Question 13.5: Analysis of the Fixed Base Bias Circuit Repeat Example 13.4 ...

First, we assume operation in the active region. This leads to

I_B=\frac{V_{CC}-0.7}{R_B}=71.5~\mu\mathrm{A}

I_C=\beta I_B=21.45\mathrm{~mA}

V_{CE}=V_{CC}-R_CI_C=-6.45\mathrm{~V}

The requirements for the active region are V_CE > 0.2 V and I_B > 0, which are not met. Thus, the transistor is not operating in the active region.
Next, we assume that the transistor is in saturation. This leads to

I_C=\frac{V_{CC}-0.2}{R_C}=14.7~\mathrm{mA}

and

I_B=\frac{V_{CC}-0.7}{R_B}=71.5~\mu\mathrm{A}

Now, we find that the conditions for saturation (I_B > 0 and βI_B > I_C) are met. Thus, we have solved the circuit, and V_CE = 0.2 V.