Help with RC Low pass filter

Discussion in 'Homework Help' started by ParaTrooperJoe, Apr 21, 2013.

  1. ParaTrooperJoe

    Thread Starter New Member

    Apr 21, 2013
    Hello everyone! I am new to this forum but am glad I found it. I have a pretty good understanding of troubleshooting and analyzing circuits, but the one I am currently working on is proving a bit difficult, especially sense my book does a terrible job at explaining how to do this.

    Anyhow, here is my assignment:

    An RC low-pass filter has the following exact values: RF = 100 Ω, RS = 50 Ω, and CF = 10 μF. The filter is connected to a load that can vary over a range of 10 to 500 Ω. Calculate:

    The range of Av for the circuit.
    The range of fC for the circuit.

    So, I know the formula for Av(max) = Req/(Req)+(Rs)
    and I also know that Req = Rp || RL

    However, I am lost as how to get started because, for one thing, I do not see a value for RL listed. Furthermore to find Rp, I would need to find Q, but to find Q I would need to calculate XL and again, no value is presented for an inductor.

    Perhaps I am just looking at this assignment the wrong way?

    Any advice would be greatly appreciated! Thanks in advance!

    - ParaTrooperJoe -


    Here is what I think is the correct way to approach this.

    Once I reread the question, I see that the load (RL) ranges from 10 to 500 ohms (DOH!)


    When RL = 10Ω -

    AV(max) = R_L/(R_L+R_F ) = 10Ω/(10Ω+100Ω) = 0.0909

    When RL = 500Ω -

    AV(max) = R_L/(R_L+R_F ) = 500Ω/(500Ω+100Ω) = 0.833


    When RL = 10Ω -
    First, I calculated the value of R while RL = 10Ω:
    R1 = 1/(1/Rf +1/RL1 +1/Rs ) = 1/(1/100+1/10+1/50) = 7.69
    fc = 1/(2π(R1)(C)) = 1/(2π(7.69)(10µF)) = 2.07 kHz

    When RL = 500Ω -
    First, I calculated the value of R while RL = 500Ω:
    R2 = 1/(1/Rf +1/RL2 +1/Rs ) = 1/(1/100+1/500+1/50) = 31.25
    fc = 1/(2π(R2)(C)) = 1/(2π(31.25)(10µF)) = 509.3 Hz

    I think that makes more sense, and hopefully I have done these calculations correct. If anyone has any advice to offer, I would greatly appreciate it!
  2. ParaTrooperJoe

    Thread Starter New Member

    Apr 21, 2013
    Hmmmm, I guess no one knows anything on these forums, I guess I will move along and hope for the best.

    Thanks for nothing!
  3. tshuck

    Well-Known Member

    Oct 18, 2012
    No problem! Not like any of us have better things to do than wait until we get the privilege of helping you. Good luck finding your answer.
  4. WBahn


    Mar 31, 2012
    Your welcome.

    Imagine, not getting a response from someone (that is NOT getting paid to help you) after a whole whopping day. That's just horrible, isn't it?

    Of course, I don't recall seeing in the ToS anything that requires me to respond to your questions in any set period of time. In fact, I don't recall seeing anything in the ToS that requires me to respond to your questions at all.

    We do so because we CHOOSE to do so and we do so WHEN it is convenient for US. And we are more likely to choose NOT to help someone that feels that they are entitled to immediate answers.

    And, for the record, my first question here went three weeks without a response.
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
    You need to post a schematic. How are we supposed to know what your circuit looks like?