y not ans 33.6 *10^3 microsecond??????????????

y The first carry and sum will be available after 4.8 us & how???????????

71 votes

A half adder is implemented with XOR and AND gates. A full adder is implemented with two half adders and one OR gate. The propagation delay of an XOR gate is twice that of an AND/OR gate. The propagation delay of an AND/OR gate is 1.2 microseconds. A 4-bit-ripple-carry binary adder is implemented by using four full adders. The total propagation time of this 4-bit binary adder in microseconds is ______.

0

y not ans 33.6 *10^3 microsecond??????????????

y The first carry and sum will be available after 4.8 us & how???????????

0

here mention xor is twice of and/or so and/or is 1.2ms so xor is 2.4 ms & 1 hald adder is 1xor+1and so= 3.6

& a full adder is implemented wt 2 half adder +1 or gate = 2*3.6+1.2= 8.4

4full adder =4*8.4= 33.6

& a full adder is implemented wt 2 half adder +1 or gate = 2*3.6+1.2= 8.4

4full adder =4*8.4= 33.6

8

okay :) If everything is done one after the other or in a serial way. But, operations can be done in parallel- you can see the figure given below. As long as input to a gate is not dependent on the output from another gate, those two can be operated in parallel.

1

Should not this be like since in the question we are given that a full adder is here implemented as 2 Half adders and an OR gate..

And we know that if we follow this constraint of circuitry then we use cascading half adders and then an OR gate..So delay should be added for each of the half adders and one OR gate for every full adder implementation.

So delay of a full adder should be : 2.4 * 2[For 2 half adders] + 1.2 [for OR gate]

= 6 ms

So delay of 4 bit ripple carry adder = 4 * 6 ms

= 24 ms

@Arjun Sir , please check this approach..

And we know that if we follow this constraint of circuitry then we use cascading half adders and then an OR gate..So delay should be added for each of the half adders and one OR gate for every full adder implementation.

So delay of a full adder should be : 2.4 * 2[For 2 half adders] + 1.2 [for OR gate]

= 6 ms

So delay of 4 bit ripple carry adder = 4 * 6 ms

= 24 ms

@Arjun Sir , please check this approach..

0

what will be the difference if i use ripple carry adder over Carry look ahead adder here?? what will be the chenge in aswer??

0

@ S Ram,

The question itself says,

"A 4-bit-ripple-carry binary adder is implemented by using four full adders"

I think it should be clear now.

The question itself says,

"A 4-bit-ripple-carry binary adder is implemented by using four full adders"

I think it should be clear now.

4

Delay of carry look ahead adder:

https://gateoverflow.in/35865/total-propagation-delay-in-carry-look-ahead-adderin-a-4-b

1

In ripple carry adder, a full adder becomes active only when its carry in is made available by its adjacent less significant full adder.

so 19.2 is the right answer.

so 19.2 is the right answer.

0

In Ripple carry adder, a stage doesn’t wait for the “full output” of previous stage. It only waits for the previous stage’s carry to reach to itself. Now, carry generated by our current stage can be written in 2 ways:-

1)Carry generated= $xy + $ $(x$$\bigoplus$$y)C_{in}$ 2)Carry generated= $xy+yC_{in} +xC_{in}$ where $C_{in}$ is carry received by our current stage

Using 1st way, delay in carry generation would be 4.8$us$ and by using 2nd way, delay in carry generation would be 2.4$us$. So, answer should be according to 2.4$us$ delay(better than 4.8$us$ delay).

So, total delay will be $(4-1)*2.4 + 4.8$ = $12us$

111 votes

$S1$ should wait for $C1$ to be ready. Delay for generating $C$ is $1$ EXOR $+ 1$ AND $+ 1$ OR $= 2.4 + 1.2 + 1.2 = 4.8$ $μs$

Delay for sum is XOR + XOR $= 2.4 + 2.4 = 4.8 μs$

But for the second adder, there the first EXOR can be done even before waiting for the previous output. So, we can get the sum in Another $2.4 μs$ and carry in another $2.4 μs$. In this way, $4$-$bit$ sum can be obtained after

$4.8 μs + 3 * 2.4 μs = 12 μs.$

But the question says we use ripple-carry adder. So, each adder must wait for the full output from the previous adder. This would make the total delay $= 4 * 4.8 = 19.2 μs$ and this is the key given by GATE, so obviously they meant this.

Reference: http://www.cs.umd.edu/class/sum2003/cmsc311/Notes/Comb/adder.html

0

plz help me out .I`m confused . I understand what is the procedure of yours .

but in the official site they have given answer 19.2 ( 4.8*4 )

see question no. 65 : http://gate.iitk.ac.in/GATE2015/AnsKey2015/CS_S06.pdf

3

I have seen it, This one and one another question regarding frame size GATE gave really "bull shit" key. Many others are debatable ones, but this is the first time really blunder ones come in GATE key.

0

Delay for generating C is 1 AND + 1 OR = 1.2 + 1.2 = 2.4 ms y 1 and here ??? shd 2 and 3 * delay for carry this 3 due to 4 bit binary adder ryt ???

1

@Pranay Sorry. There was a catch here - ripple-adder. Answer must be 19.6 only.

@Shimpy Is it clear now?

@Shimpy Is it clear now?

4

Thanks . now its cleared . Just a small correction 4.8*4=19.2 .

I compare this with the movie Race 2 . "twist k ander twist " :D .peace

I compare this with the movie Race 2 . "twist k ander twist " :D .peace

0

sir .if it were given as PARALLEL BINARy ADDER instead of ripple ,than it will take total time as 4.8msec.??

1

A 16-bit **ripple carry adder** is realized using 16 identical full adders (FA) as shown in the figure. The carry-propagation delay of each FA is 12 ns and the sum-propagation delay of each FA is 15 ns. The worst case delay (in ns) of this 16-bit adder will be _____. (The question was asked in 2014 ECE (SET-4) question no-41 .But the answer given by iit kgp (ANS-195) does not use ripple carry adder concept!! whats the concept actually?)

0

Hello Arjun Sir, one thing i didn't understand.... they said one full adder was implemented with 2 half adder and one OR gate.... so for 1 full adder delay will be 1st XOR delay (from 1st half adder) +2nd XOR delay(from 2nd half adder)+ 1 OR gate means 2.4+2.4+1.2= 6ms????? please clarify...

0

@arjun sir..how will we infer that in a question we have to calculate using n*(gate delay of 1 full adder) ,like 19.2 here or the other method which gave 12??which method to follow for what kind of adder??please explain as every year propagation delay type qustions come and they are very confusing.!!

0

what will happen if the carry and the sum generated different values of propagation delay. Then what will be the final answer ??

1

Unable to get this question clearly.

**What I understood :** when we calculate the sum and carry of first half adder ie, $AB$ and $A\oplus B$, we calculate the delay for this as max(delay for $AB$,delay for $A\oplus B$ ) = $2.4$ in this question.

After this is done we can use this result to calculate the result of 2^{nd} half adder i.e., $A\oplus B\oplus C$ and $C( A\oplus B ) + AB$ and again delay for this step is maximum of these two = $max(1.2+1.2,2.4) = 2.4$

Total delay for 1 full adder $= 2.4 + 2.4 = 4.8$

Delay for 4 full adders used in ripple carry adder = $4.8 * 4 = 19.2$

**Correct me if Iam Wrong ?**

1

Arjun Sir,

carry in is 0 for the first 1 bit adder, so we can use only 1 half adder there. and after that as you said "operations can be done in parallel-As long as input to a gate is not dependent on the output from another gate, those two can be operated in parallel." so total 2.4 x 4=9.6 .. why not 9.6 then?

carry in is 0 for the first 1 bit adder, so we can use only 1 half adder there. and after that as you said "operations can be done in parallel-As long as input to a gate is not dependent on the output from another gate, those two can be operated in parallel." so total 2.4 x 4=9.6 .. why not 9.6 then?

4

Ans is 19.2 ms

Delay for generating C is 1 EXOR + 1 AND + 1 OR = 2.4 + 1.2 + 1.2 = 4.8

Delay for sum is XOR + XOR = 2.4 + 2.4 = 4.8 ms

And for a Ripple Carry adder

T = (n-1) Tc + Ts

= 3* 4.8 + 4.8 = 19.2ms

Delay for generating C is 1 EXOR + 1 AND + 1 OR = 2.4 + 1.2 + 1.2 = 4.8

Delay for sum is XOR + XOR = 2.4 + 2.4 = 4.8 ms

And for a Ripple Carry adder

T = (n-1) Tc + Ts

= 3* 4.8 + 4.8 = 19.2ms

3

this same question was asked in one of the made easy tests and they solved it differently and their answer is 12ns. Dont know why.. According to me 19.2 is the best one. which appoach to follow in gate??

0

" Delay for generating C is 1 EXOR + 1 AND + 1 OR = 2.4 + 1.2 + 1.2 = 4.8 ms "

Sir here 2nd XOR depend on first XOR and one AND depend on 1st XOR.

Among these two, XOR has maximum delay

So, we should take 2 XOR and 1 OR gate. rt?

Cannot understand why AND gate delay is taking?

Rather 2nd level XOR gate delay should take. as it's delay is more.

Plz explain where I am doing mistake.

Sir here 2nd XOR depend on first XOR and one AND depend on 1st XOR.

Among these two, XOR has maximum delay

So, we should take 2 XOR and 1 OR gate. rt?

Cannot understand why AND gate delay is taking?

Rather 2nd level XOR gate delay should take. as it's delay is more.

Plz explain where I am doing mistake.

0

Only after getting the carry from the previous stage we can and it with the first xor output of that stage that's why and gate delay is taken right?

Also how to know whether we have to consider the first xor outputs produced paralelly or not? We can't know what the key will have. How to know in examination?

Also how to know whether we have to consider the first xor outputs produced paralelly or not? We can't know what the key will have. How to know in examination?

6

@sushmita I'm no expert in digital but if I write GATE and this question comes I'll give 12 as answer :)

1

only carry is rippled through various stages, not sum, thats why named as ripple carry adder. so parallel execution of sum is possible. so , surely answer will be 12.

15

**Sum=A⊕B⊕Cin**

**Carry Output= Cin.[A⊕B] + AB**

These are the equations which can be realized for full adder as per the description is given by the question.

XOR gate delay=2.4 microseconds and AND/OR gate delay= 1.2 microseconds.

So this way, Sum delay of 1 Full Adder= 2.4+2.4=4.8 micro-seconds.

Carry Delay= First we have to wait for the output of A⊕B, which will be available to us after 2.4 micro seconds, and then 1 AND operation + 1 OR operation. So total= 2.4+1.2+1.2=4.8 micro seconds.

The question is asking for total propagation time of 4 bit binary adder implemented using described 4 one bit full adders.

The total propagation time means when would the last Full adder give the output of Sum.

The last Full Adder can only produce sum output only when carry input from previous Full adders are available.

So, the fourth Full Adder will wait (3*4.8 microseconds) for carry from previous 3 Full adders,

and produce the sum in next 4.8 microseconds after this.

So, total delay= 3*4.8 +4.8 = 19.2 micro seconds.

0

But the carry is added with the sum right ?

so without carry how can sum be calculated

it cannot happen in parallel in case of ripple carry na?

so without carry how can sum be calculated

it cannot happen in parallel in case of ripple carry na?

0

@srestha

if we look at the diagram provided by arjun SIR

1 XOR , 2 AND and 1 OR gate is involved for carry right?

why only one AND gate is considered

if we look at the diagram provided by arjun SIR

1 XOR , 2 AND and 1 OR gate is involved for carry right?

why only one AND gate is considered

3

Do u know, how to access parallely?

For carry generation there are 2 AND gate, that is true

But for parallel access in 2 AND gates we have to take max(gate1,gate2)

So, for calculation purpose 1 AND gate is sufficient

For carry generation there are 2 AND gate, that is true

But for parallel access in 2 AND gates we have to take max(gate1,gate2)

So, for calculation purpose 1 AND gate is sufficient

0

@srestha

if the question said that the ripple adder could use only half adders provided

then what will be the answer

if the question said that the ripple adder could use only half adders provided

then what will be the answer

2

Answer should be 12 and not 19.2, just because it's given in the key we should not change our understanding of the concepts.

0

i don't understand mean of total delay?

if its (carry delay+sum delay) then answer will be different???

if its (carry delay+sum delay) then answer will be different???

0

Arjun sir ...In Morris Mano book it is given that the total propagation time in the adder would be the time in one half adder plus 2n gate levels for n bit parallel adder

3

@ayush i attempted this question in test and i marked the option with same approach that you followed but they described the solution same as Arjun sir said -

for the second adder, there the first EXOR can be done even before waiting for the previous output. So, we can get the sum in Another 2.4μs and carry in another

2.4μs. In this way, 4-bit sum can be obtained after

4.8μs+3∗2.4μs=12μs.

So finally what approach should we followed? i seen it's time complexity also which theta(n*FA) means 19.2 should be right .

for the second adder, there the first EXOR can be done even before waiting for the previous output. So, we can get the sum in Another 2.4μs and carry in another

2.4μs. In this way, 4-bit sum can be obtained after

4.8μs+3∗2.4μs=12μs.

So finally what approach should we followed? i seen it's time complexity also which theta(n*FA) means 19.2 should be right .

1

@Prince-Arjun's sir approach is correct if we want the minimum time to compute the sum using 4 full adders,sir has use concept of parallelism.But, as sir said and in my approach, we have assumed the hardware as given to be "ripple carry adder" means once the previous output is available then only the next computation starts.

However, it would be good if you post your question here on GO and comment it's link here so that we can see where the difference lies.

However, it would be good if you post your question here on GO and comment it's link here so that we can see where the difference lies.

0

The total propagation time means when would the last Full adder give the output of Sum.

It should be "When last Full added gives output of sum and output carry as well right ..at ouput we want both sum and carry ??

So it should be (N-1)Tsum + MAX(Tsum ,Tcarry)

Here it so happened that Tsum = Tcarry

0

@Jatin-No. See why we shifted from ripple carry adder to carry lookahead adder? What was the change? The carry network was implemented independently and then the adders don't have to wait for the carry to get ready from the previous adder.

We are worried about the propagation time of carry in ripple carry adder, because say if carry had no effect on the result, then why would have we bothered about designing carry look ahead adder.

Here it is very clear they are asking what is the total delay occurred in producing the carry, means when would last carry be produced.

We are worried about the propagation time of carry in ripple carry adder, because say if carry had no effect on the result, then why would have we bothered about designing carry look ahead adder.

Here it is very clear they are asking what is the total delay occurred in producing the carry, means when would last carry be produced.

0

So what is total propagation delay of this adder considering ripple carry adder

Time we get carry from last full adder ??

What IF for full adder Tsum we got 4.8 and Tcarry 5.2 ??

Time we get carry from last full adder ??

What IF for full adder Tsum we got 4.8 and Tcarry 5.2 ??

1

For this question, time we get carry from last full adder is $4*4.8=19.2\mu s$

You are asking what is total propagation delay for a 4-bit ripple carry binary adder whose sum is produced in $4.8 \mu s$ and carry in $5.2 \mu s?$

Carry to the last full adder would come after $3*5.2 \mu s$ and then output carry would be produced in another $5.2 \mu s$ and hence total propagation delay=$4*5.2 \mu s$

You are asking what is total propagation delay for a 4-bit ripple carry binary adder whose sum is produced in $4.8 \mu s$ and carry in $5.2 \mu s?$

Carry to the last full adder would come after $3*5.2 \mu s$ and then output carry would be produced in another $5.2 \mu s$ and hence total propagation delay=$4*5.2 \mu s$

0

@Arjun sir,

if they ask for how much time is going to take for generating sum then writing direct

4*4.8doesn't make good sense it should be written like[3*4.8(carry)+4.8(sum by last FA)]. which leads to definately 19.2.

0

@Arjun Sir,

The answer considered for evaluation is $12$ but as explained above it's $19.2$ please update.

0

@Devendra Thakur

This would make the total delay =4∗4.8=19.2μs and this is the key given by GATE,

so obviously they meant this.....

This would make the total delay =4∗4.8=19.2μs and this is the key given by GATE,

so obviously they meant this.....

0

@Arjun Sir

Please correct this answer.

The best way to analyse this answer is by drawing a detailed representation of the "gates."

Earlier the answer given by gate authority is 19.2 but later on after evaluation they changed it to 12.

1

Just check out my answer below. You will understand it very well as I have drawn it very neat and clean.

In the mean time I will provide you the updated key.

This is the wrong answer and I wonder how it can be the best answer considering the excellent quality of all other questions on the gateoverflow.

I am just waiting for the Arjun Sir to take a note of this question or to have a conversation.

Hopefully this will get correct soon.

0

Just check out my answer below. You will understand it very well as I have drawn it very neat and clean.

In the mean time I will provide you the updated key.

This is the wrong answer and I wonder how it can be the best answer considering the excellent quality of all other questions on the gateoverflow.

I am just waiting for the Arjun Sir to take a note of this question or to have a conversation.

Hopefully this will be corrected soon.

0

@Arjun sir which method should we go for if it is asked in exam? Isn't the formula for ripple carry adder total propagation time

(n-1)xTc + Ts ? So according to that we get 19.2 micro seconds. But according to the diagram it is clear that 12 micro seconds should be the answer. I am really confused.

0

This doesn't work all the time.

You need to have the proper analysis of such a question.

Nevertheless asking such questions is very rare.

You need to have the proper analysis of such a question.

Nevertheless asking such questions is very rare.

0

Another fastest way for answering this question is to find Tsum and Tcarry.Once u find them just apply this formula

Total Time Required =(n-1)Tcarry+maximum(Tsum,Tcarry).

Here, in one FA sum delay =XOR+XOR = 2.4+2 4 =4.8

Carry delay=XOR+OR+AND= 2.4+1.2+1.2=4.8

So apply formula,n=4(given in question)

(4-1)(4.8)+maximum(4.8,4.8)

=3×4.8+4.8

=14.4+4.8

=19.2microseconds.

Total Time Required =(n-1)Tcarry+maximum(Tsum,Tcarry).

Here, in one FA sum delay =XOR+XOR = 2.4+2 4 =4.8

Carry delay=XOR+OR+AND= 2.4+1.2+1.2=4.8

So apply formula,n=4(given in question)

(4-1)(4.8)+maximum(4.8,4.8)

=3×4.8+4.8

=14.4+4.8

=19.2microseconds.

3

@Ayush Upadhyaya, @Arjun sir I think there is no problem in applying the concept of parallelism and answer can be 12. Please tell me weather I am correct or wrong

Source:-

33 votes

Ans is 12 ns

It took me a while but here's how it is :

The first carry and sum will be available after 4.8 ns. This should be straight forward.

However, for the subsequent stages, you need to keep in mind that the output of half adders is already there at 2.4 ns. So in a sense, it is already computed. The remaining half adder for each full adder is just waiting for the previous carry, which when available from the previous stage can be processed in 2.4 ns. So each next stage will take only 2.4 ns each.

The catch here is that half of the output in each next stage is already computed, only half needs to be processed.

It took me a while but here's how it is :

The first carry and sum will be available after 4.8 ns. This should be straight forward.

However, for the subsequent stages, you need to keep in mind that the output of half adders is already there at 2.4 ns. So in a sense, it is already computed. The remaining half adder for each full adder is just waiting for the previous carry, which when available from the previous stage can be processed in 2.4 ns. So each next stage will take only 2.4 ns each.

The catch here is that half of the output in each next stage is already computed, only half needs to be processed.

0

https://electronics.stackexchange.com/questions/153650/delay-in-4-bit-ripple-carry-adder

I see what you did there :p

Oh and btw, your very first solution is the correct one, It's according to Gate's official answer key...

0

"The remaining half adder for each full adder is just waiting for the previous carry, which when available from the previous stage can be processed in 2.4 ns. So each next stage will take only 2.4 ns each."

yes..but the previous carry will be available after another 2.4ns and then second XOR will happen which will take another 2.4ns. In total it will take 4.8ns

3 votes

This should be the right procedure to get the right answer for this question.

Simply putting the formula here won't work.

Therefore, the answer comes out to be 12 microseconds.

1 vote

**b) 19.2**

**C(i+1)=(A XOR B).C(i)+A.B**

Hence Totla delay of one carry bit is

1 XOR gate + 2 AND gate + 1 OR gate

2.4 +1.2+1.2 =4.8 microsec.

1) total delay from c1 to c4 will be 4.8 x 4 =19.2 microsecond;

2) total delay from c1 to c3 + sum of A4 and B4 will be 4.8 x 3+ (2.4+2.4) =19.2 microsecond

hence for** MAX DELAY** =**MAX**( 19.2,192 )

**MAX DELAY** is 19.2 microsecond.

1 vote

Circuit diagram for the problem can be made as:

delay for XOR gate = 2*1.2 = 2.4 μs.

delay for AND/OR gate = 1.2 μs.

First XOR gate takes 2.4 μs. and meanwhile XY can be calculated in parallel. Similarly, Second XOR gate takes another 2.4 μs and in the meanwhile output of AND(1.2 μs) and OR(1.2 μs) can be calculated (Since, the output of first XOR and AND is available immediately).

So, total time taken = 2.4 + 2.4 = 4.8 μs for 1-bit calculation.

Hence, gate delay for 4-bits = $4*4.8=19.2\ μs$.