Worked example: NPN switch — active or saturated?

An NPN BJT has $\beta = 100$. A 5 V input drives the base through $R_B = 10\,\text{k}\Omega$. Take $V_{BE}(\text{on}) = 0.7\,\text{V}$. The collector connects to $V_{CC} = 12\,\text{V}$ through $R_C = 1\,\text{k}\Omega$. Find $I_B$, $\beta \cdot I_B$, the actual $I_C$, and decide whether the transistor is in the active or saturation region.

1. Given. $V_{in} = 5\,\text{V}$, $R_B = 10\,\text{k}\Omega$, $V_{BE} = 0.7\,\text{V}$, $\beta = 100$, $V_{CC} = 12\,\text{V}$, $R_C = 1\,\text{k}\Omega$, $V_{CE}(\text{sat}) = 0.2\,\text{V}$.
2. Compute base current. KVL around the base loop:
   $$I_B = (V_{in} - V_{BE}) / R_B = (5 - 0.7) / 10\,\text{k}\Omega = 0.43\,\text{mA}$$
3. Compute $\beta \cdot I_B$ (what the transistor would pass).
   $$\beta \cdot I_B = 100 \times 0.43\,\text{mA} = 43\,\text{mA}$$
4. Compute the saturation cap. The maximum current the load can carry once $V_{CE}$ bottoms out at $V_{CE}(\text{sat})$:
   $$I_{C,\text{max}} = (V_{CC} - V_{CE,\text{sat}}) / R_C = (12 - 0.2) / 1\,\text{k}\Omega = 11.8\,\text{mA}$$
5. Compare. $\beta \cdot I_B = 43\,\text{mA}$ > $I_{C,\text{max}} = 11.8\,\text{mA}$. The load resistor cannot pass that much current — the transistor saturates.
   $$I_C = I_{C,\text{max}} = 11.8\,\text{mA},\quad V_{CE} = V_{CE,\text{sat}} = 0.2\,\text{V}$$
6. Sanity-check the overdrive ratio. $\beta \cdot I_B / I_{C,\text{max}} = 43 / 11.8 \approx 3.6$. The base is overdriven by ≈3.6×, which is good engineering practice — even if $\beta$ sags from 100 to 30 (extreme temperature, low $I_C$), the transistor still saturates.