Example Programs for 6502 Microprocessor Kit
Contents
1. 020D 020D 020E 0210 0212 0214 0216 0218 021A 021A 021B 021D 021F 0221 0221 0223 OC A9 85 A9 58 4C 78 F8 E6 A5 C9 DO A9 85 18 A5 69 85 A5 8D 02 oc FE 02 FF 09 02 30 64 10 30 31 01 31 31 00 80 SEC100 SEC MAIN BLOCK 1 BLOCK 1 ORG FE WORD SERVICE IRQ ORG 200H LDA SERVICE IRQ amp SFF STA FE LDA 2 STA SFF CLI S JMP SERVICE IRQ SEI SED INC LDA SEC100 SEC100 CMP 100 BNE SKIP1 LDA 0 STA SEC100 CLC LDA SEC ADC 1 STA SEC LDA SEC STA GPIO1 ENABLE IRQ jump here DECIMAL MODE 0047 0226 0048 0226 SKIP1 0049 0226 40 RTI 0050 0227 0051 0227 END 0052 0227 tasm Number of errors 0 Main program begins with storing the service address of IRQ at location 00FE and OOFF The service address is 020C Then ENABLE the interrupt with instruction CLI And jump here waiting the trigger from 10ms tick When the CPU recognize the IRQ trigger it will save current PC and jump to interrupt service subroutine at 020C The service for IRQ will be entered every 10ms The SEC100 variable will be incremented when it is 100 it will be one second The SEC variable will then be incremented in BCD number and wrote to the GPIO1 LED Let us test the code with key GO and see what happen on the GPIO1 LED2. END The 6502 kit display and keypad use hex number Since each hex digit represents 4 bit binary number or one nibble It will make us more easy to enter the program using hex number The actual code in memory is binary number For example the hex code of instruction LDA 0 has two bytes A9 and 00 The display will show us as AQ at location 0200 The real value however in memory will be 1010 1001 Here is the number representation for decimal binary and hexadecimal number Decimal 4 bit Binary Hex 0 0000 0001 0010 0011 0100 0101 0110 0111 COIN GO Aa R WO NN 1000 COON A a KR WIN O 9 1001 9 10 1010 A 11 1011 B 12 1100 C 13 1101 D 14 1110 E 15 111 F For example the instruction hex code of LDA 0 is A9 and OO We can write the first byte in 8 bit binary as 1010 1001 and the second byte as 0000 0000 Note Writing hex number in the Assembly Program can be done with or H symbol We see that the assembler directive org 200H will be the same as org 200 The sharp symbol indicating the number is 8 bit data to be loaded If no symbol the 8 bit data will be memory location For example LDA 0 load accumulator with data 0 LDA 0 load accumulator with memory location O in zero page Program2 Bit shifting This program looks similar to Program but we use instruction that shift
3. 0206 END tasm Number of errors 0 Enter the hex code press RESET PC and GO What is result on the GPIO1 LED If we change the byte at location 201 from 00 to 01 what is the result 9 Program4 8 bit Addition Let us play with 8 bit binary addition Suppose we want to add two 8 bit binary numbers as shown below Number 1 0101 1011 5B Number 2 0101 1010 5A Result is Remember 1 1 10 Try with hand compute write the result for both binary and hex Now let us check the result from 6502 computing 0001 0000 0002 0000 GPIO1 EQU 8000 0003 0000 0004 0000 0005 0200 ORG 200H 0006 0200 0007 0200 18 CLC 0008 0201 A9 5B LDA 301011011 0009 0203 69 5A ADC 01011010 0010 0205 8D 00 80 STA GPIO1 0011 0208 00 BRK 0012 0209 0013 0209 0014 0209 END tasm Number of errors 0 The program will load number1 5B to the accumulator Then add it with 5B and then write the result to GPIO1 LED Do you have got the same result 10 Since the ADC instruction will add two numbers with carry flag so for the rightmost digit there is no carry flag We then clear it beforehand with instruction CLC CLear Carry You can try replace the first byte from 18 CLC to 38 SEC instruction that sets carry flag And test the program again What is the result Carry flag will be used for multiple bytes addition the same as we add multiple digits of decimal number We can modify above program for testing with logica
4. 200 0202 0205 0205 0207 020A 020D 0020F 0212 0215 0218 0218 021A 021B 021D 021E 021E 0220 0221 0223 0224 0224 0224 Number of errors A9 8D A9 8D 20 A9 8D 20 4C AO 88 DO 60 AO 88 DO 60 BF 02 BD 03 18 00 1E 05 01 FD C8 FD 80 80 02 80 02 02 PORT1 PORT2 LOOP DELAYON DELAY2 DELAYOFF DELAY3 0 EQU EQU 8002H 8003H ORG 200H LDA STA LDA STA JSR LDA STA JSR JMP LDY DEY BNE RTS LDY DEY BNE RTS SBF PORT1 BD PORT2 DELAYON 00000000 PORT2 DELAYOFF LOOP 1 DELAY2 200 DELAY3 END Line 6 and 7 will prevent BREAK command Let us focus at the main loop We first write BD to PORT2 and call delay for LED ON Then turn off all bits at PORT2 and call delay for LED OFF The ratio between Time ON and OFF can be approx using the loop counter in register Y We see that it is approx 1 200 or 0 5 Try enter the hex code and test it with key GO Then put the jumper J2 to pin 1 2 Check the brightness of the LED Can you make it more brighter Can you change the number to be displayed Before reset the board remove J2 to pin 2 3 24 Programl0 Testing IRQ with 10ms Tick The onboard 10ms tick generator is produced by the AT89C2051 microcontroller 5V R1 Programmable system tick 680 U3 TP1 TXD gt P3 0 RXD P1 0 AINO nea d 1 ssh P3 1 TXD P1 1 AIN1 gt y LED P3 2 INTO P1 2 N P3 3 I
5. 208 0013 0208 8E 00 80 STX GPIO1 0014 020B 4C OB 02 HERE JMP HERE 0015 020E 0016 020E END tasm Number of errors 0 Register X is loaded with 10 Main loop is to write the X register to GPIO1 LED DEX instruction will decrement X register by one Zero flag will be set if the all bits in X register are zero We can control the number of loop running by BNE instruction then 13 BNE Branch if Not Equal to ZERO has two bytes hex code i e DO and FA The second byte FA is signed number It is the OFFSET byte that will be used to add to the current program counter if the result is not equal to zero Current program counter after these two bytes were read by the CPU is 208 We see from the program that the loop location is 202 Thus the OFFSET byte is 202 208 6 Or jump backward 6 bytes We can make 6 from 6 by using 2 s complement The computer uses 2 s complement to represent the negative number Here is 6 in 8 bit binary 0000 0110 2 s complement is 1 s complement 1 This is 1 s complement 1111 1001 Then 1 1111 1010 or FA we see that the OFFSET byte FA is then 6 We can also use key REL for finding the OFFSET byte At the display 0206 with DO press REL key the address 0206 will be start address Press key for the destination enter 0202 the press key GO The OFFSET byte FA will be placed at address 0207 automatically On the hex keypad you may count backward from key F to key A You will get 6
6. Example Programs for 6502 Microprocessor Kit 0001 0000 0002 0000 GPIO1 EQU 8000 0003 0000 0004 0000 0005 0200 ORG 200 0006 0200 0007 0200 A5 00 LDA 0 0008 0202 8D 00 80 STA GPIO1 0009 0205 00 BRK 0010 0206 0011 0206 0012 0206 END tasm Number of errors 0 Build Your Own 6502 Microprocessor kit www kmitl ac th kswichit 6502 6502 html Preface The example programs for 6502 Microprocessor kit demonstrate how to enter the computer program in hex code and how to use the onboard input output devices e g GPIO LED key switch speaker 7 segment display and 10ms system tick generator Each program is simple and short The instruction hex codes are provided Students can enter the hex code to memory and test it directly No need others tools The monitor program provides single step running with user registers and zero page memory displaying Students can check the result of CPU operation with these registers easily The program can be modified at the 8 bit constants or even with the instruction itself student will learn the result after program modifications For the hardware programming e g key switch speaker 7 segment display and interrupt please study the hardware schematic in user manual Wichit Sirichote Program Program2 Program3 Program4 Program Program Program Programs Program9 Contents Binary number counting 006 4 Bit SING detest istorii ties
7. NT1 P1 3 TESE PUEN P3 4 TO P1 4 P3 5 T1 P1 5 P1 6 PHI2 P1 7 2 XTAL1 XTAL2 RESET2 TICK A RST VPP VCC AT89C2051 The 10ms tick signal is a 100Hz We will learn how the 6502 responded with this signal SW1 is the selector between 10ms tick or IRQ switch SW2 ESP switch SW 1 25 The 6502 reset and interrupt vectors are put at the last page of 64kB memory space 2106 EF1F 2107 EF1F VECTOR NMI RESET AND IRQ 2108 EF1F 2109 EF1F 2110 FFFA ORG OFFFAH 2111 FFFA 2112 FFFA 6C C8 WORD NMI NMI 2113 FFFC 00 CO WORD OCOOOH RESET 2114 FFFE 6F C8 WORD IRQ IRQ The monitor program put the vector address for IRQ at location C86F 1923 C86C 6C FA 00 NMI JMP SFA 1924 C86F 6C FE 00 IRQ JMP SFE At the location C86F we put JUMP indirect using RAM vector at location 00FE for low byte and 00FF for high byte With this method we can then test the IRQ process by placing the IRQ vector in RAM 0001 0000 0002 0000 GPIO1 EQU 8000H 0003 0000 PORTO EQU 8001H 0004 0000 PORT1 EQU 8002H 0005 0000 PORT2 EQU 8003H 0006 0000 0007 0000 0008 0030 ORG 30 0009 0030 0010 0011 0012 0013 0014 0015 0016 0017 0018 0019 0020 0021 0022 0023 0024 0025 0026 0027 0028 0029 0030 0031 0032 0033 0034 0035 0036 0037 0038 0039 0040 0041 0042 0043 0044 0045 0046 0030 0031 0032 0032 OOFE OOFE 0100 0200 0200 0200 0202 0204 0206 0208 0208 0209 020C 020C 020C 020C 020C
8. decimal number is 7 segment display We have the rightmost digit for experimenting by using jumper J2 DO Ci7 T 10uF 74HC573 We see that each bit CMOS output is capable for driving each segment of the LED Logic 1 will make the segment ON 21 To display a number says 0 1 2 3 4 5 6 7 8 9 we must convert the 8 bit value into the corresponding pattern See the table below Number SEGMENT CODE Display 0 BD 0 1 30 Y 2 9B 2 3 BA ecu A 36 4 5 AE so 6 SAF 6 7 38 va 8 BF 8 9 BE 9 If we want to show number 0 the segment code BD will be used instead By writing BD to PORT2 at location 8003 and put the jumper J2 to pin 1 2 Our segment driver using 74HC573 CMOS Latch has no current limiting resistors To produce the suitable brightness of the 7 segment display we use PWM Pulse Width Modulation method The method is to turn ON and OFF the LED at high repetition rate To make low brightness turn ON period will be smaller than turn moet i l The average DC power will then smaller and no heat dissipation If we use series current limiting resistors heat will be I R Let us have a look our program 0001 0002 0003 0004 0005 0006 0007 0008 0009 0010 0011 0012 0013 0014 0015 0016 0017 0018 0019 0020 0021 0022 0023 0024 0025 0026 0027 0028 tasm 0000 0000 0000 0200 0200 0
9. is FA We will use conditional jump instructions with relative addressing for many programs 14 Program Reading switch status The kit has one bit that ties REP switch S19 We can read the status of this switch easily by reading it then write it to GPIO1 LED Logic 1 means switch is opened and logic 0 means switch is closed U13 is tri state buffer used as the 8 bit input port Program is repeating loop read the byte from PORTO REP key is tied to bit 6 PA6 We want to check status only this bit so we remove the rest bits with instruction AND 01000000 Also we invert it with EOR Exclusive OR EOR 01000000 0001 0002 0003 0004 0005 0006 0007 0008 0009 0010 0011 0012 0000 0000 0000 0000 0200 0200 0200 0203 0205 0207 020A 020D 29 49 8D 4C 01 40 40 00 00 80 80 02 GPIO1 PORTO LOOP EQU 8000H EQU 8001H ORG 200H LDA PORTO AND 01000000 EOR 301000000 STA GPIO1 JMP LOOP 15 0013 020D 0014 020D END tasm Number of errors 0 We test this program with key GO What happen if we press key REP Note Logical AND can be used to maskout undesired bit by logic O Logical Exclusive OR can be used to invert the logic by using logic niles Logical OR can be used to set the desired bit to be 1 by using logic 1 Program7 Producing tone 800Hz We can produce the tone signal 800Hz using one bit output port A simple tone sig
10. l instructions as well By replacing ADC instruction with Logical OR with instruction ORA n Logical AND AND n Exclusive OR EOR n For example if we want to find the result of logical AND between 4A and 33 the program will be 0005 0200 ORG 200H 0006 0200 0007 0200 18 CLC 0008 0201 A9 4A LDA 4A 0009 0203 29 33 AND 33 0010 0205 8D 00 80 STA GPIO1 0011 0208 00 BRK 0012 0209 0013 0209 0014 0209 END tasm Number of errors 0 We see that at the address 0203 now the hex code is 29 the AND instruction And the 2 byte is 33 the 8 bit data to be ANDed with 4A Result will show on the GPIO1 LED directly Note Symbol shown in the program is for binary number constant Remember is for hex number The hex code for AND n is 29 ORA n is 09 EOR n is 49 Find more the 6502 hex code from http www obelisk demon co uk 6502 reference html Program5 Conditional branch instruction The 6502 CPU has the instructions that jump with flag condition The example of flag is ZERO flag The zero flag will be 1 when all bits in a given memory location or register are zero We can use such indication for changing program flowing direction Let us see the example of using BNE instruction 0001 0000 0002 0000 GPIO1 EQU 8000 0003 0000 0004 0000 0005 0200 ORG 200H 0006 0200 0007 0200 A2 10 LDX 10 0008 0202 0009 0202 8E 00 80 LOOP STX GPIO1 0010 0205 CA DEX 0011 0206 DO FA BNE LOOP 0012 0
11. nal is square wave ON and OFF We will try with 800Hz PCO PC4 PC5 PORTI gt RESETS 74HC573 TXD gt The small speaker is tied to bit 7 of PORTI It is driven by Q1 PNP transistor We can make this bit ON and OFF by writing bit 1 and bit O to this port Each time we make such bit ON or OFF a small delay is inserted The example shown below the delay will make period for 0 5 1 800Hz SPEAKER 17 0001 0000 0002 0000 GPIO1 EQU 8000H 0003 0000 PORTO EQU 8001H 0004 0000 PORT1 EQU 8002H 0005 0000 0006 0200 ORG 200H 0007 0200 0008 0200 A9 3F TONE LDA 00111111 0009 0202 8D 02 80 STA PORT1 0010 0205 20 13 02 JSR DELAY 0011 0208 A9 BF LDA 10111111 0012 020A 8D 02 80 STA PORT1 0013 020D 20 13 02 JSR DELAY 0014 0210 4C 00 02 JMP TONE 0015 0213 0016 0213 AO 79 DELAY LDY 79 0017 0215 88 DELAY2 DEY 0018 0216 DO FD BNE DELAY2 0019 0218 60 RTS 0020 0219 0021 0219 END tasm Number of errors 0 For 800Hz the half period is 625 microseconds The delay subroutine will need to delay approx a bit smaller than 625 microseconds Can you change the frequency Where is the byte to be modified 18 Program8 Simple MORSE code keyer Let us have some fun with MORSE keyer program We will combine Program 7 and Program 8 Program 7 enables us to read status of key REP and Program 8 produces 800Hz tone at the speaker We will read if key REP was pressed Tone 800Hz will be turned
12. on If key REP was released Tone 800Hz will be turned off This will make a simple MORSE code keyer 0001 0002 0003 0004 0005 0006 0007 0008 0009 0010 0011 0012 0013 0014 0015 0016 0017 0000 0000 0000 0000 0000 0200 0200 0200 0203 0206 0208 020A 020A 020D 020D 0210 0212 MORSE KEYER SPEAKER GPIO1 EQU 8000H PORTO EQU 8001H PORT1 EQU 8002H ORG 200H 20 1A 02 MORSE JSR TONE AD 01 80 LDA PORTO 29 40 AND 01000000 FO F6 BEQ MORSE 20 2B 02 JSR DELAY AD 01 80 WAIT LDA PORTO 29 40 AND 01000000 DO F9 BNE WAIT 19 0018 0214 0019 0214 20 2B 02 JSR DELAY 0020 0217 4C 00 02 JMP MORSE 0021 021A 0022 021A A9 3F TONE LDA 00111111 0023 021C 8D 02 80 STA PORT1 0024 021F 20 2B 02 JSR DELAY 0025 0222 A9 BF LDA 10111111 0026 0224 8D 02 80 STA PORT1 0027 0227 20 2B 02 JSR DELAY 0028 022A 60 RTS 0029 022B 0030 022B AO 79 DELAY LDY 79 0031 022D 88 DELAY2 DEY 0032 022E DO FD BNE DELAY2 0033 0230 60 RTS 0034 0231 0035 0231 END tasm Number of errors 0 We see that now we have two subroutines TONE and DELAY The TONE subroutine will be called from the main code only when REP key was pressed Testing key status now can be done by using logical AND bit6 of the PORTO MORSE tone is widely used between 750Hz to 900Hz Can you modify our code that produces 800Hz for higher or lower frequency 20 Program9 Seven segment display One of the generic device for displaying
13. s the bit ASL 0 instead of INC 0 Enter the hex code then test it with STEP amp REP key What is the result on the GPIO1 LED 0001 0000 0002 0000 GPIO1 EQU 8000 0003 0000 0004 0000 0005 0200 ORG 200H 0006 0200 0007 0200 A9 01 LDA 1 0008 0202 85 00 STA 0 0009 0204 0010 0204 A5 00 LOOP LDA 0 0011 0206 8D 00 80 STA GPIO1 0012 0209 06 00 ASL 0 0013 020B 0014 020B 4C 04 02 JMP LOOP 0015 020E 0016 020E END tasm Number of errors 0 We see that the GPIO1 LED is very useful for reading the 8 bit data We can see the result of internal operation by using the Accumulator register as the data carrier The instruction STA GPIO1 having hex code as 8D 00 80 will write the contents of accumulator register to GPIO1 LED Program3 LOAD and STORE Another method to test program running is to use BRK instruction We will test the code with key GO Our program has only three instructions LDA 0 STA GPIO1 and BRK BRK instruction will make the CPU to jump back to monitor program saving the CPU registers to user registers We can keep the results in user registers for program checking This program will load the accumulator with contents at memory location 0 then write it to GPIO LED and ended with BRK instruction 0001 0000 0002 0000 GPIO1 EQU 8000 0003 0000 0004 0000 0005 0200 ORG 200H 0006 0200 0007 0200 A5 00 LDA 0 0008 0202 8D 00 80 STA GPIO1 0009 0205 00 BRK 0010 0206 0011 0206 0012
14. to a tienen 7 LOAD and STORE cccccseseseeees 8 8 bit ACG ION avec eatecrdscetevindadsdledtseess 9 Conditional branch instruction 12 Reading switch statuS 6 14 Producing tone 800H2z 16 Simple MORSE code keyerv 18 Seven segment display 20 Programl0 Testing IRQ with 10ms Tick 24 Program Binary number counting This program uses memory location O in zero page as the 8 bit variable We will see the 8 bit binary number counting by writing the contents of memory location O to the GPIO1 LED at location 8000 Logic 1 will make LED ON and logic 0 for LED OFF 8 B38855885 LEREERE T4HC573 p7 _ D8 pe D1 E D12 I RR KR 4 4 D13 1N5236A We can enter the instruction hex code to the memory started from location 200 to the last byte at location 20D xe Pa K5 4 tl To test the program press key RESET PC then press STEP and REP together Did you see the 8 bit binary number counting Line Addr hex code Label Instruction 0001 0000 0002 0000 GPIO1 EQU 8000 0003 0000 0004 0000 0005 0200 ORG 200H 0006 0200 0007 0200 A9 00 0008 0202 85 00 0009 0204 0010 0204 A5 00 LOOP 0011 0206 8D 00 80 0012 0209 E6 00 0013 020B 0014 020B 4C 04 02 0015 020E 0016 020E tasm Number of errors 0 LDA STA LDA STA INC 0 0 0 GPIO1 0 LOOP
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