Why does a Darlington saturate less completely than a single transistor?
Answer: C
The output device's base sits at the driver's emitter, so its collector cannot go below one base-emitter drop plus the driver's own saturation voltage, leaving perhaps 0.8 V rather than 0.2. Drive current is ample by construction, differing temperatures are a secondary matter, and the gain fall-off is a separate effect. Tracing a limit to the topology rather than to either device explains why no better transistor would fix it. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.
Please log in or register to report this question.