Memory Address and Capacity MCQ Quiz - Objective Question with Answer for Memory Address and Capacity - Download Free PDF

Explanation:

A memory chip has a capacity of 4 kB. To determine the number of address lines required, we need to understand the relationship between memory capacity and address lines. The capacity of a memory chip is the product of the number of addressable memory locations and the size of each memory location.

Firstly, 4 kB (kilobytes) needs to be converted into bytes. Since 1 kB = 1024 bytes, we have:

4 kB = 4 × 1024 bytes = 4096 bytes

Each memory location in a typical memory chip stores 1 byte of data. Therefore, the memory chip with a capacity of 4096 bytes has 4096 addressable memory locations.

The number of address lines required (n) can be determined using the formula:

Number of addressable memory locations = 2ⁿ

Setting this equal to the number of memory locations we have:

2ⁿ = 4096

To solve for n, we take the logarithm base 2 of both sides:

n = log₂(4096)

Since 4096 is a power of 2 (specifically, 2¹²), we have:

n = 12

Therefore, 12 address lines are required to address 4096 memory locations.

Hence, the correct answer is option 2, which is 12.

The correct answer is Programmable read-only memory

Key Points

• PROM stands for Programmable Read-Only Memory.
• Read-only memory (ROM) is a type of memory that is used in electronics and computers.
• It is a non-volatile memory, that is, information stored in such memory is permanent and cannot be changed.
• The other two types of ROM are
  • Electrically Erasable Programmable ROM (EEPROM)
  • Erasable Programmable ROM (EPROM)

Additional Information

• Primary memory can be divided into RAM (Random Access Memory) and ROM (Read Only Memory)

Explanation:

The Remaining Address Line of Bus:

Definition: In a computer system, the address bus is responsible for carrying the addresses from the CPU to the memory or other devices. This bus is used to specify the location where data is to be read from or written to. The remaining address line of the address bus, after decoding, is used to generate the chip select signal. This signal is crucial for enabling the specific memory chip or peripheral device that the CPU intends to communicate with.

Working Principle: The address bus consists of multiple lines, each line representing a bit in the address. The decoding process involves interpreting the binary address to select a specific memory location or device. The remaining address line, after decoding, is utilized to generate the chip select signal, which activates the appropriate device for communication.

This option correctly identifies that the remaining address line of the address bus is used to generate the chip select signal. The address bus carries the address information, and the decoding process interprets this information to generate the necessary control signals, including the chip select signal.

Explanation:

Microprocessor 8085

Introduction: The 8085 microprocessor is an 8-bit microprocessor designed by Intel in the mid-1970s. It is widely used in various computing applications and known for its simplicity and ease of understanding. One of the fundamental aspects of any microprocessor is its addressing capability, which determines how much memory it can access.

Address Lines: The number of address lines in a microprocessor determines the range of addresses it can generate. Each address line can be in one of two states (0 or 1), and with 'n' address lines, the total number of unique addresses is 2ⁿ. Therefore, if a microprocessor has 16 address lines, it can generate 2¹⁶ unique addresses.

Memory Access: The total memory that a microprocessor can access is directly related to the number of unique addresses it can generate. In the case of the 8085 microprocessor, it has 16 address lines. Thus, it can generate 2¹⁶ = 65536 unique addresses. Since each address corresponds to one byte of memory, the 8085 microprocessor can access 65536 bytes, or 64 kilobytes (KB) of memory.

This option correctly states that the 8085 microprocessor has 16 address lines, allowing it to access 64 kilobytes of memory

The correct option is 3

Concept:

To determine the minimum number of bits needed to address 2000 memory locations, we can calculate the binary logarithm (log base 2) of 2000 and round it up to the nearest integer. This will give us the minimum number of bits required to represent 2000 unique memory locations.

Calculating the binary logarithm of 2000:

log₂(2000) ≈ 10.965784

Rounding up to the nearest integer:

⌈10.965784⌉ = 11

Therefore, the correct answer is 11.

The Correct Answer is "Cache Memory".

Important Points

Cache Memory :

• Cache Memory is a special very high-speed memory.
• It is used to speed up and synchronizing with a high-speed CPU. Cache memory is costlier than main memory or disk memory but economical than CPU registers.
• Cache memory is an extremely fast memory type that acts as a buffer between RAM and the CPU.
• It holds frequently requested data and instructions so that they are immediately available to the CPU when needed.
• Cache memory is used to reduce the average time to access data from the Main memory.

Additional Information

Secondary Memory​ :

• It is non-volatile, i.e. it retains data when power is switched off.
• It is large capacities to the tune of terabytes.
• It is cheaper as compared to the primary memory.
• Depending on whether the Secondary memory device is part of the CPU or not, there are two types of secondary memory – fixed and removable.

 Auxiliary Memory :

• Auxiliary memory is the non-volatile memory lowest-cost, highest-capacity, and slowest-access storage in a computer system.
• It is where programs and data kept for long-term storage or when not in immediate use.
• Such memories tend to occur in two types-sequential access (data must access in a linear sequence) and direct access (data may access in any sequence).
• The most common sequential storage device is the hard disk drives, whereas direct-access devices include rotating drums, disks, CD-ROMs, and DVD-ROMs.
• It used as permanent storage of data in mainframes and supercomputers.

Virtual Memo :

• A computer can address more memory than the amount physically installed on the system.
• This extra memory is actually called virtual memory and it is a section of a hard disk that's set up to emulate the computer's RAM.
• The main visible advantage of this scheme is that programs can be larger than physical memory.
• Virtual memory serves two purposes.
  • First, it allows us to extend the use of physical memory by using the disk.
  • Second, it allows us to have memory protection because each virtual address is translated to a physical address.

In a DRAM, a capacitor is used to store a bit of data along with a MOSFET (transfer device) which acts as a switch.

The circuit is as shown:

In a DRAM:

• Periodic refreshing is required.
• The information is stored in a capacitor.
• Both read and write operations cannot be performed simultaneously.

For an SRAM, the storage element is a  flip-flop.

A comparison of SRAM and DRAM is as shown:

Concept:

1 kB = 1024 bytes

One memory location of computer occupies 1 byte of memory.

So, to occupy 32 k of memory = 32 × 1024 = 32768 bytes are required.

∵ 1 byte = 1 memory location

∴ 32 k memory unit of computer will require 32768 memory locations.

Concept:

• The maximum number of bytes directly addressable in a computer depends on the number of bits in the memory address.
• For an n-bit address bus, the memory can store:

          M ≤ 2n Bytes, since an 8-bit represents a byte.

Calculation:

Given Address lines, n = 9

The maximum number of addressable locations in memory using 9 address lines

M = 2⁹ 

M = 512 addressable locations

Given the size of memory = 4k

1k represents 1024 memory locations represented as:

1024 = 210

So, 12 bits are needed to address 4k memory locations.

Concept:

Address line: An address line is basically refers to a physical connection between a CPU/Chipset and memory. They specify which addresses to access in memory. When there are k address lines, then 2k memory word can be accessed.

Data line: Data lines provides the information to be stored in memory. It represents the number of bits in the word.

Calculation:

Here, it is given that memory unit size = 256 bytes

2k = 256

k = 8

Concept:

An address line usually refers to a physical connection between a CPU and memory.

It specifies which address to access in the memory.

For an ‘n’ bit address line, we can access 2ⁿ memory locations.

Application:

Given, the number of memory locations = 16 kB

i.e. 2ⁿ = 16 kB

2ⁿ = 16 × 1 kB

2ⁿ = 2⁴ × 2¹⁰

2ⁿ = 2¹⁴

Calculation:

Required memory system size = 16K bytes

Memory chip size = 2¹² × 4 (Given)

Required size = 16K bytes = 2⁴ x 2¹⁰× 8 bits

Concept of memory:

Any memory size is given by = 2^K × m

K = address line

m = data line

Eg: 1 KB memory = 2¹⁰ × 8

Concept of decoder:

For n × 2ⁿ decoder no. of AND gates required are 2ⁿ.

Eg: 2:4 decoder,  4 AND gates are required.

Analysis:

Given

For this memory 15 × 2¹⁵ decoders required

Since n = 15

So no. of AND gate are required = 2¹⁵

The correct option is 1

Concept:

DRAM:

•  In a DRAM, a capacitor to store a bit of data is used along with a MOSFET (transfer device) which acts as a switch.
• It requires only one transistor to store one-bit data whereas SRAM requires 6 transistors.
• The information is stored in a capacitor.
• Both read and write operations cannot be performed simultaneously.
• To store one bit of data, DRAM needs one transistor and one capacitor.
• Since the capacitor stores the data in the form of charge.
• To keep the charge stored for a long duration, the capacitor needs to constantly be refreshed. So DRAM is slower and used for main memory.