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.

 

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).

 

Considering notations in http://www.ele.uri.edu/courses/ele343/lab/lab1/quescient.pt/quescient.html, 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.

 

Great reply eduard!! Thanks alot for the help.