#testbook_digital

Question

explain ?

Comments

@praveen_saini sir

Answer 1

In carry lookahead adder, the carry is calculated based on initial carry and hence it is not dependent on previous carry.So here we have 4 bit carry lookahead adder , we calculate C₄ in terms of C₀ without involvement of C₃ using the carry lookahead equations .To use the carry looahead equations , we need the generator functions G_i and propogator functions P_i for each bit pair of operands a_i and b_i.They are calculated as :

G_i   =  a_i . b_i

P_i =  a_i  ⊕ b_i

So all these operations(calculation of G_i 's and P_i 's takes place simulatnoeusly.After that :

we calculate C₄ from the C₀ using the carrylookahead equations :

C_i+1   =  G_i   +  P_i . C_i

So calculating C₁ in terms C_o  , we get :

C₁   =  G₀  +  P₀. C₀

Similarly calculating C₂ in terms C₁  , we get :

C₂   =  G₁ +  P₁. C₁

Having calculated the C₁ earlier , we can find C₂ in terms of C₀ by substitution :

C₂  =  G₁ + G₀ . P₁ + C₀ . P₀ . P₁

Proceeding in the same manner , substituting repeatedly , we get :

C₄   =  G₃ + G₂ . P₃ + G₁ . P₂ . P₃ + G_{0 .} P_{1 .} P₂ . P₃  + C₀ . P₀ . P₁ . P₂ . P₃

Thus we can see final carry C₄ is caluclated in terms of C₀ by using generator and propogator functions

As we can see for C₄ , no of product terms containing more than 1 literal = 4

So no of AND gates required  = 4 for C₄ 

and to combine them i.e.to find C₄ , we need to OR them , so we 

No of OR gates for C₄ = 1

Similarly we require 3 AND gates and 1 OR gate for C₃

                            2 AND gates and 1 OR gate for C₂   and 

                            1 AND  and  1 OR gate for C₁

This carry lookaheads C₁ , C₂ , C₃ and C₄ that we have calculated in terms of C₀ are stored in carry lookahead circuit.But prior to calculating them as we mentioned earlier , we calculate G_i and P_i where i ranges from 0 to 3 here since it is a 4 bit adder.

So in calculating G_i , we saw G_i  = a_i . b_i  so here 1 AND gate is required as well .As i ranges from 0 to 3 , so need 4 AND gates.

Therefore, total number of AND gates = 4 + 3 + 2 + 1(from carry lookahead circuit) + 4(in calculating G_i)

                                                      =  14

Total no of OR gates                        =  4 [For finding C₁ to  C₄]

Therefore , x = 14 and y = 4

Hence , 2x + 2y  =  28 + 8

                         =  36

Hence the correct option is B

Comments

@habib

for calculating pi we need n EXOR gate  or 2n EXOR gate (mentioned in testbook solution)?

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For calculating P_i , we need n Ex OR gates and n Ex OR gates to calculate the sum bits S_i.So total no of XOR gates = 2n