Question 5.18: Calculate the dc bias currents and voltages for the circuit ...
Calculate the dc bias currents and voltages for the circuit of Fig. 5.89 to provide V_o at one-half the supply voltage (9 V).
I_{B_{1}}=\frac{18 V -0.7 V }{2 M \Omega+(140)(180)(75 \Omega)}=\frac{17.3 V }{3.89 \times 10^{6}}=4.45 \mu A
The base Q_2 current is then
I_{B_{2}}=I_{C_{1}}=\beta_{1} I_{B_{1}}=140(4.45 \mu A )=0.623 mAresulting in a Q_2 collector current of
I_{C_{2}}=\beta_{2} I_{B_{2}}=180(0.623 mA )= 1 1 2 . 1 ~ m Aand the current through R_C is then
Eq. (5.119):
I_{C}=I_{E_{1}}+I_{C_{2}} \approx I_{B_{2}}+I_{C_{2}} (5.119)
I_{C}=I_{E_{1}}+I_{C_{2}}=0.623 mA +112.1 mA \approx I_{C_{2}}=112.1 mA
V_{C_{2}}=V_{E_{1}}=18 V -(112.1 mA )(75 \Omega)
= 18 V – 8.41 V
= 9.59 V
V_{B_{1}}=I_{B_{1}} R_{B}=(4.45 \mu A )(2 M \Omega)= 8.9 V
V_{B C_{1}}=V_{B_{1}}-0.7 V =8.9 V -0.7 V= 8.2 V
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