Common Emitter Coupling

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).

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.

• 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. – Bimpelrekkie Sep 17 '16 at 21:00
• There's a schematic button on the editor toolbar. Click it! It's easy to use. Double click a component to edit its properties. – Transistor Sep 17 '16 at 21:01
• @FakeMoustache : Sorry about that.. – Lu Ka Sep 17 '16 at 21:56
• @Transistor: I have copied a picture from my source. I hope now it is more understandingly. – Lu Ka Sep 17 '16 at 21:58

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