The values in the circuit shown in <image 1> are: $R_1$ = 33 k$\Omega$, $R_2$ = 10 k$\Omega$,  $R_C$ = 2.2 k$\Omega$, $R_E$ = 1 k$\Omega$, and $V_{CC}$ = 18 V. A silicon NPN transistor with $\beta $ = 50 is used. Determine the operating point $V_{CEQ}$.
[]
We first calculate the base equivalent resistance and open-circuit

                        voltage.

                                    RB = R1 || R2 = [{(33) (10)} / (33 + 10)]k\Omega = 7.7 k\Omega

                                    VBB = VCC {R2 / (R1 + R2)} = (18) {10 / (33 + 10)}V = 4.2 V

                        The quiescent base current is now calculated. We write the base-loop

                        equation:

                                                VB - VBE = R2IB + REIE

                        Since IE = IB + IC = IB + \betaIB = (\beta + 1)IB we can solve for IB.

                                                IBQ = [(VBB - VBE) / {RB + (\beta + 1)RE}]

                        In this active mode, VBE = 0.6 V Thus,

                                                IBQ = [(4.2 - 0.6) / {7.7 + (51)1}] mA = 61 \muA

                        With \beta = 50, the collector current is:

                                                ICQ = \betaIBQ = 50(0.061 mA) = 3 mA

                        The collector voltage is determined to complete the cal\xadculation of the

                        operating point:

                                                  VCEQ  = VCC - RCICQ - REIEQ

                                                            \approx VCC - (RC + RE)ICQ= 18 - (2.2 + 1)3 = 8.4 V