Which statement is correct about RAM and ROM?

RAM is volatile; ROM is non-volatile

RAM is non-volatile; ROM is volatile

In a PROM implementation, what is programmed?

The OR array (links from minterm lines to outputs)

Why does the PROM method need no simplification?

The decoder already provides every minterm

Boolean laws are applied automatically

The functions are always single-minterm

2076 Chaitra · 2+4 marks

RAM vs ROM; functions using PROM

Short answer

RAM is volatile and allows read/write; ROM is non-volatile and read-only. Implement F1=Σ(1,2,4,6) and F2=Σ(0,2,3) with an 8×2 PROM: a 3-to-8 decoder generates all minterms, and the OR array connects m1,m2,m4,m6 to F1 and m0,m2,m3 to F2.

Differentiate between RAM and ROM. Implement F1 = Σ(1,2,4,6) and F2 = Σm(0,2,3) using PROM.

RAM vs ROM

Point	RAM	ROM
Full form	Random Access Memory	Read Only Memory
Operation	Read and write	Read only (written once/rarely)
Volatility	Volatile: data lost on power off	Non-volatile
Use	Temporary data and running programs	Fixed programs (BIOS, firmware), tables
Types	SRAM, DRAM	Mask ROM, PROM, EPROM, EEPROM
Speed	Fast read and write	Fast read; writing slow or impossible

F1 = Σ(1,2,4,6) and F2 = Σ(0,2,3) using PROM

Three inputs ( $A$ , $B$ , $C$ ) and two outputs, so an 8 × 2 PROM is used: a fixed 3-to-8 decoder (AND array) produces all minterms $m_0$ – $m_7$ , and the programmable OR array connects the needed minterms to each output.

PROM truth table (contents):

Address $ABC$	Minterm	$F_1$	$F_2$
000	$m_0$	0	1
001	$m_1$	1	0
010	$m_2$	1	1
011	$m_3$	0	1
100	$m_4$	1	0
101	$m_5$	0	0
110	$m_6$	1	0
111	$m_7$	0	0

A 1 means the fuse/link between that minterm line and the OR gate is kept; a 0 means it is blown.

A B C
| | |
[3-to-8 decoder] (fixed AND array)
 m0 m1 m2 m3 m4 m5 m6 m7
 |  x  x  |  x  |  x  |   --> OR1 --> F1
 x  |  x  x  |  |  |  |   --> OR2 --> F2
 (x = intact link)

$F_1 = m_1 + m_2 + m_4 + m_6$
$F_2 = m_0 + m_2 + m_3$

No simplification is needed with a PROM, because the decoder already gives every minterm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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