Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Coupling capacitor's value

  1. Sep 7, 2016 #1
    How do I determine the value of a blocking capacitor for say this circuit:
    What I mean is, sure Cin must behave like a short circuit at working frequency, but something tells me there is more to the choice of a coupling capacitor than this.
  2. jcsd
  3. Sep 7, 2016 #2


    User Avatar
    Homework Helper
    Gold Member

    IMO not a lot! Price, physical size, reliability - which tend to favour small, non-electrolytic devices.
    You don't need (and can't get) a short cct, so you can calculate how low you need to go to get the frequency response you need. If you make the capacitance too large, you may get irritating issues with it charging to quiescent DC levels at switch on. Low enough is low enough.
  4. Sep 7, 2016 #3


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    circuit operating frequency is the main criteria
    audio freqs -- 1uF to ~ 15uF electrolytics would be very commonly used

    At RF, a whole different ball game and cap values at different freq ranges ( bands) becomes more critical
    could be anything from 100's of pF @ HF 3 - 30MHz to a few pF at freq's over 1000 MHz (1GHz)

  5. Sep 7, 2016 #4
    Are there any particular formulas?
  6. Sep 7, 2016 #5
    I found this in a site called "Learningaboutelectronics" and it says I can use this table to choose a coupling cap for my frequency needs. This sure solves some of my problems, but I need to know how they got these values.
  7. Sep 7, 2016 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If you replace the transistor and all it's biasing components with a resistor equal to the input impedance then you essentially end up with a high pass circuit like this..


    Choose C to set the corner frequency.
  8. Sep 7, 2016 #7
  9. Sep 7, 2016 #8
    Yes, there is. The corner frequency is equal to Fc = 1/(2*Π*R*C)≈ 0.16/(R*C). The frequency at which R = Xc
    C is the capacitance of a capacitor and the R is the resistance "seen" by capacitor The effective resistance that will discharge the capacitor.
    So for Cin we have Cin ≈ 0.16/(Rs + Rin*Fc), Ce ≈ 0.16/(1/gm * Fc), Cout ≈ 0.16/(Rc+RL*Fc)

    Rs - is a signal source resistance
    Rin - the amplifier input resistance ≈ R1||R2||(β*1/gm )
    gm - the BJT transconductance gm ≈ Ic/26mV ≈ 40*Ic
  10. Sep 7, 2016 #9
    Thanks! And the input resistance is given by R1||R2 + the transistor's input resistance(Ube/Ib)?
  11. Sep 7, 2016 #10
  12. Sep 7, 2016 #11
    Thanks once again!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted