Chip Select

Consider a system that has four of the 4x3 memory chips we discussed in class.

• Each chip only needs two address lines, since there are only four words in each chip. Inside each chip, the addresses would be:
  

  00

  01

  10

  11

  00, 01, 10, 11.

• The number of words for the entire system is 16: 4 in each chip. How do the chips know when a word is for them? The addresses for the entire system are:
  

  0000	0001	0010	0011
  0100	0101	0110	0111
  1000	1001	1010	1011
  1100	1101	1110	1111

• We will design the system so that the each row of addresses will correspond to a different memory chip. The row addresses can be viewed as 00XX, 01XX, 10XX, 11XX. The X's represent the address inside the chip, the other two bits are used to select one of the chips. Using a 2-to-4 decoder, the high two address lines can be used to select the appropriate chip, the low two address lines go to each memory chip.
  

A more complicated example

The example in the book in section 3.7.2 also shows how to configure chip select for a larger address space, using different size chips.

1. The first step is to do address line calculations
   • The address space is 64K.
     64K = 64 * 1K = 64 * 2^10 = 2^6 * 2^10 = 2^16
     16 address lines are needed to access all of the 64K addresses.
   • Next is to determine how many of these address lines are needed for the chip. Consider the EPROM. It is 2k in size.
     2K = 2 * 1K = 2^1 * 2^10 = 2^11
     11 address lines are needed to address all the addresses inside the EPROM.
   • A similar calculation reveals that the 2K RAM also needs 11 address lines.
   • The PIO chip only has 4 bytes inside, so it only needs 2 address lines.
2. The second step is do starting location calculations. The idea behind placing the chips is to place them at multiples of there size. This keeps the circuitry at a minimum (and the math). Therefore, a chip of size 2K will only be placed at 0K, 2K, 4K, 6K, 8K, etc.
   • Consider the EPROM
     • Calculate how many multiples of the chip size there are in the address space.
       64K/2K = 32
       There are 32 possible multiples.
       32 = 2^5
       There will be 5 bits for the chip select lines.
     • Calculate which multiple corresponds to the starting location. For the EPROM, the starting location is 0K.
       0K / 2K = 0
       Write the multiple as a binary number, using 5 places (one for each chip select line).
       00000
     • For the EPROM, the 16 bit address can be broken into 5 chip select lines, and 11 lines for inside the EPROM
       00000XXXXXXXXXXX
   • Consider the RAM
     • It is the same size as the EPROM, so it will also have 5 chip select lines.
     • Calculate the multiple that corresponds to 32K (the starting location for the RAM)
       32K/2K = 16
       Write 16 as a binary number in 5 places.
       10000
     • For the RAM, the 16 bit address can be broken into 5 chip select lines, and 11 lines for inside the RAM
       10000XXXXXXXXXXX
   • Consider the PIO.
     • Calculate how many multiples of the chip size there are in the address space.
       64K/4 = 16K
       There are 16K possible multiples.
       16K = 2^4 * 2^10 = 2^14
       There will be 14 bits for the chip select lines to the PIO.
     • Calculate the multiple that corresponds to the last 4 bytes before 64K (the starting location for the PIO). Theoretically, the PIO could be placed at any multiple of 4. To keep the math simple, we will place the PIO either at the start, or at the end of a 1K block in the address space.
       A 1K block has 10 address lines. The PIO only needs 2 of those address lines. The remainder are for chip select.
       If the PIO is placed at the start of the 1K block, there would be 8 chip select lines and 2 address lines as follows:
       00000000XX
       If the PIO is placed at the end of the 1K block, there would be 8 chip select lines and 2 address lines as follows:
       11111111XX
       This example places the PIO at the end of the 63K block.
       64K/1K = 64
       There are 64 possible multiples of 1K.
       63K/1K = 63
       Write 63 as a binary number in 6 places.
       111111
       Combining the 8 chip select lines with these 6, results in the following
       11111111111111XX
       (If this chip were placed at the start of the 63K block, its chip select lines would be
       1111100000000XX)

Chips of different sizes

The same calculations work for chips of different sizes as well. Consider a second EPROM of size 4K that starts at 40K. The EPROM now requires 12 address lines for inside the chip.

• Calculate how many multiples of the chip size there are in the address space.
  64K/4K = 16
  There are 16 possible multiples.
  16 = 2^4
  There will be 4 bits for the chip select lines.
• Calculate which multiple corresponds to the starting location. For the EPROM, the starting location is 40K.
  40K / 4K = 10
  Write the multiple as a binary number, using 4 places (one for each chip select line).
  1010
• For the EPROM, the 16 bit address can be broken into 4 chip select lines, and 12 lines for inside the EPROM
  1010XXXXXXXXXXX