A transistor inverter has a supply of 5.0 V, a collector resistor of 1.0 kΩ, a base resistor of 47…

Step 1. Input of $0.5$ V. This is below $0.7$ V, so the transistor is off. There is no drop in $R_C$, so $V_o = 5.00$ V.

Step 2. Input of $2.0$ V. $I_B = \dfrac{2.0\ \text{V} - 0.7\ \text{V}}{47\ \text{k}\Omega} = 27.7\ \mu\text{A}$. Then $I_C = 100 \times 27.7\ \mu\text{A} = 2.77$ mA, and $V_o = 5.0\ \text{V} - 2.77\ \text{mA} \times 1.0\ \text{k}\Omega = 2.23$ V. The transistor is active.

Step 3. The saturation input. At saturation, $I_C = \dfrac{5.0\ \text{V} - 0.2\ \text{V}}{1.0\ \text{k}\Omega} = 4.8$ mA, so $I_B = \dfrac{4.8\ \text{mA}}{100} = 48\ \mu\text{A}$. The input is $V_i = 0.7\ \text{V} + 48\ \mu\text{A} \times 47\ \text{k}\Omega = 2.96$ V.