Difficulty: Easy
Correct Answer: Very high
Explanation:
Introduction / Context:
One of the main reasons to use an emitter follower is to avoid loading a preceding stage. The base input resistance of an emitter follower is large, so only a small input current is required to develop the needed input voltage. This permits accurate voltage transfer from high-impedance sources to lower-impedance loads downstream.
Given Data / Assumptions:
Concept / Approach:
The input resistance looking into the base is approximately Rin ≈ (beta + 1) * RE_eff. Because beta is sizable and RE_eff can be in the kilo-ohm range, Rin frequently lands in the hundreds of kilo-ohms to mega-ohm range. This is not infinite, but is “very high” relative to many signal sources. The high Rin, together with low output resistance, makes the emitter follower an excellent buffer and impedance transformer with a voltage gain near unity.
Step-by-Step Solution:
Verification / Alternative check:
Measured with a small AC test source at the base and observing current draw, the calculated Rin aligns with the (beta + 1) scaling. Real designs also include bias resistor dividers that may lower the net input resistance; even then, the base itself presents a very high input compared to many other stages.
Why Other Options Are Wrong:
Very low: contradicts the buffer role of the emitter follower.
Shorted to ground: only in fault conditions; not a design feature.
Open (infinite): Rin is high but not infinite; base currents and bias networks exist.
Approximately equal to RC: RC is not central in a common-collector stage.
Common Pitfalls:
Ignoring the effect of a bias divider that can reduce overall input resistance; assuming infinite Rin and miscalculating source loading; forgetting RE_eff includes the external load.
Final Answer:
Very high
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