Stairstep-ramp ADC (4-bit) — With clock frequency = 100 kHz (period = 10 microseconds), full-scale = 10 V, and an input of 6 V applied, estimate the conversion time for the counter/stairstep process (time to reach or exceed the input level).

Introduction / Context:
A stairstep-ramp (counter-type) analog-to-digital converter generates a digital count from 0 upward while a DAC creates a corresponding staircase. A comparator stops the count when the DAC staircase reaches the analog input. The conversion time therefore depends on the input amplitude (larger inputs take more steps) and the clock rate (how fast steps occur).

Given Data / Assumptions:

• Resolution: 4 bits ⇒ 2^4 = 16 staircase steps across 0–10 V.
• Clock frequency: f_clk = 100 kHz ⇒ T_clk = 10 microseconds per step.
• Input voltage: V_in = 6 V; full-scale: V_FS = 10 V.
• Each step height: V_step = V_FS / 16 = 10 / 16 = 0.625 V.

Concept / Approach:
Conversion time is approximately the number of steps needed to reach (or just exceed) V_in multiplied by the clock period. Steps required ≈ (V_in / V_FS) * 16. Because the comparator halts on the first step at or above V_in, practical answers round to the nearest whole step based on implementation specifics.

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