Given figure shows a silicon transistor connected as a common emitter amplifier. The quiescent collector voltage of the circuit is approximately.

Analysis:

The quiescent Base voltage (V_b) is determined by the potential divider network formed by the two resistors R₁, R₂ (connected to the base of the transistor), and the power supply voltage V_cc as shown, with the current flowing through both the resistors.

\({V_B} = \frac{{{V_{CC}}{R_2}}}{{{R_1} + {R_2}}}\)

Calculation:

\({V_B} = \frac{{20}}{{10\; + \;5}} \times 5 = \frac{{20}}{3}\)

The Thevenin equivalent resistance connected to the base of the circuit will be:

\({R_{BB}} = \frac{{5\; \times \;10}}{{15}} = \frac{{10}}{3}\)

With I_E = (β + 1)I_B ≈ β I_B:=\(I_C\)

\({I_B} = \frac{{{I_E}}}{\beta }\)

Now \(V_{BE}=0.7\)

 \(V_B -V_E =0.7V\)

\(V_E=\frac{20}{3}-0.7\)

\(V_E=5.966 V\)

\(V_E=I_ER_E\)

\(I_E=0.596 mA\)

Assuming I_E ≈ I_C, the output voltage will be:

\({V_o} = 20 - {I_E}{R_C} = 14\;V\)