BJT operating regions — identifying cutoff in an NPN transistor Which condition best indicates that an NPN BJT is in cutoff (no conduction) between collector and emitter?

Introduction / Context:Understanding BJT operating regions—cutoff, active, and saturation—is essential for analog biasing and digital switching. This question asks you to identify the signature condition of cutoff in an NPN transistor.

Given Data / Assumptions:

• NPN transistor in a typical biasing configuration.
• Cutoff means no base drive sufficient to forward-bias the base–emitter junction.
• Voltages are referenced in standard silicon BJT terms.

Concept / Approach:

In cutoff, the base–emitter junction is not forward biased; as a result, collector current (IC) is approximately zero and the device behaves like an open switch between collector and emitter. VCE typically rises toward the supply because there is no conduction path to drop voltage across the collector resistor.

Step-by-Step Solution:

Verification / Alternative check:

Compare to saturation: in saturation, VCE(sat) is low (typically 0.1–0.3 V) and IC is relatively large for the circuit’s load. In active region, VBE ≈ 0.7 V and IC = β * IB with VCE moderate. Neither matches cutoff behavior.

Why Other Options Are Wrong:

• Collector current is maximum: describes saturation, not cutoff.
• VCE = 0 V: also indicates saturation (short-like behavior), not cutoff.
• VBE = 0.7 V: implies a forward-biased base–emitter junction (active or saturation), not cutoff.
• Base current very high: opposite of cutoff.

Common Pitfalls:

• Confusing high VCE in cutoff with a fault; high VCE is normal when no current flows.
• Assuming VBE must be exactly 0.7 V; it varies with device and current.

Final Answer:

collector to emitter appears to be open