It’s organized as x8 (bytes), which means there are only
256 unique addresses
, as the question states. 2^n=256, where n is the number of address lines required, so 8 address bits are needed. The width of the memory bus will only tell the maximum flat memory space a particular processor can address.
How many address lines are needed for memory?
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 addresses are in memory?
With 32 bits, you can store
2^32 distinct
numbers, ranging from 0 to 2^32 – 1. “Byte addressing” means that each byte in memory is individually addressable, i.e. there is an address x which points to that specific byte.
How many address lines are required by 8kB memory?
It is an 8kB and his question uses a 32kB RAM. Both simply have 8k of addressable words and therefore need
13 address lines
.
What are memory address lines?
An address line usually refers to
a physical connection between a CPU/chipset and memory
. They specify which address to access in the memory. So the task is to find out how many bits are required to pass the input number as an address.
How many address lines will a 4K memory have?
12 address lines
are require for 4k memory.
What is the memory address size?
As I mentioned before, it’s the
size of a machine word
. Machine word is the amount of memory CPU uses to hold numbers (in RAM, cache or internal registers). 32-bit CPU uses 32 bits (4 bytes) to hold numbers. Memory addresses are numbers too, so on a 32-bit CPU the memory address consists of 32 bits.
How many memory locations will be available in a 4 GB 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
.
How are memory addresses stored?
They are typically
stored inside the code of the program
. Sometimes the program will want to calculate addresses as well, off of some base value. The addresses are not stored in memory, unless you want them to be.
How many address lines will a 2K memory have *?
It is 2k in size.
11 address lines
are needed to address all the addresses inside the EPROM. A similar calculation reveals that the 2K RAM also needs 11 address lines.
How many address lines are needed for 64Kb memory?
Sixteen address lines
will address 64K bytes. If you count in binary (which computers always do) and limit yourself to 16 binary columns, you can count from 0 to 65,535. (The colloquial ”64K” is shorthand for the number 66,536.)
How many address lines are required by a microprocessor to address 256kb of memory locations?
In total, it works out to be
10 address
bits and therefore, 10 address signals. If static RAM or flash, the size of that memory will be 2^[number of address lines].
How many address lines does a 64Kb memory device have?
The memory you state has a size of 64Kb. A byte is 8 bits and has been since around 1962. So you have 64 Kilobyte memory addressable using
16 address lines
.
How many address lines are required by the memory that contain 16k words?
There are 1024 bytes in one k byte. A 16k ram would therefore need 1024*16 =
16384 addresses
.
How many memory locations are addressed using 18 address bits?
How many memory locations are addressed using 18 address bits? Explanation: For n address bits, the memory location will consist of 2
n
bits. Using 18 address bits, 2
18
=
262,144 (= 256 K)
words are addressed.
How many address and data lines will be there for a 16M 32 memory system Mcq?
How many address and data lines will be there for a 16M x 32 memory system? d. None of the above Correct solution is (c). Since there are 16M words, the number of address lines will be
24
, since 224 = 16M.
How many address lines are necessary on the chip of 4K byte memory?
Detailed Solution. So,
12 bits
are needed to address 4k memory locations.
How many address lines are necessary to address two gigabytes of memory write all the steps followed to reach your answer?
Answer: It requires
21 address lines
to address two megabytes of memory.
How many address lines are required to represent 32K memory?
4K bytes) has 12 address lines. A JEDEC 27256 EPROM (32K bytes) has
15 address lines
. A JEDEC 27512 EPROM (32K words, aka. 64K bytes) also has 15 address lines, but holds twice as much as a 27256.
How many memory cells does a 4 Gbit RAM has?
Over time the size of the actual DRAM cells (a modern 4 gigabit DDR3-SDRAM chip has 4 * 2^30 of these, and a single-sided 4 gigabyte DIMM will have
eight
of these chips) has been dwarfed by the control electronics that are necessary to provide the external interface.
How many memory locations are identified by a processor with 8 bits address bus at a time?
The facile answer is “
256
” because 8 bits can only have 256 different values.
What type are memory addresses?
The data type of a memory address is
a pointer
, which is denoted by the type that it points to, followed by an asterisk ( * ).
How many memory locations does 8k memory capacity have?
8k= 2^13. So 2^13 address lines are required. So, number of 8k x 4 memory required=2^16÷2^13=2^3=
8
.
How many address lines are needed for a memory size of 1K locations?
To see why this is natural first ask yourself how many memory locations can be addressed using a single address line? 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.
How many address lines a memory chip of 128k capacity will have?
for 128kb, you need 10 bits for each k and then 7 more bits. and thats for the address and if has
8 lines
means its 128 kBytes.
How many address lines are needed for 16kb ROM?
→ it means
20 address lines
and 16 data lines. therefore, general formula to find out ROM memory size is 2^m * n, where m is address lines and n is data lines.
What is 64Kb memory?
64 Kb means =
2 ^16 bytes
. Addressability concerns the different number of bytes we can point to. So 8 bytes = 2 ^3. If we can address each 8 bytes out of 64Kb, that means= 2^16 / 2^3= 2^13 is your answer.
What is the maximum number of memory locations that can be addressed by a chip with 16 address pins?
And that limits the amount of memory that can be accessed. For example, with a 16-bit address register, the address bus has 16 bits to address RAM and ROM. With 16 bits, a maximum of
2
16
= 65,536 words
can be addressed.
How many words memory can accommodate if address space is of 16 bit?
You can address 2^16 words and each word is 8 bit (= 1 byte). Therefore it is 64 KB. If the word size was 16 bit. The answer would be
128 KB
.
How many bits of storage does a memory system contain if it has a 16K address space and is 8 bit addressable?
If each memory location holds one byte, the addressable memory space is 4 GB. Based on your input, 16k is the memory location with 8 bits(1byte) of data storage at each memory location. Hence the address bus length is
14 bits
.
How many address lines will be required to access all memory locations of a 16 KB memory chip?
A 16-bit microprocessor has
twenty address lines
(A
0
to A
19
) and 16 data lines. The higher eight significant lines of the data bus of the processor are tied to the 8-data lines of a 16 Kbyte memory that can store one byte in each of its 16K address locations.
How many address pins are needed to address 256K DRAM?
Thus with
9 pins
we can address 256K locations. NVRAM is random-access memory that retains its information when power is turned off (non-volatile). This is in contrast to dynamic, random-access memory(DRAM) and static random-access memory (SRAM), which both maintain data only for as long as power is applied.
How many address lines and input output data lines are needed in each case?
How many address lines and input-output data lines are needed in each case? (a) 32 x 8 32 = 25, so 32 x 8 takes
5 address lines
and 8 data lines, for a total of 5 + 8 = 13 I/O lines.
Is 256 a special number?
A byte represents 256 different values
. So that’s it. A byte is a unit of storage in a computer which contains 8-bits and can store 256 different values: 0 to 255. … In any case, 256 is special since it represents the most common base unit in a computer.
How many address lines are required to address each memory location in 16384 * 16?
The total number of addresses in use in the system described would be 16384+16*2^4 for a total of
16640 addresses
. In order to address 16640 addresses, you need an address bus with at least 15 address lines, which could address a maximum of 32768 addresses, so 16128 addresses would be unused.
How many address lines are needed for memory?
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 are needed to address each machine location in a 4096 4 memory chip?
ANSWER:
11
It means that a memory of 2048 words, where each word is 4 bits.
