BJT as a Switch

          +12 V
            │
          Load (relay/LED strip)   ← flyback diode across inductive loads
            │
MCU ──[ Rb ]── B   NPN
               E
               │
              GND

1. Load current, e.g. $I_C$ = 200 mA.
2. Required base current with margin: use a "forced β" of 10–20 → $I_B$ ≈ 10–20 mA (check the pin can supply it).
3. $R_B = \dfrac{V_{pin} - V_{BE}}{I_B} = \dfrac{3.3 - 0.7}{0.013} \approx 200$ Ω.

The transistor then saturates and dissipates only $V_{CE(sat)} \times I_C$. For larger currents a logic-level MOSFET is usually better.