Question 13.7: Analysis of the Four-Resistor Bias Circuit Find the values o...

Substituting into Equations 13.20 and 13.21, we find that

R_B=\frac{1}{1/R_1+1/R_2}=R_1\|R_2             (13.20)

R_B=\frac{1}{1/R_1+1/R_2}=3.33\mathrm{~k}\Omega

V_B=V_{CC}\frac{R_2}{R_1+R_2}=5\mathrm{~V}

Then, substituting into Equation 13.23 and using β = 100, we have

I_B=\frac{V_B-V_{BE}}{R_B+(\beta +1)R_E}=41.2~\mu\mathrm{~A}

For β = 300, we find that I_B = 14.1 μA. Notice that the base current is significantly smaller for the higher β.
Now, we can compute the collector current by using I_C = βI_B. For β = 100, we find that I_C = 4.12 mA, and for β = 300, we have I_C = 4.24 mA. For a 3:1 change in β, the collector current changes by less than 3 percent. The emitter current is given by I_E = I_C + I_B. The results are I_E = 4.16 mA for β = 100 and I_E = 4.25 mA for β = 300.
Finally, Equation 13.24 can be used to find V_CE. The results are V_CE = 6.72 for β = 100 and V_CE = 6.51 for β = 300.

V_{CE}=V_{CC}-R_CI_C-R_EI_E      (13.24)