CE Amplifier Design question

Thread Starter


Joined Nov 25, 2011

I'm currently studying electronics and came across this diagram for the design of CE amplifiers.

I'm just trying to figure out why Rc has been chosen to be 20k?

In order for Ic to be 0.5mA shouldn't Rc be 40k? ( R = v/i = 20/0.5*10^-3 )

Also, how was the value for Re chosen?

Any help would be greatly appreciated!



Joined Aug 10, 2008
For a CE amp the voltage at the collector needs to be around 1/2 of VCC.

So 10v. / 0.5mA will be the 20K ohm resistance.

The DC bias voltage at the emitter is 1v. so 1v/ 0.5mA will give the 2K value.

R3 is not part of the biasing voltage at the emitter, it is used for gain calculations.
Due to the bypass cap.