Questions

How many chips are needed and how should their address line be connected to provide a memory capacity of 1024 bytes?

How many chips are needed and how should their address line be connected to provide a memory capacity of 1024 bytes?

1024 x 4 means 2^10 address lines. 2K bytes= 2048 x 8 means 2^11 address lines. 2048 = 2^11. Therefore number of memory chip needed= 2^11/2^10=2.

How many chips are needed to provide a memory capacity of?

Since 8 bits = 1 byte, Each RAM chip has 64 x 1 byte = 64 bytes. Thus the number of chips to address a memory capacity of 2048 bytes will be, 2048/64 = 32 chips….Subscribe to GO Classes for GATE CSE 2022.

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How many chips are needed to provide a memory capacity of 16 Kb explain in words how the chips are to be connected to the address bus?

How many chips will be required to obtain a memory capacity of 16 𝐾 bytes? How many chips will be required and how many address lines will be connected to provide a capacity of 1024 bytes? We have to obtain a memory whose capacity is 16 K bytes or ( 16 × 1024 × 8 ) (16 \times 1024 \times 8 ) (16×1024×8) bits.

How many chips are needed to provide a memory capacity of 16k bytes?

A) 16K bytes = 16 x 1024 x 8 => 128 chips. 16 groups of 8 chips which have same…

How many chips are needed to provide a memory capacity of 16 K bytes of a computer uses RAM chips of 1024 1 capacity?

A RAM chip has a capacity of 1024 words of 8 bits each. The number of 2- to-4 decoders with enable lines needed to construct a 16K x 16 RAM system will be ………………….. Correct answer is 5. To construct a memory system with 16K words using 1024×8 RAM chips, we have to connect 16 such chips in parallel.

How the chips are to be connected to the address bus?

RAM and ROM chips are connected to a CPU through the data and address buses. The low-order lines in the address bus select the byte within the chips and other lines in the address bus select a particular chip through its chip select inputs. The connection of memory chips to the CPU is shown in Figure 5.3.

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How many 128 16 RAM chips are needed to provide?

Answer: 16K bytes = 16 x 1024 x 8 => 128 chips.

How many lines must be decoded for chip select specify the size of the decoders?

How many lines must be decoded for the chip select inputs? Specify the size of the decoder. So, we need 4 lines to select the chip. and one for a register address.

How many address lines does a 256K memory chip need?

2^n=256, where n is the number of address lines required, so 8 address bits are needed.

What is number of bits required for address lines to access 512 bytes of RAM?

The connection of memory chips to the CPU is shown in Figure 5.3. This configuration gives a memory capacity of 512 bytes of RAM and 512 bytes of ROM. Each RAM receives the seven low-order bits of the address bus to select one of 128 possible bytes.

How many address lines are needed for a 1024×8 microcontroller?

(i) As given available chips = 1024 x 1 capacity and Required capacity = 1024 x 8 capacity Number of Chips= (1024X8)/ (1024X1) = 8 Number of address lines are needed = 10 (that is 1024 = 210) Here the word capacity is similar ( 1024 ) so similar address lines will be connected to all chips.

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How many address bits are required for 1 RAM chip?

Memory is usually designed to store or retrieve data in word-length quantities. Thus, one RAM chip of 128 ∗ 8 size has 128 words and 8 bits of each word. Thus, 7 address bits are required for 1 RAM chip. We are given 4 RAM chips.

How many chips do I need to reduce the address pin count?

This allowed us to reduce the address pin count from the processor and the memory chip pin count. As to “how many” chips, you have to select a chip. You might select a 1024×1 chip and you will need 8 chips, or a 1024×8 chip, and you will need 1 chip, or a 512×1 chip, and you will need 16, or 512×8 and you will need 2.

How many chips are there in a 16K memory?

Computer Science & Engineering • Solu 4:a) • Memory size is 1024 bytes = 8 x 1024 x 1 RAM => 8 chips • All has same address lines and output is one bit from every chip. • b) • 16K bytes = 16 x 1024 x 8 => 128 chips. • 16 groups of 8 chips which have same address/chip select lines.

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