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Microcomputers questions

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Microcomputers questions part 2

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Microcomputers questions

  1. 1. ozahmetdemir@gmail.com 2014 Reference : Microprocessors and Peripherals Hardware, Software, Interfacing and Applications Second Edition Barry B. Brey INTRODUCTION to MICROCOMPUTERS in ESOGU Questions about 8085 Microprocessor with Solutions Solved by Ahmet ÖZDEMİR Chapter-3/Q-5 What is the clock cycle time whwnever a 4-MHz crystal is attached to the 8085A? The clock frequency of an 8085A is one half the crystal frequency. The period of one T –cycle is the inverse of the clock frequency. At a crystal frequency of 4-MHz, the clock is 2-MHz and clock cycle time equal to 500ns. Chapter-3/Q-8 How many bytes of memory can the 8085A address directly? The 8085A MPU with 16 address lines capable of addressing 65536 (generally knowns as 64K) memory locations. Chapter-3/Q-12 What is the purpose of the signal? The control signal Read ( ) enables the output buffer , and data from the selected register are made available on the output lines. This is an active low input control signal used to read data from the memory location whose address is available on address lines whenever chip select signal is enable. This signal is available on system control bus and generated by the microprocessor or the other master in the system such as DMA controller or co-processor.
  2. 2. ozahmetdemir@gmail.com 2014 Reference : Microprocessors and Peripherals Hardware, Software, Interfacing and Applications Second Edition Barry B. Brey Chapter-3/Q-13 What is the purpose of the signal? The control signal Read ( ) enables the input buffer , and data on the input lines are written into memory cells. This is an active low input control signal used to write data to the memory location whose address is available on address lines if chip select is enable . This signal is available on system control bus and generated by the microprocessor or the other master in the system such as DMA controller or co-processor. Chapter-5/Q-16 Develop a memory system using two 2716 EPROMs and three 2732 EPROMs located in memory location 0000H through and including memory 3FFFH. D E C O D E R G 2716 0000H to 07FFH CE 2716 0000H to 0FFFH CE 2732 3000H to 3FFFH CE OE 2732 2000H to 2FFFH CE OE 2732 1000H to 1FFFH CE OE M E M O R Y 8 0 8 5 A 16 K 2 K 4 K 2 K 4 K 4 K 0000H - The 2716 EPROM is 2K and similarly 2732 EPROM is 4K. We known what the 8085A has 64K memory locations. I shown 16K memory in 8085A. 2x2716  4K 3x2732 12K -Means that 16 K memory locations are enought to illustrate memory locations 0000H through and including memory 3FFFH.
  3. 3. ozahmetdemir@gmail.com 2014 Reference : Microprocessors and Peripherals Hardware, Software, Interfacing and Applications Second Edition Barry B. Brey Chapter-3/Q-10 What is the purpose of the ALE signal? ALE is the address latch enable signal. This is a positive going pulse generated every time the 8085 begins an operation(machine cycle); it indicates that the bits on are the address bits. This signal is primarily used to latch the low order address bus, its generate a separate set of eight address lines . Means that these lines contain address bits whenever ALE is a logic 1, and data bus connections when ALE is a logic 0. Chapter-3/Q-19 What memory access time does the 8085A allow if operated at its maximum clock frequency? In the 8085A the amount of time allowed for the memory or the I/O to access data is time, which amounts to 575 ns at the highest allowable clocking rate 3 MHz.
  4. 4. ozahmetdemir@gmail.com 2014 Reference : Microprocessors and Peripherals Hardware, Software, Interfacing and Applications Second Edition Barry B. Brey Chapter-5/Q-18 Develop a memory system that uses one 2716 EPROM and one 4016 RAM. Locate them anywhere you wish. (HINT: Use incompletely specified decoding.) 2716 EPROM and 4016 RAM are 2K. We can find this dividing the last two term by two. Means that they are required totaly 4K memory locations. Therefore 8085A have 64K memory space, we must divide by 32 field. And we can take into consideration minimum field 2K in this separation. So we can use 5 digit address bits to declare our devices’ selection bits. Designing a decoder We can use 74138 for decoding 60K RAM (2K) ROM (2K) ROM RAM 0 1 2 3 4 5 6 7 ROM RAM C B A 7 4 1 3 8 Means that inverter
  5. 5. ozahmetdemir@gmail.com 2014 Reference : Microprocessors and Peripherals Hardware, Software, Interfacing and Applications Second Edition Barry B. Brey Chapter-5/Q-30 From the flowchart of figure 5-29 develop a program that will test a 1K-byte RAM residing at locations 1000H through 13FFH. We can find any error with this code about working RAM at starting microprocessors. If RAM correctly works then there is no indicate failure. But there is any indicate failure at any time, means that we have some problem about working RAM. SOLUTION Start: Clear: Set: Indicate Failure: LXI LXI XRA MVI ORA JNZ INX DCX JNZ LXI LXI MVI MVI CMP JNZ INX DCX JNZ JMP H,1000H B,03FFH A M,00H M Indicate Error H B Clear H,1000H B,03FFH M,FFH A,FFH M Indicate Failure H B Set Start START Clear all locations All clear? No Yes Set all locations All set? No Yes RETURN Indicate failure
  6. 6. ozahmetdemir@gmail.com 2014 Reference : Microprocessors and Peripherals Hardware, Software, Interfacing and Applications Second Edition Barry B. Brey Chapter-6/Q-10 Develop a decoder that will generate a logic one for page EDH. This decoder generate a logic one when the address lines have the number EDH. We can illustrate the hexadecimal number ED to 1110 1101. The other bits are not important, means that they are don’t care. My decoder is: or ED strobe
  7. 7. ozahmetdemir@gmail.com 2014 Reference : Microprocessors and Peripherals Hardware, Software, Interfacing and Applications Second Edition Barry B. Brey Chapter-6/Q-12 Design a circuit that will develop eight I/O strobes at memory-mapped I/O locations 10XXH. Make certain to label the address ranges of your output strobes. In this memory range there are 256 bytes. If we have eight I/O strobes , this range will be divided into eight field. Therefore every field has 32 bytes memory space. Address bits are illustrate this range are and are illustrate in this range when after division. The other address bits are seletion pins our devices. My decoder is : 10FFH 1000H 0001.0000.1111.1111 0001.0000.0000.0000 0 1 2 3 4 5 6 7 C B A 7 4 1 3 8 Means that inverter without
  8. 8. ozahmetdemir@gmail.com 2014 Reference : Microprocessors and Peripherals Hardware, Software, Interfacing and Applications Second Edition Barry B. Brey Chapter-6/Q-19 Develop initialization dialog for the 8155 in problem 15 if ports A and C were to function as inputs, port B as an output, and the timer were to produce a series of pulses at 1/374 of the input rate. Reference Question 15: Interface an 8155 to function at isolated I/O space CXH. Firstly we will draw the interface to understand solution. Isolated I/O space CXH tells us what the memory range is starting from C000H to C0FFH of the 256x8 static RAM in the 8155. And the interface of this function can drawn as ; A B C RAM TIMER 8155 8085A Firstly initialized my timer with count 374; 1 0 0 0 10100 110 0 1 1 0 MSB LSB Cont. Square Wave Our input rate 374 41H 76H
  9. 9. ozahmetdemir@gmail.com 2014 Reference : Microprocessors and Peripherals Hardware, Software, Interfacing and Applications Second Edition Barry B. Brey Chapter-6/Q-25 Connect seven solenoids to an output port of the 8155 and develop the hardware and software to control these seven 10-ms solenoids. Your subroutine should fire the solenoids in the pattern passed to it in the accumulator register. Firstly we have interface an 8155 to function at isolated I/O space CXH. Because of this reason we can also use this one in new question. But this time, seven solenoids will be connected my output port B. START: MVI A,76H ;Load LSB of timer OUT C4H MVI A,41H ;Load MSB og timer and its mode(as continuous square wave) OUT C5H MVI A,C2H ;Timer is started, Ports A and C choose inputs, Port B is output OUT C0H CMD EQU C0 PA EQU C1 PB EQU C2 PC EQU C3 LSB EQU C4 MSB EQU C5 A B C RAM TIMER 8155 8085A Not connected anywhere connected solenoids CMD EQU C0 PA EQU C1 PB EQU C2 PC EQU C3 LSB EQU C4 MSB EQU C5 INIT : MVI A,02H OUT C0H XRA A OUT C2H FIRE: OUT C2H MVI A,10H FIRE2: CALL DELL1 DCR A JNZ FIRE1 XRA A ;Port B selected output port ;Accumulator will reset why turn off solenoids ;Send to pattern to solenoids ;Waste 10 milisecond ;Turn off solenoids
  10. 10. ozahmetdemir@gmail.com 2014 Reference : Microprocessors and Peripherals Hardware, Software, Interfacing and Applications Second Edition Barry B. Brey If there are any error or you have some questions you can contact me. dept. of computer engineering in esogu ozahmetdemir@gmail.com

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