How many address bits are needed to operated a 8k 8 ROM?
8K∗16=213∗16, thus 13 address lines and 16 data lines.
How many address lines are required to decode 8k memory?
Both simply have 8k of addressable words and therefore need 13 address lines.
How many address bits are required for a 2k * 8 memory?
There are k address lines and therefore 2k addresses. Each of these addresses contains an n-bit word. In this instance, 2k = 2048 = 211. You need 11 address lines.”
How many 8k x 4 memory are required to construct a 64K x 8 memory?
How many 8k x 4 memory are required to construct a 64K x 8 memory? So, number of 8k x 4 memory required=2^16÷2^13=2^3= 8. Similarly for number of bits: 2^3÷2^2=2. So, total memory required=8+2=10.
How many number of bits must be stored in a RAM of 8K?
RAM with 8K memory can store 8192 bytes or 65536 bits.
How do I calculate the number of address lines?
If n=2, you can address 2 locations (0, 1, 2, and 3). As you can see, number of addressable locations = n^2. This means that n=log(1024) to the base 2. Thus, n=10.
What are the number of memory required of size 16 x 4 to design a memory of size 64 x 8?
124. What are the number of memories required of size 16 x 4 to design a memory of size 64 x 8? 125.
How many 1024 * 1 RAM chips are required to construct a 1024 * 8 memory System 2 points?
RAM chip size = 1k ×8[1024 words of 8 bits each] RAM to construct =16k ×16 Number of chips required = (16k x 16)/ ( 1k x 8) = (16 x 2) [16 chips vertically with each having 2 chips horizontally] So to select one chip out of 16 vertical chips, we need 4 x 16 decoder.
How many address lines are necessary to access 64K x 8 memory system?
It does not matter what size the memory chips are. If the memory system is 64K x 8, then you need 16 address lines to access it. Perhaps you meant to ask how many address lines per chip. If so, there would be 8 chips and you would use 3 address lines to select which chip, with the remaining 13 address lines being common to all 8 chips.
How many address lines are needed for 8K words?
Let’s suppose computer’s memory is composed of 8k words of 32 bits each. How many bits are required for memory address? Click to expand… I know for 1k we need 10 address lines. So for 2k it would 11. For 4k it would be 12. And for 8k, it should be 13.
How many address lines are needed for dynamic RAM?
For example, 16 address lines would be 64k bytes, assuming 8 data lined. For dynamic RAM, you multiply the number of address lines by 2. The formula is: Also, in the past, DRAM chips were 1 bit. A 64k bit DRAM was using 8 address lines. For 1M×16 memory chips, how many address lines and data lines are required?
How many address lines are needed to access 4MB?
Binary, power of two’s yields the answer: 1 address line = 2 bytes. 2 address lines = 4 bytes. 3 address lines = 8 bytes. 4 address lines = 16 bytes. 5 address lines = 32 bytes. 6 address lines = 64 bytes. 7 address lines = 128 bytes.