Biasing: find Rc and Rb

Biasing means applying suitable DC voltages and currents to a transistor to fix its operating point (Q-point) in the active region, so that the signal is amplified without distortion. The operating point should be stable against changes in temperature and $\beta$.

Given: collector-feedback bias, $V_{CC}=12$ V, $I_C=1.2$ mA, $V_{CE}=6$ V, $\beta=100$. Assume $V_{BE}=0.7$ V (silicon).

Base current:

$$I_B=\frac{I_C}{\beta}=\frac{1.2\ \text{mA}}{100}=12\ \mu\text{A}$$

Finding $R_C$. The current through $R_C$ is $I_C+I_B=1.212$ mA. KVL on the output loop:

$$V_{CC}=(I_C+I_B)R_C+V_{CE}$$ $$R_C=\frac{12-6}{1.212\ \text{mA}}=4.95\ \text{k}\Omega\ (\approx5\ \text{k}\Omega)$$

Finding $R_B$. KVL through $R_B$ and the base-emitter junction:

$$V_{CE}=I_BR_B+V_{BE}$$ $$R_B=\frac{6-0.7}{12\ \mu\text{A}}=441.7\ \text{k}\Omega$$

Answer: $R_C\approx4.95\ \text{k}\Omega$ (about 5 kΩ), $R_B\approx441.7\ \text{k}\Omega$ (nearest standard values 430 kΩ or 470 kΩ).

If $I_B$ is neglected in the $R_C$ drop, $R_C=6/1.2\ \text{mA}=5\ \text{k}\Omega$, which is the same to standard-value accuracy.

(If the figure is instead a fixed-bias circuit, with $R_B$ connected from $V_{CC}$ to the base, then $R_C=6\ \text{V}/1.2\ \text{mA}=5\ \text{k}\Omega$ and $R_B=(12-0.7)/12\ \mu\text{A}=941.7\ \text{k}\Omega$.)