BJT switching in cutoff: when a transistor switch is cut off (no base drive), what is the approximate collector-to-emitter voltage VCE across the device relative to the supply?

Introduction / Context:
Understanding the two extreme states of a BJT used as a switch—cutoff and saturation—is crucial for estimating power dissipation and verifying logic levels. In cutoff, the transistor is off, so there is essentially no collector current other than leakage. The collector circuit then behaves as an open switch.

Given Data / Assumptions:

• Transistor base drive removed (IB ≈ 0), device in cutoff.
• Normal silicon device at room temperature; leakage currents are negligible for first-order analysis.
• Collector connected through a load to a positive supply VCC.

Concept / Approach:
With no conduction path through the transistor, the load line forces the collector voltage to rise to the supply. Thus the collector-to-emitter voltage VCE is approximately VCC (minus tiny leakage-induced drops). This is the complement of saturation, where VCE collapses to a small value. Hence, in digital switching the off-state channel blocks current and supports nearly the full supply voltage across the device.

Step-by-Step Solution:

Verification / Alternative check:
Review typical switching waveforms: turning the transistor off makes VCE rise to near the rail. Datasheets quote ICBO and ICEO leakage currents in microamps or less, confirming that the drop across the load is negligible and VCE remains nearly VCC.

Why Other Options Are Wrong:
Approximately equal to VB: VB is undefined in cutoff and not directly indicative of VCE.
Approximately 0.2 V or 0.7 V: these are characteristic forward drops (VCE(sat), VBE(on)), not applicable in cutoff.
Approximately equal to zero: would imply a short circuit, which is the saturation state, not cutoff.

Common Pitfalls:
Confusing cutoff with saturation; assuming VBE(on) relates to VCE in the off state; forgetting the role of leakage and the load.

Final Answer:
Approximately equal to VCC