Coupling capacitor's value

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How do I determine the value of a blocking capacitor for say this circuit:
common-emitter-RC-coupled-amplifier.png

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.
 

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  • #2
Merlin3189
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...something tells me there is more to the choice of a coupling capacitor than this.
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.
 
  • #3
davenn
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How do I determine the value of a blocking capacitor for say this circuit:
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)


Dave
 
  • #4
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Are there any particular formulas?
 
  • #5
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Screenshot_2016-09-07-15-57-19.png

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.
 
  • #6
CWatters
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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.
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..

High-pass-filter-diagram2.png


Choose C to set the corner frequency.
 
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  • #7
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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..

High-pass-filter-diagram2.png


Choose C to set the corner frequency.
Thanks!
 
  • #8
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Are there any particular formulas?
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
 
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  • #9
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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
Thanks! And the input resistance is given by R1||R2 + the transistor's input resistance(Ube/Ib)?
 
  • #10
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And the input resistance is given by R1||R2 + the transistor's input resistance(Ube/Ib)?
Yes.
 
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