hello students
today's video is again about construction of bus system
today we wil see how bus system can be constructed with the help of multiplexers
here i got one question to design a common bus system using multliplexers which consist of 8 registers
and each register have 3 bits stored in them. Show the transfer of data from any of these register to another 3 bit register with name as M
So i have draw a diagram which i will discuss later
first we will discuss the digram given in your Maris Mano book
so here bus system for 4-bits is constructed
so here you should focus that there are four registers of 4-bit each
how many bits are there in register , that many bits of common bus system can be constructed
the capacity of coomon bus system is equal to the capacity of the registers
that much lines are there in bus
here we have taken  four multiplexers because we are having four registers
these four registers will give input to four multiplexers
and  the multilpexer will produce any one output
next we will see selection lines, these are two in number
because we are having four registers, and to select between 4  choices we should have 2 selection lines
because 2 to power 2 is equal to 4. if we have to select from 4 choices we should have 2 selection lines
if we have 8 choices then we require 3 selection lines
so we sould understand this logic
what multiplexer does here
here we are having 4 multipleser and these are 4 by 1 multiplexers
because its taking 4 inputs and gives one output
inputs are from register and output is to produce one  line for common bus
the name of the registers are A,B,C and D
so A0,A1,A2,A3, B0,B1,B2,B3,C0,C1,C2,C3 AND D0,D1,D2,D3 are data inside the registers A,B,C and D respectivly
this data is going as a input to multiplexers
so if we talk about MUX0, so here we are having A0,B0,C0 and D0 as a inputs
that means MUX0 is having first bit of each register as a input
MUX1 is having 2nd bit of each register as a input  means  A1,B1,C1 and D1 as a input
Third multiplexer MUX2 is having 3rd bit of each register that is A2,B2,C2 and D2 in input
and 4th multiplexer MUX3 is having A3,B3,C3 and D3 as inputs
now i have draw diagram according to requested question
Design a common bus system using multiplexer which consists of 8 registers and each register have 3 bits each
so we need to take 8 registers
so i have taken register A,B,C,D,E,F,G, and H
Each register is having 3 bits 0,1,and 2
A0,A1 and A2 are bits of register A
like it B0,B1,B2,C0,C1,C2, are bits of different registers
we need three multiplexers , ok, because we are having 3-bits only in registers
and we can construct 3-line common bus system
we are having 8 registers so we need 3 selection lines because 2 to power 3 is 8
so slection lines are S0,S1 and S2
these selection lines will go to in each multiplexer
which tells each 8by1 multiplexer that which input should  go as a output of that multiplexer
so it will decided by the selection lines
and  this will generate three line coomon bus system
its was an easy construction
now we will discuss one general  instructions to construct common bus system using multiplexer
so that you can understand any question related to  this topic
so its also given in book
in general a bus system will multiplex k registers of n bits each  to produce an n-line common bus
so you have to construct common bus with n-bit registers , so your common bus bill contain n lines only
as we have seen that we are having 3-bit registers so we constructed 3 line common bus
if we have 4-bit register than we can construct 4 line bus system
and the number of multilexers needed to constrcuct the bus is equal to n, the number of bits in each register.
so when we have 3-bit register then we need 3 multiplexers
if you have 4 -bit registers then you need 4 multiplexers
number of multiplexer is equal to number of  registers
so multiplexer size kby1 where k is the number of registers
so if we are having 8 registers then we need 8by1 multiplexers
when we have 4 resisters then we need 4by1 multiplexers
because data lines or inputs of multiplexers are coming from registers
so if 8 registers are there then 8 lines will go to the multiplexer as input
for example , a coomon bus for eight register of 16 bits each require 16 multiplexers
one for each line in the bus and 16 line bus will be constructed
each multiplexer will have 8 data lines because  we are having 8 different registers here
and size of multiplexer is 8by1
so you can solve any question like this
thanks for watching
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