A certain ROM has a capacity of 16 K times 4 and an internal

A certain ROM has a capacity of 16 K times 4 and an internal structure like that shown in figure below How many registers are in the array? How many bits are there per register? What size decoders does it require? For A3=1, A2=1, A1=0 and A0=0 which register will be selected

Solution

(a) Decoder-1 has a total of 4*4=16 registers and similarly Decoder-2 has 16 registers. Therefore, there are a total of 32 registers (16+16). Horizontally(row wise), it has 4 registers in the array and Vertically(column wise), it has 4 registers in the array.

(b) As it is 16KB*4 ROM, so each row wise registers must contain 16 bits.(16KB*4=64KB=(2^6)*(2^10)=(2^16)).so 16/4=4bits------->each register must contain 4 bits.

(c) Two decoders are required. Each of size 2*4 are needed

(d)For A3=1,A2=1,A1=0,A0=0 Decoder-1 line zero is activated ,Decoder-2 line three is activated. So, register 15 is selected.