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Interfacing 2732 EPROM and 6116 RAM

Short answer

EPROM 4K at 0000H–0FFFH, spare 4K at 1000H–1FFFH, RAM 2K at 2000H–27FFH. Decoder: 74LS138, C,B,A = A13,A12,A11, G1=+5V, G2A'=IO/M', G2B'=A15 OR A14. Y0' and Y1' ANDed to CE' of 2732, Y4' to CS' of 6116. A11–A0 to EPROM, A10–A0 to RAM, RD' to OE', WR' to WE' of RAM.

Design an interfacing circuit for following problem. i) 74LS138: 3 to 8 Decoder ii) 2732 (4K8): EP-ROM address range should begin at 0000H and additional 4K memory space should be available for future explanation iii) 6116 (2K8): CMOS R/W memory

Chips: 2732 is a 4K×8 EPROM (A11–A0, pins CE' and OE'/Vpp). 6116 is a 2K×8 CMOS static RAM (A10–A0, pins CS', OE', WE').

Memory map (my choice)

BlockDeviceRangeA15 A14 A13 A12 A11
12732 EPROM (4K)0000H – 0FFFH0 0 0 0 x
2Spare (future 4K)1000H – 1FFFH0 0 0 1 x
36116 RAM (2K)2000H – 27FFH0 0 1 0 0

The EPROM starts at 0000H as required (the 8085 begins execution at 0000H after reset). The next 4K is kept free for future expansion.

Decoder connections (74LS138)

  • C, B, A = A13, A12, A11. The EPROM covers 000 and 001 → Y0', Y1' (combined by an AND gate → CE' of 2732).
  • Future block: 010 and 011 → Y2', Y3' (left free, or ANDed and taken to a connector for a future 4K chip).
  • 6116: 100 → Y4' → CS'.
  • G1 = +5 V; G2A' = IO/M'; G2B' = output of an OR gate (A15 + A14), so A15 = A14 = 0 is needed.
  • A11–A0 to 2732; A10–A0 to 6116.
  • RD' → OE' of 2732 and 6116; WR' → WE' of 6116.
 A13 ---- C \
 A12 ---- B  | 74LS138   Y0' --+
 A11 ---- A /                  [AND]-- CE' 2732 (0000-0FFF)
 IO/M' -- G2A'           Y1' --+
 A15 --+                 Y2' --+
       [OR]-- G2B'             +------ spare (1000-1FFF)
 A14 --+                 Y3' --+
 +5V ---- G1             Y4' ---------- CS' 6116 (2000-27FF)
 RD' -> OE' of both;  WR' -> WE' of 6116

The lower address byte is multiplexed with data on AD7–AD0, so a 74LS373 latch enabled by ALE is first used to separate A7–A0. A15–A8 are available directly. Signals marked with a prime (RD', WR', IO/M', CE', OE') are active low.

✓ Answer checked by Claude Opus 5.5 (high effort)

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