ECE 265 – Lecture 8

1. ECE265 ECE 265 – Lecture 8 The M68HC11 Basic Instruction Set The remaining instructions

2. Lecture Overview • The M68HC11 Basic Instruction Set • Stack and index register instructions • The remaining instructions classes are for program flow control • Branch Instructions • Jump Instructions • Subroutine Calls • No Operation • Stop • REF: Chapter 3 and the appendix that details the instructions. ECE265

3. Stack &Index register instructions • These instructions allow manipulation of the index registers, X and Y, and exchange with the D accumulator. Additionally, they provide the ability to push and pop X and Y to/from the stack. • This class of instructions can be further subdivided into sub-classes of the type of operation. ECE265

4. Modify value index reg or SP • To modify the value of the index register • Add B accumulator to X or Y: ABX ABY • Decrement the register: DES (stack pointer) • DEX DEY • Increment the register: INS (stack pointer) • INX INY • INX and INY (DEX,DEY) are only ones that affect CC register and then only the Z bit. ECE265

5. Data manipulation of contents • Add the B accumulator to and index register • ABX ABY • Does not affect the CC register • Data testing of contents of index register • CPX CPY • There are compares to a 16-bit value in memory ECE265

6. Move the X and Y index registers • Push them on the stack – copy the value in the register to the top of the stack. The register stays the same value. These are 2 byte push and pop operations. • PSHX PSHY • Pull the top 2 bytes from the stack into X or Y • PULX PULY • Store the X or Y index register or the SP • STX STY STS • Load the X or Y index register or the SP • LDX LDY LDS ECE265

7. Register interchange instructions • Exchanges • D with X: XGDX • D with Y: XGDY • Transfers • Stack pointer to X or Y: TSX TSY • X or Y to the Stack Pointer: TXS TYS • On a transfer what is the value of the register transferred from after the operation? • What values is transfrred? ECE265

8. Register interchange instructions • Values before execution • SP -- $C100 • X -- $000C • Execute a TSX instruction • Values now • SP -- $C100 • X -- $C101 • X is loaded with 1 plus the contents of the stack pointer so that it points to the last value that was stored on the stack. ECE265

9. Ended here on Fri Jan 27, 12 ECE265

10. Program flow control instructions • When a task is sequential simply increment through the steps of tasks will accomplish it. • However, in computer programming for most applications, and especially embedded control applications, the programs typically have actions based on the status of the various inputs or internal counters and timers. • These instructions control what part of the program will execute next. They control the flow of execution and are often referred to as program flow control instructions. ECE265

11. Conditional Branches • These are instructions to the control the flow of execution based on a value in the condition code register. • Their general execution is: • If the condition tested is false execution continues with the next instruction. • If the condition tested is true then execution continues at the relative location indicated in the argument of the instruction. ECE265

12. The effect of a branch • A branch controls the flow of execution ECE265

13. A table of the HC11 branches • The • Instructions • Note the • mode is • relative ECE265

14. Relative addressing mode • Relative means the address is related to something. • It is related to the current instruction address which is in the program counter (PC). The instruction argument is the offset to be taken if the condition is TRUE. • The offset can be positive of negative. It is an 8 bit quantity that is treated as a 2’s complement value and allows offsets of -128 to +127. ECE265

15. An example • This is a timing delay loop • DECB takes 1usec • BNE takes 1.5 usec • LDAB immediate mode takes 1us • Coding • LDAB #$count • DELAY: DECB • BNE DELAY • Time delay = 1us+(1us+1.5us)*count ECE265

16. More calculation • Most times you will know to time delay that you need and must calculate what count will create a loop with the correct delay. • Time Delay – LDAB time • Count = DECB time + BNE time • Taking into account the MCUs clock frequency which then tells us the time for the instructions. • Count = (Time Delay – 1us) / (2.5 us) ECE265

17. Simple Branches summary • Branch on Carry Set of Carry Clear • BCS BCC • Branch on Zero • Equal to zero – BEQ Not Equal to zero – BNE • Branch on Minus – BMI • Branch on Plus – BPL • Branch on Overflow clear or set • BVC BVS ECE265

18. Unsigned Binary Branches • Branch if higher BHI • CMPA $D380 compares A to the contents of $D380 • BHI will branch if A > $D380 i.e. A > M • Branch on higher or same BHS • BHI is > • BHS is a ³ test • Branch if lower BLO < • Branch if lower or equal BLS £ ECE265

19. Signed Branches • These branches treat the data as 2’s complement • Branch if greater than BGT • Branch is greater than or equal BGE • Branch is less than BLT • Branch is less than or equal BLE ECE265

20. Jump and Branch always • Sometimes called unconditional branches • Transfers control to the location specified in the argument of the instruction. • Jump, JMP, uses direct, extended, or indexed addressing for the target address • Branch always – BRA – us a relative transfer ECE265

21. Subroutines • High level languages and good programming practice use subroutines. • The 68HC11 has two branches to subroutine • and one return from subroutine • The BSR is a relative branch to subroutine ECE265

22. The effect of the JSR,BSR & RTS • Instruction and their effect • JSR shows the effect of the addressing mode ECE265

23. Lecture summary • Have covered the remaining instructions • Now knowing the instruction set, the processor may be used in applications • To do that you need a program. • How do you design a program? • How do you design a well documented, understandable and maintainable program? ECE265

24. Assignment • Problems Chapter 3 page 87 • Problem 28 ECE265