Electronics and Communication Engineering - Digital Electronics - Discussion

44. 

A memory system of size 16 k bytes is to be designed using memory chips which have 12 address lines and 4 data lines each. The number of such chips required to design the memory system is

[A]. 2
[B]. 4
[C]. 8
[D]. 18

Answer: Option C

Explanation:

.


Anilkumar said: (Jun 3, 2014)  
What is the exact formula used?

Athi said: (Jun 18, 2014)  
How did the answer come?

Daleep said: (Aug 3, 2014)  
Logic behind this calculation?

Krishna said: (Jan 15, 2015)  
16K bytes = 16*1024*8.

12 address lines = 2^12 bytes.

4 data lines.

Now the simplification is direct.

Sai said: (Aug 19, 2015)  
16k means 16*1000 but you used 1024 why?

Jayasankar said: (Dec 9, 2015)  
16k means 16*1024 but how 8 will comes?

Shikha said: (Feb 6, 2016)  
Because in digital byte means 8.

= 2^10 = 1026 is called kilo.

Kumar said: (Aug 4, 2016)  
2^16 = 2^10 * 2^6.
1024 * 64 BYTE.
1024 * 64 * 8 BIT.

Preethi said: (Aug 30, 2016)  
The numerator is in bits and denominator is in bytes. How could it be possible?

Dharwendra Pandit said: (Mar 22, 2017)  
1 kilobyte = 1024 byte.
1 byte = 8 bit.

So 16 * 1024 * 8 bit.

Aswathy said: (Mar 25, 2017)  
Number of chips required= Required size/given size.

Required size = 16 Kb
1kb = 1024 bytes
1 byte = 8 bits
therefore 16 KB = 16 * 1024 * 8 bits.

Now, we have 12 address lines and 4 data lines,
12 address lines means 2^12 bytes which is equal to 4096 bytes.

Generally,
1024 * 8 means that 1024 locations are there and each location can store 8 bits.
1024 * 2 means that 1024 locations are there and each location can store 2 bits.

Data lines means no of bits.

Therefore as per the question given above only 4 data lines are there which means only 4 bits can be stored.

So, 4094 * 4 indicates 4096 location are there and each location can store 4 bits.

therefore the answer to the question is

16*1024*8bits/4096*4bits

Twinkle said: (May 11, 2017)  
Thank you @Aswathy

Kapil said: (Sep 3, 2017)  
With the help of 12 address lines, we can create 2^12 data locations.
Each data location has 4 bits.

So the size of memory in each chip = no. of data locations * size of each data location
= 2^12 * 4 bits.

The required memory system is 16KB i.e. 16 kilo bytes = 16 * 1024 * 8 bits.

The Number of chips required = Total memory required / memory of each chip.
= 16 * 1024 * 8 bits / 2^12 * 4 bits
= 8 chips.

Guru said: (Aug 3, 2019)  
Thank you @Kapil.

Manivannan said: (May 5, 2020)  
Thank you @Aswathy.

Sridevi said: (May 7, 2020)  
Hello all,

The memory system size to be designed is 16kb in total.
Given memory, chip specs is 12 address line and 4 data lines.
Ques: How many memory chips with given specs is required to design a 16 kb system?
Ans: No.of mem chips =
System size/ given specs size.

= 16 * 1024 * 8 bits / 2^12 * 4 bits.

Numerator: 16 kb can be written as 16* (1k= 1024)*(1byte= 8 bits)
Denominator: (address lines in powers of 2)*(4-> Data lines refer to no of bits).
Final answer= 8 chips.

Aaryan said: (Jul 26, 2020)  
Required Total Size of memory = 16KByte = 16 *1024 *8 = 2^17
Available memory chip with 12 address line and 4 data lines so available memory size = [2^12] *4 = 2^14.

No of chips required = required total size of memory/available chip size = [2^17]/[2^14] = 2^3 = 8.

Deepa said: (Sep 16, 2021)  
Thank you @Aswathy.

Deepa said: (Sep 16, 2021)  
Thank you @Aswathy.

Post your comments here:

Name *:

Email   : (optional)

» Your comments will be displayed only after manual approval.