I can't understand the current flow in a CE amplifier with bypass capacitor. If there is no capacitor, then the AC signal will pass through the emitter resistance and cause additional voltage drop there, making the emitter more positive. This will decrease the base and the collector currents and the voltage at the collector will increase (when it is supposed to decrease). This is the reason to bypass the Ac part of the signal with an emitter capacitor.
But, since the capacitor is supposed to have very little reactanse at the working frequency, in order to be like short circuit for the ac signal, wont that mean, that for that part of the amplification process, when the AC source is positive at the base, the emitter will be grounded at 0volts and the whole bias current through the emitter resistance will decrease since now it is approximately 0,1v/RE, non like the previous 1v/RE? I dont understand this, I ask for your help.
But, since the capacitor is supposed to have very little reactanse at the working frequency, in order to be like short circuit for the ac signal, wont that mean, that for that part of the amplification process, when the AC source is positive at the base, the emitter will be grounded at 0volts and the whole bias current through the emitter resistance will decrease since now it is approximately 0,1v/RE, non like the previous 1v/RE? I dont understand this, I ask for your help.