Discussion in 'Homework Help' started by Joe24, Sep 9, 2007.

  1. Joe24

    Thread Starter Active Member

    May 18, 2007
    Hello all,

    I have a simple Common-Emmitter BJT Amplifier circuit in a Voltage Divider Bias configuration with a Source and Load Resistor. The question is:

    1) Show that the ICQ that produces the largerst AC output voltage swing is given by ICQ = VCC / Rac + Rdc, where Rac is the AC resistance and Rdc is the DC resistance, and -1/Rac = slope of ac load line
    -1/Rdc = slope of dc load line

    The circuit does not have a specified value for VCC.

    The way I was thinking of going about this problem was to draw the AC and DC load lines on the characteristics graph and show the best ICQ that gives the largers signal swing. But, since there is no VCC given, I can't go about this method.

    Does anyone know how I can go about solving this problem.
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    Possibly you could select some arbitrary values for Vcc and run the equation to see if the answers give sensible results. If a trend develops, you might find there is a best value for Vcc with the resistor values given (you don't mention what they are).
  3. Eduard Munteanu

    Active Member

    Sep 1, 2007
    Considering notations in, we have: R_{DC} = R_C + R_E \\ R_{AC} = R_C (since the emitter resistance is bypassed by the capacitor at AC)

    IMO, this is plainly wrong.

    Choosing V_C = {1 \over 2} V_{CC} will give the maximum output voltage swing. Going for the quiescent current, we get:
    V_C = V_{CC} - I_C R_C \Leftrightarrow -{1 \over 2} V_{CC} = - I_C R_C \Leftrightarrow I_C = {V_{CC} \over {2 R_C}} \Leftrightarrow I_C = {V_{CC} \over {2 (R_{DC} - R_E)}}\\
    But R_{DC} - R_E = R_{AC}, so I_C = {V_{CC} \over {R_{DC} + R_{AC} - R_E}}, which is just different from what you asked.

    Putting this a bit differently, I don't see how the quiescent point (i.e. DC analysis output voltage/current) can be affected by AC conditions. The actual output swing depends on the input, but this naive model assumes R_{E (AC)}=0, which predicts infinite AC gain (since G_v = -{R_C \over R_{E (AC}}). But such a gain would only require the quiescent point to be set at half the power supply's voltage.
  4. Joe24

    Thread Starter Active Member

    May 18, 2007
    Great reply eduard!! Thanks alot for the help.