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How many 128 x 4 ram chips are needed to provide a memory capacity of 640 bytes?
To determine how many 128 x 4 RAM chips are needed for a memory capacity of 640 bytes, first calculate the capacity of one chip. A 128 x 4 RAM chip has 128 words, each 4 bits. This equals 128 x 4 bits = 512 bits, or 64 bytes (since 8 bits = 1 byte). To find the number of chips needed for 640 bytes, divide 640 bytes by 64 bytes per chip, which equals 10 chips. Therefore, 10 chips are required to provide a memory capacity of 640 bytes.
Ram memory is measured in bytes how many bytes are there in 2gb of ram memory?
2147483648 bytes
What is the address of the last byte of RAM in the Pep8 memory?
2 bytes or 16 bits
Which are internal bit addressable RAM locations of 8051?
The 16 bytes (128 bits) at internal RAM locations 0x20-0x2F are bit-addressable.
How many 1288RAM memory chips are required to contruct RAM memory system of 2 bytes?
To construct a RAM memory system of 2 bytes using 1288 RAM memory chips, first, determine the capacity of a single 1288 chip. Each 1288 chip typically has 128 bits (or 16 bytes). Therefore, to achieve 2 bytes, you would need 2 bytes / 16 bytes per chip = 0.125 chips. Since you cannot use a fraction of a chip, you would need at least 1 chip to construct the RAM memory system of 2 bytes.
How many bytes in NASA computer that landed on the moon?
See the related link. According to that info, you could say about 2048 bytes RAM (it had 1K of 16 bit words, but a byte is 8 bits). And 12K ROM.
Consider a 256x8 RAM chip a- How many 256x8 Ram chip are needed to provide a memory capacity of 4096 bytes b- How many bits will each memory address contain and address line must go to each chip?
a) To provide a memory capacity of 4096 bytes using 256x8 RAM chips, you need 4096 bytes / 256 bytes per chip = 16 chips. b) Each memory address for the 256 locations in a chip will require 8 bits (since 2^8 = 256). Therefore, each chip will require 8 address lines to select one of the 256 locations.
How many bits are bit addressable in the internal RAM and what is the range of the address?
Of the 128-byte internal RAM of the 8051, only 16 bytes are bit-addressable. The rest must be accessed in byte format. The bit-addressable RAM locations are 20H to 2FH.
What is bit addressable memory in microcontroller 8051?
The bit addressable memory in 8051 is compose from 210 bits: - bit address space: 20H - 2FH bytes RAM = 00H - 7FH bits address; - SFR registers; The following addresses are NOT bit addressable, only 1-byte addressable: - 32 bytes RAM from 00H to 1FH (R0 - R7 registers in all four banks); - 80 bytes RAM general user from 30H to 7FH.
How many ram chips of 256kb required to realize 1 GB?
A hypothetical 32x1 RAM chip provides storage for 32 bits or 4 bytes. 256k bytes require 256 * 1000 * 8 = 2048000 bits (or 256 * 1024 * 8 = 2097152 bits, if the k is interpreted to mean kibibyte rather than kilobyte, using the IEC nomenclature).Because 2048000 / 32 = 64000, you'd need 64000 chips.
How many Mega-bytes are in a RAM?
3 Gigabytes
How many bytes are in 2 BGof RAM?
2.00000000.000000
