| ??? 02/22/07 13:55 Modified: 02/22/07 13:55 Read: times |
#133501 - Depends on buffer... Responding to: ???'s previous message |
Mike Stegmaier said:
Andy, do the diodes represent logic high?
Yes. Follow the link, there's a truth table for the example. 8 bytes per '138, so you'll need 9 '138's, creating 8 rom banks, and another to select among them by activating the output of one of 8 tristate inverting buffers. If you're going to buy chips, you should be able to find 74HCT154's, which will cut your logic count. You can still use the '138 you already have to decode which buffer to turn on, you'll just have 4 unused lines. You're going to need as many as 512 1N914 diodes to store all 1's. Your program will be less than this of course. Also... You need a slower xtal. The 18Mhz one will not work with a diode board, even if you know it works with your MCU. You're looking at drilling over 700 pin holes in your PCB, pick up a couple spare drill bits. While you're there, ask if they have any AT89S52's. They'll cost about the same as 2 74HCT154's. |
