Memory Unit

What is the Memory Unit?

i. The memory unit is part of the computer that holds data and instructions for processing.
ii. It may also be defined as the storage space in the computer where data to be processed and instructions required for processing are stored.
Although closely associated with the central processing unit, the memory unit is separated from it.

Types of Memories in Computer

Memory is of two types:
i.Primary memory/main memory/internal memory
ii.Secondary memory/auxiliary memory/external memory
Primary Memory:
The primary memory is the memory that can be directly accessed by the CPU, the CPU constantly interact with it, reads instructions stored there and executes them as required.

Types of Primary memory

There are two types of primary memory:
i. RAM
ii. ROM
RAM (Radom Access Memory):
Random access memory also called the Read/Write memory, is the temporary memory of a computer. It is said to be ‘volatile’ since its contents are accessible only as long as the computer is on. The contents of RAM are cleared once the computer is turned off or if there is a power cut.
Types of RAM
1. Dynamic RAM: Dynamic RAM (DRAM) is a type of physical memory used in most personal computers. The term dynamic indicates that the memory must be constantly refreshed (reenergized) or it will lose its contents.
2. Static RAM: Static RAM (SRAM) is a type of RAM that holds its data without external refresh, for as long as power is supplied to the circuit. This is contrasted to dynamic RAM (DRAM), which must be refreshed many times per second to hold its data contents

ROM (Read Only Memory)
Read Only Memory is a special type of memory which can only be read and the contents of which are not lost even when the computer is switched off or if there is a power cut. It typically contains the manufacturer’s instructions.
Types of ROM
i. Programmable Read-Only Memory (PROM): This type of ROM can be re-programmed by using a special device called a PROM programmer. Generally, a PROM can only be changed/updated once.
ii. Erasable Programmable Read-Only Memory (EPROM): This type of ROM can have its contents erased by ultraviolet light and then reprogrammed by a PROM programmer. This procedure can be carried out many times; however, the constant erasing and rewriting will eventually render the chip useless.
iii. Electrically Erasable Programmable Read-Only Memory (EEPROM): These are also erasable like EPROM, but the same work of erasing is performed with electric current. Thus, it provides the ease of erasing it even if the memory is positioned in the computer. It stores the computer system’s BIOS. Unlike EPROM, the entire chip does not have to be erased for changing some portion of it. Thus, it even gets rid of some biggest challenges faced by using EPROMs.

Differences between RAM and ROM

RAM	ROM
It is volatile (It loses data when power is turned off)	It is non-volatile (retains data even when power is off)
Data in RAM can be changed or deleted	ROM is fixed or data cannot be modified.
RAM chip are bigger in size	ROM chip are smaller in size
RAM chips are relatively expensive	ROM chips are relatively less expensive
RAM stores all the applications and data when the computer is running	ROM usually stores instructions that are required for booting the computer

Secondary Memory

This type of memory is a non-volatile memory. It is slower than the main memory. These are used for storing data/Information permanently. CPU directly does not access these memories instead they are accessed via input-output routines. Contents of secondary memories are first transferred to the main memory, and then the CPU can access it.
Types of Secondary Memory
Secondary memory can be of the following types:
Magnetic: Data and information are stored and retrieved using magnetism
Examples
Magnetic disk
Magnetic tape
Hard disk
floppy disks

Optical: Optical storage devices employ light (laser beams) to store and retrieve data and information
Examples
CD
VCD
DVD

Electronic: This holds data and information in the form of electric voltage
Example
Flash Memory

Others include Punch card

Secondary Storage Devices

1. Floppy Diskette
There are two standard sizes used these days which are 5 □((1 )/4) inches and 3 □(1/2) inches, in size, commonly referred to as the mini-floppy and microfloppy. The capacity of 5 □((1 )/4) inches floppy is 1.2 MB and that of 3 □(1/2) inches is 1.44 MB.
2. CD-ROM drives
They are of three types
a. CD-ROM
CD-ROM stands for (Compact Disc Read Only Memory), and it is mainly used to mass produce audio CDs and computer games. Computer users can only read data and music from the discs, but they cannot burn their information onto the discs, from their personal computers.
b. CD-R
CD-R (Compact Disc Recordable) also known as WORM (Write Once Read Many) is a blank disc that users can put into a CD-ROM drive to burn or make a copy of their personal data, music, videos and information. CD-Recordable discs are designed for one-time recording only.
3. CD-RW
The CD-RW (Compact Disc Rewritable) can be erased and returned to its original blank state. New files can then be copied onto the rewritable disk. CD-RW never became as popular as the CD-R because they are not compatible with most disc players to listen to music. They are primarily used to move data from one computer to another or to copy files that are only needed a few times.
3. DVD (Digital Versatile Disk)
DVD is very similar to a CD but it has a much larger capacity. A standard DVD can hold 4.7 GB of data. DVDs are of the same types as CDs along with different formats as explained about CDs.
4. Combo Drives
These drives combine the function of a DVD drive and a CD-RW drive.
5. Hard disk: Largest in capacity

Differences between Primary and Secondary Memory

Primary Memory	Secondary Memory
These devices are temporary (Volatile)	These devices are permanent (Non-volatile)
These devices are expensive	These devices are cheaper
They have less storage capacity	They have storage capacities
They are usually faster	They are slower
Directly accessed by the CPU	Not directly accessed by the CPU
Internal memory	External Memory

Units of Storage in Computer

i. Bit (b): Bits is an acronym that stands for Binary digITS. It is the Smallest Unit of data on a binary computer. A single bit consists of 0 (zero) or 1 (one).
ii. Nibble (nybble, nyble, or nybl): A nibble is a collection of four bits.
iii. Bytes (B): A byte consists of eight bits. It is the smallest item that can be individually accessed by a program. It is the fundamental unit of storage on a binary computer.
iv. Word: A word is a collection 32 bits.
v. Kilobyte (KB): A Kilobyte consists of 2¹⁰ (1,024) Bytes which is approximately 1000 Bytes.
vi. Megabyte (MB): One Megabyte consists of 2²⁰ (1,048,576) Bytes.
vii. Gigabyte (GB): A Gigabyte is a collection of 2³⁰ (1,073,741,824) Bytes.
viii. Terabyte (TB): A Terabyte consists of 2⁴⁰ (1,099,511,627,776) Bytes.
ix. Petabyte (PB): A Petabyte consists of 2⁵⁰ (1,125,899,906,842,624) Bytes.
x. Exabyte (EB): An Exabyte consists of 2⁶⁰ (1,152,921,504,606,846,976) Bytes.
xi. Zettabyte (ZB): A Zettabyte consists of 2⁷⁰ (1,180,591,620,717,411,303,424) Bytes.
xii. Yottabyte (YB): A Yottabyte consists of 2⁸⁰ (1,208,925,819,614,629,174,706,176) Bytes.

Conversion from one Unit to another

The conversion process from one unit to another can be done using the following relationships.
1 bit = 0 or 1
1 nibble = 4 bits
1 bytes = 8 bits
1 word = 16 bits
1 KB = 1024 bytes
1 MB = 1024 KB
1 GB = 1024 MB
1 TB = 1024 GB
1 PB = 1024 TB
1 EB = 1024 PB
1ZB = 1024 EB
1 YB = 1024 ZB

Example 1
Convert 1208bits to byte
Solution
We are required to convert 1204 bits to byte
Let the unknown byte be x
It implies that;
1208bits = xbyte
The relationship between bits and byte is
8bits = 1 byte
Cross multiplying we have
xbyte × 8bits = 1208bits × 1byte
Divide both sides by the coefficient of the unknown unit of storage which is the 8bits
(xbyte × 8bits)/(8bits) = (1208bits × 1byte)/(8bits)
xbyte = (1208 byte)/8
xbyte = 151bytes
Therefore 1208bits = 151bytes

Example 2
Convert 330KB to byte
Solution
We are required to convert 330KB to byte
Let the unknown byte be x
It implies that;
330KB = xbyte
The relationship between kb and byte is
1KB = 1024byte
Cross multiplying we have
xbyte × 1KB = 330KB × 1024byte
Divide both sides by the coefficient of the unknown unit of storage which is the 1 kb

(xbyte × 1KB)/(1KB) = (330KB × 1024byte)/(1KB)
xbyte = 330 × 1024bytes
xbyte = 337920bytes
Therefore 330kb = 337920bytes

Example 3
Convert 14200 KB to MB
Solution
We are required to convert 14200 to kilobyte
Let the unknown kilobyte be x
It implies that;
14200KB = xMB
The relationship between KB and MB is
1024KB = 1MB
Cross multiplying we have
xMB × 1024KB = 14200KB × 1MB
Divide both sides by the coefficient of the unknown unit of storage which is the 1024 byte
(xMB × 1024KB)/(1024KB) = (14200KB × 1MB)/(1024KB)
xMB = (14200 × 1MB)/1024
xKB = 13.87MB
Therefore 1420KB =13.87MB

Example 4
Convert 1.44 MB to Byte (B)
Solution
We are required to convert 1.44MB to B
Let the unknown B be x
It implies that;
1.44MB = xB
The relationship between MB and B is
1MB = 1024 × 1024B
1MB = 1048576B
Cross multiplying we have
xB × 1MB = 1.44MB × 1048576B
Divide both sides by the coefficient of the unknown unit of storage which is the 1MB
(xB × 1MB)/(1MB) = (1.44MB × 1048576B)/(1MB)
xB = (1.44 × 1048576B)/1
x B=1509949.44B
Therefore 1.44MB = 1509949.44B

Example 5
Two storage devices have capacities of 2KB and 2MB respectively:
i. Calculate the number of bits each can hold
ii. putting the two storage together, calculate the total capacity in bytes
Solution
2 KB to bits
Let the unknown bits be x
2KB = xbits
but 1KB = 8 × 1024bits
1KB = 8192bits
cross multiply
xbits × 1KB = 2KB × 8192bits
Divide both sides by 1 kb
(xbits × 1KB)/(1KB) = (2KB × 8102bits)/(1KB)
xbits = 16384bits
2KB = 16384bits

2 MB to bits
Let the unknown bits be y
2MB = ybits
but 1MB = 8 × 1024 × 1024bits
1MB = 8388608bits
cross multiply
ybits × 1MB =2MB × 8388608bits
Divide both sides by 1MB
(ybits × 1MB)/(1MB) = (2MB × 8388608bits)/(1MB)
ybits=16777216bits
2MB = 16777216bits

ii. Putting the two storages together
(16384 + 16777216)bits= 16793600bits
Next is to convert 16793600bits to bytes
Let the unknown bytes be z
16793600bits = zbytes
but 8bits = 1bytes
cross multiply
zbytes × 8bits = 16793600bits × 1bytes
Divide both sides by 8 bytes
(z bytes × 8 bits)/(8 bits) =(16793600 bits× 1 bytes)/(8bits)
z bytes=2099200 bytes
Therefore the two storage capacity in bytes is 2099200 bytes