Question

3. (15 Pts.) In this problem you are given a microprocessor with 24-bit address bus and 8-bit data bus. a. What is the addressing space of this microprocessor? 113 16 MB b. how many bytes are contained in the sub-space starting at address C00 000h and ending at address DFF FFFH? Express your answer in KB and MB. DFFFEE CoO DOU - TFFFFF in ke 20 4SKe c. You would like to interface a single 4 MB memory IC to this microprocessor so that the last address of the IC is at FFE EFEh Assume the IC to have CS1 (active-low) and CS2 (active high) in addition to the OE and WE (both active-low). What is the starting address of this IC's space ? Draw your interface. No foldback it to occur. Show all details. Ic : start : ooo00 END: FFFFFF An MEMB MEMW A. ir d. Unfortunately you check your store and it has 1 MB memory ICs only Repeat your interface using multiple ICs instead. Indicate the starting and ending of each IC. Use the 3- to-8 decoder (LS 138) for a more efficient design Show your interface at the paper back

Answer 1

3' (a) 24 bit adduo Ans: 24 20 4 2 X2 16 X 2620. %3D 16 MB (b) oddut-starting Juime oddrs = undimg d-starting addsus DEF FFF -CO0000 DF FFFF CO0000 I.FF FFF is Ne: of Memory uze = 22 = 2MB 21 2.2 = 2048KB : %3D 20 4MB = A23 Ao1 A21 Ao 2 A22 dcsi A23 CS22MB. D4 Do. addus:- tlanting A23 A2 Azı Az0 A19 Ag- A - A16 A15 A, tositing addrus-800000H

A23 A22 A21 A20 Slart Ao lait .F AzI A20 A A22 10 = (A23t A22t Azı + Azo) (A23+Ag2+ Azit A20) F A2 A22 A21O A2o Day zveing this funetion at chip salaet com get the suquisud addut