Topic: Electronic Devices and Circuits
If for a silicon npn transistor, the base-to-emitter voltage $(V_{BE})$ is 0.7 V and the collector-to-base voltage $(V_{CB})$ is 0.2 V, then the transistor is operating in the
- A. normal active mode
- B. saturation mode
- C. inverse active mode
- D. cut-off mode
Answer: A. normal active mode
Explanation: $V_{BE} = 0.7\text{ V}$, meaning the base-emitter junction is forward biased. $V_{CB} = 0.2\text{ V}$. Since $V_{C} - V_{B} = 0.2$, it implies $V_{C} > V_{B}$. This means the collector-base junction is reverse biased. Thus, the transistor is in normal active-mode.