How many address lines are needed for a 2K 4 memory chip?
11 address lines
How many lines of address bus must be used to access 2048 bytes of memory how many of these lines will be common to all chips?
How many lines of address bus must be used to access 2048 bytes of memory when available RAM chips 128 × 8. How many lines of these will be common to each chip? 128 locations is 7. Since seven address lines can address 27 locations so all seven lines are common to all chips.
How many address lines are required to address each memory location in 16.384 * 16 Memory?
In a system each byte is addressed individually. There are 1024 bytes in one k byte. A 16k ram would therefore need 1024*16 = 16384 addresses.
How many address lines does 256 * 4 memory have?
four data lines
How many address bits are required for a 512 * 4 memory?
210 = 1024, so you need 10 bits to address every byte in a kilobyte. Likewise, you need 20 bits to address every byte in a megabyte, and 30 bits to address every byte in a gigabyte. 232 = 4294967296, which is the number of bytes in 4 gigabytes, so you need a 32 bit address for 4 GB of memory.
How do you 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.
How many address lines would be required to address 64 MB directly?
You have 8 bits for every location, therefore your memory needs a data bus with 8 lines.
How many unique addresses would be required to access 2 to the 16 memory cells?
So 2 to the power of 16 is 65,536, commonly referred to as 64K. The number of memory words addressed on a bus is 2 to the power of the width of the address bus, in this case 16. So 2 to the power of 16 is 65,536, commonly referred to as 64K.
How many address lines are needed for a memory size of 1K locations?
Notice that with ten address lines you can address the magic number of 1024 memory locations, and this is the closest you can get to 1000 using this system – hence 1K = 1024.
What will be the size of memory if address bus size is 22 bits?
13.6 Memory addressing size
| Address bus size | Addressable memory (bytes) |
|---|---|
| 21 | 2M |
| 22 | 4M |
| 23 | 8M |
| 24 | 16M |
How many locations can be selected by 16 address lines?
65536 different addresses
What is the smallest and highest address?
Indeed, if the “smallest addressable unit” is a “4 byte word”, then it would follow that in 2^10 bytes there are 2^8 different addresses, which means that the highest address is 2^8-1 or 255.
What is the highest address written in binary?
99,999
How do you read HEX memory address?
Memory addresses are displayed as two hex numbers. An example is C800:5. The part to the left of the colon (C800) is called the segment address, and the part to the right of the colon (5) is called the offset. The offset value can have as many as four hex digits.
How do you access a value from a pointer?
Steps:
- Declare a normal variable, assign the value.
- Declare a pointer variable with the same type as the normal variable.
- Initialize the pointer variable with the address of normal variable.
- Access the value of the variable by using asterisk (*) – it is known as dereference operator.
How long does it take to read one address of memory?
— The time to read from one address is eight cycles or 80ns. — You can do 12.5 million reads per second, for an effective bandwidth of (12.5 x 106 reads/second) x (8 bytes/read) = 100MB/s. Example Bus Problems, cont. 2) Assume the following system: — A CPU and memory share a 32-bit bus running at 100MHz.