Coupling capacitor's value

1. Sep 7, 2016

Grim Arrow

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. Sep 7, 2016

Merlin3189

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. Sep 7, 2016

davenn

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. Sep 7, 2016

Grim Arrow

Are there any particular formulas?

5. Sep 7, 2016

Grim Arrow

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. Sep 7, 2016

CWatters

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.

7. Sep 7, 2016

Thanks!

8. Sep 7, 2016

Jony130

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

9. Sep 7, 2016

Grim Arrow

Thanks! And the input resistance is given by R1||R2 + the transistor's input resistance(Ube/Ib)?

10. Sep 7, 2016

Jony130

Yes.

11. Sep 7, 2016

Grim Arrow

Thanks once again!