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

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Discussion Overview

The discussion focuses on the design of an active low pass filter with specific parameters: a cutoff frequency of 1kHz, a low frequency gain of 20, and an input impedance of 22kΩ. Participants are exploring the theoretical and practical aspects of the circuit design, including calculations for resistor and capacitor values.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents a starting equation for the filter design, relating output and input voltages to resistances and capacitance.
  • Another participant questions the input impedance of the op-amp filter and prompts for clarification on whether a single stage is being used.
  • A participant confirms that the input impedance is indeed the specified 22kΩ and states that only a single stage will be used.
  • One participant calculates resistor and capacitor values, asserting that R1 equals the input impedance and deriving Rf based on the gain requirement.
  • Another participant challenges the calculated capacitance value, suggesting that the resistor used in the calculation may need to be reconsidered.

Areas of Agreement / Disagreement

Participants generally agree on the values for R1 and Rf, but there is disagreement regarding the correct value for the capacitance, indicating that the discussion remains unresolved.

Contextual Notes

There are limitations regarding the assumptions made about the circuit configuration and the specific values used in calculations, which may affect the results. The discussion does not resolve the correct capacitance value.

kimandrew20
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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

Homework Equations





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!

http://upload.wikimedia.org/wikipedia/commons/thumb/5/59/Active_Lowpass_Filter_RC.svg/300px-Active_Lowpass_Filter_RC.svg.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 ...
 

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