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Practice English Speaking&Listening with: Lecture 11 Arithmetic Circuits

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So today we will look at arithmetic circuits, computers of course are the most important

digital systems at least it is the most visible digital system. Whenever you say digital immediately

what comes in your mind is the computer and computer consists of all these arithmetic

units, of course there is a lot of control also, arithmetic units have to be controlled,

when does an arithmetic unit work, where does it take the input from, where does it write

the output, where does it send the output to, so all those things are there, the control

aspect of computing.

We are not discussing that here but it will be in a future course. But in this course

we will see some of the basic circuits which perform arithmetic in computers. When we say

computer we need not imagine a big Pentium P3 P4 but it can be anything even a much simpler

computing circuit.

So far we used logic gates in order to implement some of those circuits like parity generator

for example code converters etc. whatever is the circuits definition input output specification

we finally reduce it a Boolean equation minimum sum of products or product of sums implemented

using gates and gates are basically logic devices, logic circuits wherein you have your

true or false inputs and true or false outputs. So when some combination of true or false

inputs gives rise to a specific output being a true or false.

How you translate this logical operation into arithmetic operation is what is required in

a computer. In computer we give numbers as inputs and we want numbers as the outputs

so it is how you convert or how you transform these logical gates into arithmetic circuits.

Again it is simple because of the binary nature of the numbers we are representing. Because

we talk about binary representation in numbers, however big the number is we have to translate

it into binary format binary representation 0s and 1s. So as long as of 0s and 1s are

inputs of circuits we can have an output which is also 0 or 1.

You can interpret it as logical signals and logical inputs logical outputs or interpret

it as arithmetic circuits whose values can be 0 and 1 and the output as 0 or 1. So it

is a question of interpretation of the same logic gates whether it is an arithmetic circuit

or a logical circuit the simple example will be a gate in which I have two inputs have

0 and 1 if I want to sum of these I want to sum of 0 and 1 so the output can be 1 and

if both are 0 the output has to be 0. So this is the type of arithmetic operation we will

perform using the same logic gate. So the simplest of arithmetic operation is an adder,

of course all of us to know that.

Let us say we have two numbers A and B and we want C which is sum of A and B. this is

the sum operator not the OR operator, the arithmetic sum. Since the numbers are binary

each of them can only take the value of 0 or 1. So I have four values if both are 1

what should be the output? The output sum is 2 which cannot be represent with one binary

digit because binary digit can only represent 0 or 1 so I know the number will be 0 sum

is 0, it is truncating it is something like I am trying to add two single digit numbers

and I have only a single digit to represent my result and if the number exceeds the single

digit number naturally it will only show the first digit and the second digit will be discarded.

Suppose I have a single digit decimal representation both inputs and outputs as long as the number

is less than 9 I can get the actual value but if the sum is more than 9 then the first

digit only will appear the second digit will be discarded. Suppose I have two numbers added

5 and 5 and the answer is 10 but 10 cannot accommodate my single digit output but only

represent the first digit 0 and the second digit 1 will be discarded. so that way when

we look at it that the sum is 0 but we need to take care of fact that the sum is 0 not

because both were 0 but because both were 1 so I need to have one more output so I will

not even call this C now I will call it S as a sum.

So S is a sum and I will call this C as the carry so S stands for sum and C for carry.

The carry is when you have a digit which is overflowing that overflow has to be accommodated

elsewhere in the next computation stage. The next time you compute you have to take into

account this overflow before you can proceed with the addition. So I need a carry which

is only one in this case but in all other cases it is 0. It is a simple adder so I have

two inputs and two outputs so this is going to be my adder circuit with two inputs A and

B, two outputs S and C adder and you can very easily see that sum is Exclusive OR of A and

B and C is AND of A and B. So without having to go through the Karnaugh Map and Boolean

algebra, reduction and everything I can look at the truth table and write the sum as A

Exclusive OR B C as this. So basically this circuit is nothing but this, and the carry

is nothing but a gate like this.

So an Exclusive OR takes this A and B and gives the sum and the AND gate takes this

A and B and gives a carry. Now this is one bit addition, suppose no number is small enough

to be represented by one bit in that case you dont need a computer, you dont need

a calculator; you dont need an electronic circuit.

Suppose all you can do with a computer is add 1 and 1 and nothing more than that so

you dont need a computer so I need to naturally represent the numbers by larger number of

bits so the whole range of decimal number I want to operate from 0 to infinity as infinity

is only a fixatious quantity any large number you want to deal with you want to add only

then you need a computer you need a calculator you punch these numbers to get the output

when the number is large and not when the number is very small. somebody asks to get

three and four do you need a calculator for that you dont, may be after sometime but

right now at least people can add single digit without having to resolve through calculator,

earlier it used to be large numbers.

But now we have to depend on computers and calculators so much but for single digits

we still dont need calculators but then we dont need these circuits. So naturally

this one bit is only one of the several bits that will be in your number of representation.

Hence this A and B will be really multi-bit representation of a decimal number. we have

gone through the decimal to binary representation so when I have a given number A and B are

both decimal numbers of several digits the first thing you do is convert them into binary

and then add those numbers so reasonably assume that A and B are both multi-bit numbers. That

means when I add A and B the first digit let us say I have multi-bit A and B, A and B are

multi-bit numbers so to start with I will call them 4-bits. Now I represent A as A0

the lowest significant bits A1 A2 A3 and B as B0 B1 B2 B3 and I will have the sum as

S0 S1 S2 S3 and then I may have a carry from A3 and B3 so there will be a carry out also

in addition to this.

Therefore this is sum, A, B, sum and a carry Co is for output carry. So when I consider

A0 and B0 first two bits of A and B first bit of A and first bit of B naturally there

is no carry because it is the very first bit from A and very first bit from B so you can

use this hardware. But when I try do the addition of A1 and B1 I need to consider not only A1

B1 but I need to also consider the carry from A0 B0. So the carry from here has to flow

into this and carry from here has to flow into this carry from here has to flow into

this and the final carry is what we say here. So this is thecarry inand this is

thecarry out’. So this carry is a carry out the output carry which has to be fed as

the input carry in the next bit addition. In the next stage of the binary arithmetic

I need to take the output carry from the previous stage and feed it as the input carry. So C

out would be Cin for the next stage so this adder cannot do the job this is having limitation

and it can only take two numbers and not the third number and such an adder is called a

half adder.

A half adder is the one which can only take two digits and add the sum and get the carry

but not take into account the carry from the previous position of the addition. I need

to expand my circuit into a circuit in which I will take my A B and Cin the Cin being the

carry from the previous bit position and get my sum Co Cin is the input carry and Co is

the output carry. Input carry comes from the previous position if this is the nth bit this

is n minus 1th carry out of the previous bit. So, if I generalize it Ci Ai and Bi are the

ith bit of A and ith bit of B this is Cin which is nothing but Co Ci minus 1 from the

previous position. I get Ci minus 1 and I get sum ith bit carry ith bit and this Ci

will become in the next stage carry in. Such an adder which can take into account the carry

from the previous position is called a full adder

which is only of practical use. This is not a practical use except for the very first

bit position. The very first bit position in a large multi-bit number the very first

bit can be added using half adder but all other bits have to be added with a full adder.

For a full adder you can connect the carry to Co so I dont want to have two types

of circuits suppose I have 16-bit numbers I will have sixteen As and sixteen Bs so instead

of having fifteen full adders and one half adder I can have sixteen full adders for uniformity

sake, the very first full adder I will make the carry input as 0 then I will have uniformity.

So what is the behavior of this full adder? How does it look like? Full adder truth table:

We will call A and B C input sum and C output. Sum is 0 1 1 0, 1 0 0 1, carry is 0 0 0 1,

0 1 1 1. You have look at this and write it very easily. For example, if you take this

there are two bits 1 and the third bit is 0, two bits will lead to 2 and 2 is 1 0 so

sum is 0 carry is 1. all you have to do this count the number of 1s A B C. in A B Ci count

the number of 1s if it is a 0 you put a 0 if it is 1 you put 1 and if it is 2 you put

a 0 and the one gets the next one, if it is 3 you put 1 and 1 in the next one thats

all. So the number can be all the three can be 0s all the three can be 1s, this only a

binary representation of the number of 1s.

The output is nothing but the binary representation of the number of 1s in the truth table. The

only thing is this if all are 0s we have 0 0, take this for example there are two 1s

and the output is 1 0 binary representation of 2 is 1 0. This is 3 and the binary representation

of 3 is 1 1. This is how we write the truth table, it is very easy number of 1. If it

is 0 you put a 0, if it is 1 you put 1, if it is 2 you put a 0 1, if it is 3 you put

1 1.

Therefore now I can write the truth table for this. If the truth table is there I can

put the Karnaugh Map for this and get the simplification and draw the logic for my full

adder. I am going to give this an exercise, I am go to draw the Karnaugh Map, simplification

I am going to leave it as an exercise to you which is very easy it is there in all the

books, we have the truth table A B Ci this is sum so first row is 0, Ci is 0 1 0 1 and

1 0 1 0 this is the truth table and the Karnaugh Map. This is the Kmap for S, the first

row is 0, second row is 1, third row is 1, fourth row is 0, fifth is 1, sixth is 1, seventh

is 0 and eight is 1.

We know how to draw. We can simplify it apparently there are no prime implicants but remember

we had this type of pattern in our parity generator, we had this pattern also in one

of those converters gray code converters so this is the Exclusive OR solution. I will

leave it as an exercise to you to get this expression. I will give the expression to

you it is A Exclusive OR B Exclusive OR Ci please get this expression as a homework,

this is very easy, the pattern is there. The pattern has been identified earlier in our

parity checker parity generator. And the carry out is 0 0 0 1 there are four 1s. If you try

to join them it will be three prime implicants all essential three essential prime implicants

we can ADD them or OR them really not ADD them. Co would be AB ACi BCi or AB Ci A OR

B this is the carry output. If Ci is the carry input Co is the carry output and A and B are

the bits. it can be applied to any bit Ai or Bi it can be the third bit or fourth bit

or fifth bit and if it is the fifth bit then it is the fifth bit input of A, fifth input

of B, fourth bit output of C, and this will be the fifth bit output of C. Ci is the one

bit before the current bit, Co is that current bit output for carry.

I want to give this as an assignment. I can also write it as Ci A Exclusive OR B. We will

use it sometimes not now. Occasionally we will use this expression for carry and not

this expression. That means this should be equivalent that means this should also be

right. I can tell you intuitively without even going through Boolean algebra how this

can be written, when you have A B you are taking these two terms into account, this

one and this one have been considered in getting this expression, A B is same as this and this,

I have this one extra 1 and this extra one and in order to make my prime implicants smaller

my sum of product expressions smaller what did I was to combine this 1 with A1 and combine

this 1 with the 1. Remember this one is already been considered this expression so what I

am saying again I will do it here one more time. Consider this carefully.

I have already considered this as AB BCi, AB is what? Co is 0 0 0 1, BCi is 0 1 1 1

so when we consider this as BCi AB we will first take AB which is AB? AB let us say this

has already been considered as AB I have these two 1s left I combine this 1 with the 1 2

make my product terms less by one literal. Whenever I combine two 1s my product term

gets reduced by one literal. Every time I combine a factor of 2 one literal goes out,

4 if I combine two literals go out, if I combine eight terms three literals go out. So that

was the reason why I have to make this and this even though this was not necessary to

have been combined this with this and this because these has already been take into account,

now if I consider these two terms separately what is this? This is A B bar Ci and this

is A bar B Ci. So, if I take Ci out it becomes AB bar or A bar B these two terms is this

which is equal to Ci times AB bar A bar B which is nothing but A Exclusive OR B. So

this is same as Ci A Exclusive OR B that is this. so I can either write it this way by

grouping the 1s more than once, grouping this one three times actually, if I group this

one three times I get this, if I do not do it if I only group this here for which I get

AB and these two 1s I separately write as a sum of product term it reduces to an Exclusive

OR.

In other words my Co can be AB or Ci and A OR B or A Exclusive OR B. I will use this

to my advantage again as I said hardware advantage. Suppose I am asked to give an Exclusive OR

solution I want more Exclusive OR gates and less OR gates for some reason may be I have

an Exclusive OR array and I want to use as many Exclusive ORs as possible in my solution

I can use this. Sometimes I may say give me the simplest possible solution then in that

case I have do go for an OR solution. Exclusive OR gate is definitely a more complex gate

than a simple OR gate. So this is the philosophy of minimum sum of product expression. Minimum

sum of product expression not always give you the most optimum solution in terms of

practicality. We saw it in the case of gray code converter, we saw it in the case of parity

generator, if I did not use this Exclusive OR solution I will have four min terms each

min term with three literals, there will be four AND gates with three inputs each and

one OR gate with four inputs and each of these A B C is inverted so there are three invertors.

Three invertors for A B Ci, four AND gates with three inputs each, one OR gate with four

inputs. As again said I am now having two Exclusive OR gates with no invertors; A need

not be inverted, B need not inverted, Ci need not be inverted, all I need is two Exclusive.

Suppose I have two Exclusive OR spare I am having a big hardware, board in which I am

doing several things and I happened to have a couple of these Exclusive ORs unutilized

so I will use it. So after the minimum sum of product expression all these Karnaugh Map

mapping and minimum sum of product expression, trying to get as many 1s as possible to be

grouped together is all fine I am not saying you should not do it but after that also you

be a sort of little bit vigilant after that some little manipulation can lead to a very

simple or elegant circuitry, thats what we did in the case of Exclusive OR gate, in

parity check generator, that is what we did here in design gray code and that is exactly

what I am doing here now.

Sometimes A OR B can be replaced by Exclusive OR if necessary I am not saying do it every

time. Suppose an Exclusive OR solution is needed for some practical reason because probably

you have stocked Exclusive OR gates a large number you want to use it rather than go to

a market and buy an OR gate that is a good enough reason or I have an Exclusive OR gate

unused in my board I use it rather than going and finding an another OR gate and finding

a place for it and connecting it together and all that.

All I am trying to say is these two expressions are equivalent. Depending on whichever you

want you can use it but generally people stop with this. And this you can see but this we

will use later on, we are going to use this later in our consideration for simplification

of circuitry. We will use that some other time but meanwhile I thought it is worthwhile

to say this to you.

What we have done so far is to do a half adder two bits both A and B to be added with the

carry, two input and two outputs with a half adder, not of much practical use except in

the least significant bit where there is no carry input but generally when you are doing

in a multi-bit addition one bit not having a carry is not significant at all so you can

as well forget about the half adder and do a full adder in which there are three inputs

A and B and previous input, previous carry output as the input of the carry, and two

outputs sum and carry and sum is coming in the Exclusive OR format and carry can come

in this format and this format. This is for a single bit. As I said we have to do multi-bit

addition. Suppose I have 4-bit number as I assumed here A and B are multi-bit additions

so how will it look like? A 4-bit adder will look like, I have 4-bits Ai A0 B0, A1 B1 A3

B3 now the carry from the previous input, let us assume all of them to be full adders

for the sake of uniformity this carry in I will ground it make it 0 because in the very

first bit there is no carry.

This carry out goes into carry in, this carry out goes into this carry in, this carry out

goes into this carry in so we will have sum 0, sum bit 1, sum bit 2, sum bit 3 and the

final carry bit will be my C3 so my sum will be two 4-bits adder and I can have a sum of

five bits because the last bit can produce a carry. So two 4-bits when I add the sum,

when I say sum that includes carry the total number I get at the output will be 4 or 5

depending on whether there is an output carry or not so I should allow for five bits. Whether

I have it or not depends on the numbers, if both the numbers are 0s the output is 0 so

there is no need for a carry out B, if I knew the numbers A and B why should I add, so if

I dont know the numbers A and B it is a possibility that can generate a carry at the

output so when you add two 4-bit numbers the result can have up to five bits.

The fifth bit is nothing but the carry out of the fourth stage. So when you are going

to the shop to buy a 4-bit adder you have only the As and Bs to be fed in A0 B0, A1

B1, A2 B2, A3 B3 outcomes S0 S1, S2, S3 and C3 and you have to give the Cin as the input

as 0. This is called a 4-bit full adder as one package. Whatever I have drawn inside the rectangle is a one package of IC integrated

Circuit wherein I can feed four A bits, 4-bits of B and carry from the input. Why do we need

the carry input? Of course in the first stage the carry is 0 but suppose I want an 8-bit

adder what will I do for 8-bit adder, I will use these 4-bits full adder.

I will use another 4-bit full adder, this is A0 to 3, B0 - 3 this arrow with a cross

means there is more than one bit. I write here 0 to three bits that means it is 4. A0

to 3 means 4-bits of A, B0 to 3 means 4-bits of B but one 4-bits of B a single arrow cannot

show 4-bits so you put a cross across it. If you know this means it is more than one

input thats what it means. An arrow cross in a digital drawing indicates it is more

than one bit input. If you know the number of inputs you can write 4. You know more than

one but you are not sure how many then you put a cross and stop.

Similarly I get here A5, A4 to 7 4-bits this is my 4-bit, B4 to 7 so I have two numbers

now A is equal to A0 to A7, B is equal to B0 to B7 8-bit numbers. A is an 8-bit number

B is an 8-bit number the numbers are A0 to A7 B0 to B7 these are the bits. So the sum

can be 8-bits or 9-bits, the eighth bit can have a carry so I will feed the output Co

which is actually C3, C3 of the previous block as the input here. So dont assume that

every time the first bit will have a 0 input, dont assume that the first bit of the 4-bit

adder will always have a carry input as 0.

If it is an intermediate stage in a multi-bit addition then my input carry would be the

output of the previous block. This block of 4-bits, the output feeding into a block of

another 4-bits, the output carry C3 goes as input carry Cin and output of this is C is

8 C7 so this would be sum 0 to 4, sum 0 to 3 4-bits, 4 to 7 4-bits so this is four, 4-bits,

4-bits, C7. So C7 S4 to S7 S3 to S0 if you want to write S7 to S4 S3 to S0 first 4-bits,

second 4-bits and the last bit the carry bit is the ninth bit. Hence like that you can

add. You can cascade any number of 4-bits.

The 4-bit is generally available as a slice. The 4-bit adder is the component which is

available as readymade circuits, sometimes 8-bit adders are available but not 16-bit

adder you cant buy a 16-bit adder, you cant buy a 32-bit adder. if you want to

build a 32-bit adder you have to buy either 4-bits or 8-bits, if 8-bits are available

you buy 8-bits otherwise buy 4-bit adders and cascade them, first 4-bit feeding into

the second 4-bit carry, carry of the second 4-bit feeding into the carry in of the third

4-bit etc each so each of these will give you 4-bit sum, 4-bit sum, 4-bit sum, 4-bit

sum and the last carry would also be considered and intermediate carries will be fed into

the next stage and the very first carry in is 0 and four As and Bs are fed into each

of these blocks as usual. This is a multi-bit adder in 8-bit full adder. Like that you can

go on building any number of 4-bit adders to make higher bit adders like 32-bits, 64-bits

but let us examine whether it is a very efficient way of doing it.

Any questions so far on the half adder the full adder, how do you combine full adders

to 4-bit adders and 4-bit adders into multi-bit adders of any number of bits, the concept

of carry in and carry out, sum. So individually each of these blocks will have A Exclusive

OR B Exclusive OR C as the sum and AB or C A Exclusive OR B as the carry circuits so

there will be so many gates inside.

Take it as an exercise; find out how many gates will be there in a 16-bit full adder.

Assuming the first stage also is a full adder. Let us not assume a half adder. Let us assume

for uniformity sake there are 16-bits, I am adding using this scheme that means I will

have four of this 4-bit adder and in turn a 4-bit adder will be like this full adder,

full adder, full adder, full adder that means there is going to be sixteen full adders and

you know each full adder is having the sum circuit like this and carry circuit like this.

So find out and list using this equation and using this equation. For sum use this equation

always for carry you can use both these equations first this and this. In this case list the

number of gates, whenever you are asked to give the number of gates you should give the

type of the gates you cant say seventeen gates, you will not to say eight gates of

three input AND gates, eight three input AND gates, six two input OR gates, four four input

OR gates like that we should give a list of types and the number of inputs only then it

becomes complete. As if we can go to a market with that and to a shop with this and then

buy those. Make a shopping list of the number of gates required to build a 16-bit full adder.

Now having said that let us quickly see whether it is the most efficient way of doing the

job. When I say efficiency we talk of many things, I already said in the beginning. The

speed is one thing, the power consumption, the number of gates so that the size is reduced

so you can make all in one IC it will be very fine so I can save on the size. For a moment

you forget about that size assume that it is all that one integrated circuit. We will

not even make 4-bit, 4-bit, 4-bit, 4-bit and in a 16-bit gate we will try to integrate

sixteen of those units inside one small IC block so that size can be saved we will not

talk about power saving now because I have not told you how to compute the power requirement

of a gate. But one of the parameter I am stressing on from the beginning is the speed.

How much time it takes to compute a 16-bit addition? Do you have any idea how much time

does it take? Suppose I have A and B, A is 16-bit, B is 16-bit all of them are available

to start with. I am giving you a 16-bit number for A, another 16-bit number for B and say

go and we are asked to do the addition. Can you do it instantaneously? No, because the

gate themselves will take time. For example, even if a single bit adder or let us take

this half adder, the moment I gave A and B the output S and C are not available because

however fast these circuits work the gates work there is a finite time delay.

So, there is a time delay involved in propagation of these values for the switching operation

Exclusive OR gates and switching operation of this AND gates before we get the result.

Since we dont have an idea of the gate delays of individual Exclusive OR gate and

individual AND gate the delay will depend on the type of the gate and number of inputs

both decides the delay. The gate delay depends on the number of inputs of the gate as well

as the type of the gate. Since we dont know anything of all those things we will

only simply say the time it takes for the output to come after the input is given we

will call it propagation delay.

The delay of propagation of the inputs to the outputs is called propagation delay. There

is only one Exclusive OR gate and one AND gate. When you go to a full adder there are

two Exclusive OR gates and a two level gate one AND gate feeding into a NOR gate, whether

you use this model or this model let us use this model it is one AND gate feeding into

another gate so there is going to be some delay there.

There is going to be a delay into these two stages of gate thing, here also there are

two stages of a gate thing A Exclusive OR B has to be fed into this C, first Exclusive

OR gate will find A Exclusive OR B, second Exclusive OR gate will take that and put it

into C so two levels. Here is one level of gating some propagation delay, here there

are two levels of gating propagation delay. Since we have taken the full adder as a unit

of our consideration because half adder is used as only the first stage so we will forget

about it for the moment for all purposes of calculation and standardization we will assume

full adder as the basic building block.

A single bit full adder 1-bit full adder is a basic building block we will call the propagation

delay of the full adder TP that consists of two parts the carry and the sum whether the

sum comes first or carry comes first because one path carries another path whichever path

activates first will come out first. We will not worry about, we are only interested in

the carry now because the carry has to come from here go here, go here. Even if the sum

comes with a slightly more delay sum and carry may not have equal propagation delays. we

dont know but we will have to look at the gates structure to see which is faster, sum

can be faster carry can be faster let us not talk about it for a moment.

For a moment we will only consider on carry propagation because carry propagation is more

important the carry has to go from here to here, here to here, here to here, here to

here and then from here. The 16-bit full adder which I asked to do as an exercise carry has

to propagate through sixteen stages, first stage has to go to the second stage, second

has to go to the third and finally it has to go to the sixteenth stage so sixteen carry

delays have to be accounted for. Meanwhile the sum will come because as soon as A and

B are known and once the carry comes the carry triggers the sum, this carry trigger the sum.

It is this something like the domino effect. As soon as the carry comes sum comes but then

carry goes on, carry goes from here to here leaving sum behind, goes from here to here

leaving sum behind. The sum may be slightly delayed than the carry but then before the

carry goes to the next stage the sum could have been settled. So you dont worry about

the minor difference between the sum delay and the carry delay let us only worry about

carry delay because it is more significant of the two because it has to travel a wide

long distance so the total propagation delay is sixteen carry propagation delays, carry

of the full adder is carry propagation delay sixteen of them has to be used so it is a

16-bit adder, it could be a 32-bit adder, 64-bit adder you can imagine.

Therefore we have to think of, if you want to talk about one Giga hertz, Pentium II Giga

hertz processor, what is a Giga hertz in terms of clock frequency? It is nano seconds. Giga

is 10 power 9 reciprocal of 10 power 9 is nano. We are talking of nano seconds, the

whole operation. When you say computer works at nano seconds the whole thing has to be

done, some addition process, some multiplication process or some other process and we are talking

about full adder here and how many such things will be there in a computer, adding two numbers

is the least complicated thing you can think of in computing.

In a computing environment there are so many other more complex things that need to be

done. That means we have to definitely try to do something to reduce the carry delay.

There are several techniques and to exhaust all of them will be beyond the scope of this

lecture series. But I thought at least some of those concepts are important it has to

be drilled in the beginning. When you are learning the first course in digital you need

understand to certain parameters.

At least you should know a problem exists. Later on you have opportunity to learn about

solutions of these. So we will take one technique of improving the speed of carry. One technique

of reducing the propagation delay of a 4-bit full adder we will see in the next lecture.

But that is not the only way to do it there are several other methods. We will not be

talking about all those methods. We will take one simple method of reducing the propagation

delay of the 4-bit full adder in the next lecture.

The Description of Lecture 11 Arithmetic Circuits