Memory Organization IIMicroprocessor and Interfacing

261214

PIC Flash Memory Segments

2K2K2K2K

01012

PIC’s RAM Banking

Instruction Format

f = 7 bitMaximum memory = 2^7 = 128 Bytes

PIC 16F877 has 512 bytes of RAM: How do we access all of it?

STATUS (F# 0x03)

Bit 5-6 (RP0,RP1) in STATUS are used for RAM Page selection

07

Total Memory becomes 2^9 = 512 Bytes

Creating a 9 bit RAM address

STATUS

07

068

f (File#)

06713

9 Bit RAM Address

OPCODE

void main(){

int i;i = 1;i++;

}

.................... int i;

....................

.................... i=1; 000D: MOVLW 01000E: BCF 03.5000F: MOVWF 21.................... i++; 0010: INCF 21,F

This code should now make sense

RAM location

First 32 bytes of every pageare reserved. Except on pages 3 and 4

User RAM space.Last 16 Bytes are mirrored

The big picture: where the bits are

Bit 5-6 (RP0,RP1) in STATUS are used for RAM Page selection

07

Bit 3-4 in PCLATH are usedfor memory access

07 4 3

6 5

2. เร่�ง Memory Operations ด้�วย Indirect Addressing

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int value[8];

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With indirect addressing

For (i=0 ; i<8 ; i++) {value[i] = 0;

}

Without indirect addressing

Value[0] = 0;Value[1] = 0;Value[2] = 0;Value[3] = 0;Value[4] = 0;Value[5] = 0;Value[6] = 0;Value[7] = 0;

PIC’s Indirect Addressing

Opcode File

Creating a 9 bit indirect RAM address

STATUS

07

078

8-bit address

07

9 Bit RAM Address

FSR (File #4)

Example: clear RAM locations 0x20 – 0x27 (8 Bytes)

Instruction Comment

BCF 0x03, 7 ;select page

MOVLW 0x20

MOVWF 0x04 ;Set FSR

Next CLRF 0x00 ;clear RAM

INCF FSR,F ;increase counter

BTFSS FSR,3 ;all done?

GOTO Next ; no, clear next

0x20 = 0b0010 00000x27 = 0b0010 0111

The big picture: where the bits are

Bit 5-6 (RP0,RP1) in STATUS are used for RAM Page selection

07

Bit 3-4 in PCLATH are usedfor memory access

07 4 3

6 5

Bit 7 (IRP) in STATUS is used for indirect Addressing

Indirect Addressing ExFind the sum of values in RAM address 0x22-0x25 and storeat address 0x21

Address Function

0x21 Sum

0x22 Value1

0x23 Value2

0x24 Value3

0x25 Value4

End Result: Sum = Value1+Value2+Value3+Value4

; 0x21 = sum result variableclrf 0x21; 0x22 - 0x25 = input numbers to be summedmovlw 4movwf 0x22movlw 3movwf 0x23movlw 2movwf 0x24movlw 1movwf 0x25