ET355 Microprocessors Thursday 6:00 pm 10:20 pm

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1 ITT Technical Institute ET355 Microprocessors Thursday 6:00 pm 10:20 pm Unit 4 Chapter 6, pp Chapter 7, pp

2 Unit 4 Objectives Lecture: BCD Programming Examples of the 805x Microprocessor Mathematics Programming Examples of the 805x Microprocessor Signed and Unsigned Numbers Programming Examples of the 805x Microprocessor Logic Programming Examples of the 805x Microprocessor Lab:. Reinforce Lecture Topics through Practical Experience. Assignment: Reinforce both Lecture & Lab Topics through Practical Experience.

3 Unit 4 Agenda Lecture: Chapter 6, pp Chapter 7, pp Lab: Refer to ET355 Agenda and Assignment document provided by your instructor or obtained on Your instructor must sign-off lab and you must submit with Lab Questions answered. Assignment: Refer to ET355 Agenda and Assignment document provided by your instructor or obtained on

4 Unit 4: Chapter 6, Section 6.1 Arithmetic Instructions Addition of Unsigned Numbers ADD A, SOURCE ; A = A + <source>: Example 6.1 Show how the flag register is affected by the instructions. MOV A, #0F5H ADD A, #0BH F5H BH H Reg A contains 00H and flags are: CY = 1 (carry from D7) P = 0 (even) AC = 1 (carry from D3 to D4)

5 Unit 4: Chapter 6, Section 6.1 Example 6.2 Assume that RAM locations have the following values: 40 = 7D, 41 = EB, 42 = C5, 43 = 5B, 44 = 30 Write a program to find the sum of the values. At the end of the program, register A should contain the low byte and R7 the high byte. All values are in hex. MOV R0, #40H ; load pointer MOV R2, #5 ; load counter CLR A ; A = 0 MOV R7, A ; clear R7 AGAIN: ADD ; add the byte pointer to A by R0 JNC NEXT ; if CY=0 don t accumulate carry INC R7 ; keep track of carries NEXT: INC R0 ; increment pointer DJNZ R2, Again ; repeat until R2 is zero

6 Unit 4: Chapter 6, Section 6.1 ADDC instruction (add with carry) When adding two 16-bit data operands, we are concerned about the carry from the lower byte to the higher byte. When adding 3CE7H + 3B8DH 1 (note the carry is propagated to the high byte) 3C E7 + 3B 8D 78 74

7 Unit 4: Chapter 6, Section 6.1 BCD instruction (binary coded decimal) Two terms for BCD: Unpacked lower 4 bits of the number represent the BCD number and the upper bits are zero = = 5 Packed A single byte has two BCD numbers = 95

8 Unit 4: Chapter 6, Section 6.1 Problem: after adding packed BCD numbers, the result is no longer BCD Example: MOV A, #17H ADD A, #28H 17H H H should be 45H but is 3FH Remember, BCD cannot contain values of: 1010, 1011, 1100, 1101, 1110, or 1111 This is an invalid BCD representation! To correct this problem, a programmer must add 6 (0110) to the low digit

9 Unit 4: Chapter 6, Section 6.1 Problem: after adding packed BCD numbers, the result is no longer BCD Example: MOV A, #17H ADD A, #28H 17H H H should be 45H but is 3FH This problem is so pervasive, that most microprocessors have an instruction to deal with it. The 8051 instruction is DA A which is designed to correct the BCD addition problem.

10 Unit 4: Chapter 6, Section 6.1 DA instruction (decimal adjust for addition) this instruction will add 6 to the lower nibble or higher nibble if needed, otherwise, it will leave the result alone. When adding 47H + 25H MOV A. #47H MOV B, #25H ADD A, B DA, A 47H H H ; A=47 first BCD operand ; B=25 second BCD operand ; hex (binary) addition (A=6CH) ; adjust for BCD addition (A=72H) The DA instruction will work only after an ADD instruction; it will not work after an INC instruction.

11 Unit 4: Chapter 6, Section 6.1 SUBB instruction (Subtract with borrow) Many microprocessor have two instructions, SUB and SUBB. The 8051 has only SUBB. SUBB A, source ; A = A source CY Example: SUBB when CY = 0 1. Take the 2 s complement of the subtrahend (source) 2. Add it to the minuend (A) 3. Invert the carry Refer to: Example 6-5 on page 145

12 Unit 4: Chapter 6, Section 6.1 SUBB instruction (Subtract with borrow) Example: SUBB when CY = 1 1. Take the 2 s complement of the subtrahend (source) 2. Add it to the minuend (A) 3. Invert the carry Refer to: Example 6-7 on page 146

13 Unit 4: Chapter 6, Section 6.1 MUL instruction (multiplication) One of the operands MUST be in register A After the multiplication, the result is in the A and B registers; a total of 16 bits (lower byte is in A and upper byte is in B) Example: MOV A, #25H ; load 25H to reg. A MOV B, #65H ; load 65H in reg. B MUL AB ; 25H * 65H = E99 ; where B = 0EH and A = 99H

14 Unit 4: Chapter 6, Section 6.1 DIV instruction (division) When dividing a byte by a byte, the numerator must be in register A and the denominator must be in register B. A divide by Zero sets the OV flag to 1 Example: MOV A, #95 ; load 95H to reg. A MOV B, #10H ; load 10H in reg. B DIV AB ; now A = 09 (quotient) and ; B = 05 (remainder)

15 Unit 4: Chapter 6, Section 6.2 Signed Number Concepts and Arithmetic Operations Signed 8-bit operands: In signed byte operands, D7 (MSB) is the sign and D0 to D6 are set aside for the magnitude of the number. If D7 = 0, the operand is positive, if D7 = 1, the operand is negative. Positive numbers:

16 Unit 4: Chapter 6, Section 6.2 Signed Number Concepts & Arithmetic Operations Negative numbers: 1. Write the magnitude of the number in 8-bit binary (no sign) 2. Invert each bit 3. Add 1 to it Example 6-12: show how the 8051 would represent in 8-bit binary invert each bit add 1 (becomes 80 in hex) h h h FEh FFh

17 Unit 4: Chapter 6, Section 6.2 Overflow problem in signed number operations: The 8051 indicates the existence of an error by raising the OV (overflow flag), but it is up to the programmer to take care of the erroneous error. The OV is set to 1 if either of the following two conditions occurs: 1. There is a carry from D6 to D7 but no carry out of D7 (CY=0) 2. There is a carry from D7 out (CY= 1) but no carry from D6 to D7

18 Unit 4: Chapter 6, Section 6.2 Overflow problem in signed number operations: Example 6-14: observe the following, noting the role of the OV flag MOV A, #-128 ; A = (80H) MOV R4, #-2 ; R4 = (FEH) ADD A, R4 ; A = (7EH = +126) Solution: and OV = 1 Error should not be positive

19 Unit 4: Chapter 6, Section 6.2 Overflow problem in signed number operations: Example 6-15: observe the following, noting the OV flag MOV A, #-2 MOV R1, #-5 ADD A, R1 ; A = (FEH) ; R1 = (FBH) ; A = (F9H = -7, correct, OV=0) Solution: and OV = 0 correct

20 Unit 4: Chapter 6, Section 6.2 Overflow problem in signed number operations: Example 6-16: observe the following, noting the OV flag MOV A, #+7 MOV R1, #+18 ADD A, R1 ; A = (07H) ; R1 = (12H) ; A = (19H = +25, correct, OV=0) Solution: and OV = 0 correct

21 Unit 4: Chapter 6, Section 6.2 Overflow problem in signed number operations: We can conclude that in any signed number addition, OV indicates whether the result is valid or not. If OV = 1, the result is erroneous; if OV = 0, the result is valid. The programmer must look at the OV flag after an addition of signed numbers. An instruction similar to the JNC or JC for the carry bit would be a nice jump to handle the error, but the 8051 does not have such an instruction; we can use the JB PSW.2 or JNB PSW.2 to look at the bit within the PSW to make our jumps.

22 Unit 4: Chapter 6, Section 6.3 Logic and Compare Instructions AND OR XOR CPL A CJNE

23 Unit 4: Chapter 6, Section 6.3 AND Instruction (logical ANL) Both bits must be 1 to obtain a one output. Example 6-17 MOV A, #35H ; A = 35H ANL A, #0FH ; A AND 0FH ( A = 05 ) 35H FH H

24 Unit 4: Chapter 6, Section 6.3 OR Instruction (ORL) Either bit must be 1 to obtain a one output. Example 6-18 MOV A, #04H ; A = 04 ORL A, #30H ; A OR 30H ( A = 34H ) 04H H H

25 Unit 4: Chapter 6, Section 6.3 XOR Instruction (XRL) Either bit must be 1 (but not both) to obtain a one output. Example 6-19 MOV A, #54H ; A = 54H XRL A, #78H ; A XOR 78H ( A = 2CH ) 54H H CH

26 Unit 4: Chapter 6, Section 6.3 CPL A Instruction (CPL A) Compliments the A register 1 s compliment) Example 6-22 MOV A, #85H ; A = 85H CPL A ; 1 s compliment ADD A, #1 ; 2 s compliment 85H AH BH

27 Unit 4: Chapter 6, Section 6.3 CJNE Instruction (Compare and jump if not equal) also sets the CY flag to indicate whether the destination is larger or smaller Destination >= source; CY = 0 Destination < source; CY = 1 Does the jump occur? What does A contain? Example 6-23 MOV A, #55H CJNE A, #99H, NEXT NEXT: Yes - #55H

28 Unit 4: Chapter 6, Section 6.4 Rotate Instructions and Data Serialization Rotating the bits of A right or left RR A ; rotate right A RL A ; rotate left A MOV A, #36H ; A = RR A ; A = RR A ; A = RR A ; A = RR A ; A =

29 Unit 4: Chapter 6, Section 6.4 Rotate Instructions and Data Serialization Rotating through the carry RRC A RLC A CLR C ; CY = 0 MOV A, #26H ; A = RRC A ; A = CY = 0 RRC A ; A = CY = 1 RRC A ; A = CY = 1

30 Unit 4: Chapter 6, Section 6.4 Serializing a byte of data one of the most widely used applications of the rotate instruction transfer of a bit within a register one at a time Example pg 163

31 Unit 4: Chapter 7, Section 7.1 Data Types and Time Delay in 8051 C The goals of 8051 C programming is to create smaller hex files; therefore, it is worth time to re-examine C-language data types. Unsigned Char 8-bit data type that takes a value in the range of (00H FFH) EXAMPLE 7-1: Write a program to send values 00 - FF to P1 #include <reg51.h> void main (void) { unsigned char z; for (z=0;z<=255;z++) P1=z; }

32 Unit 4: Chapter 7, Section 7.1 Signed Char 8-bit data type that uses the most significant bit (D7) to represent the or + value. Using only 7 bits for data, we have the range of -128 to EXAMPLE 7-4: Write a C program to send values of -4 to +4 to P1 #include <reg51.h> void main (void) { char mynum[] = {+1, -1, +2, -2, +3, -3, +4, -4}; unsigned char z; for (z=0;z<=8;z++) P1=mynum [z]; }

33 Unit 4: Chapter 7, Section 7.1 Unsigned int 16-bit data type that that takes the value in the range of (0000 FFFF). Do not use unless required use a unsigned char type to save memory and performance. Signed int 16-bit data type that that uses the most significant bit (D15) to represent the or + value. Using only 15 bits provides a range of -32,768 to +32,767. Sbit designed specifically to access single-bit addressable registers (SFR registers of the 8051). Bit & sfr while the sbit data type is used for bit-addressable SFRs, the Bit data type is used for the bit-addressable section of RAM space 20-2F.

34 Unit 4: Chapter 7, Section 7.1 Table 7-1: Some Widely Used Data Types for 8051 C

35 Unit 4: Chapter 7, Section 7.1 EXAMPLE 7-5: Write a C program toggle bit D0 of P1 50,000 times #include <reg51.h> sbit MYBIT = P1^0 // notice that the sbit is // declared outside of the main void main (void) { unsigned char z; for (z=0; z<=50000; z++) { MYBIT = 0; MYBIT = 1; } }

36 Unit 4: Chapter 7, Section 7.1 Time Delay Created by: 1. Using a simple for loop 2. Using the 8051 timers Chapter 9 discusses 8051 Timers in detail Factors that can affect the accuracy of the delay when using a simple for loop: 1. Machine cycles and the number of clock periods per machine cycle vary among different versions of the 8051/52 microprocessor. 2. The duration of the clock period is a function of the crystal oscillator frequency. 3. C compile itself when selecting code

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