Channel multiplexing – when several analog sources must be digitized, can a single ADC be time-shared via an analog multiplexer that selects one input at a time for conversion?

Difficulty: Easy

Correct Answer: Correct

Explanation:


Introduction / Context:
Many embedded systems must digitize multiple sensor channels but can afford only one ADC due to cost or space. A standard approach is to time-division multiplex channels using an analog multiplexer (MUX) and a sample-and-hold or track-and-hold front end as needed for accuracy and speed.


Given Data / Assumptions:

  • ADC has adequate sampling rate to service all channels within required update times.
  • An analog MUX selects which sensor connects to the ADC input.
  • Settling and source impedance are compatible with ADC input requirements.


Concept / Approach:
Time-sharing a single ADC works by cycling through inputs: select channel i, allow settling and hold if necessary, start conversion, read result, then proceed to channel i+1. Provided the ADC's throughput meets the sum of per-channel rates and the analog front end is designed for proper settling and isolation, one ADC can serve many sources effectively.


Step-by-Step Solution:

1) Use a MUX to connect one sensor at a time to the ADC input.2) Provide adequate acquisition time and buffering to meet input capacitance/impedance specs.3) Trigger conversion; read and store the result with channel ID.4) Repeat for all channels within the sampling schedule.


Verification / Alternative check:
Microcontrollers with built-in ADCs commonly expose multiple channels through an internal MUX; discrete designs replicate this externally with 74HC4051-class devices or instrumentation multiplexers.


Why Other Options Are Wrong:
Current-mode only is false; strict 1 mV ground match is unnecessarily tight (use proper grounding/guarding). A dedicated S/H per channel is optional, not always required.


Common Pitfalls:
Insufficient settling time, high source impedance causing gain errors, and crosstalk between channels if MUX isolation is poor.


Final Answer:
Correct

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