GATE 2020 IN Q.23: solution and answer

GATE 2020 · INQ.23 · 1 mark · NAT

A 200 mV full-scale dual-slope analog to digital converter (DS-ADC) has a reference voltage of 100 mV. The first integration time is set as 100 ms. The DS-ADC is operated in the continuous conversion mode. The conversion time of the DS-ADC for an input voltage of 123.4 mV (in ms, rounded off to one decimal place) is ____

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Answer and solution

ANSWER

The conversion time is $T_{c} = T_{1} + T_{2} = 100 + 123.4 = 223.4 ms$ .

Official answer: 223.3 to 223.5Source: Official answer key, GATE 2020 IN (IIT Delhi)

Phase 1 (run-up, fixed time). The integrator integrates the input $V_{in}$ for a fixed time $T_{1} = 100 ms$ . At the end of this phase its output has magnitude $V_{o} = \frac{V _{in} T _{1}}{R C}$ .
Phase 2 (run-down, variable time). The integrator input is switched to the reference, which has the opposite polarity, and the output ramps back to zero in a time $T_{2}$ . The two ramps cover the same voltage, so $\frac{V _{r e f} T _{2}}{R C} = \frac{V _{in} T _{1}}{R C}$ .
Charge balance of the dual-slope ADC. $R C$ cancels, which gives $T_{2} = T_{1} \cdot \frac{V _{in}}{V _{r e f}}$
Decisive computation. $T_{2} = 100 ms \times \frac{123.4 mV}{100 mV} = 123.4 ms$ .
Check against full scale. The longest run-down happens at the full-scale input: $T_{2, m a x} = 100 ms \times \frac{200 mV}{100 mV} = 200 ms$ . Since $123.4 ms < 200 ms$ , the input is within range.
Conversion time. In continuous conversion mode the next run-up starts as soon as the run-down ends, so one conversion takes the fixed run-up plus the variable run-down: $T_{c} = T_{1} + T_{2} = 100 + 123.4 = 223.4 ms$

KEY IDEA

A dual-slope ADC integrates the input for a fixed time $T_{1}$ , then integrates the opposite-polarity reference until the integrator output returns to zero. Because both ramps cover the same voltage, $T_{2} = T_{1} V_{in} / V_{r e f}$ , and $R C$ drops out. In continuous mode, one conversion lasts the whole cycle $T_{1} + T_{2}$ , so the conversion time depends on the input voltage.

WATCH OUT FOR

Giving only the run-down time $T_{2} = 123.4 ms$ , or only the fixed run-up $T_{1} = 100 ms$ , as the conversion time. A conversion needs both phases, so the answer is $T_{1} + T_{2} = 223.4 ms$ . A second error is using the full-scale value (200 mV) in place of $V_{r e f}$ (100 mV) in $T_{2} = T_{1} V_{in} / V_{r e f}$ , which gives $T_{2} = 61.7 ms$ and a wrong total of 161.7 ms. Another is taking the worst case, $T_{1} + T_{2, m a x} = 300 ms$ , when the question asks about one particular input.

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GATE 2020 IN Q.23: solution and answer

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