Question

A main memory unit with a capacity of 4 megabytes is built using 1M ×1-bit DRAM chips. Each DRAM chip has 1K rows of cells with 1K cells in each row. The time taken for a single refresh operation is 100 nanoseconds. The time required to perform one refresh operation on all the cells in the memory unit is 
1. 100 nanoseconds
2. 100*2¹⁰ nanoseconds
3. 100*2²⁰ nanoseconds
4. 3200*2²⁰ nanosecond

Answer 1

Correct Answer - Option 2 : 100*2¹⁰ nanoseconds

The correct answer is option 2

Data:

The capacity of the main memory unit = 4 MB =4× 2²⁰× 8 bit

Size of each DRAM chips = 1M ×1-bit =1× 2²⁰ bit

Time for one refresh operation refreshes one row of memory chip =100ns

Number of rows = 1K =1× 2¹⁰

Cells in each row = 1K = 1× 210

Explanation:

Number of chips required for 4MB =\({{ 4\times 2^{20}\times8 bit}\over{1\times2^{20}}}\)=2⁵ = 32

Now all chips can be refreshed in parallel

It is to be noted that a row in all chips in series can be refreshed in one refresh cycle. This makes the total time to refresh the 4MBytes of memory as same as that of one chip.

In a chip is 1K =1× 210

Given time for one refresh operation refreshes one row of memory chip =100ns

total 2¹⁰ rows in a chip

all chips can be refreshed in parallel

So, the time required to refresh the rows in one chip = time required to refresh all the 32 chips

the time required to refresh the rows in one chip = time required to refresh the main memory

So, the time required to refresh the main memory = 100 * 2¹⁰ ns