When the bipolar transistor is in the so-called saturation state, the voltage VCE between the collector and the emitter is ideally

Question

BlackTom AI · Accepted Answer

To tackle this question, we first recall what saturation means for a bipolar transistor: both the base-emitter and base-collector junctions are forward-biased, causing the collector-emitter voltage to drop to a very small value. 
Option a: 'n. 0.7 V' suggests a fixed drop similar to a diode drop; however, in saturation the voltage across the collector-emitter is not determined by a single diode drop but by the device being in deep conduction, and while it is small, it is not consistently about 0.7 V. This makes this choice misleading as a general statement.
Option b: 'in its maximum value' implies the transistor is not saturated but rather in cutoff or active mode where VCE can rise toward Vcc; saturation pushes VCE downward, not upward, so this is incorrect.
Option c: '0 V' captures the idealized view of saturation where the collector-emitter path behaves like a short circuit; in real devices VCE(sat) is a small finite voltage (a few tens of millivolts to a few hundred millivolts depending on device and current), but the statement '0 V' communicates the essential idea that VCE is driven to a very small value, which aligns with the saturation concept.
Option d: 'equal to supply voltage, Vcc' would place the transistor in a condition where the collector-emitter path carries no increased conduction relative to supply; this contradicts saturation, where VCE collapses toward near-zero, not toward Vcc.
In summary, the core idea of saturation is that the collector-emitter voltage is driven down to a minimal value, so the description in option c best matches the saturation state concept, while the other statements mischaracterize the voltage behavior across VCE in saturation.

When the bipolar transistor is in the so-called saturation state, the voltage VCE between the collector and the emitter is ideally单项选择题

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