Chapter 3 BRANCH, CALL, AND TIME DELAY LOOP
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1 Islamic University Gaza Engineering Faculty Department of Computer Engineering ECOM 3022: Embedded Systems Discussion Chapter 3 BRANCH, CALL, AND TIME DELAY LOOP Eng. Eman R. Habib February, 2014
2 2 Embedded Systems Discussion Looping in PIC - Repeat a sequence of instructions or an operation a certain number of times. - Two ways to do looping o Using DECFSZ instruction o Using BNZ\BZ instructions DECFSZ instruction Decrements file register; skip the next instruction if the result is equal 0. AGAIN DECFSZ 0x10, F GOTO AGAIN BNZ\BZ instructions These instructions check the status flag; zero flag. Back DECF 0x10, F BNZ Back Other conditional jumps Find the sum of the values 79H, F5H, and E2H. Put the sum in filereg loc. 5H and 6H.
3 3 Embedded Systems Discussion Calculating the short branch address - The address of the target must be within 256 bytes of the program counter (PC). - [Target address = (2nd byte of instruction x 2) + PC]. - If the second byte is positive, the jump is forward. If the second byte is negative, then the jump is backwards. - The second byte can be a value from -127 to Unconditional branch instruction - Control is transferred unconditionally to the target location (at ROM). - Tow unconditional branches: o GOTO o BRA GOTO( LONG JUMP) - It can go to any memory location in the 2M address space of the PIC Its 4-Byte (32 bit) instruction o 12 bit Opcode o 20 bit Address BRA (branch) - This is 2-byte (16 bit ) instruction o First 5 bit Opcode o Lower 11 bit relative address of the target The address to bytes relative to the address of the current PC - If the jump is forward, then target address is positive. - If the jump is backward then the target address is negative. - The 2-byte instruction is preferred because it takes less ROM space. GOTO to itself using $ sign - HERE GOTO HERE - GOTO$ - OVER BRA OVER - BRA $ Call instructions and Stack - CALL instruction is used to call a subroutine. - Subroutines are often used to perform tasks the need to performed frequently. - In the PIC18 there are two instruction for call o CALL (long call) o RCALL (relative call)
4 4 Embedded Systems Discussion CALL - This is 4-Byte (32-bit) instruction o 12 bit Opcode o 20 bit target subroutines address. - To make sure the PIC knows where to come back after execution of the called subroutines, the microcontroller automatically saves on the stack the address of the instruction immediately below the CALL. - After finishing execution of the subroutine, the instruction RETURN transfers control back to caller. Stack and stack pointer in the PIC18 - The Stack is read/write memory (RAM) used by the CPU to store some very critical information temporarily. - The stack wide is 21-bit. - The Stack Pointer (SP) is register used to access the stack - SP is 5-bit, this give use 32 location each 21bit wide - When the PIC18 is powered up, the SP register contains value 0. - The stack location 1 is the first location used to stack, because SP point to last-used location. - The location 0 of the stack is not available and we have only 31 stack location in the PIC18 - The storing of CPU information such as PC on the stack is called PUSH, and loading back the contents into CPU register is called POP. - As data is pushed onto the stack, the stack pointer is incremented. - When the RETURN instruction at the end of the subroutine is executed, the top location of the stack is copied back to the program counter, the stack pointer is decremented. - The Stack is (LIFO) memory.
5 5 Embedded Systems Discussion RCALL (Relative Call) - 2-Byte instruction - The target address must be within 2K - 11 bits of the 2 Byte is used - Save a number of bytes. Instruction cycle time for the PIC - One instruction cycle consists of four oscillator periods. - To calculate the instruction cycle for the PIC, we take 1/4 of the crystal frequency, then take its inverse. - Unconditional branch takes 2 instruction cycles. - Conditional branch takes 2 instruction cycles if it jumps, and takes 1 when not jumps. PROBLEMS 27. Find the oscillator frequency if the instruction cycle = 1.25 µs. Instruction frequency = 1/1.25μs = 800 KHz; Oscillator frequency = 800 KHz * 4 = 3.2 MHz 28. Find the instruction cycle if the crystal frequency is 20 MHz. 20MHz / 4 = 5 MHz; Instruction Cycle = 1 / 5 MHz = 200 ns 30. Find the instruction cycle if the crystal frequency is 16 MHz. 16MHz / 4 = 4 MHz; Instruction Cycle = 1 / 4 MHz = 250 ns 35. Find the time delay for the delay subroutine shown below if the system has a PICI8 with a frequency of 10 MHz: MOVLW D'200' 1 MOVWF REGA 1 BACK MOVLW D'100' 1 MOVWF REGB 1 BACK HERE DECF REGB, F 1 BNZ HERE 2 HERE DECF REGA, F 1 BNZ BACK 2 10MHz / 4 = 2.5 MHz; Instruction Cycle = 1 / 2.5 MHz = 400 ns HERE loop: (1+2)*100 1 = 299 instruction cycles. Overall delay = [ *200 + ( )*200-1] * 400 ns = ms. Best Wishes
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