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How many programmable fuses are required in a PLA which takes $16$ inputs and gives $8$ outputs? It has to use $8$ OR gates and $32$ AND gates.

  1. $1032$
  2. $776$
  3. $1284$
  4. $1536$
in Digital Logic 3.9k views
0
Is the answer c)

Mine is coming 1280...
0
@Kapil

Your answer is correct

But can please explain how you got 1280.
0

diagram of PLA

 

now read the answer 

1 Answer

12 votes
 
Best answer

answer is  C)

Total programmable fuses= fuses required by AND gates + fuses required by OR gates

(Fuses are attached to and- or gate inputs to allow inputs to reach the and - or gates and If fuses don't work , then no input can reach to these gates of PLA)

Fuses required by AND gates = 2* no. Of inputs * no. Of and gates = 2*16*32= 1024 fuses

Fuses required by OR gates = no. Of outputs * no. Of and gates  = 8* 32 = 256 ( total outputs should be equal to no. Of OR gates and inputs have to cross AND gates and then goto OR gates)

Total fuses = 1024+ 256= 1280


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1
from where u got these formulas?
1
Google search .

Read about programmable fuses and got above definitions .
0
But option C is 1284
0
That is wrong. Right answer is 1280 only.
Answer:

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