Which statement correctly differentiates PROM and PLA?

PROM has a fixed AND array (all minterms); PLA has a programmable AND array

PROM has a programmable AND array; PLA has a fixed AND array

Which arrays are programmable in a PAL?

Programmable AND array, fixed OR array

Fixed AND array, programmable OR array

Why does the unused sixth product term not change output B?

Both x and x' stay connected so the term is always 0

It is connected to Vcc so it is always 1

All its fuses are blown so it gives 1

2078 Kartik · 2+2×2 marks

PROM vs PLA; functions using PAL

Short answer

PROM has a fixed AND array (full decoder) and programmable OR; PLA has both arrays programmable. For PAL, A(x,y,z)=Σ(2,4,5,7) simplifies to xy'+xz+x'yz', B=Σ(0,2,6) to x'z'+yz'. Program these product terms in the AND array; fixed ORs combine them.

Differentiate between PROM and PLA. Implement the following boolean functions using PAL. a) A(x,y,z) = Σ(2,4,5,7) b) B(x,y,z) = Σ(0,2,6)

PROM vs PLA

Point	PROM	PLA
AND array	Fixed (full decoder, all $2^n$ minterms)	Programmable
OR array	Programmable	Programmable
Functions in form	Sum of minterms (no simplification)	Simplified SOP, shared terms
Size for many inputs	Grows as $2^n$ , wasteful	Only needed product terms
Typical use	Look-up tables, code converters	Multi-output logic with common terms

Functions using PAL

In a PAL the AND array is programmable and the OR array is fixed, so each function is first simplified to SOP.

a) $A(x,y,z) = \Sigma(2,4,5,7)$

       yz=00 01 11 10
x=0      0   0  0  1
x=1      1   1  1  0

$m_4, m_5$ give $xy'$ ; $m_5, m_7$ give $xz$ ; $m_2$ alone gives $x'yz'$ .

A = xy' + xz + x'yz'

b) $B(x,y,z) = \Sigma(0,2,6)$

       yz=00 01 11 10
x=0      1   0  0  1
x=1      0   0  0  1

$m_0, m_2$ give $x'z'$ ; $m_2, m_6$ give $yz'$ .

B = x'z' + yz'

PAL programming table (each OR gate has 3 AND inputs; 1 = true, 0 = complement, – = not connected):

Product term	x	y	z	Output
1: $xy'$	1	0	–	A
2: $xz$	1	–	1	A
3: $x'yz'$	0	1	0	A
4: $x'z'$	0	–	0	B
5: $yz'$	–	1	0	B
6: (unused)	–	–	–	B

x y z --[prog. AND]-- P1,P2,P3 --[fixed OR]--> A
               \----- P4,P5,P6 --[fixed OR]--> B

The unused term 6 has all its fuses intact (both $x$ and $x'$ connected) so it gives 0 and does not affect B.

✓ Answer checked by Claude Opus 5.5 (high effort)

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