GATE 2020 IN Q.41: solution and answer

GATE 2020 · INQ.41 · 2 marks · NAT

The address lines $A_{9} \dots A_{2}$ of a 10 bit, 1.023 V full-scale digital to analog converter (DAC) is connected to the data lines $D_{7}$ to $D_{0}$ of an 8-bit microprocessor, with $A_{1}$ and $A_{0}$ of the DAC grounded. Now, $D_{7} \dots D_{0}$ is changed from 1010 1010 to 1010 1011. The corresponding change in the output of the DAC (in mV, rounded off to one decimal place) is ____

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Answer and solution

ANSWER

The DAC output changes by 4.0 mV.

Official answer: 3.5 to 4.5Source: Official answer key, GATE 2020 IN (IIT Delhi)

Step size of the DAC. A 10-bit DAC has codes $0$ to $2^{10} - 1 = 1023$ . The full-scale (largest) output is $1023$ steps, so the step is $Δ = \frac{V _{F S}}{2 ^{n} - 1} = \frac{1.023 V}{1023} = 1 mV$ .
How the processor drives the DAC. $D_{7} \dots D_{0}$ go to $A_{9} \dots A_{2}$ , and $A_{1} = A_{0} = 0$ . So the 10-bit DAC code is the 8-bit data word followed by two 0s: $N_{D A C} = (D_{7} \dots D_{0} 00)_{2} = 4 \times N_{d a t a}$ .
Change in the data word. $(10101010)_{2} = 170$ and $(10101011)_{2} = 171$ , so $Δ N_{d a t a} = 1$ .
Change in the DAC code. The DAC codes are $(1010101000)_{2} = 680$ and $(1010101100)_{2} = 684$ , so $Δ N_{D A C} = 4 \times 1 = 4$ . Each LSB of the processor's data word is worth four DAC steps.
Output change. $Δ V_{o} = Δ N_{D A C} \times Δ = 4 \times 1 mV = 4.0 mV$
Check with the other convention. If you take the step as $V_{F S} / 2^{n} = 1.023/1024 V \approx 0.999 mV$ , you get $Δ V_{o} = 4 \times 0.999 mV \approx 3.996 mV$ . That still rounds to $4.0 mV$ at one decimal place, so the answer is the same either way.

KEY IDEA

The two DAC LSBs ( $A_{1}$ , $A_{0}$ ) are grounded and the processor drives the upper eight bits, so the 10-bit code is always $4 \times$ the data word. A one-count change in the data therefore moves the DAC by $2^{2} = 4$ steps of $1.023/1023 V = 1 mV$ each.

WATCH OUT FOR

Treating the change as one DAC step (1 mV), which ignores that $D_{0}$ drives $A_{2}$ , not $A_{0}$ . Another slip is to work out the step from an 8-bit DAC ( $1.023/255 V \approx 4.01 mV$ ). That lands inside the key range by luck, but it describes the wrong hardware.

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GATE 2020 IN Q.41: solution and answer

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