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I have a problem understanding the following schematic, specifically the "Glue Logic(Memory Decoding)" section in the lower half middle:

circuit

This is a simple breadboard computer based on a 6502 CPU with RAM, ROM, and a D-Latch that serves as IO(seen on the lower right corner driving a single LED). The memory map is

  • $0000 - $3FFF: RAM (16K)
  • $4000 - $7FFF: I/O (16K)
  • $8000 - $FFFF: ROM (32K)

The one signal I don't understand is /RAMSEL. As shown in the schematic A15 and A14 are used to address the three chips(U2, U3, and U7). /RAMSEL feeds into /CE(chip enable) of the RAM chip(U3) which is active low, so if I what to read or write from RAM /RAMSEL has to be low(0).

Suppose I want to read the value located at $1000 in RAM(e.g., LDA $1000 in 6502 assembly). I expect the address bus to be $1000 which in binary is b0001000000000000. As can be seen the signals A15 and A14 are both low(0) making /RAMSEL high(1) through the glue logic. Therefore the signal on /CE is high, making the RAM chip not readable since it will be in high-impedance mode.

On the other hand, if I want to read from a ROM location say $F000/b1111000000000000, A15 and A14 are both high(1) and through the glue logic /ROMSEL will be low(0) (so the ROM chip can be read since its /CE pin is low) and /RAMSEL will be high(1) rendering it unreadable. Same applies for the IOSEL signal.

So as I understand this schematic, the signals for reading/writing to ROM and I/O works fine, but not for the RAM chip. Am I reading the glue logic wrong?

EDIT: This schematic is taken from this book, in case you wonder.

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  • \$\begingroup\$ I agree. The /RAMSEL is wrong for the given memory map. U5B needs to be an OR gate \$\endgroup\$
    – Icy
    Dec 1, 2015 at 10:29
  • \$\begingroup\$ I had guessed so, but I was wondering since this comes from a book and he uses the same glue logic for another project - I wouldn't expect for such an error to happen twice. So I figured I might simply be misunderstanding the schematic but seems I'm right after all. \$\endgroup\$
    – koalag
    Dec 1, 2015 at 10:41
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    \$\begingroup\$ It wouldn't be a Black Art if you didn't mislead the uninitiated... \$\endgroup\$
    – user16324
    Dec 1, 2015 at 10:58

2 Answers 2

Reset to default
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You are correct, The /RAMSEL is wrong for the given memory map. U5B needs to be an OR gate.

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    \$\begingroup\$ Thanks, that is the right answer. Using the 74HC04 to invert A15 and A14 before feeding them int U5B will do the trick. Thanks a lot! \$\endgroup\$
    – koalag
    Dec 1, 2015 at 11:01
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Instead of address lines, the output of U5-A and output of U5-C should be fed into U5-B to get the proper ram select without using another chip or gate. (Romselect NAND ioselect = ram select)

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  • \$\begingroup\$ Thanks, that would work too I guess, I might even try this once I come back to this project. Currently I'm working on a ROM emulator to go with it, looking good so far, but need to order the components. \$\endgroup\$
    – koalag
    Aug 31, 2016 at 22:04

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