BJT Common Emitter AC Analysis

Discussion in 'General Electronics Chat' started by SiegeX1, Mar 10, 2009.

  1. SiegeX1

    Thread Starter Member

    Mar 10, 2009
    I was watching this video on Troubleshooting Transistor Circuits which goes through circuit analysis on a common-emitter amplifier circuit. I have a few questions that didnt get answered in the video:

    1) Why is it that an AC signal will see a power supply as a ground? I'll assume this doesn't apply to DC?

    2) This is a multi-part question. When he calculated the AC collector current, he had to first calculate the *parallel* resistance of R4||R6 which are the "pull-up" resistor from collector to power supply and the load resistance respectively. The reason for it being a parallel calculation is the heart of Question #1 above.

    • Did he ignore the series capacitance C3 (10uF) as a 1Khz AC signal will only see an additional reactance of only 16.91Ohm, basically a short?
    • Did he neglect C2 for the same reason when calculating the emitter current?
    3) What method(s) does one use to come up with the the coupling and bypass capacitor values?

    Thank you for your help.
  2. mik3

    Senior Member

    Feb 4, 2008
    You use the superposition theorem to analyze the circuit. First you do the DC analysis and then the AC analysis. For the DC analysis you short the AC sources and for the AC analysis you short the DC sources, that is why you short the battery.

    AC analysis!

    • Yes.

    These values depend on the desired frequency response of the circuit.