Difficulty: Easy
Correct Answer: VOH = 8.5 V, VOL = 0.1 V
Explanation:
Introduction / Context:The bipolar NE555 output stage uses bipolar transistors that cannot swing exactly to the rails under load. Consequently, the HIGH level is typically below VCC by about 1.2 V to 1.7 V, and the LOW level is a small saturation voltage above ground. This question checks recognition of realistic 555 output levels versus ideal rail-to-rail assumptions.
Given Data / Assumptions:
Concept / Approach:Because the bipolar totem-pole cannot reach the rails, VOH ≈ VCC − about 1.5 V. VOL is on the order of a few hundred millivolts. These typical values align with many datasheet examples given specified sink or source currents.
Step-by-Step Solution:
1) Start with VCC = 10 V.2) Apply typical headroom for VOH: around 1.5 V below VCC → about 8.5 V.3) Assume VOL near saturation, roughly 0.1 V to 0.2 V.4) Therefore, VOH ≈ 8.5 V and VOL ≈ 0.1 V match practical expectations.Verification / Alternative check:Review the NE555 datasheet curves showing VOH vs. load current and VOL vs. sink current; the numbers cluster around these values at common loads.
Why Other Options Are Wrong:
Common Pitfalls:Confusing CMOS 555 (closer to rail-to-rail) with the bipolar version; ignoring load dependence.
Final Answer:VOH = 8.5 V, VOL = 0.1 V
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