Mine is coming 1280...

5 votes

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.

- $1032$
- $776$
- $1284$
- $1536$

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