GATE 2016 EE – Question 47

Analog and Digital Electronics · Schmitt triggers, Sample and hold circuits, A/D and D/A converters · 2 marks · Multiple choice

A 2-bit flash Analog to Digital Converter (ADC) is given below. The input is $0 \leq V_{IN} \leq 3$ Volts. The expression for the LSB of the output $B_0$ as a Boolean function of $X_2$, $X_1$, and $X_0$ is

A 3 V reference feeds a resistor ladder of 100 Ω, 200 Ω, 200 Ω and 100 Ω to ground. Three comparators compare $V_{IN}$ with the three tap voltages, giving outputs $X_2$ (top tap), $X_1$ (middle tap) and $X_0$ (bottom tap). A digital circuit turns $X_2, X_1, X_0$ into the output bits $B_1$ and $B_0$.

$X_0 [\overline{X_2 \oplus X_1}]$
$\bar{X}_0 [X_2 \oplus X_1]$
$X_0 [X_2 \oplus X_1]$
$\bar{X}_0 [\overline{X_2 \oplus X_1}]$

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: (A) $X_0 [\overline{X_2 \oplus X_1}]$

Explanation

The ladder has 600 Ω in total, so the taps are at $3 \times \frac{500}{600} = 2.5$ V, 1.5 V and 0.5 V. The comparator outputs $X_2 X_1 X_0$ are 000 for $V_{IN} < 0.5$, 001 for $0.5$ to 1.5, 011 for 1.5 to 2.5, and 111 above 2.5. These give $B_1 B_0 = 00, 01, 10, 11$. So $B_0 = 1$ for the inputs 001 and 111 and 0 otherwise. This is $X_0$ AND ($X_2$ equals $X_1$), which is $X_0 [\overline{X_2 \oplus X_1}]$.