I was about to go and learn common emitter amplifier configuration in details and the problem occurred right at the beginning of it. I have found this equation of calculating the C1 (capacitor that blocks DC component at the input of the junction transistor) for its value (it shouldn't be to high or too low (the capacitance) because it also puts on a high/low-pass filter on the input of the transistor).

enter image description here

So, I almost perfectly understand the whole equation but what I don't understand is how should I get the required capacitance for C1 from this calculation.

 - low frequency break point of the filter

- resistance from R1, R2 resistors (from voltage divider) and the input resistance of transistor - in parallel manner

For C2 equation is the same except there is resistance of load instead of those parallel resistances mentioned before.

enter image description here

  • \$\begingroup\$ Blabla C1, blabla R1, R2. Uhm why don't you include a schematic so that we know what you talk about instead og having to guess. Yes we know what a CE circuit looks like but still we want to see that schematic. There's a tool to draw a schematic, use it or include a pic from a Google search. \$\endgroup\$ Sep 17, 2016 at 21:00
  • \$\begingroup\$ There's a schematic button on the editor toolbar. Click it! It's easy to use. Double click a component to edit its properties. \$\endgroup\$
    – Transistor
    Sep 17, 2016 at 21:01
  • \$\begingroup\$ @FakeMoustache : Sorry about that.. \$\endgroup\$
    – Lucenzo97
    Sep 17, 2016 at 21:56
  • \$\begingroup\$ @Transistor: I have copied a picture from my source. I hope now it is more understandingly. \$\endgroup\$
    – Lucenzo97
    Sep 17, 2016 at 21:58

1 Answer 1


I think you understand the theory but have missed something simple.

  • As far as the AC signal \$ V_{IN} \$ is concerned R1 and R2 are in parallel as the ends away from C1 are at fixed potential. So, work that out and call it \$ R_{||} \$ as you have done.
  • C1 is the input decoupling capacitor. It prevents the bias from R1 and R2 being leaked away or shunted out through the input.
  • The bigger you make C1 the lower the cut-off frequency becomes and you will get extended bass response in an audio circuit. (C1, R1 and R2 form a high-pass filter.) For cost, size and because we're engineers we choose one that's appropriate for the application.
  • Decide your cut-off frequency and calculate the required value using your formula

$$ C1 = \frac {1}{2 \pi f_{LOW} R_{||}} $$

The capacitance shouldn't be to high or too low because it also puts on a high/low-pass filter on the input of the transistor.

It's a high-pass filter only. The sanity check on this is to take the extreme low-frequency test case - DC. It obviously blocks DC so it's high-pass. (You can use this technique to do a quick circuit analysis in lots of applications.)


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