equal to R1, which is also the value of the resistors used in the circuit. The gain of the circuit can be calculated using the formula A = Rf/R1, where Rf is equal to 440 kΩ. The capacitance can be found using the formula C = 1/[2∏ * R * f], where R is equal to 22kΩ and f is equal to 1kHz. Therefore, the correct value for C is 7.234 x 10-
kimandrew20

## Homework Statement

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

## 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!

#### Attachments

• active-low-pass-filter.png
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Hi kimandrew20! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

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

Yes, just a single stage.

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

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

## 1. What is an active low pass filter?

An active low pass filter is a type of electronic circuit that allows low frequency signals to pass through while attenuating or reducing higher frequency signals. It is designed using active components such as operational amplifiers to achieve this filtering effect.

## 2. What is the significance of a cutoff frequency of 1kHz in this filter design?

The cutoff frequency of a low pass filter is the frequency at which the output signal is reduced by 3dB (half of its original amplitude). In this case, a cutoff frequency of 1kHz means that signals above 1kHz will be attenuated by 3dB or more.

## 3. Why is the gain of 20 chosen for this filter design?

The gain of a filter is the factor by which the output signal is amplified or attenuated. In this design, a gain of 20 means that the output signal will be 20 times larger than the input signal. This specific gain value may have been chosen to match the desired amplification for the specific application of the filter.

## 4. What is the role of the 22kΩ input impedance in this filter design?

The input impedance is the resistance that is presented to the input signal by the filter. In this design, the 22kΩ input impedance helps to ensure that the input signal is not affected or distorted by the filter. It also helps to match the impedance of the input signal to the impedance of the active components in the filter.

## 5. Can this filter design be used for other cutoff frequencies and gains?

Yes, this filter design can be modified for different cutoff frequencies and gains by adjusting the values of the components, such as the resistors and capacitors, used in the circuit. There are also different types of active low pass filters, such as Butterworth, Chebyshev, and Bessel filters, that can be used for different applications and specifications.

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