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![Stack Related Instructions
• PUSH B
Let BC=3010, B=30h, C=10h
suppose SP initialized at FFFF h
after execution of instruction PUSH B
SP=SP-1=FFFF-1=FFFE
B
[FFFE] =30h
again SP=SP-1=FFFE-1=FFFD
C
[FFFD]=10h
SP=[FFFD]
5-Mar-14
Prof.Nitin Ahire
73](https://image.slidesharecdn.com/microprocessor-140304211733-phpapp01/85/Microprocessor-73-320.jpg)
![Stack Related Instructions
• POP B
initially B=20, C=40h
SP at=[FFFD]=10h
at=[FFFE]=30h
After execution of POP B
SP=[FFFD]=10h
[C]
SP=SP+1=[FFFE]=30h
[B]
Again SP=SP+1=[FFFF]
Now B=30h, C=10h and SP=[FFFF]
5-Mar-14
Prof.Nitin Ahire
74](https://image.slidesharecdn.com/microprocessor-140304211733-phpapp01/85/Microprocessor-74-320.jpg)
![Delay using 8 –bit counter
• Formula for delay value
Td=[ M + {(count) X N} -3 ] T
M=No. of T state out side the loop
N=No. of T state inside the loop
M=10+7=17 T; N=10+4=14T
5-Mar-14
Prof.Nitin Ahire
93](https://image.slidesharecdn.com/microprocessor-140304211733-phpapp01/85/Microprocessor-93-320.jpg)
![Delay using 8 –bit counter
• Td max=
=
=
=
[17+[ {255} X 14 ] -3] T
3584 T
3584 X 0.5 microsecond
1792 microsecond
(FF)=(255)
5-Mar-14
Prof.Nitin Ahire
95](https://image.slidesharecdn.com/microprocessor-140304211733-phpapp01/85/Microprocessor-95-320.jpg)
![Delay using 16 bit counter
Td=[ M + {(count) X N} -3 ] T
M=No. of T state out side the loop
N=No. of T state inside the loop
M=20 T; N=24T
5-Mar-14
Prof.Nitin Ahire
97](https://image.slidesharecdn.com/microprocessor-140304211733-phpapp01/85/Microprocessor-97-320.jpg)
![Delay using 16 bit counter
• Td max= [20+[ {65535}X 24 ] -3] T
= 1572857 T
=1572857 X 0.5 microsecond
= 78642 microsecond
(FFFF)=(65535) largest count.
5-Mar-14
Prof.Nitin Ahire
98](https://image.slidesharecdn.com/microprocessor-140304211733-phpapp01/85/Microprocessor-98-320.jpg)
Uploaded byNitin Ahire
Microprocessor
The document provides an overview of microprocessors and microcontrollers. It discusses the functional blocks of a microprocessor system including the CPU, memory, and I/O devices. It then covers various number systems used in microprocessors like binary, hexadecimal, and their advantages. The document reviews logic devices like tri-state buffers and latches. It provides details on the 8085 microprocessor including its features, registers, interrupts, pin diagram, instruction format and classification.
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Microprocessor
- 1. Microprocessor & Microcontroller- I T.E Sem V (Rev) Prof. Nitin Ahire XIE, Mahim
- 2. Overview of Microprocessor MEMORY MICROPROCESSOR (CPU) INPUTOUTPUT (I/O) DEVICE 5-Mar-14 Prof.Nitin Ahire 2
- 3. Functional block Diagram •INPUT OUTPUT (I/O) DEVICE I/P :Key board, scanner, card reader etc O/P : Display, printer LED etc • MEMORY RAM, ROM • MICROPROCESSOR Central Processor Unit ( CPU ) include ALU, Timing & control unit for synchronizations 5-Mar-14 Prof.Nitin Ahire 3
- 4. Number System • Decimalnumber system (DNS)(10) 0,1,2 ……,9,10 • Binary number system(2) 0,1,10,11,100 • Hexadecimal number system (16) 0,1,2,…..,9,A,B,C,D,E,F,10,11 • Advantages of Hex No over BCD No system (1111 1111)2 (FF)16 (255)10 5-Mar-14 Prof.Nitin Ahire 4
- 5. Review for LogicDevices • Tri State Devices : 3 States are logic 1, logic 0 & high impedances state ( Z ) Enable Active high Enable Active Low 5-Mar-14 Prof.Nitin Ahire 5
- 6. Tri-State Buffers • Animportant circuit element that is used extensively in memory. • This buffer is a logic circuit that has three states: – Logic 0, logic1, and high impedance. – When this circuit is in high impedance mode it looks as if it is disconnected from the output completely. The Output is Low The Output is High High Impedance 6
- 7. The Tri-State Buffer •This circuit has two inputs and one output. – The first input behaves like the normal input for the circuit. – The second input is an “enable”. • If it is set high, the output follows the proper circuit behaviour. • If it is set low, the output looks like a wire connected to nothing. Input Output Enable OR Input Output Enable 7
- 8. Review for LogicDevices • Buffer e.g. 74LS244(unidirectionl) & 74LS245(Bidirection) • Buffer is a logic CKT that amplifies the current or power • It has one I/P line and one O/P line • The logic level of O/P is the same as that of the I/P • Basically used as to increase the driving capacity of logic CKT simple buffer 5-Mar-14 Active low buffer Prof.Nitin Ahire 8
- 9. D – F/F(Latch) Q clk I/P I/P D F/F clk Q 5-Mar-14 O/P Prof.Nitin Ahire 9
- 10. Introduction to 8085 •CPU built into a single semiconductor chip is called as microprocessor • The microprocessor work as a brain of a computer • It consist of ALU, registers and control unit • The microprocessor are usually characterized by speed, word length (bit), architecture, instruction set Etc 5-Mar-14 Prof.Nitin Ahire 10
- 11. 8085 Features • 8085is a 8-bit processor • Frequency of operation a) 8085 --- 3Mhz b) 8085-2 --- 5Mhz c) 8085-1 --- 6Mhz • 8085 has 16 bit address bus to access memory • 8 bit address bus to access I/O location 5-Mar-14 Prof.Nitin Ahire 11
- 12. 8085 Features • Itrequired only single +5V power supply • 8085 has following registers a) 8 bit accumulator b) six 8- bit general purpose registers c) 8-bit flag register d) 16 –bit PC and SP • It has 5 hardware and 8 software interrupt • 8085 required 6 Mhz crystal • It can transmit and receive serial data 5-Mar-14 Prof.Nitin Ahire 12
- 13. 1 X1 2 X2 3RESET OUT 4 SOD 5 SID 6 TRAP 7 RST 7.5 8 RST 6.5 9 RST 5.5 10 INTR 11 INTA 12 AD0 13 AD1 14 AD2 15 AD3 16 AD4 17 AD5 18 AD6 19 AD7 20 VSS 5-Mar-14 8085 PIN DIG 40 VCC 39 HOLD 38 HLDA 37 CLOCK (OUT) 36 RESET IN X1 Crystal 6 MHz 35 READY 34IO/M 33 S1 32 RD X2 8085 31 WR (3 MHz ) 30 ALE 29 S0 28 A15 27 A14 26 A13 25 A12 24 A11 PIN DIG 23 A10 8085 22 A9 21 A8 Prof.Nitin Ahire 13
- 14. X1 Serial I/O ports X2 vcc SID CLKCKT & P.S. H.O.A.B A8-A15 SOD AD0- AD7 TRAP Externally Initiated Signal RST 7.5 RST 6.5 RST 5.5 INTR READY ALE 8085 Functional Pin Diagram HOLD 5-Mar-14 S0 S1 Control & Status Signal IO/M RD WR RESET IN External Acknowledge Signal Multiplexed A/D Bus INTA CLK OUT HLDA Prof.Nitin Ahire RESET OUT 14
- 15. INTA RST 7.5to 5.5 TRAP SID SOD INTR P.S +5V GND Serial I/O Control Interrupt control 8 bit Internal BUS MUX W8 I.R. 8 F/F 5 ALU 8 Inst. Decoder & M/C Encoder DECODRE Accumulator 8 Temp. Reg Z8 B C D E H L SP 16 PC 16 Internal latch CLK OUT RESET IN RESET OUT X1 Timing and control unit A/D. Buffer Add. Buffer X2 READY WR RD ALE S0 S1Prof.Nitin Ahire HOLD IO/M HLDA 5-Mar-14 AD0-AD7 A15-A8 15
- 16. INTA RST 7.5to 5.5 TRAP SID SOD INTR P.S +5V GND Serial I/O Control Interrupt control 8 bit Internal BUS MUX I.R. 8 F/F 5 ALU 8 Inst. Decoder & M/C Encoder DECODRE Accumulator 8 Temp. Reg 8 W8 B Z8 C D E H L SP 16 PC 16 Internal latch CLK OUT RESET IN RESET OUT X1 Timing and control unit A/D. Buffer Add. Buffer X2 READY WR RD ALE S0 S1Prof.Nitin Ahire HOLD IO/M HLDA 5-Mar-14 AD0-AD7 A15-A8 16
- 17. Registers • The registercontains a set of binary storage cells/Flip Flop • 6 general purpose 8 bit Reg. B,C,D,E,H&L (or can be used as pair of 16 bit reg. like BC,DE,HL) • W & Z (Temp reg.) • 16 bit Reg are PC And SP • 8 bit flag register 5-Mar-14 Prof.Nitin Ahire A F B C D E H L SP PC 17
- 18. Interrupts • Hardware interrupt Trap(Non Mask able) (vectored) RST 7.5(Mask able) (vectored) RST 6.5 (Mask able) (vectored) RST 5.5(Mask able) (vectored) INTR (Mask able) (Non vectored) • Software interrupt RST 0 to RST 7 All are vectored interrupt 5-Mar-14 Prof.Nitin Ahire 18
- 19. Interrupts • 8085 has5 hardware interrupts 8 software interrupts • All software interrupt are vectored • Out of 5 hardware interrupt 4 are vector and 1 is non vector also 4 are maskable and one is non mask able 5-Mar-14 Prof.Nitin Ahire 19
- 20. De multiplexing OfAD0-AD7 ALE Latch AD0-AD7 A0-A7 8085 D0-D7 IO/M A8-A15 5-Mar-14 Prof.Nitin Ahire 20
- 21. De multiplexing OfAD0-AD7 ALE AD0-AD7 Latch A0-A7 8085 IO/M D0-D7 5-Mar-14 Prof.Nitin Ahire 21
- 22. Differentiate between IO/M ALE AD0-AD7 Latch A0-A7 IOdevice 8085 D0-D7 A0-A7 IO/M Memory A8-A15 5-Mar-14 Prof.Nitin Ahire 22
- 23.
- 24. Flags Register (8 bit ) S D7 Z D6 -- AC D5 D4 • S –sign flag P D3 D2 -- D1 C D0 (for signed number) if D7=1 the number in accumulator will be –ve number D7=0 the number in accumulator will be +ve number • Z – zero flag 5-Mar-14 if D6=1The zero flag is set if the result in accumulator is zero Prof.Nitin Ahire 24
- 25. Flags Register (8 bit ) S D7 Z D6 AC D5 D4 P D3 D2 C D1 D0 AC – Auxiliary carry in the arithmetic operation, when the carry is generated digit D3 and passed on digit D4 the AC flag is set P– parity flag after an arithmetic and logical operation, if the result has even number of ones the flag is set if it has odd numbers of ones, the flag is reset CY – Carry flag if an arithmetic operation results in carry, the carry flag is set otherwise it is reset. The carry flag also serves as a barrow flag for subtraction 5-Mar-14 Prof.Nitin Ahire 25
- 26. Subtraction process in8085 • 1 : find 1‟s complement of the subtrahend • 2 : find 2‟s complement of the subtrahend • 3 : Adds 2‟s complement of the subtrahend to the minuend • 4 : complements the CY flag. These steps are invisible to the user, only the result is available to the user. For unsigned number if CY is reset the result is positive and if CY is set the result is negative(2‟complement) 5-Mar-14 Prof.Nitin Ahire 26
- 27. Sign flag (usedonly for sign No.) • Sign flag: This flag is used with signed numbers in the arithmetic operation. With sign number, bit D7 is reserved for indicating the sign and the remaining 7 bit are used to represent the magnitude of a number • Sign flag is irrelevant for unsigned number 5-Mar-14 Prof.Nitin Ahire 27
- 28. Instruction, Data format andstorage • Part of instruction each instruction has two parts 1 opcode: one is the task to be perform (operational code) 2 operand: data to be operated on (data) The data can be specified in the various form it may in the memory or I/O or in the instruction it self. 5-Mar-14 Prof.Nitin Ahire 28
- 29. Opcode • Opcode :operational code • Operand : Data • Mnemonics : Instructions Memory Locations Opcode 2000 2001 2002 2003 2004 3E 20 67 12 4F 5-Mar-14 Mnemonics Operand MVI A, 20 MVI B, 12 MOV C, A Prof.Nitin Ahire 29
- 30. DATA BUS Internal DataBUS MEMORY LOCATION B C (A) D E INST. DECODER H L SP CONTROL LOGIC DECODER ALU 3E 20 67 12 4F 2000 2001 2002 2003 2004 PC (2000) ADD BUS MERD 3E 5-Mar-14 Prof.Nitin Ahire 30
- 31. DATA BUS Internal DataBUS MEMORY LOCATION B C (A) D E INST. DECODER 3E H L SP CONTROL LOGIC DECODER ALU 3E 20 67 12 4F 2000 2001 2002 2003 2004 PC (2001) ADD BUS MERD 20 5-Mar-14 Prof.Nitin Ahire 31
- 32. DATA BUS Internal DataBUS MEMORY LOCATION B C (A) D E INST. DECODER H L 20 SP CONTROL LOGIC DECODER ALU 3E 20 67 12 4F 2000 2001 2002 2003 2004 PC (2002) ADD BUS MERD 67 5-Mar-14 Prof.Nitin Ahire 32
- 33. DATA BUS Internal DataBUS MEMORY LOCATION B C (A) D E INST. DECODER H L 20 SP CONTROL LOGIC DECODER ALU 3E 20 67 12 4F 2000 2001 2002 2003 2004 PC (2003) ADD BUS MERD 12 5-Mar-14 Prof.Nitin Ahire 33
- 34. DATA BUS Internal DataBUS MEMORY LOCATION B C (A) D E INST. DECODER H L 20 SP CONTROL LOGIC DECODER ALU 3E 20 67 12 4F 2000 2001 2002 2003 2004 PC (2004) ADD BUS MERD 4F 5-Mar-14 Prof.Nitin Ahire 34
- 35. Instruction classification • “Instruction”is a command to the microprocessor to perform a given task on specified data”. • The instruction can be classified into following fundamental categories 1 Data transfer 2 Arithmetic & Logical operation 3 Branching operation 4 Machine control operation 5-Mar-14 Prof.Nitin Ahire 35
- 36. Instruction classification • 1Data transfer (copy) basically used to copies data from source to destination without modifying the content of the source like, Opcode operand MOV rd, rs MVI r, 8-bit IN 8 bit port add. OUT 8 bit port add. 5-Mar-14 Prof.Nitin Ahire 36
- 37. Instruction classification • 1Data transfer (copy) • LXI Rp, 16-bit add. • MOV R,M • MOV M,R • LDA 16-bit add. • STA 16-bit add. • LDAX R* • STAX R* • LHLD 16-bit add • SHLD 16-bit add *R – Register pair 5-Mar-14 Prof.Nitin Ahire 37
- 38. Instruction classification • Arithmeticoperation These instruction perform arithmetic operation such as addition subtraction, increment, decrement. • ADD R • ADI data • ADC R • ADC M • ACI data • DAD Rp 5-Mar-14 Prof.Nitin Ahire 38
- 39. Instruction classification • • • • • • • SUB R SUBM SBB R SBB M SUI Data SBI Data DAA 5-Mar-14 Prof.Nitin Ahire 39
- 40. Instruction classification • INRR • DCR R • INR M • DCR M • INX Rp • DCX Rp 5-Mar-14 Prof.Nitin Ahire 40
- 41. Instruction classification • Logicalinstruction. These instruction perform various logical operation with the content of the accumulator e.g. 1) AND,OR,EX-OR(ANA R,ANI Data) 2) Rotate (RAL,RAR,RLC,RRC) 3) Compare (CMP B,CPI Data) 4) Complement (CMC, CMA,STC) 5-Mar-14 Prof.Nitin Ahire 41
- 42. Instruction classification • Branchingoperation These group of instruction alter the sequence of program execution either conditionally or unconditionally e.g. JUMP (conditionally or unconditionally) CALL & RET (conditionally or unconditionally) 5-Mar-14 Prof.Nitin Ahire 42
- 43. Instruction classification • Machinecontrol instruction These instruction control the machine function such as Halt (HLT), interrupt (RST 1) or do noting (NOP) 5-Mar-14 Prof.Nitin Ahire 43
- 44. Addition of two8 bit number (30H + 40H) LXI H, C200H; load HL pair by C200h MOV A,M; Move 1st No. in the Reg. A INX H ; increment the HL pair by 1 ADD M; Add A+(M)=A INX H; increment the HL pair by 1 MOV M,A; move the result in M HLT; Stop 5-Mar-14 Prof.Nitin Ahire 44
- 45. Subtraction of two8 bit number (10H-05H) LXI H, C200H; load HL pair by C200h MOV A,M ;Move 1st No. in the Reg. A INX H; increment the HL pair by 1 SUB M ; Subtract A-(M)=A INX H ; increment the HL pair by 1 MOV M,A ; move the result in M HLT ; Stop 5-Mar-14 Prof.Nitin Ahire 45
- 46. Addition of two16 bit number ( 1234H+4321H ) LHLD C200H ; XCHG ; LHLD C202H ; DAD D ; SHLD C204H ; HLT ; 5-Mar-14 Here L=34, H=12 exchange HL with DE Here L= 21, H=43 DE+HL = HL store the result stop Prof.Nitin Ahire 46
- 47. Write a programto transfer a block of data from C550H to C55FH. Store the data from C570H to C57FH . LXI H ,C550H LXI B ,C570H MVI D,0FH UP MOV A,M STAX B INX H INX B DCR D JNZ UP RST1 5-Mar-14 Prof.Nitin Ahire 47
- 48. Minimum mode system(8085) VCC A0-A15 X1 A8-A15 ALE X2 RESETOUT AD0-AD7 RESET IN TRAP RST 7.5 RST 6.5 RST 5.5 INTR 74 LS 373 D0-D7 A0-A7 8 0 8 5 Memory INTA READY SOD SID 5-Mar-14 MERD IO/M RD WR I/O 74 LS 138 MEWR IORD IOWR Prof.Nitin Ahire 48
- 49. Instruction classification • The8085 instruction set is classified into the following 3 group according to word size or byte size 1) 1- byte instruction 2) 2- byte instruction 3) 3 –byte instruction 5-Mar-14 Prof.Nitin Ahire 49
- 50. 1- byte instruction •A 1- byte instruction includes the opcode and the operand in the same byte e.g. Opcode operand hex code 1 MOV C, A (4F) (opcode) 2 ADD B (80) (Data) each instruction required 1 memory location ( 8-bit) 5-Mar-14 Prof.Nitin Ahire 50
- 51. 2- byte instruction •In the 2- byte instruction the first byte specifies the operation code and the second byte specifies the operand e.g. Opcode operand hex code MVI A, 12H 3E (opcode) 12 (Data) These instruction required 2 memory location 5-Mar-14 Prof.Nitin Ahire 51
- 52. 3- byte instruction •In 3-byte instruction the first byte specifies the opcode and the following 2 bytes specify the 16bit address e.g. Opcode operand hex code LDA 2050 3A (Opcode) 50 (Data) 20 (Data) Note the second byte is the lower address and the third byte is the high order address These instruction required 3 memory location 5-Mar-14 Prof.Nitin Ahire 52
- 53. Addressing mode • Thedifferent methods (mode) to select the operands (address) are called addressing mode • For 8085 they are 1 Immediate addressing 2 Register addressing 3 Direct addressing 4 Indirect addressing 5 Implied addressing 5-Mar-14 Prof.Nitin Ahire 53
- 54. Addressing mode 1 Immediateaddressing In the immediate addressing mode the data is specified in the instruction it self. The immediate addressing mode instruction are either 2- byte or 3- byte long. The instruction contain the letter “I” indicate the immediate addressing mode. e.g. 1 MVI A,12h 2 LXI H,2000h 5-Mar-14 Prof.Nitin Ahire 54
- 55. Addressing mode 2 Registeraddressing mode In register addressing mode the source and destination operands are general purpose registers The register addressing mode instructions are generally of 1 –byte e.g. 5-Mar-14 1 MOV A,B 2 ADD B 3 PCHL Prof.Nitin Ahire 55
- 56. Addressing mode 3 Directaddressing In the direct addressing mode the 16 bit address of the data or operand is directly specified in the instruction These instruction are 3 –byte instruction. e.g. 1 LDA 2000h 2 STA 2060h 5-Mar-14 Prof.Nitin Ahire 56
- 57. Addressing mode 4 Indirectaddressing In the Indirect addressing mode the instruction reference the memory through the register pair i.e. the memory address where the operand/data is located is specified by the register pair e.g.1 MOV A,M 2 LDAX B 5-Mar-14 Prof.Nitin Ahire 57
- 58. Addressing mode 5 Impliedaddressing • The Implied mode of addressing does not required any operand • The data is specified within the opcode itself • Generally these instructions are 1-byte instruction • The data is supposed to be present in the Accumulator e.g. 1 RAL 2 CMC 5-Mar-14 Prof.Nitin Ahire 58
- 59. Timing diagram • Forbetter understanding of each instruction it is very essential to understand the Timing diagram of each instruction. • The graphical representation of each instruction with respective to time i.e. CLOCK is called “Timing Diagram” 5-Mar-14 Prof.Nitin Ahire 59
- 60. Timing diagram • Instructioncycle (IC) : The essential step required by CPU to fetch and execute an instruction is called IC IC=FC+EC • Machine cycle (MC) : Time required by microprocessor to complete the operation of accessing memory or I/O device is called MC. • T –state :Each clock cycle is called T-state 5-Mar-14 Prof.Nitin Ahire 60
- 61. Timing diagram • • • 1. 2. 3. 4. 5. 5-Mar-14 The µPoperates with reference to clock signal. The rise and fall of the pulse of the clock gives one clock cycle. Each clock cycle is called a T state and a collection of several T states gives a machine cycle. Important machine cycles are : Op-code fetch. Memory read. Memory write. I/O-read. I/O write. Prof.Nitin Ahire 61
- 62.
- 63. Timing diagram • Step1 (State T1) In the state, 8085 sends the status signals, IO/M=0, S1=1 and S0=1 • The 8085 send a 16 bit address on A8-A15 and AD0-AD7 • The high order bytes of PC is placed on the A8-A15 lines, and it remain there upto T3 state. The low order bytes of PC placed on the AD0-AD7,lines which remain there only for T1 • During this state, ALE gives a positive pulse signal is used to latch the add A0-A7. • No control signal is generated in state. 5-Mar-14 Prof.Nitin Ahire 63
- 64. Timing diagram • Step2(T2):The content of PC lower will disappear on AD0-AD7 lines, so the same line can be used as data line . The contents of A0-A7 are still available for memory from external Latch. • The control signal RD is made low by the processor which enables the read ckt of addressed memory device. • Then the memory device send the content on the data bus AD0-AD7 • In addition to these the processor increments PC content by one 5-Mar-14 Prof.Nitin Ahire 64
- 65. Timing diagram • Step3(T3):during this cycle the data from memory (opcode) is transfer in the instruction Reg. and RD control signal made HIGH 5-Mar-14 Prof.Nitin Ahire 65
- 66. Timing diagram • Step4 (T4): the microprocessor perform only internal operation. The opcode decoded by the CPU and 8085 decide 1) Whether it should enter T5 and T6 states or not 2) How my bytes of instruction it is? If instruction doesn‟t required T5 &T6 states, it go to the next MC 5-Mar-14 Prof.Nitin Ahire 66
- 67. Timing diagram • Step5 (T5 &T6): T5 and T6 states, states are required to complete decoding and some operations inside the 8085 it depend on the type of instruction 5-Mar-14 Prof.Nitin Ahire 67
- 68. Timing diagram • 1) 2) 3) 4) 5) 6) 7) 8) Following instructionrequired the T5 & T6 States CALL CALL Conditionally DCX RP INX RP PCHL PUSH RP SPHL RET Conditionally 5-Mar-14 Prof.Nitin Ahire 68
- 69. Timing diagram • Machinecycle (type) 1 opcode fetch 2 operand fetch 3 Memory read 4 Memory write 5 I/O read 6 I/O write 7 Interrupt Ack M-cycle 8 Ideal M-cycle 5-Mar-14 Prof.Nitin Ahire 69
- 70. Timing diagram • Followinginstruction required T5 & T6 states for the opcode fetch MC 1 CALL 2 CALL conditional 3 DCX Rp 4 INX Rp 5 PCHL 6 SPHL 7 PUSH Rp 8 RET conditional All other instruction except the above instruction required opcode fetch of T1 to T4 states only. 5-Mar-14 Prof.Nitin Ahire 70
- 71. Stack control andbranching group • Stack is the reserved area of the memory in RAM where temporary information may be stored • Stack pointer (SP): an 16-bit SP is used to hold the address of the most recent stack entry. It work on the principle of LIFO or FILO. 5-Mar-14 Prof.Nitin Ahire 71
- 72. Stack Related Instructions •LXI SP,16-bit address • PUSH Rp • POP RP • SPHL ( HL SP) • XTHL ( HL SP) • PCHL ( HL PC) 5-Mar-14 Prof.Nitin Ahire 72
- 73. Stack Related Instructions •PUSH B Let BC=3010, B=30h, C=10h suppose SP initialized at FFFF h after execution of instruction PUSH B SP=SP-1=FFFF-1=FFFE B [FFFE] =30h again SP=SP-1=FFFE-1=FFFD C [FFFD]=10h SP=[FFFD] 5-Mar-14 Prof.Nitin Ahire 73
- 74. Stack Related Instructions •POP B initially B=20, C=40h SP at=[FFFD]=10h at=[FFFE]=30h After execution of POP B SP=[FFFD]=10h [C] SP=SP+1=[FFFE]=30h [B] Again SP=SP+1=[FFFF] Now B=30h, C=10h and SP=[FFFF] 5-Mar-14 Prof.Nitin Ahire 74
- 75. Subroutines • Whenever weneed to use a group of instruction several times throughout a program there is a way we can avoid having to write the group of instructions each time we want to use them. • one way is to write the group of instruction separately, Called Subroutines • whenever we want to execute that group of instruction we can call that Subroutine. • The return address has to be stored back on the stack memory 5-Mar-14 Prof.Nitin Ahire 75
- 76. Subroutines e.g 6FFF 31 LXI SP, FFFFh 7000 CD CALLC200h 7001 00 7002 C2 7003 Next instruction When this instruction is executed PC contents 7003h (next instruction) will stored on to the stack and microprocessor will load PC with C200h and start executing instruction from C200h 5-Mar-14 Prof.Nitin Ahire 76
- 77. Subroutines • If SP=FFFF h • CALL C200. (SP-1)=Pc H (SP-2)=Pc L SP=SP-2 PC=new C200 5-Mar-14 PC 70 Prof.Nitin Ahire stack (memory) FFFF 03 70 FFFE 03 FFED 77
- 78. Subroutines • Conditional Callinstructions When condition is true, then CALL at NEW address else execute the next instruction of the program 1) CZ Add 2) CNZ Add 3) CP Add 4) CM Add 5) CPO Add 6) CPE Add 7) CC Add 8) CNC Add If condition false 9T True16T 5-Mar-14 Prof.Nitin Ahire 78
- 79. Subroutines • RET unconditionally 1SP (PC L) PC Stack 2 SP+1 (PC H) 70 03 3 SP+2=SP 70 03 FFFF FFFE FFFD Initially SP at FFFD After execution of RET SP=FFFF 5-Mar-14 Prof.Nitin Ahire 79
- 80. Subroutines • RET conditionally Whencondition is true, then RET at the main program 1) RZ 2) RNZ 3) RP 4) RM 5) RPO 6) RPE 7) RC 8) RNC 5-Mar-14 Prof.Nitin Ahire 80
- 81. Nested Subroutines • Wheneverone subroutine calls another subroutine in order to complete a specific task, the operation is called as nesting. The First subroutine may call the second subroutine and in turn the second subroutine may called first or third subroutine such routines called NESTED subroutines 5-Mar-14 Prof.Nitin Ahire 81
- 82. Nested subroutines . . call sub1 . . . 5-Mar-14 sub 1 . call sub 2 . . RET Prof.Nitin Ahire sub 2 . . . . RET 82
- 83. Nested Subroutines • Thereare two kinds of subroutines 1) Re-entrant subroutines. 2) Recursive subroutines. 5-Mar-14 Prof.Nitin Ahire 83
- 84. Re entrant Subroutines 1)Re-entrantSubroutine It may happened a Subroutine „1‟ is called from main program and „2‟ Subroutine is called from Subroutine „1‟ and Subroutine „2‟ may called Subroutine „1‟ then the program re entrant in Subroutine „1‟ this is called re-entrant Subroutine 5-Mar-14 Prof.Nitin Ahire 84
- 85. Re entrant subroutine • MainProgram SUB 1 SUB 2 CALL CALL • • Next line RET 5-Mar-14 RET Prof.Nitin Ahire 85
- 86. Recursive subroutine • Aprocedure which called it self is called a recursive subroutine • The recursive subroutine loop takes long time to execute • In this type of subroutine we normally define N ( recursive depth) it is decrement by one after each subroutine is call until N=0 5-Mar-14 Prof.Nitin Ahire 86
- 87. Recursive subroutine • Main ProgramN=3 SUB 1 N=2 N=1 SUB 2 SUB 3 CALL CALL • • CALL Next line RET 5-Mar-14 RET RET Prof.Nitin Ahire 87
- 88. Software Delay • Delay: operating after an some time interval. • Microprocessor take fixed amount of time to execute each instruction • Microprocessor driven by fixed frequency (crystal) • So using instruction we can generate a Delay. 5-Mar-14 Prof.Nitin Ahire 88
- 89. Software Delay • E.g.delay using NOP instruction NOP ( 1-byte instruction- 4T state) assume crystal freq= 4Mhz.. CLK freq = 2Mhz (T=0.5 microsecond) Delay using NOP = 4 X 0.5 microsecond = 2.0 microsecond 5-Mar-14 Prof.Nitin Ahire 89
- 90. Software Delay • Ifwe want the more delay than 4T,then we go on increasing NOP after NOP. • Impractical (size of program increase) 5-Mar-14 Prof.Nitin Ahire 90
- 91. Delay using 8–bit counter Delay Initializes 8-bit counter Decrement counter No yes RET 5-Mar-14 Prof.Nitin Ahire 91
- 92. Delay using 8–bit counter • MVI C, Count 7T up: DCR C 4T JNZ :up 10T/7T RET 10T The loop is executed ( count-1 ) times 5-Mar-14 Prof.Nitin Ahire 92
- 93. Delay using 8–bit counter • Formula for delay value Td=[ M + {(count) X N} -3 ] T M=No. of T state out side the loop N=No. of T state inside the loop M=10+7=17 T; N=10+4=14T 5-Mar-14 Prof.Nitin Ahire 93
- 94. Delay using 8–bit counter JNZ instruction required 10 or 7 T state based on the condition ( Z=0 or 1) (when condition is satisfied it take 10T state and if not satisfied it take 7T state) so 3 T must be subtracted from total value 5-Mar-14 Prof.Nitin Ahire 94
- 95. Delay using 8–bit counter • Td max= = = = [17+[ {255} X 14 ] -3] T 3584 T 3584 X 0.5 microsecond 1792 microsecond (FF)=(255) 5-Mar-14 Prof.Nitin Ahire 95
- 96. Delay using 16bit counter • LXI B, (count)H 10T up: DCX B 6T MOV A,C 4T ORA B 4T JNZ :up 10/7T RET 10T DCX not affect the zero flag. 5-Mar-14 Prof.Nitin Ahire 96
- 97. Delay using 16bit counter Td=[ M + {(count) X N} -3 ] T M=No. of T state out side the loop N=No. of T state inside the loop M=20 T; N=24T 5-Mar-14 Prof.Nitin Ahire 97
- 98. Delay using 16bit counter • Td max= [20+[ {65535}X 24 ] -3] T = 1572857 T =1572857 X 0.5 microsecond = 78642 microsecond (FFFF)=(65535) largest count. 5-Mar-14 Prof.Nitin Ahire 98
- 99. Memory and I/Ointerfacing • Types of memory 1 ROM (EPROM) 2 RAM 5-Mar-14 Prof.Nitin Ahire 99
- 100. Memory structure &it‟s requirement's I/P decoder • The read write memories consist A0 of an array of A1 registers, where in each register has a unique address • M=No. of register AM • N=No. of bits Data i/Ps I/P Buffer WR MXN O/P Buffer RD CS 5-Mar-14 Prof.Nitin Ahire Data o/p s 100
- 101. Memory structure &it‟s requirement's • If the memory having 13 address line and 8 data lines, then it can access 213 address lines = 8192 and N= 8 bit or 1-byte • No of address lines of the „up‟ is to be used to find how much memory array can be access by that processor. 5-Mar-14 Prof.Nitin Ahire 101
- 102. No. of addressline required to accessed the memory • No of lines 1 2 3 4 10 11 16 5-Mar-14 size of memory 21= 2 22= 4 23= 8 24 =16 210= 1K= 1024 211 =2K= 2048 216 =64K= 65536 Prof.Nitin Ahire 102
- 103. EPROM IC availablein the market IC NO 2716 2732 2764 27128 27256 27512 5-Mar-14 size 2k X 8 4k X 8 8k X 8 16k X 8 32K x 8 64k X 8 Prof.Nitin Ahire 103
- 104. RAM IC availablein the market IC NO 6116 6264 62512 2114 5-Mar-14 size 2k X 8 8k X 8 64k X 8 1k X 4 Prof.Nitin Ahire 104
- 105. Comparison between full andpartial decoding • full decoding • partial decoding 1) Also referred to be as absolute decoding 2) All address lines are consider 3) More hardware required 1) Also referred to be as liner decoding 2) Few address lines are ignored 3) Decoder hardware is simple 5-Mar-14 Prof.Nitin Ahire 105
- 106.
- 107. Memory and I/Ointerfacing • Types of memory 1 ROM (EPROM) 2 RAM 5-Mar-14 Prof.Nitin Ahire 107
- 108. Memory structure &it‟s requirement's I/P decoder • The read write memories consist A0 of an array of A1 registers, where in each register has a unique address • M=No. of register AM • N=No. of bits Data i/Ps I/P Buffer WR MXN O/P Buffer RD CS 5-Mar-14 Prof.Nitin Ahire Data o/p s 108
- 109. Memory structure &it‟s requirement's • If the memory having 13 address line and 8 data lines, then it can access 213 address lines = 8192 and N= 8 bit or 1-byte • No of address lines of the „up‟ is to be used to find how much memory array can be access by that processor. 5-Mar-14 Prof.Nitin Ahire 109
- 110. No. of addressline required to accessed the memory • No of lines 1 2 3 4 10 11 16 5-Mar-14 size of memory 21= 2 22= 4 23= 8 24 =16 210= 1K= 1024 211 =2K= 2048 216 =64K= 65536 Prof.Nitin Ahire 110
- 111. EPROM IC availablein the market IC NO 2716 2732 2764 27128 27256 27512 5-Mar-14 size 2k X 8 4k X 8 8k X 8 16k X 8 32K x 8 64k X 8 Prof.Nitin Ahire 111
- 112. RAM IC availablein the market IC NO 6116 6264 62512 2114 5-Mar-14 size 2k X 8 8k X 8 64k X 8 1k X 4 Prof.Nitin Ahire 112
- 113. Comparison between full andpartial decoding • full decoding • partial decoding 1) Also referred to be as absolute decoding 2) All address lines are consider 3) More hardware required 1) Also referred to be as liner decoding 2) Few address lines are ignored 3) Decoder hardware is simple 5-Mar-14 Prof.Nitin Ahire 113
- 114. 8085 Memory Interfacing • GenerallyµP 8085 can address 64 kB of memory . • Generally EPROMS are used as program memory and RAM as data memory. • We can interface Multiple RAMs and EPROMS to single µP . • Memory interfacing includes 3 steps : 1. Select the chip. 2. Identify register. 3. Enable appropriate buffer. 5-Mar-14 Prof.Nitin Ahire 114
- 115. 8085 Memory Interfacing •Example: Interface 2Kbytes of Memory to 8085 with starting address 8000H. Initially we realize that 2K memory requires 11 address lines (2^11=2048). So we use A0-A10 . • Write down A15 –A0 A15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 ADD 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8000H 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 87FFH 5-Mar-14 Prof.Nitin Ahire 115
- 116. 8085 Memory Interfacing •Address lines A0-A10 are used to interface memory while A11,A12,A13,A14,A15 are given to 3:8 Decoder to provide an output signal used to select the memory chip CS¯or Chip select input. • MEMR¯ and MEMW¯are given to RD¯and WR¯pins of Memory chip. • Data lines D0-D7 are given to D0-D7 pins of the memory chip. • In this way memory interfacing can be achieved. 5-Mar-14 Prof.Nitin Ahire 116
- 117. 8085 Memory Interfacing •The diagram of 2k interfacing is shown below: A15- A11 3:8DECODER 8085 CS A15-A8 ALE A10- A0 AD7-AD0 WR RD IO/M Latch A7- A0 2K Byte Memory Chip D7- D0 RD 5-Mar-14 Prof.Nitin Ahire WR 117
- 118. Microprocessor & Microcontrolle -I T.ESem V (Rev) Prof. Nitin Ahire XIE, Mahim
- 119. Connection of I/Odevices. • Polling method • Interrupt method 5-Mar-14 Prof.Nitin Ahire 119
- 120. Interrupt system of8085 • Definition: “It is a mechanism by which an I/O device ( Hardware interrupt) or an instruction (software interrupt) can suspend the normal execution of the processor and get it self serviced.” 5-Mar-14 Prof.Nitin Ahire 120
- 121. Types of interrupt •1) Hardware interrupt • 2) Software interrupt 5-Mar-14 Prof.Nitin Ahire 121
- 122. Hardware interrupt • Interrupt: “It is an external asynchronous input that inform the „up‟ to complete the instruction that it is currently executing and fetch a new routine in order to offer a service to that I/O devices. Once the I/O device is serviced, the „up‟ will continue with execution of its normal program.” 5-Mar-14 Prof.Nitin Ahire 122
- 123. Hardware interrupt • 8085has „5‟ hardware interrupt 1)Trap 2)RST 7.5 3)RST 6.5 4)RST 5.5 5)INTR 5-Mar-14 Prof.Nitin Ahire 123
- 124. Types of Hardwareinterrupt • NMI( non maskable) • Maskable 1) It can‟t be masked or 1) It can be masked or made pending made pending 2) Highest priority 3) This interrupt disable 2) Lower priority all maskable interrupts 3) These interrupt dose not disable non 4) Used for emergency maskable interrupt purpose like power failure, smoke detector 4) Used to interface e.g. TRAP peripherals. e.g. RST 7.5 5-Mar-14 Prof.Nitin Ahire 124
- 125. Hardware Interrupt Priority 1 2 3 4 5 5-Mar-14 interrupt TRAP RST 7.5 RST6.5 RST 5.5 INTR ISR address 0024h 003Ch 0034h 002Ch No specific location Prof.Nitin Ahire trigger edge +level edge level level level 125
- 126. Software interrupt • 8085has „8‟ software interrupt 1)RST0 2)RST1 3)RST2 4)RST3 5)RST4 6)RST5 7)RST6 8)RST7 5-Mar-14 Prof.Nitin Ahire 126
- 127. Software interrupt • Theseinstruction ( RST0-RST7) allow the „up‟ to transfer the program control from main program to the subroutine program (i.e. ISR) ISR: interrupt service routing 5-Mar-14 Prof.Nitin Ahire 127
- 128. Software interrupt Interrupt RST 0 RST1 RST 2 RST 3 RST 4 RST 5 RST 6 RST 7 5-Mar-14 Restart locations 0 X 8 = 0000h 1 X 8 = 0008h 2 x 8 = 0010h 3 X 8 = 0018h 4 X 8 = 0020h 5 X 8 = 0028h 6 X 8 = 0030h 7 X 8 = 0038h Prof.Nitin Ahire 128
- 129. Software interrupt /hardware interrupt • Hardware interrupt • Software interrupt 1)It is an asynchronous 1)It is as synchronous event event 2)This interrupt is requested 2)This interrupt is requested by external by executing instruction device 3)PC is incremented 3)PC is not incremented 4)The priority is highest 4)The priority is lower than softer interrupt 5)It can‟t be ignored 6)It is not used to interface 5)Can be masked 6)It is used to interface the peripheral peripheral devices Used in debugging 5-Mar-14 Prof.Nitin Ahire 129
- 130. Interrupt related instructions 1) EI: it is used to enable the all maskable interrupt. It required 1byte, one MC (4T). It does not affect on TRAP 2) DI : it is used to disable all maskable interrupt. 1-byte (4T). It does not affect on TRAP 5-Mar-14 Prof.Nitin Ahire 130
- 131. Interrupt related instructions SIM :(set interrupt mask) 1-byte (4T) state. • Used to enable or disable RST 7.5, RST 6.5, RST 5.5 interrupts. • It does not affect on TRAP & INTR . • It is used in serial data transmission • It also transfer serial data bit „D7‟of „A‟ to the SOD pin • Hence the CWR format must be load in the „A‟ before execution of SIM 5-Mar-14 131 Prof.Nitin Ahire instruction.
- 132. SIM (bit pattern) •SOD pin SOD SDE X R 7.5 MSE M’ 7.5 M’ 6.5 M’ 5.5 • • • • • D7= SOD D6= serial data enable 1=enable, 0=disable D5= Don‟t care D4= Reset R7.5 F/F, 1=Reset 0=no effect D3=MSE Mask set enable 1=D2,D1,D0 bit are effective 0=D2,D1,D0 bit are ignored • D2= M‟7.5 Mask RST 7.5 1= Mask or disable R7.5 0= Enable RST 7.5 • D1=M‟6.5 Mask RST 6.5 1= Mask or disable R6.5 0= Enable RST 6.5 • D0=M‟5.5 Mask RST 5.5 1= Mask or disable R5.5 0= Enable RST 5.5 5-Mar-14 Prof.Nitin Ahire 132
- 133. Interrupt related instructions RIM :( read interrupt mask) 1-byte (4T) state. • It gives the status of the pending maskable interrupt (RST 7.5 – RST 5.5) • It does not affect on TRAP & INTR • It can also transfer the contents of the serial input data on the SID pin into the accumulator („D7‟ bit.) • Hence after execution of this instruction serial data get load in to the accumulator 5-Mar-14 Prof.Nitin Ahire 133
- 134. RIM (bit pattern) •SID pin SID • • • • I 7.5 I 6.5 I 5.5 IE M 7.5 M 6.5 M 5.5 D7= SID D6= D5= if 1 respective interrupt is pending D4= 0 respective interrupt is not pending • D3=IE interrupt enable • D2= • D1= • D0= 5-Mar-14 if 1 respective interrupt is Masked 0 respective interrupt is unmasked Prof.Nitin Ahire 134
- 135. 8155 (PPI) Prof. NitinAhire XIE, Mahim
- 136. Features of 8155(PPI) • • • 1) 2) 3) 4) 5) 6) It is a multifunction device designed to use in minimum mode system It contain RAM, I/O ports and Timer Features 2k static RAM cell organized as 256 bytes 2 programmable 8 bit I/O ports (A,B) 1 programmable 6 bit I/O port (c) 1 programmable 14 bit binary down counter/timer An internal address latch to de multiplex AD0AD7 using ALE Internal selection logic for memory and I/O. using command register 5-Mar-14 Prof.Nitin Ahire 136
- 137. 8155 VCC GND IO/M A AD0-AD7 256x8 Static RAM B CE ALE RD PA0-PA7 C Timer PB0-PB7 PC0-PC5 WR CE –8155 CE - 8156 5-Mar-14 RESET TIMER IN TIMER OUT Prof.Nitin Ahire 137
- 138. Pin out • ADo-AD7: address and data lines internally de multiplex by using internal latch and ALE signal address lines are used to access the memory or I/O port depending on the status of IO/M^ pin i/p • D0-D7 lines act as data bus 5-Mar-14 Prof.Nitin Ahire 138
- 139. Pin out • ALE: used to de multiplex the AD0AD7 • IO/M^ : used to differentiate between IO or memory • CE^ : used to select the 8155 • RD^ : used for to read the data from memory or I/O • WR^: used for to write the data from memory or I/OProf.Nitin Ahire 5-Mar-14 139
- 140. Pin out • Reset: used to reset the 8155 IC • PA0-PA7,PB0-PB7 : Port A and Port B I/P 8 bit pins they can be programmed either i/p or o/p port using command register • PC0-PC5 : these are 6 bit I/O pins they can be used as simple Input output port or control port when PA and/or PB are used in handshake mode • Timer in: this is an i/p to the timer • Timer out :This is an o/p pin depending on the mode of the timer o/p can be either a square wave or pulse. • VCC, GND : +5 v resp. to GND 5-Mar-14 Prof.Nitin Ahire 140
- 141. I/O port ortimer selection IO/M^ 1 1 1 1 1 1 0 5-Mar-14 A2 0 0 0 0 1 1 A1 0 0 1 1 0 0 Prof.Nitin Ahire A0 0 1 0 1 0 1 CWR PORT A PORT B PORT C Timer LSB Timer MSB Memory option 141
- 142. Control Word of8155 D7 • • • • • • D6 D5 D4 D3 D2 D0=Port A 0=Input D1=Port B 1=Output D2 &D3 used with port C D4 (IEA=interrupt Enable Port A) D5 (IEB=interrupt Enable Port B) D6&D7 used in timer mode 5-Mar-14 Prof.Nitin Ahire D1 D0 1=Enable 0=Disable 142
- 143. D7 D6 0 0 01 1 1 5-Mar-14 0 1 timer commands NOP Stop counting if timer is running Stop after TC (stop after at the count) Start timer if is not running Prof.Nitin Ahire 143
- 144. Timer mode • MSB •LSB M2 M1 D13 D12 D11 D10 D9 D8 D7 D6 D5 D4 D3 D2 D1 D0 • Mode 0 In this mode the timer o/p remains high for half the count and goes low for the remaining count, thus providing the single square wave. The pulse width is determined by count and clk freq • Mode 1 In this mode the initial count is automatically reloaded at the end of each count. Provide the continuous square wave. • Mode 2 In this mode single clock pulse is provided at the end of count • Mode 3 This is similar to mode 2 except the initial count is reloaded to provided a Prof.Nitin Ahire wave form continuous 5-Mar-14 144
- 145. • For portC D3 D2 0 0 = ALT 1 1 = ALT 0 1 = ALT 1 0 = ALT 5-Mar-14 1(port C as Input mode) 2(Port C as Output Mode) 3 used in handshake mode 4 along with Port A &B Prof.Nitin Ahire 145
- 146. Example 1 • Designa square wave generator with a pulse width of 100 us by using the 8155 timer if clock freq is 3MHz. Sol : T=1/F=1/3MHz=330ns Timer count=pulse period/CLK period = 200us/330ns=606 count = 025Eh 5-Mar-14 Prof.Nitin Ahire 146
- 147. • Assuming theport addresses CWR=20h Timer LSB=24h Timer MSB=25h Count =025Eh Therefore 5Eh must be load in the LSB timer Select mode 1 for square wave. Therefore 42h (01000010)must be load in the MSB timer 5-Mar-14 Prof.Nitin Ahire 147
- 148. • Control word Tostart the timer D7 and D6 bit must be 1 set the bit of CWR and send to address 20h therefore C0h (11000000) must be load in CWR register. 5-Mar-14 Prof.Nitin Ahire 148
- 149. Program for squarewave • MVI A,5Eh OUT 24H MVI A,42H OUT 25H MVI A,C0H OUT 20H HLT 5-Mar-14 Prof.Nitin Ahire 149
- 150.
- 151. Feature of 8255 •It is programmable parallel I/O device • It has 3,8 bit I/O Ports: Port A, Port B, Port C, which are arranged in two groups of 12 pins. • TTL compatible • Direct bit set/reset capability is available for Port C 5-Mar-14 Prof.Nitin Ahire 151
- 152. 8255 PIN DIAGRAM PA0-PA7 PB0-PB7 PC0-PC7 D0-D7 RESET RD¯ WR¯ A0-A1 CS ¯ 5-Mar-14 Vcc , Gnd I/O I/O I/O I/O I I I I I I Port A Pins Port B Pins Port C Pins Data Pins Reset pin Read input Write input Address pins Chip select Prof.Nitin Ahire +5volt supply 152
- 153.
- 154. 8255 BLOCK DIAGRAM Data Bus Buffer: It is an 8 bit data buffer used to interface 8255 with 8085. It is connected to D0-D7 bits of 8255. Read/write control logic: It consists of inputs RD¯,WR¯,A0,A1,CS¯ . RD¯,WR¯ are used for reading and writing on to 8255 and are connected to MEMR¯,MEMW¯ of 8085 respectively. A0,A1 are Port select signals used to select the particular port . CS ¯ is used to select the 8255 device . It is controlled by the output of the 3:8 decoder 5-Mar-14 154 Prof.Nitin Ahire used to decode the address lines of 8085.
- 155. 8255 BLOCK DIAGRAM A0,A1decide the port to be used in 8255. A1 A0 Selected port 0 0 1 1 0 1 0 1 Port A 5-Mar-14 Port B Port C Control Register Prof.Nitin Ahire 155
- 156. 8255 BLOCK DIAGRAM GroupA and Group B Control: Group A control consists of Port A and Port C upper. Group B control consists of Port B and Port C lower. Each group is controlled through software. They receive commands from the RD¯, WR¯ pins to allow access to bit pattern of 8085. The bit pattern consists of : 1. Information about which group is operated. 2. Information about mode of Operation. 5-Mar-14 Prof.Nitin Ahire 156
- 157. 8255 BLOCK DIAGRAM •PORT A,B: These are bi-directional 8 bit ports each and are used to interface 8255 with CPU or peripherals. • Port A is controlled by Group A while Port B is controlled by Group B Control. • PORT C: This is a bi-directional 8 bit port controlled partially by Group A control and partially by Group B control . • It is divided into two parts Port C upper and Port C lower each of a nibble. • It is used mainly for control signals and interfacing with peripherals. 5-Mar-14 157 Prof.Nitin Ahire
- 158. 8255 Operating Mode •8255 provides one control word register • It is selected when A0=1,A1=1,CS^=0,WR^=0 • The read operation is not allowed for CWR • There are two CWR formats (mode) 1)BSR mode 2)I/O mode • The two basic modes are selected by D7 bit of control register • when D7=1 it is a I/O mode & D7=0 it is BSR mode 5-Mar-14 158 Prof.Nitin Ahire
- 159. 8255 MODES • BSR(Bit Set Reset) Mode • Only C is available for bit mode access. • Allows single bit manipulation for control applications • Mode 0 : Simple I/O • Any of A, B, CL and CH can be programmed as input or output • Mode 1: I/O with Handshake • A and B can be used for I/O • C provides the handshake signals • Mode 2: Bi-directional with handshake • A is bi-directional with C providing handshake signals • B is simple I/O (mode-0) or handshake I/O 5-Mar-14(mode-1) Prof.Nitin Ahire 159
- 160. BSR mode D7 D5 D4 D3 D2 D1 D0 0 • • • D6 X x x b b b S/R Bit D7must be zero for BSR mode The BSR mode is a port C bit set/reset mode. The indivisible bit of port C can be set or reset by writing a control word in CWR. • Port C bit set/reset ; if D0=0 reset D0=1 set • The port pin of port C is selected using bit D3,D2,D1 • The BSR mode affect only one bit of port C at a time 5-Mar-14 160 Prof.Nitin Ahire
- 161. Port C bitselection D3/b 0 0 0 0 1 1 1 1 5-Mar-14 D2/b 0 0 1 1 0 0 1 1 D1/b 0 1 0 1 0 1 0 1 Prof.Nitin Ahire Bit Bit Bit Bit Bit Bit Bit Bit Bit 0 1 2 3 4 5 6 7 161
- 162. Example • Write aset of instruction to perform the following 1)Set bit 4 of port C 2)Reset bit 4 of port C( Assume Port C Address =12 h) Sol: 1) MVI A, 09h OUT 12h 2) MVI A,08h OUT 12h 5-Mar-14 Prof.Nitin Ahire 162
- 163. I/O mode (CWR) D7 D6 D5 I/O,BSRMode A Mode A D4 PA D3 D2 PCU Mode B D1 D0 PB PCL • When the bit D7=1 then I/O mode is selected • The bit D6 and D5 used for mode selection of Group A • If the D4 =1port A act as I/P port D4 = 0 port A act as O/P Port • If the D3 = 1 Port C Upper act as I/P Port D3 = 0 port C Upper act as O/P Port • The bit D2 used for mode selection of Group B • If the D1 = 1 port B act as I/P Port • D1 = 0 port B act as O/P Port • If the D0 = 1 Port C Lower act as I/P Port D0 = 0 Port C Lower act as O/P Port 5-Mar-14 Prof.Nitin Ahire 163
- 164. Example • Write aprogram to initialize 8255 in the configuration given below: 1 Port A : Simple I/P 2 Port B : Simple O/P 3 Port CL: O/P 4 Port CU: I/P Assume CWR address is 83h 5-Mar-14 Prof.Nitin Ahire 164
- 165. solutions I/O,BSR Mode AMode A 1 0 0 PA 1 PCU Mode B 1 0 PB PCL 0 0 • MVI A,98h OUT 83h 5-Mar-14 Prof.Nitin Ahire 165
- 166. INTERFACING 8085 &8255 • Here 8255 is interfaced in I/O Mapped I/O mode. Initially we write down the addresses and then interface it . A15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 Port 1 0 0 0 0 X X X X X X X X X 0 0 A 1 0 0 0 0 X X X X X X X X X 0 1 B 1 0 0 0 0 X X X X X X X X X 1 0 C 1 0 0 0 0 X X X X X X X X X 1 1 CW 5-Mar-14 Prof.Nitin Ahire 166
- 167. INTERFACING 8085 &8255 • Thus we get addresses ,considering don‟t cares to be zero as Port A =80H Port B =81H Port C =82H CWR =83H • Then, we give A11,A12,A13 pins to A,B,C inputs of Decoder to enable 8255 or Chip Select. • A15 is logic 1 so it is given to active HIGH G1 pin& A14 ,IO/M ¯ are given to active low G2B ¯,G2A ¯ pins. 5-Mar-14 167 • Output from Latch Prof.Nitin Ahire as A0,A1 pins to 8255 is given while D0-D7 are given as data inputs.
- 168. INTERFACING 8085 &8255 A15 A14 IO/M A13 A12 A11 A G2A G2B G1 B PA C /CS 8085 8255 3:8 decoder (AD0-AD7) O0 74373O1 A0-A7 PB PC O7 RD ¯ WR ¯ RD¯ WR¯ 5-Mar-14 A0 A1 D7-D0 ALE Prof.Nitin Ahire 168
- 169. INTERFACING 8085 &8255 Example: Take data from 8255 port B. Add FF H . Output result to port A. 5-Mar-14 Prof.Nitin Ahire 169
- 170. Solution MVI A,82H OUT 83H LDA81H ADI FFH OUT 80H. RST1. 5-Mar-14 Initialize 8255. Take data from port B Add FF H to data OUT Result to port A. STOP. Prof.Nitin Ahire 170
- 171. INTERFACING STEPPER MOTOR with8255 5-Mar-14 Prof.Nitin Ahire 171
- 172. Stepper motor • Hardware: A stepper motor is a digital motor. It can be driven by digital signals motor shown in the figure ( ckt ) has two phases, with center tap winding. • The center taps of these winding are connected to the +5Volt supply. • Due to this, motor can be excited by grounding 4 terminals of the 2 5-Mar-14 172 Prof.Nitin Ahire winding.
- 173. Locking system forstepper motor • In the stepper motor it is not desirable to excite both the ends of the same winding simultaneously. • This cancel the flux and motor winding may damage. • To avoid this digital clocking system must be design. 5-Mar-14 Prof.Nitin Ahire 173
- 174. Data bit passto the port A PA0 1 1 0 0 5-Mar-14 PA1 PA2 PA3 0 1 0 = 0A 0 0 1 = 09 1 0 1 = 05 1 1 0 = 06 Prof.Nitin Ahire 174
- 175. Data bits • 5000H •5001H • 5002H • 5003H 5-Mar-14 0AH 09H 05H 06H Prof.Nitin Ahire 175
- 176. Initialized Port Aas O/P Port • Program for stepper motor LXI SP,FFFFH MVI A,80H ; to make port A as o/p port OUT CWR BACK: LXI H,5000H ;HL act as memory pointer MVI C,04H ;counter for the steps UP: MOV A,M ;data bits transfer on the lower nibble of port A OUT PORT A CALL DELAY ; delay for the steps INX H ; increment the HL pair for the next data bits DCR C ; decrement the counter JNZ UP ;check zero flag JMP BACK ; jump back for continuous loop ( motor rotation) 5-Mar-14 Prof.Nitin Ahire 176