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Dimas [21]
2 years ago
10

Consider a 512 Mb DRAM with 8 banks, 256k subarrays, and 8 bit word. How are the address bits allocated? How many subarrays are

there and how are they arranged?
Computers and Technology
1 answer:
Softa [21]2 years ago
4 0

Answer:

26 address bits are needed to address 2∧26 locations.

There are 24 = 16 subarrays.

Explanation:

Considering a 512Mb DRAM, it means it can store 512*2∧20 bits = 2∧9X2∧20 = 2∧29. Since it is a 8 bit word, at a time, 8 bits are accessed. Therefore, the capacity can be re-written as 2∧26X 2∧3 = 2∧26X8

It simply means 26 address bits are needed to address 2∧26 locations, since there are 8 banks, 3 address bits are needed to select a bank ( 8 = 2∧3 )

Each subarray is 256K which means 19 bits are needed to select one address from this 256K subarray ( since 256K = 2∧19 )

22 addresses are used up out of a total of 26 addresses (19 for locating a location in 256K subarray and 3 for banks). The remaining addresses are 4 ( 26 - 22 ). These 4 addresses are used to locate a subarray. thus there are 2∧4 = 16 subarrays

There are a total of 8 banks  and within a bank, there are 32 subarrays.

Within each subarray there are 256K locations each of 8 bit size.

Note: " ∧" means raise to power

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Answer:

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// C program to read nth byte of a file and

// copy it to another file using lseek

#include

#include

#include

#include

void func(char arr[], int n)

{

// Open the file for READ only.

int f_read = open("start.txt", O_RDONLY);

// Open the file for WRITE and READ only.

int f_write = open("end.txt", O_WRONLY);

int count = 0;

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{

// to write the 1st byte of the input file in

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if (count < n)

{

// SEEK_CUR specifies that

// the offset provided is relative to the

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lseek (f_read, n, SEEK_CUR);

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// After the nth byte (now taking the alternate

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{

count = (2*n);

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write(f_write, arr, 1);

}

}

close(f_write);

close(f_read);

}

// Driver code

int main()

{

char arr[100];

int n;

n = 5;

// Calling for the function

func(arr, n);

return 0;

}

Explanation:

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