Coupling Capacitors

[zpyder]

How does one calculate the values for coupling/decoupling capacitors for a transistor/op amp?

Is that a simple enough question?

thanks,
zpyder

[brett]

Good question.
First, estimate the input impedance.
For transistors, it is a bit complicated, because the resistance through the base is determined by the current flow and HFE.

Assuming that the bias resistors are large, and have minimal effect, the input impedance, seen by a source looking at the base is Rin = B.re  (B is the hFE, and re is the emiiter resistance).  If there is actual no emitter resistor, or it is very small (a few ohms), the resistance is equal to the diode ac emitter resistance, = 25 mV/the emitter current)

Here's some calculated examples:
hFE    Ie (mA)    re (ohms)  Rin (ohms)
400    0.01          2500      1000k
400    0.1            250        100k
400    1               25          10k
100    0.01         2500       250k
100    0.1            250         25k
40     0.01         2500        100k
40     0.1            250         10k

For op-amps, estimating the input impedance is a LOT easier:
For inverting op-amps, the input impedance is equal to the input resistor (connecting the input to the - input and feedback loop)
For non-inverting op-amps, the + input has almost infinite impedance, so the impedance is set by the bias network.  Check the resistor connecting the + input to the voltage divider (or LEDS or whatever else is producing the bias voltage).  The input impedance is equal to that resistor (usually very large, like 1M).

Now that you've got the input impedance, choose a cutoff frequency (fc, usually about 30Hz), and use the formula C= 1/(2.pi.R.fc) to calculate C in farads (R in ohms, pi=3.14, fc in Hz).

But here's a shortcut I use all the time.    
I have a guess at the input impedance (in kohms), and divide 10 by that number to get the answer in microfarads.  e.g. 100k input impedance = 0.1uF, 50k = 0.2 uF.  This method gives an very useful fc of 16 Hz if the guess at the input impedance is perfect.  But even if the guess is wrong by half or double, the fc is still acceptable.

cheers

[zpyder]

Thanks so much brett!!

I'll be printing this out 

zpyder