GATE 2020 EC Q.19: solution and answer

GATE 2020 · ECQ.19 · 1 mark · NAT

In an 8085 microprocessor, the number of address lines required to access a 16 K byte memory bank is ______.

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Answer and solution

ANSWER

14 address lines are needed, because a 16 K byte memory has $16 \times 1024 = 2^{14}$ byte locations.

Official answer: 14Source: Official answer key, GATE 2020 EC (IIT Delhi)

The 8085 memory is byte-organised: each address selects one 8-bit location, because the data bus is 8 bits wide. So a $16 K byte$ bank has $16 K$ separate locations to address.
In memory sizes, $1 K = 2^{10} = 1024$ . So the number of locations is $N = 16 \times 1024 = 2^{4} \times 2^{10} = 2^{14} = 16384$ .
$n$ address lines can select $2^{n}$ different locations, so we need the smallest $n$ with $2^{n} \geq N$ : $n = lo g_{2} N = lo g_{2} 2^{14} = 14$ .
Check: the 14 lines $A_{0}$ to $A_{13}$ give addresses from $(0000)_{16}$ to $(3 FFF)_{16}$ , which is exactly $16384$ locations. The 8085's other two address lines ( $A_{14}$ , $A_{15}$ ) are free for chip-select decoding. The answer is 14 address lines.

KEY IDEA

In a byte-addressable memory, $n$ address lines select $2^{n}$ locations. So the number of address lines is $lo g_{2} (number of byte locations)$ , with $1 K = 2^{10}$ . The number of lines depends only on how big the memory bank is. The 8085's total of 16 address lines does not change it.

WATCH OUT FOR

Answering 16 because the 8085 has 16 address lines. That is the whole address space ( $64 K$ ), not what a $16 K$ bank needs. Two other mistakes: counting bits instead of bytes ( $16 K \times 8$ bits, giving 17), and taking $1 K = 1000$ instead of $1024$ .

From the GATE 2020 EC paper. Every question of this paper · Download the full paper · All 9 Processors & memory questions, with notes

GATE 2020 EC Q.19: solution and answer

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