PIC Discussion. By Eng. Tamar Jomaa

Transcription

1 PIC Discussion By Eng. Tamar Jomaa

2 Chapter#2 Programming Microcontroller Using Assembly Language

3 Quiz#1 : Time: 10 minutes Marks: 10 Fill in spaces: 1) PIC is abbreviation for 2) Microcontroller with..architecture called RISC microcontroller, which is the abbreviation for 3) The various circuits inside a typical computer are connected to one another through three busses which are..,,.. 4) Write the size of each memory: program memory.. RAM.. EEPROM.. 5) The program must start from location..

4 Bonus : Time: 10 minutes Marks: +2 Write the function for each pin:

5 Outlines for part#1: 2.1 Machine language and assembly language. 2.2 Assembler statements (instructions/ directives) and number representation. 2.3 Addressing modes Opcode The configuration Hexa file Direct and indirect addressing

6 2.1 Machine language and assembly language: We as humans express our ideas in complex and often loosely defined linguistic forms. A computer reads and understands binary, and responds in a precise way to precise instructions. It is ruthlessly logical and does exactly what it is told.

7 2.1 Machine language and assembly language: How can a programmer write programs for a computer? The human learns machine code. Use a high-level language (HLL). Use Assembler.

8 2.1 Machine language and assembly language: Assembly language: Every one of the computer s instructions set is given a mnemonic. This is usually a three- or four-letter word that can be used to represent directly one instruction from the instruction set. The programmer then writes the program using the instruction mnemonics. The programmer has to think at the level of the computer, as he/she is working directly with its instructions, but at least the programmer has the mnemonics to use, rather than actually working with the computer machine code. A special computer program called a Cross-Assembler, usually these days running on a PC, converts the code written in mnemonics to the machine code that the computer will see.

9 2.1 Machine language and assembly language:

10 2.1 Machine language and assembly language: The PIC 16 Series instruction set, with a little more on the ALU:

11 2.1 Machine language and assembly language: The PIC 16 Series instruction set, with a little more on the ALU: The ALU can operate on data from two sources. One is the W (or Working ) register. The other is either a literal value or a value from a data memory (whose memory locations Microchip call register files ). A literal value is a byte of data associated with a particular instruction that the programmer embeds in the program. The data that the instruction operates on, or uses, is called the operand. Operands can be data or addresses. Always need an operand to be specified with them, others do not.

12 2.1 Machine language and assembly language: Once an instruction has been executed, where is the result stored? For many instructions Microchip offer a choice, whereby the result can either be held in the W register or stored back in data memory. Which one is used is fixed by certain instructions; in others it is determined by the state of a special d bit, which is specified within the instruction.

13 The PIC 16 Series instruction

14 2.1 Machine language and assembly language: You can see that the table is divided into six columns, and each of the 35 instructions gets one line. The first column gives the actual mnemonic, together with the code specifying the type of operand it acts on. There are four such operand codes: f for file (i.e. memory location in RAM), a 7-bit number b for bit, to be found within a file also specified, a 3-bit number d for destination, as described above, a single bit k for literal, an 8-bit number if data or 11-bit if address.

15 2.1 Machine language and assembly language: The second column summarises what the instruction does. In some cases this gives adequate information. The third column shows how many instruction cycles the instruction takes to execute. The fourth column gives the actual 14-bit opcode of each instruction. This is the code that the Cross- Assembler produces, as it converts the original program in Assembler language to machine code. The fifth column shows which bits in the Status register are affected by each instruction.

16 2.1 Machine language and assembly language: Five example instructions: 1) clrw : This clears the value in the W register to zero. There are no operands to specify. Column 5 tells us that the Status register Z bit is affected by the instruction. As the result of this instruction is always zero, the bit is always set to 1. No other Status register bits are affected. 2) clrf f : This clears the value of a memory location, symbolised as f. It is up to the programmer to specif a value for f. Again, because the result is zero, the Status register Z bit is affected.

17 2.1 Machine language and assembly language: 3) addwf f,d: This adds the contents of the W register to the contents of a memory location symbolised by f. It is up to the programmer to specify a value for f. There is a choice of where the result is placed. This is determined by the value of the operand bit d. Because of the different values that the result can take, all three condition code bits, i.e. Z, the Carry bit C, and the Digit Carry bit DC are affected by the instruction. 4) bcf f,b : This instruction clears a single bit in a memory location. Both the bit and the location must be specified by the programmer. The bit number b will take a value from 0 to 7, to identify any one of the 8 bits in a memory location. No Status register flags are affected, even though it is possible to imagine that the result of the instruction could be to set a memory location to zero.

18 2.1 Machine language and assembly language: 4) addlw k: This instruction adds the value of a literal, whose value k must be specified by the programmer, to the value held in the W register. The result is stored in the W register; there is no choice. Like addwf, all condition code bits can be affected by this instruction. Question: What is the meaning of the following instructions:

19 Solution:

20 2.1 Machine language and assembly language: Exercise: What does the following code do?

21 2.1 Machine language and assembly language: Solution:

22 2.1 Machine language and assembly language: To make the previous code more easier we use equ: All the identifiers for memory locations find in P16F84a.INC, instead of writing the name of each location by equ

23 2.1 Machine language and assembly language: Another important instructions for condition case:

24 2.1 Machine language and assembly language: Instructions for subroutine and stack:

25 2.1 Machine language and assembly language: Instructions for subroutine and stack:

26 2.1 Machine language and assembly language: Important note: If you call a sub-program within another sub-program called nested subroutine. When you do so, you must remember that each time the program is called sub-program or book a place in the stack, which is free when you return from sub program. If you call a sub-program of another branch within the program we use two stack sites may be three if there is another internal call. Because a microcontroller PIC16 stack owns eight sites (levels) must caution that no rash or flood or increase the stack is what is known as the stack overflow.

27 2.2 Assembler statements (instructions/ directives) and number representation An assembly language source cod consists of a set of statements. The two types of statements are : Instruction : which are translated into machine instruction by the assembler. Directives : which tell the assembler to perform a specific action during the assembling process. The format of the source code is shown below. The maximum column width is 255 characters. Example:

28 2.2 Assembler statements (instructions/ directives) and number representation Some common MPASM Assembler directives: Number representation in MPASM Assembler:

29 2.3 Addressing modes: General format for instructions: The format for instructions of PIC16 series is the following three kinds: Byte-oriented file register operations. Bit-oriented file register operations. Literal and control operations. The instructions are written in the program memory and one instruction is composed of 14 bits. These 14 bits are called a word.

30 2.3 Addressing modes: 1. Byte-oriented file register operations: The instructions of this format are the instructions which process a byte unit.

31 2.3 Addressing modes: Note: f: file register, it is the address of GPR or SFR. GPR and SFR are 8 bit register but in the instruction it has only 7 bits, the last bit is dropped.

32 2.3 Addressing modes: Example: porta 05h address trisa 85h address 05h: h: These bits are dropped. These bits only appear in the instruction set. Note: The dropped bit was defined according to the value of RP0 in the status register.

33 2.3 Addressing modes: Complete example: ADDWF PORTA,W From table the opcode like this: The opcode: dfff fff opcode operand PORTA has 05h address so, the instruction will be as the following:

34 2.3 Addressing modes: 2. Bit-oriented file register operations: The instructions of this forma are the instruction which processes a bit unit.

35 2.3 Addressing modes: Think : Why does b available in 3 digits but the f in 7 digits? Solution: Because b is from 0 to 7 so we can act it in 3 digits But f is from 0 to 7f so we can act it in 7 bits. Example#1: BSF portb,2 From table the opcode is: f : portb has address 06h b: 2 So the opcode is:

36 2.3 Addressing modes: Example#2: BSF trisb,2 From table the opcode is: f : trisb has address 86h b: 2 So the opcode is: Note: bsf trisb and bsf portb have the same machine language.

37 2.3 Addressing modes: Literal and control operations: The instructions of this format do the processing which used the fixed number (k) which was written in the instruction. There are two instruction types and fixed number (k) is 11 bits about GOTO and CALL instruction. In case of GOTO and CALL instruction

38 2.3 Addressing modes: In the first form k is a constant and since all registers are 8 bit, registers maximum value that can be stored in the register is 255(ffh). In the second form k is address of program memory. It s used with CALL and goto instructions. It s 11 bit because max range from 0 to 2047(7FFh).

39 2.3 Addressing modes: In the first form k is a constant and since all registers are 8 bit, registers maximum value that can be stored in the register is 255(ffh). In the second form k is address of program memory. It s used with CALL and goto instructions. It s 11 bit because max range from 0 to 2047(7FFh). Example: XORLW 15h The opcode: kkkk kkkk

40 2.3 Addressing modes: Write the machine language for the following code: Remember : org and end are directives that haven t machine code, because machine code for instructions only.

41 2.3 Addressing modes: Solution : The instruction The opcode rule The opcode ORG 0X000 directive GOTO START ORG 0X004 directive RETFIE BSF STATUS,RP0 BCF TRISA,0 BSF TRISB,0 BCF STATUS,RP0 MOVF PORTB,W MOVWF PORTA BSF STATUS,RP0 END directive

42 2.3 Addressing modes: Configuration: The configuration bits may be change to 0 or left it to default value which is 1. This bits exist in address 2007h in program memory. Bit0 & Bit1: to define the type of crystal, as the following: Bit2: active or not the watch do timer. Watch dog active Watch dog not active 1 default 0

43 2.3 Addressing modes: Bit3: PWRTE to active or not the power timer after 72ms during reset. Power timer not active Power timer active 1 default 0 Bit4: CP to prevent the program code The program code is not Protected The program code is Protected 1 default 0 The configuration bits: FOSC0 FOSC1 WDTE PWRTE CP CP CP CP CP CP CP CP CP CP CP Example: = 1FF3

44 2.3 Addressing modes: The configuration is written by CONFIG directive. Summary: all in the green are defaults

45 2.3 Addressing modes: Hex file: The hexa file contains the machine language that deals with PIC microcontroller. The hexadecimal object file format for PIC Microcontroller General Record Format: Record Mark: each record begins with a Record Mark field containing 03AH,the ASCII code for the colon ' : ' character. Record Length: The number of bytes of information or data Example: 02 2 bytes byte=8 bits so 2 byte=2x8=16 bits --4bit bit bit bit hexadecimal digits. 08 8byte 8x8=64 bits 64/4=16 hexadecimal digits.

46 2.3 Addressing modes: Offset: By the offset we can arrive to the address, the address for the byte not for the word. Example: bits Byte=8bit offset Record Type: It is used to interpret the remaining information within the record. The encoding for all the current record types are:

47 2.3 Addressing modes: Chick Sum: This field contains the check sum on the Record length, Load Offset, Record Type, and Information or Data. Therefore, the sum of all the ASCII pairs in a record after converting to binary, from the Record length field to and including the Chick Sum field, is zero. An important question for Mid Term Exam : Write the program which has the following hexa file: : FA : D1 : FF :0C B2803 : FF

48 2.3 Addressing modes: SOLUTION: Step(1): distribute the bits as following: the last byte is Check sum : FA : D1 : FF : 0C B23 D3 : FF Step(2): convert the data that have only 00 type. Step(3):distribute all 2 bytes separately, because each 2 bytes act one instruction. Step(4):reverse LSB with MSB then convert.

49 Instruction Reverse ins. The opcode The code Goto Goto BSF STATUS,RP0 FF30 30FF MOVLW FF MOVWF PORTA MOVLW 0X MOVWF PORTB BCF STATUS,RP MOVF PORTA,W MOVWF PORTB 0B23 280B GOTO 0B

50 2.3 Addressing modes: After organize the previous instruction we got the following code:

51 2.3 Addressing modes: Direct and indirect addressing: Addressing mode: it is the way in which an operand is specified. There are 3 types of addressing modes: 1. Immediate addressing mode. 2. Register operand addressing mode. 3. Memory operand addressing mode, which classified into to groups: A. Direct addressing B. Indirect addressing

52 2.3 Addressing modes: Direct and indirect addressing: 1. Immediate addressing mode: The operand is a number or constant not an address. Example: movlw 43h 2. Register operand addressing mode: Deals with registers Example: CLRW W:work register. 3.Memory operand addressing mode: A) Direct : deals with address or memory location Example: CLRF 13H or MOVWF TRISA B) Indirect : uses INDF and FSR.

53 2.3 Addressing modes: Direct and indirect addressing: A)Direct: Direct Addressing is done through a 9-bit address. This address is obtained by connecting 7th bit of direct address of an instruction with two bits (RP1, RP0) from STATUS register as figure. Any access to SFR registers can be an example of direct addressing.

54 2.3 Addressing modes: Direct and indirect addressing: Quick Question We observe that SFR is classified into 2 banks, each bank consist from 8 bit but we need only 7 bit because the last bit for select the bank. Now observe the picture, the GPR does not classified into 2 banks, so shall I need 16 bit?? the solution is No, because each register in GPR consist from 2 register each one has 8 bit and any thing you write it in first register is directly copy to the other register, but we can reach to any register from any bank, so we also need 7 bit.

55 2.3 Addressing modes: Direct and indirect addressing: B) Indirect: It does not take an address from an instruction, but it derives it from IRP bit of STATUS and FSR registers. Addressed location is accessed through INDF register which in fact holds the address indicated by the FSR. Indirect addressing is very convenient for manipulating data arrays located in GPR registers. In this case, it is necessary to initialize FSR register with a starting address of the array, and the rest of the data can be accessed by incrementing the FSR register.

56 2.3 Addressing modes: Direct and indirect addressing: Figure below shows the indirect addressing concept. we put the address of the register which we want to deal with it in FSR, which the FSR work as pointer. We deal with the location which the FSR points to it from INDF register.

57 Example#1: Filling 0C-0F GPR registers with 55h Indirect addressing examples

58 Example#2: Filling All GPR registers with 00h Indirect addressing examples

59 Indirect addressing examples

60 Homework Q1) Extract the assembly language instruction from the following hex file : B E5 Q2) Implement a program that copy the input data from portb to all GPR. Note: date delivery is next lecture.

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