What should be the size of Mar to address 4GB memory?
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What should be the size of Mar to address 4GB memory?
At first sight, 32 bit address registers seem to perfectly match 2^32 byte = 4 GB of physical memory. However, with 32 Bit you can also address more than 4 GB, like Physical Address Extension (PAE) does. The opposite way round means that you can access 4 GB with less than 32 bit address register.
How many bytes is a memory address?
Each address identifies a single byte (eight bits) of storage.
How many memory locations will be available in 4GB memory?
There is no difference: on a typical system RAM and ROM share the same address space, and each byte needs to have a unique address, so regardless of whether you have 4 GB of RAM or ROM you will need to have 32 address lanes.
What memory is 4 byte?
A collection of 8 bits is called a byte and (on the majority of computers today) a collection of 4 bytes, or 32 bits, is called a word. Each individual data value in a data set is usually stored using one or more bytes of memory, but at the lowest level, any data stored on a computer is just a large collection of bits.
What is the size of the Mar?
At its equator, Mars has a diameter of 4,222 miles (6,794 km), but from pole to pole, the diameter is 4,196 miles (6,752 km). Mars’ radius is, of course, half of planet’s diameter.
What is the size of Mar register?
The Memory Address Register (MAR) in a simple microprocessor needs enough bits for the address. For example, if the address requires 8 bits then the The size of the register needs to be 8 bits wide.
How do I find the memory size of an address?
Step 1: calculate the length of the address in bits (n bits) Step 2: calculate the number of memory locations 2^n(bits) Step 3: take the number of memory locations and multiply it by the Byte size of the memory cells.
How many memory locations are addressed using 18 address bits how many address bits are needed to operate a 2 K 8 bit ROM memory?
8. How many memory locations are addressed using 18 address bits? Explanation: For n address bits, the memory location will consist of 2n bits. Using 18 address bits, 218 = 262,144 (= 256 K) words are addressed.