Question bank — BJT operating regions (cutoff, active, saturation)

False. Saturation is the fully ON state — collector current is at its ceiling $I_{C(sat)}=(V_{CC}-V_{CE(sat)})/R_C$ while $V_{CE}$ collapses to ~0.2 V. The word misleads; cutoff is the near-zero-current state.

True — but only while still active. As long as $\beta I_B<I_{C(sat)}$, $I_C=\beta I_B$ tracks $I_B$ linearly. Once $\beta I_B$ would exceed the ceiling, the device saturates and the proportionality breaks.

False. It holds only in the active region. In saturation the collector can't supply $\beta I_B$, so $I_C$ is fixed by the external circuit $(V_{CC}-V_{CE(sat)})/R_C$; in cutoff both are ≈0.

True. $I_E=I_C+I_B$ and $I_B>0$ whenever the device conducts, so $I_E>I_C$ always. Equivalently $I_C=\alpha I_E$ with $\alpha <1$.

False. A forward B–E junction only rules out cutoff — it could be active or saturation. You must also check the B–C junction (equivalently, whether $V_{CE}$ has hit $V_{CE(sat)}$).

True. Since $\alpha =\beta /(\beta +1)$, $\alpha$ rises toward 1 as $\beta$ grows. But the change is tiny: $\beta =99\to \alpha =0.99$, $\beta =199\to \alpha =0.995$.

False. Power is $V_{CE}\times I_C$. In cutoff $I_C\approx 0$, so power ≈ 0 despite the large $V_{CE}$. Both OFF (cutoff) and hard-ON (saturation) are low-power states — see BJT as a Switch.

False. Beyond $I_{B(sat)}$ the collector is already at its ceiling; extra base drive is "overdrive" that only ensures a solid, robust ON — it does not raise $I_C$.

Error: applying $I_C=\beta I_B$ without checking the ceiling. If $I_{C(sat)}$ (from $V_{CC}$ and $R_C$) is below 20 mA, the device saturates and $I_C=I_{C(sat)}$, not 20 mA. Always compute both and take the smaller.

Error: wrong current. $\alpha$ multiplies the emitter current: $I_C=\alpha I_E$. The base relation uses $\beta$: $I_C=\beta I_B=99I_B$. Mixing them swaps a gain of ~99 for a loss.

Error: that describes the active region. In saturation the B–C junction is forward-biased, so it stops helping and $V_{CE}$ drops to ~0.2 V. The reverse-biased B–C that "sucks carriers across" is precisely what makes active mode a clean amplifier.

Error: that's the boundary, not a safe ON. At exactly $I_{B(sat)}$ the device only just reaches the edge of saturation. Real switches use $I_B$ a few times larger to stay firmly saturated across temperature and $\beta$ spread.

Error: that's cutoff. In active mode current flows, so $V_{CE}=V_{CC}-I_C{R}_C<V_{CC}$. Only in cutoff ($I_C\approx 0$) does $V_{CE}\approx V_{CC}$.

Error: dropped the big term. $I_C$ is the large term ($\beta$ times $I_B$), so $I_E\approx I_C$, not $I_B$. $I_B$ is the small one being neglected — the opposite of the claim.

Error: that's the reverse-active (inverse) region, the fourth combination, not standard active. Its gain is poor because the collector was never doped to be an efficient emitter; it's normally avoided, not equivalent.

So few injected carriers recombine in the base, pushing $\alpha$ close to 1. Since $\beta =\alpha /(1-\alpha )$, a tiny loss fraction $(1-\alpha )$ in the denominator yields a large $\beta$ — that's where the amplification comes from. See BJT Structure and Doping.

As $I_C$ rises, $V_{CE}=V_{CC}-I_C{R}_C$ drops; once $V_{CE}$ approaches $V_{BE}$ the B–C junction becomes forward-biased and clamps further fall. That clamp voltage is $V_{CE(sat)}\approx 0.2$ V.

The forward-biased B–C junction injects carriers backward, and the external resistor already limits current to $(V_{CC}-V_{CE(sat)})/R_C$. The collector circuit physically cannot supply $\beta I_B$, so $I_C$ is set by the resistor, not $\beta$.

A saturated transistor has $V_{CE}\approx 0.2$ V, so power loss $V_{CE}\times I_C$ is tiny and the output is a clean "low." Active mode wastes power as heat and gives an ill-defined intermediate voltage — great for amplifying, bad for switching.

We usually know $I_B$, $\beta$, $V_{CC}$, $R_C$ but not $V_{CE}$ up front. Comparing the active prediction $\beta I_B$ to the ceiling $I_{C(sat)}$ tells us the region, from which $V_{CE}$ then follows — this is exactly Load Line Analysis in numbers.

The emitter is heavily doped to flood the base with carriers; the collector is lightly doped and physically large to sweep them out under reverse bias. Swapping roles (reverse-active) works poorly because the collector was never optimized to inject.

Because some injected carriers always recombine in the base or fail to cross, so the collector receives strictly fewer than the emitter emits: $I_C<I_E$. If $\alpha$ hit 1, $\beta$ would be infinite — physically impossible.

Cutoff. No forward B–E means no injection, so $I_B\approx I_C\approx 0$ regardless of $V_{CE}$ — the OFF state, even with the supply fully across it.

The device sits exactly at the edge of saturation: $I_C=I_{C(sat)}$ and $V_{CE}=V_{CE(sat)}$, with the active law just satisfied. It is the crossover point where increasing $I_B$ further gains no extra $I_C$.

With no resistor $I_{C(sat)}=(V_{CC}-V_{CE(sat)})/0\to \infty$, so the circuit imposes no ceiling and the device stays active as $I_C=\beta I_B$ — until some real limit (power, internal resistance) intervenes. This is why a switch needs a collector resistor.

Recover $\beta =\alpha /(1-\alpha )=0.995/0.005=199$, then proceed normally. The region test never truly needs $\alpha$ directly — convert to $\beta$ first.

Saturation. Both junctions forward → $V_{CE}\approx 0.2$ V and $I_C<\beta I_B$ (collector maxed out by the external resistor, not by base drive).

Cutoff → (B–E turns forward at ~0.7 V) → Active → (at $I_B=I_{B(sat)}$) → Saturation. $I_C$ climbs linearly through active, then flattens at $I_{C(sat)}$ once saturated.

A MOSFET's cutoff mirrors BJT cutoff (OFF), and its ohmic/triode region plays the switch-ON role like BJT saturation; its saturation region is the amplifying one — opposite naming to the BJT, a classic trap. Compare MOSFET Operating Regions.

No — it's at the active/saturation edge only if $\beta I_B$ equals $I_{C(sat)}$. If the active prediction is genuinely below the ceiling, the transistor is active and $V_{CE}$ would actually sit above $V_{CE(sat)}$; the two descriptions must agree.

Cutoff = reverse+reverse; Active = B–E forward, B–C reverse; Saturation = forward+forward.

Compute $\beta I_B$ and $I_{C(sat)}=(V_{CC}-V_{CE(sat)})/R_C$; the smaller one is the real $I_C$ — smaller-is-$\beta I_B$ means active, smaller-is-ceiling means saturation.