Z80 Microprocessor Kit

Programming Lab Book

Student:_________________________ID:____________________

Study program:_________________________________________
 PREFACE

This lab book is designed for self-learning how to program the Z80 microprocessor in
machine language with the Z80 Microprocessor Kit. The demonstration programs were
written in assembly program using Z80 instructions. The program listings are provided with
instruction hex code. Students can enter the program by using hex code to the memory and
test it directly. Illustrations of program flow were also provided for easy understanding.
After program testing the exercise will ask questions and program modification only at the
hex code, no need the assembler program.

LAB1 to LAB 6 are focusing on software programming. LAB7 to LAB12 are for hardware
interfacing using the on-board I/O devices.
 Z80 MICROPROCESSOR KIT
PROGRAMMING LAB BOOK
CONTENTS

LAB 1 RUNNING DOT LED...........................................................................4

LAB 2 FILL MEMORY WITH CONSTANT......................................................6

LAB 3 ADDING 16-BIT NUMBER..................................................................8

LAB 4 ADDING BCD NUMBER....................................................................11

LAB 5 COMPARISON...................................................................................14

LAB 6 STOP-WATCH....................................................................................17

LAB 7 INTERRUPT........................................................................................21

LAB 8 7-SEGMENT DISPLAY.......................................................................26

LAB 9 KEYBOARD.......................................................................................30

LAB 10 DIGITAL TIMER...............................................................................34

LAB 11 LCD MODULE INTERFACING.......................................................37

LAB 12 SERIAL COMMUNICATION............................................................43

APPENDIX A

MONITOR SUBROUTINES..........................................................................50

APPENDIX B

ASCII CODE CHART...................................................................................51

DECIMAL, BINARY and HEXADECIMAL
LAB 1 RUNNING DOT LED

Your first program will get you familiar with how to enter the hex code and test run quickly.

The main code is repeat writing accumulator register to GPIO1 with delay.

line addr hex code label instruction comment

---- ---- -------- ----- –------------ –--------------
0001 1800 .ORG 1800H
0002 1800
0003 1800 GPIO1 .EQU 40H
0004 1800
0005 1800 3E 01 MAIN LD A,1
0006 1802
0007 1802 D3 40 LOOP OUT (GPIO1),A
0008 1804 CD 0C 18 CALL DELAY
0009 1807 CB 07 RLC A
0010 1809 C3 02 18 JP LOOP
0011 180C
0012 180C
0013 180C 11 FF FF DELAY LD DE,-1
0014 180F 21 00 10 LD HL,1000H
0015 1812 19 LOOP2 ADD HL,DE
0016 1813 38 FD JR C,LOOP2
0017 1815 C9 RET
0018 1816
0019 1816 .END
0020 1816
tasm: Number of errors = 0
0 0 0 0 0 0 0 1 Accumulator

GPIO1
Delay
0 0 0 0 0 0 1 0 Rotate left A

Repeat write to GPIO1

Procedure

1. Enter the hex code from address 1800 to 1815.

2. Run the program with key PC, and key GO.

3. Stop program running with key RESET.

4
Exercise

1. Suppose we want new initial value to be loaded into register A, where is the byte to be
modified? Try change it and run the program again.

__________________________________________________________________________
2. Let us make running speed slower. Where is the byte to be modified? Test it.

__________________________________________________________________________

3. Can you change the direction of LED running? How to do that?

__________________________________________________________________________

Another example is to show the useful of delay subroutine for fun.

0001 1800 .ORG 1800H

0002 1800
0003 1800 GPIO1 .EQU 40H
0004 1800 TONE2K .EQU 05E2H
0005 1800
0006 1800 21 64 00 MAIN LD HL,100
0007 1803 CD E2 05 CALL TONE2K
0008 1806 CD 0C 18 CALL DELAY
0009 1809 C3 00 18 JP MAIN
0010 180C
0011 180C 11 FF FF DELAY LD DE,-1
0012 180F 21 00 10 LD HL,1000H
0013 1812 19 LOOP2 ADD HL,DE
0014 1813 38 FD JR C,LOOP2
0015 1815 C9 RET
0016 1816
0017 1816 .END
0018 1816
tasm: Number of errors = 0

The new subroutine called TONE2K is built-in code that produces 2kHz tone with period
set by HL. The delay subroutine is the same as LED running. Students may try enter the hex
code and test run again. You may learn how the code running and modify it.

Summary

The delay subroutine is a useful code for demonstration computer program running and
even for complicated applications as well. Since the microprocessor cycle time is quite short
compare to our response. If students change the load value of HL register to a small number,
and try the LED running. We will see all LED will be lit, why? 5
LAB 2 FILL MEMORY WITH CONSTANT

This lab demonstrates how to fill block of memory using indirect addressing mode.

line addr hex code label instruction comment

---- ---- -------- ----- –------------ –--------------
0001 1800 .ORG 1800H
0002 1800
0003 1800
0004 1800 06 10 MAIN LD B,16 ; loop counter
0005 1802 21 00 20 LD HL,2000H ; HL = 2000H
0006 1805 3E 00 LD A,0 ; A=0
0007 1807 77 LOOP LD (HL),A ; [HL]=A
0008 1808 23 INC HL ; HL=HL+1
0009 1809 10 FC DJNZ LOOP ; jump if B!=0
0010 180B FF RST 38H ; jump back monitor
0011 180C
0012 180C .END
0013 180C
tasm: Number of errors = 0

Before After

2000H 89H A=0 2000H 00H

AEH 00H
[HL]=A
57H 00H
HL=HL+1

B not eq. 0
200FH 9FH 200FH 00H

RST 38H

Procedure

1. Enter the hex code from address 1800 to 180B.

2. Write down the content of memory locations 2000 to 200F before press key GO.

3. Press key PC to set address to 1800 then key GO to run the program.

4. Write down the content of memory locations 2000 to 200F after press key GO. 6
Results

Address Data (before) Data (after)

2000H
2001H
2002H
2003H
2004H
2005H
2006H
2007H
2008H
2009H
200AH
200BH
200CH
200DH
200EH
200FH

Exercise

1. Now we want to clear 256 bytes from 2000 to 20FF. How to do that? Try and show the
result to TA.

_______________________________________________________________________

2. Instead of clearing the memory, let us fill the block of data with FF, 256 bytes, from 2000
to 20FF. How to do that? Try and show the result to TA.

_______________________________________________________________________________

Summary

We can use register B together with DJNZ instruction to make loop running. The load value
to register B will be loop counter. The body of loop can be any code or subroutine to be
repeated. HL register pair forms a 16-bit pointer. We use it to point the memory address with
indirect addressing mode.

7
LAB 3 ADDING 16-BIT BINARY NUMBER

We will learn how to add two 16-bit binary numbers using indirect addressing mode.

line addr hex code label instruction comment

---- ---- -------- ----- –------------ –--------------
0001 1800 .ORG 1800H
0002 1800
0003 1800 NUMBER1 .EQU 2000H
0004 1800 NUMBER2 .EQU 2002H
0005 1800
0006 1800 MAIN
0007 1800 21 00 20 LD HL,NUMBER1
0008 1803 11 02 20 LD DE,NUMBER2
0009 1806
0010 1806 06 02 LD B,2 ; loop counter
0011 1808 AF XOR A ; clear carry flag
0012 1809
0013 1809 1A LOOP LD A,(DE)
0014 180A 8E ADC A,(HL)
0015 180B 77 LD (HL),A
0016 180C
0017 180C 13 INC DE
0018 180D 23 INC HL
0019 180E
0020 180E 10 F9 DJNZ LOOP
0021 1810
0022 1810 FF RST 38H
0023 1811
0024 1811 .END
0025 1811
tasm: Number of errors = 0

Number1 and number2 are memory address that stores two 16-bit data. The program uses
HL and DE registers as a pointer. Register B is loop counter. The result will be saved to
number1 location. Before After

NUMBER1 2000H 89 2000H XX

2001H AE 2001H YY
NUMBER2 2002H 57 2002H 57
2003H 9A 2003H 9A

AE 89+
CY 9A 57
YY XX 8
Procedure

1. Enter the hex code from address 1800 to 1810.

2. Edit the data to be added for number1 and number2 as shown above.

3. Compute by hand with binary addition, keep the result.

4. Run the program, press key PC and key GO.

5. Write the result that saved in number1 location and carry flag after adding. Compare the
result with hand calculation. Carry flag can be viewed with key REG, D. The display will
show the low nibble of flag register. Carry flag is the right most bit.

Results

Address Data (before) Hand compute Data (after)

2000H
2001H
2002H
2003H
Carry Flag= --

Exercise

1. Let us try with new value of number1 and number2 by editing them in location 2000H to
2003H. Then ask your friend add it by hand calculation. Check the your friend result with
Z80 running, correct or not?

Address Data (before) Hand compute Data (after)

2000H
2001H
2002H
2003H
Carry Flag= --

2. If we want to add two 32-bit number, how to do that?

__________________________________________________________________________

__________________________________________________________________________

9
Summary

Adding binary number can be done with 8-bit addition instruction. For multi-byte adding,
we can use ADC instruction with loop running. The number of loop will be the number of
byte to be added. If there is a carry bit from lower significant byte adding, the carry bit will
be added to the next higher significant byte automatically.

Important note: adding instruction is for binary number. The hexadecimal representation are
for shorter written and code entering.

10
LAB 4 ADDING BCD NUMBER

For some applications that use BCD number, Z80 also has instruction DAA used to adjust or
correct the result of addition in accumulator.

line addr hex code label Instruction comment

---- ---- -------- ----- –------------ –--------------
0001 1800 .ORG 1800H
0002 1800
0003 1800 NUMBER1 .EQU 2000H
0004 1800 NUMBER2 .EQU 2002H
0005 1800
0006 1800 MAIN
0007 1800 21 00 20 LD HL,NUMBER1
0008 1803 11 02 20 LD DE,NUMBER2
0009 1806
0010 1806 06 02 LD B,2
0011 1808 AF XOR A ; clear carry flag
0012 1809
0013 1809 1A LOOP LD A,(DE)
0014 180A 8E ADC A,(HL) ; Binary add
0015 180B 27 DAA ; adjust result to BCD
0016 180C 77 LD (HL),A
0017 180D
0018 180D 13 INC DE
0019 180E 23 INC HL
0020 180F
0021 180F 10 F8 DJNZ LOOP
0022 1811
0023 1811 FF RST 38H
0024 1812
0025 1812 .END
0026 1812
tasm: Number of errors = 0

Before After

NUMBER1 2000H 89 2000H XX

2001H 19 2001H YY
NUMBER2 2002H 88 2002H 88
2003H 90 2003H 90

19 89+
CY 90 88
YY XX
11
We see that above program is the same as LAB 3. Only, DAA instruction is followed with
ADC instruction. For now number1 and number2 must be BCD, all digits must be 0-9 only.
As we know that BCD is a decimal representation by 4-bit binary number.

Procedure

1. Enter the hex code from address 1800 to 1811.

2. Edit the data to be added for number1 and number2 as shown above.

3. Compute by hand, keep the result.

4. Run the program, press key PC and key GO.

5. Write the result that saved in number1 location and carry flag after adding. Compare the
result with hand calculation. Carry flag can be viewed with key REG, D. The display will
show the low nibble of flag register. Carry flag is the right most bit.

Results

Address Data (before) Hand compute Data (after)

2000H
2001H
2002H
2003H
Carry Flag= --

Exercise

1. Let us try with new value of number1 and number2 by editing them in location 2000H to
2003H. Then ask your friend add it by hand calculation. Check the your friend result with
Z80 running, correct or not?

Address Data (before) Hand compute Data (after)

2000H
2001H
2002H
2003H
Carry Flag= --

2. Suppose we want to add 10 digits BCD number. For example,

12
Number1 = 1234567890
Number2 = 8976543254

Find the unknown values, xx, yy, jj, kk and nn in the program below.

line addr hex code label Instruction comment

---- ---- -------- ----- –------------ –--------------
0001 1800 .ORG 1800H
0002 1800
0005 1800
0006 1800 MAIN
0007 1800 21 xx yy LD HL,NUMBER1
0008 1803 11 jj kk LD DE,NUMBER2
0009 1806
0010 1806 06 nn LD B,nn
0011 1808 3F CCF
0012 1809
0013 1809 1A LOOP LD A,(DE)
0014 180A 8E ADC A,(HL)
0015 180B 27 DAA ; Adjust to BCD
0016 180C 77 LD (HL),A
0017 180D
0018 180D 13 INC DE
0019 180E 23 INC HL
0020 180F
0021 180F 10 F8 DJNZ LOOP
0022 1811
0023 1811 FF RST 38H
0024 1812
0025 1812 .END

3. Test your program running by checking the result of addition.

Summary

Z80 provides DAA instruction for correcting the result after using ADD or ADC
instructions. DAA must be placed after ADD or ADC instruction. The values to be added
also must be BCD number, 0-9. DAA is not the instruction that converts binary to decimal
number. It is designed for correcting the result of BCD number adding.

13
LAB 5 COMPARISON

Number comparison can be used to control flow or decision of program running. Z80
provides 8-bit compare instruction. The carry flag and zero flag will be affected according
to the result of comparison.

Suppose we have two 8-bit number to be compared, A and B. The result would be,

1. A < B, A is less than B,

2. A = B, A is equal to B,
3. A > B, A is larger than B.

0001 1800 .ORG 1800H

0002 1800
0003 1800 NUMBER1 .EQU 2000H ; A
0004 1800 NUMBER2 .EQU 2001H ; B
0005 1800
0006 1800 MAIN
0007 1800 3A 00 20 LD A,(NUMBER1)
0008 1803 21 01 20 LD HL,NUMBER2
0009 1806 BE CP (HL) ; A-B
0010 1807 FF RST 38H
0011 1808
0012 1808 .END
0013 1808
tasm: Number of errors = 0

Memory

2000 5FH Number1 Compare Number1 Carry flag,

2001 30H Number2 Number2 Zero flag

Procedure

1. Enter the hex code from address 1800 to 1807.

2. Edit the data to be compared for number1 and number2 at location 2000 and 2001.

4. Run the program, press key PC and key GO. Write down the result of carry and zero flags
with key REG, C, or REG D.

14
Results table

2000H (A) 2001H (B) Zero flag Carry flag comment

2AH 2FH CASE 1: A < B
2FH 2FH CASE 2:A = B
35H 2FH CASE 3: A > B

Check your results with below table.

2000H (A) 2001H (B) Zero flag Carry flag comment

2AH 2FH 0 1 CASE 1: A < B
2FH 2FH 1 0 CASE 2: A = B
35H 2FH 0 0 CASE 3: A > B

We may combine the result for A<=B, Zero and carry flags will be set.

For A > = B, Zero flag will be set and carry will be cleared.

We can use instructions conditional jump after carry or zero flags are affected. So the
decision can be made. The example below shows how to use the conditional jump.

0001 1800 .ORG 1800H

0002 1800
0003 1800 GPIO1 .EQU 40H
0004 1800 NUMBER1 .EQU 2000H
0005 1800 NUMBER2 .EQU 2001H
0006 1800
0007 1800 MAIN
0008 1800 3A 00 20 LD A,(NUMBER1)
0009 1803 21 01 20 LD HL,NUMBER2
0010 1806 BE CP (HL)
0011 1807 DA 12 18 JP C,LT
0012 180A
0013 180A CA 17 18 JP Z,LTEQ
0014 180D
0015 180D 3E 03 GT LD A,3
0016 180F D3 40 OUT (GPIO1),A ; CASE 3
0017 1811 FF RST 38H
0018 1812
0019 1812 3E 01 LT LD A,1
0020 1814 D3 40 OUT (GPIO1),A ; CASE 1
0021 1816 FF RST 38H
0022 1817
0023 1817 3E 02 LTEQ LD A,2
0024 1819 D3 40 OUT (GPIO1),A ; CASE 2
0025 181B FF RST 38H
0026 181C
0027 181C .END
0028 181C
tasm: Number of errors = 0

Procedure

1. Enter the hex code from address 1800 to 181B.

2. Edit the data to be compared for number1 and number2 at location 2000 and 2001.

4. Run the program, press key PC and key GO. Write down the result of GPIO1 LED for
three cases (1 for LED lit, 0 for LED off)

2000H (A) 2001H (B) GPIO1 LED comment

2AH 2FH CASE 1: A < B
2FH 2FH CASE 2: A = B
35H 2FH CASE 3: A > B

Summary

Comparison instruction can be used to make decision of program flow. After two numbers
are compared, the zero and carry flags will be affected. Program flow can be controlled with
conditional jump instructions.

16
LAB 6 STOP-WATCH

This lab demonstrates using monitor subroutines to make a simple stop-watch. The timebase
1/100 second is done by display scanning loop period. For more accurate, we will use timer
interrupt for the next lab.

0001 1800 .ORG 1800H

0002 1800
0003 1800 BUFFER .EQU 2000H
0004 1800 SCAN1 .EQU 0624H
0005 1800 HEX7SEG .EQU 0678H
0006 1800
0007 1800 DD 21 00 20 MAIN LD IX,BUFFER
0008 1804 11 00 00 LD DE,0
0009 1807
0010 1807 CD 24 06 LOOP CALL SCAN1
0011 180A 30 FB JR NC,LOOP
0012 180C
0013 180C 7B LD A,E
0014 180D C6 01 ADD A,1
0015 180F 27 DAA
0016 1810 5F LD E,A
0017 1811 7A LD A,D
0018 1812 CE 00 ADC A,0
0019 1814 27 DAA
0020 1815 57 LD D,A
0021 1816 7B LD A,E
0022 1817
0023 1817 21 00 20 LD HL,BUFFER
0024 181A CD 78 06 CALL HEX7SEG
0025 181D 36 02 LD (HL),2
0026 181F 23 INC HL
0027 1820 7A LD A,D
0028 1821 CD 78 06 CALL HEX7SEG
0029 1824 36 00 LD (HL),0
0030 1826
0031 1826 C3 07 18 JP LOOP
0032 1829
0033 1829 .END
0034 1829
tasm: Number of errors = 0

Register pair DE holds SEC and SEC/100. We see that incrementing was done by two bytes
adding with DAA for BCD adjustment. Result of BCD addition will be 00 to 99 for both
register D and E. When user press any key, the carry flag will be cleared, so jump
instruction at line 11 will repeat scanning without incrementing DE. This makes as a STOP
key. When key was released, carry flag will be set, the incrementing will be continued.
17
 SCAN DISPLAY

key pressed
No key
pressed

DE=DE+1
Convert DE to
7-segment pattern

Procedure

1. Enter the hex code from address 1800 to 1828.

2. Run the program, press key PC and key GO.

3. Try press any key, it will stop counting.

Suppose we want to clear the count to zero with a given key. How to do that?

0001 1800 .ORG 1800H

0002 1800
0003 1800 BUFFER .EQU 2000H
0004 1800 SCAN1 .EQU 0624H
0005 1800 HEX7SEG .EQU 0678H
0006 1800
0007 1800 DD 21 00 20 MAIN LD IX,BUFFER
0008 1804 11 00 00 LD DE,0
0009 1807
0010 1807 CD 24 06 LOOP CALL SCAN1
0011 180A 38 0B JR C, SKIP
0012 180C FE 18 CP 18H ; KEY CBR=18H
0013 180E 20 F7 JR NZ,LOOP
0014 1810
0015 1810 11 00 00 LD DE,0
0016 1813 3E 00 LD A,0
0017 1815 18 09 JR SKIP2
0018 1817
0019 1817 7B SKIP LD A,E
0020 1818 C6 01 ADD A,1
0021 181A 27 DAA
0022 181B 5F LD E,A
0023 181C 7A LD A,D
0024 181D CE 00 ADC A,0
0025 181F 27 DAA
0026 1820
0027 1820 57 SKIP2 LD D,A
0028 1821 7B LD A,E
0029 1822
0030 1822 21 00 20 LD HL,BUFFER
0031 1825 CD 78 06 CALL HEX7SEG
0032 1828 36 02 LD (HL),2
0033 182A 23 INC HL
0034 182B 7A LD A,D
0035 182C CD 78 06 CALL HEX7SEG
0036 182F 36 00 LD (HL),0
0037 1831
0038 1831 C3 07 18 JP LOOP
0039 1834
0040 1834 .END
0041 1834
tasm: Number of errors = 0

Not CBR key SCAN DISPLAY

key pressed
Carry flag=0
A=key position No key
pressed Carry flag=1
If key= CBR then
DE=0, A=0
DE=DE+1

Convert DE to
7-segment pattern

Since SCAN1 subroutine also scans the keyboard. It returns hardware wiring position of the
key being pressed. We will learn how to use SCAN1 more in later LAB. Let us try detecting
key CBR having position key of 18H to be a reset counter key. We use compare instruction
to check if key CBR has been pressed or not. It is was pressed, register DE and A are cleared
to zero.

Procedure

1. Enter the hex code from address 1800 to 1833.

2. Run the program, press key PC and key GO. 19

3. Try press any key, it will stop counting.

4. Try press key CBR, did you see the counter is zero?

Exercise

1. We can try change the control key to a given key. The example uses CBR for resetting the
counter to zero. Try to change another key for stop function.

2. Have the real stop watch for checking time error, with the accumulating time for one
minute, our program is run faster or slower in seconds unit? Why?

3. Can you change the counting from incrementing to decrementing using SUB and SBC
instructions.

Summary

We can use monitor subroutine scan display for displaying counter running. The count is
updated every approx. 1/100 sec. Registers DE was used for four digits BCD counting.
SCAN1 subroutine also returns hardware wiring position of the key. We can have a given
key by detecting its position for the STOP function.

20
LAB 7 INTERRUPT

Hardware interrupt

We will learn a maskable interrupt process and write the code for testing interrupt. The Z80
Kit has two sources of hardware interrupt. One for INTR key and a 10ms tick signal
produced by timer chip, AT89C2051 microcontroller. SW4 is a slide switch. We can select
which one will be the interrupt source.

Z80 CPU

INTR

INT
SW4

Timer chip 10ms tick

0001 1800 .ORG 1800H

0002 1800
0003 1800 ADDISP .EQU 1FDEH
0004 1800 PORT2 .EQU 02H
0005 1800 GPIO1 .EQU 40H
0006 1800
0007 1800 ; main code
0008 1800
0009 1800 21 0E 18 MAIN LD HL,SERVICE
0010 1803 22 FF 18 LD (18FFH),HL ; insert vector
0011 1806 3E 18 LD A,18H
0012 1808 ED 47 LD I,A ; high order byte
0013 180A
0014 180A ED 5E IM 2
0015 180C FB EI
0016 180D
0017 180D FF RST 38H
0018 180E
0019 180E ; interrupt service subroutine
0020 180E
0021 180E E5 SERVICE PUSH HL
0022 180F F5 PUSH AF
0023 1810
0024 1810 AF XOR A
0025 1811 D3 02 OUT (PORT2),A
0026 1813 2A DE 1F LD HL,(ADDISP)
0027 1816 7E LD A,(HL) 21
0028 1817 D3 40 OUT (GPIO1),A
0029 1819
0030 1819 F1 POP AF
0031 181A E1 POP HL
0032 181B
0033 181B FB EI
0034 181C ED 4D RETI
0035 181E
0036 181E .END
0037 181E
tasm: Number of errors = 0

Above program demonstrates how the maskable interrupt mode 2 is responding to the
trigger signal by key INTR.

When Z80 was triggered by low level logic at INT pin, it will acknowledge by requesting
the low order byte from device. The memory address that stores service vector is formed by
register I for high byte and low order byte from the device. Z80 will save current program
counter to stack memory, get the vector address then jump to interrupt service routine. When
interrupt process was finished, the program counter that saved in stack memory will be
retrieved and then return to the main program.

Device
INTR (1) Interrupt MODE 2

INT

(2) Supply low order byte

(3)
Register I=18H 18 FF

(4) PC STACK
18FFH 0E
(5) get vector 180E 1900H 18
1901H XX
1902H YY

(6) INTERRUPT SERVICE

(8) Return to main program

(7) RETI STACK PC

22
Our Kit has no circuit that supplies the low order byte, however with the pull-up resistor at
the data bus, the CPU will read byte as FF. We then load the register I with 18H, thus the
location that stores interrupt vector will be 18FFH.

Procedure

1. Slide SW4 to select INTR key. The INTR key will be used as the interrupt source.

2. Enter the hex code from address 1800 to 181D.

3. Press RESET key, PC key. Try press INTR key. Did you see any change at GPIO1?

_____________________________________________________________________
4. Now press key RESET, PC then GO. Did you see any change?

_____________________________________________________________________
5. Press key INTR again, what value will be displayed on GPIO1?

6. Change address to 0000, press key INTR again, write down the hex content and draw
GPIO1 LED. Press key + for next address and key INTR to make interrupt..

Address Data (HEX) GPIO1 LED (binary)

0000
0001
0002
0003
0004
0005
0006
0007
0008
0009
000A
000B
000C
000D
000E
000F

23
10ms Tick Timer interrupt

Let us do experiment with 10ms tick using timer chip. The 89C2051 microcontroller is used
to produce 10ms tick signal. Slide SW4 to 10ms tick position, we can trigger the Z80 CPU
to jump to interrupt service routine every 10ms. The demonstration program will display
BCD counting at GPIO1 every one second while the monitor program is functioning.

Z80 CPU

INTR

INT
SW4

Timer chip 10ms tick

0001 1800 .ORG 1800H

0002 1800
0003 1800 GPIO1 .EQU 40H
0004 1800 TICK .EQU 2000H
0005 1800 COUNTER .EQU 2001H
0006 1800
0007 1800 ; main code
0008 1800
0009 1800 21 0E 18 MAIN LD HL,SERVICE
0010 1803 22 FF 18 LD (18FFH),HL ; insert vector
0011 1806 3E 18 LD A,18H
0012 1808 ED 47 LD I,A ; high order byte
0013 180A
0014 180A ED 5E IM 2
0015 180C FB EI
0016 180D
0017 180D FF RST 38H
0018 180E
0019 180E ; interrupt service subroutine
0020 180E
0021 180E E5 SERVICE PUSH HL
0022 180F F5 PUSH AF
0023 1810
0024 1810 21 00 20 LD HL,TICK
0025 1813 34 INC (HL)
0026 1814 3A 00 20 LD A,(TICK)
0027 1817 FE 64 CP 100
0028 1819 38 10 JR C,SKIP
0029 181B 3E 00 LD A,0
0030 181D 32 00 20 LD (TICK),A
0031 1820
0032 1820 3A 01 20 LD A,(COUNTER)
0033 1823 C6 01 ADD A,1
0034 1825 27 DAA
0035 1826 32 01 20 LD (COUNTER),A
0036 1829 D3 40 OUT (GPIO1),A
0037 182B
0038 182B F1 SKIP POP AF
0039 182C E1 POP HL
0040 182D
0041 182D FB EI
0042 182E ED 4D RETI
0043 1830
0044 1830 .END
0045 1830
tasm: Number of errors = 0

Main code is the same as hardware interrupt. When Z80 is triggered by logic low signal at
pin INT, it will get the vector at location 18FFH, where we insert the interrupt service
address 180EH. The same as hardware interrupt, but now we use 10ms tick as the interrupt
source.

The code for interrupt service may called timer interrupt, will increment the tick variable, if
it is equal or greater than 100, then increment the counter variable.

Procedure

1. Slide SW4 to select TICK. 10ms Tick will be used as the interrupt source.

2. Enter the hex code from address 1800 to 182F.

3. Press RESET key, PC key. Then key GO. Did you see any change at GPIO1? Explain.

4. Try press key ADDR, + or – to change the address. Can you hear the beep sound that
slightly change in frequency? Explain why?

__________________________________________________________________________
25
LAB 8 7-SEGMENT DISPLAY

The kit has 6-digit 7-segment LED display. We will learn how to use it for displaying hex or
BCD data.

Display buffer

IX+0 BDH
IX+1 30H
IX+2 9BH
IX+3 BAH
IX+4 36H
IX+5 AEH

Display buffer uses 6 bytes of RAM for storing the data to be scanned. We can use IX
register to point to buffer memory. Example above shows the rightmost digit pattern, BDH
for number 0, and the left most digit pattern, AEH for number 5.

Bit pattern for LED display and bit position for each segment is shown below. For example
number 1, segment b and c are logic one , the data will be 30H.

a
f g b dpcb afge
e c 8-bit segment designation, e.g. 30H for '1'

d p

26
Let us try this program for testing the display.

0001 1800 .ORG 1800H

0002 1800
0003 1800 SCAN .EQU 0624H
0004 1800
0005 1800 ; main code
0006 1800
0007 1800 DD 21 00 20 MAIN LD IX,DISPLAYBUF
0008 1804 CD 24 06 LOOP CALL SCAN
0009 1807 18 FB JR LOOP
0010 1809
0011 1809
0012 2000 .ORG 2000H
0013 2000
0014 2000 DISPLAYBUF .BLOCK 6
0015 2006
0016 2006 .END
0017 2006
tasm: Number of errors = 0

We use IX register for pointing the display buffer memory begin at 2000H. And then call
the subroutine that scans display.

Procedure

1. Enter the hex code from address 1800 to 1808.

2. Press PC and GO, Did you see any display on the 7-segment LED? How it looks like?

3. Edit the memory address 2000 to 2005 with these value, 30, 02, 02, 0F, 1F, A1. Press PC
and GO. Draw the pattern on the segment below. Try also your pattern.

27
Exercise

1. Edit the bit pattern in display buffer, then show TA with the your name on the LED
display.

Now let us try display the 16-bit data and 8-bit data using monitor subroutines.

0001 1800 .ORG 1800H

0002 1800
0003 1800 SCANDISP .EQU 0624H
0004 1800 DATADISP .EQU 0671H ; A=nn
0005 1800 ADDRDISP .EQU 0665H ; DE=nnnn
0006 1800 DISPLAYBUF .EQU 1FB6H
0007 1800
0008 1800 ; main code
0009 1800
0010 1800 3A 00 20 MAIN LD A,(2000H)
0011 1803 CD 71 06 CALL DATADISP
0012 1806
0013 1806 ED 5B 01 20 LD DE,(2001H)
0014 180A CD 65 06 CALL ADDRDISP
0015 180D
0016 180D DD 21 B6 1F LD IX,DISPLAYBUF
0017 1811 CD 24 06 LOOP CALL SCANDISP
0018 1814 18 FB JR LOOP
0019 1816
0020 1816 .END
0021 1816
tasm: Number of errors = 0

Subroutine DATADISP will convert the 8-bit data in register A to two digits 7-segment
pattern and save to display buffer memory at DATA field.

Subroutine ADDRDISP will convert the 16-bit data in register DE to four digits 7-segment
pattern and save to display buffer memory at ADDRESS field.

We use memory address 2000H for storing 8-bit data and 2001H-2002H for 16-bit data.

Procedure

1. Enter the hex code from address 1800 to 1815.

2. Edit memory contents at location 2000H to 2002H with 12H, 34H and 56H.

3. Press PC and GO. Did you see the number on the display?
28
Summary

The onboard 6-digit LED can be used to display 8-bit or 16-bit data. Each digit uses one
byte memory which contains the pattern for a given number to be displayed. Z80 kit
provides monitor subroutines for use with applications program by calling them with proper
preset registers.

The example of code for displaying number and A to Z letters. The code is hex number.

Letter CODE (hex) Letter CODE (hex)

0 BD I 89
1 30 J B1
2 9B K 97
3 BA L 85
4 36 M 2B
5 AE N 23
6 AF O A3
7 38 P 1F
8 BF Q 3E
9 BE R 03
A 3F S A6
B A7 T 87
C 8D U B5
D B3 V B7
E 8F W A9
F 0F X 07
G AD Y B6
H 37 Z 8A

29
LAB 9 KEYBOARD

The Z80 kit has 36 keys for hex numbers, function keys and CPU control keys. We can do
experiments that demonstrates the use of keyboard. The monitor subroutine for keyboard
scanning is also provided for program testing.

Each key has its position by hardware wiring. But to use it with applications program, we
may arrange it to provide proper code that suitable for logical programming. For example
after key scanning, we will get key position code 12H for hex key '0'. We then can change it
to binary data 0 for key '0' by table indexing.

Let us do the first experiment to find the position code for each key.

0001 1800 .ORG 1800H

0002 1800
0003 1800 SCAN1 .EQU 0624H
0004 1800 HEX7SEG .EQU 0678H
0005 1800
0006 1800 DD 21 13 18 MAIN LD IX,BUFFER
0007 1804
0008 1804 CD 24 06 LOOP CALL SCAN1
0009 1807 30 02 JR NC,SKIP
0010 1809 3E FF LD A,0FFH
0011 180B
0012 180B 21 13 18 SKIP LD HL,BUFFER
0013 180E CD 78 06 CALL HEX7SEG
0014 1811 18 F1 JR LOOP
0015 1813
0016 1813 00 BUFFER .BYTE 0
0017 1814 00 .BYTE 0
0018 1815 00 .BYTE 0
0019 1816 00 .BYTE 0
0020 1817 00 .BYTE 0
0021 1818 00 .BYTE 0 30
0022 1819
0023 1819 .END
0024 1819
tasm: Number of errors = 0

Procedure

1. Enter the hex code from address 1800 to 1812.

2. Press PC and GO, write down the position key in the box or close to the key when the key
was pressed (except CPU control and user keys).

To make the key position useful and proper functioning, we change it to internal code then.

The monitor subroutine SCAN, the upper level subroutine uses KEYTABLE with offset
byte (key position) for pointing to the internal code. The IX register will load start address
of the table.

For example position code is 0. It will be used as the offset byte. The instruction LD A,
(IX+0) will get value of 03 for hex key 3.

KEYTABLE:
2519 077B 03 K0 .BYTE 03H ;HEX_3
2520 077C 07 K1 .BYTE 07H ;HEX_7
2521 077D 0B K2 .BYTE 0BH ;HEX_B
2522 077E 0F K3 .BYTE 0FH ;HEX_F
2523 077F 20 K4 .BYTE 20H ;N/A
2524 0780 21 K5 .BYTE 21H ;N/A
2525 0781 02 K6 .BYTE 02H ;HEX_2
2526 0782 06 K7 .BYTE 06H ;HEX_6
2527 0783 0A K8 .BYTE 0AH ;HEX_A
2528 0784 0E K9 .BYTE 0EH ;HEX_E
2529 0785 22 K0A .BYTE 22H ;N/A
2530 0786 23 K0B .BYTE 23H ;N/A
2531 0787 01 K0C .BYTE 01H ;HEX_1
2532 0788 05 K0D .BYTE 05H ;HEX_5
2533 0789 09 K0E .BYTE 09H ;HEX_9
2534 078A 0D K0F .BYTE 0DH ;HEX_D
2535 078B 13 K10 .BYTE 13H ;STEP
2536 078C 1F K11 .BYTE 1FH ;DOWNLOAD
2537 078D 00 K12 .BYTE 00H ;HEX_0
2538 078E 04 K13 .BYTE 04H ;HEX_4
2539 078F 08 K14 .BYTE 08H ;HEX_8
2540 0790 0C K15 .BYTE 0CH ;HEX_C
2541 0791 12 K16 .BYTE 12H ;GO
2542 0792 1E K17 .BYTE 1EH ;UPLOAD
2543 0793 1A K18 .BYTE 1AH ;CBR
2544 0794 18 K19 .BYTE 18H ;PC
2545 0795 1B K1A .BYTE 1BH ;REG
2546 0796 19 K1B .BYTE 19H ;ADDR
2547 0797 17 K1C .BYTE 17H ;DEL
2548 0798 1D K1D .BYTE 1DH ;RELA
2549 0799 15 K1E .BYTE 15H ;SBR
2550 079A 11 K1F .BYTE 11H ;-
2551 079B 14 K20 .BYTE 14H ;DATA
2552 079C 10 K21 .BYTE 10H ;+
2553 079D 16 K22 .BYTE 16H ;INS
2554 079E 1C K23 .BYTE 1CH ;COPY

Let us try the second experiment to find the internal code for each key.

0001 1800 .ORG 1800H

0002 1800
0003 1800 SCAN .EQU 05FEH
0004 1800 HEX7SEG .EQU 0678H
0005 1800
0006 1800 DD 21 0F 18 MAIN LD IX,BUFFER
0007 1804
0008 1804 CD FE 05 LOOP CALL SCAN
0009 1807
0010 1807 21 0F 18 SKIP LD HL,BUFFER
0011 180A CD 78 06 CALL HEX7SEG
0012 180D 18 F5 JR LOOP
0013 180F
0014 180F 00 BUFFER .BYTE 0
0015 1810 00 .BYTE 0
0016 1811 00 .BYTE 0
0017 1812 00 .BYTE 0 32
0018 1813 00 .BYTE 0
0019 1814 00 .BYTE 0
0020 1815
0021 1815 .END
0022 1815
tasm: Number of errors = 0

Procedure

1. Enter the hex code from address 1800 to 1814.

2. Press PC and GO, write down the internal key in the box or close to the key when the key
was pressed (except CPU control and user keys).

Summary

The Z80 kit has keyboard for inputting hex number or set the functions. The monitor
subroutine scans the key switch in sequential manner. When it was pressed, the low level
subroutine, SCAN1 will return hardware wiring position code. The upper level SCAN
subroutine returns the internal code that suitable for logical programming. Students may try
using the keyboard for many applications.

33
LAB 10 DIGITAL TIMER
This lab will use timer interrupt to be timebase for making a digital timer.

0001 1800 .ORG 1800H

0002 1800
0003 1800 SCAN .EQU 05FEH
0004 1800 DATADISP .EQU 0671H ; A=nn
0005 1800 ADDRDISP .EQU 0665H ; DE=nnnn
0006 1800 DISPLAYBUF .EQU 1FB6H
0007 1800
0008 1800 TICK .EQU 2000H
0009 1800 SEC .EQU 2001H
0010 1800
0011 1800
0012 1800
0013 1800 ; main code
0014 1800
0015 1800 21 20 18 MAIN LD HL,SERVICE
0016 1803 22 FF 18 LD (18FFH),HL ; insert vector
0017 1806 3E 18 LD A,18H
0018 1808 ED 47 LD I,A ; high order byte
0019 180A
0020 180A ED 5E IM 2
0021 180C FB EI
0022 180D
0023 180D ; CLEAR DISPLAY BUFFER
0024 180D
0025 180D 21 B6 1F LD HL,DISPLAYBUF
0026 1810 06 06 LD B,6
0027 1812 36 00 CLEAR LD (HL),0
0028 1814 23 INC HL
0029 1815 10 FB DJNZ CLEAR
0030 1817
0031 1817 DD 21 B6 1F LD IX,DISPLAYBUF
0032 181B CD FE 05 LOOP CALL SCAN
0033 181E 18 FB JR LOOP
0034 1820
0035 1820
0036 1820 ; interrupt service subroutine
0037 1820
0038 1820 E5 SERVICE PUSH HL
0039 1821 F5 PUSH AF
0040 1822
0041 1822 21 00 20 LD HL,TICK
0042 1825 34 INC (HL)
0043 1826 3A 00 20 LD A,(TICK)
0044 1829 FE 64 CP 100
0045 182B 38 11 JR C,SKIP 34
0046 182D 3E 00 LD A,0
0047 182F 32 00 20 LD (TICK),A
0048 1832
0049 1832
0050 1832 3A 01 20 LD A,(SEC)
0051 1835 D6 01 SUB 1
0052 1837 27 DAA
0053 1838 32 01 20 LD (SEC),A
0054 183B
0055 183B CD 71 06 CALL DATADISP
0056 183E
0057 183E F1 SKIP POP AF
0058 183F E1 POP HL
0059 1840
0060 1840 FB EI
0061 1841
0062 1841 ED 4D RETI
0063 1843
0064 1843
0065 1843
0066 1843 .END
0067 1843
tasm: Number of errors = 0

Repeat scan
display

SCAN buffer
memory 10ms Interrupt

If tick=100 then sec=sec-1

Convert sec to
7-segment pattern

We use interrupt mode 2 and have the interrupt service routine now with timer function.
Every 10ms, the CPU will jump to interrupt service routine at location 1820H. The tick
variable will be incremented by one and checked if it was equal or larger than 100, i.e.
100x10ms = 1 second, the variable sec will be decremented by one. Again we use DAA to
adjust the result from subtraction to be BCD number. The subroutine DATADISP will
convert the accumulator A and save the LED pattern to the display buffer in data field. Main
code is repeat scan the display. So we will see the counting down every one second. 35
Procedure

1. Ensure SW4 is set to 10ms tick position.

2. Enter the hex code from address 1800 to 1842.

3. Set the variable sec at 2001 to 99.

4. Press PC and GO, Did you see any display on the 7-segment LED?

_______________________________________________________________________

Exercise

1. The counting is decremented every one second. Can you modify the program to make
counting up every one second.

2. Try change the code to make counting every 0.1 second.

3. If we want to stop count when time is zero, give the idea how to do that?

________________________________________________________________________

Summary

We see that the main code is only scan the display. It reads the display buffer memory and
writes to the LED. But when the CPU was interrupted by 10ms tick, Z80 stop running the
main code and jump to interrupt service routine. We use variable to count the number of
interrupt, if it is equal to 100 ticks, time has passed for one second. We can use another
variable for counting 60 seconds for one minute as well.

36
LAB 11 LCD MODULE INTERFACE

In the applications that need more details readout, the LCD module is more suitable. One of
the example is LCD text module having HD44780 controller chip. Our kit prepares the 16-
pin header for easy interface. We can learn how to use it without the need of extra
interfacing circuit. The LCD module will need only female 16-pin header for direct
plugging into the on-board 16-pin header. Left hand is pin 1, and right hand is pin 16.

The example is 16 characters x 2 lines or

16x2 text LCD.

Each character's position has its location

depending on the size of LCD module.

The kit provides a hardware driver for the

LCD module. The hardware driver will set
the LCD mode, clear display, set cursor
position, write the ASCII data to display registers. User may try many size of LCD, the
driver can set the position for up to 4 lines LCD module.

The list of hardware driver is shown below.

Subroutine Address Parameters Function

INITLCD 0B64H none Initialize LCD
PRINTCHAR 0CCEH L=ASCII Print letter on LCD
PRINTTEXT 0CC8H HL=pointer Print text
GOTOXY 0CC2H H=X, L=Y Set cursor position
CLEAR 0B06H none Clear screen
PRINTHEX 0CD4H A=nn Print hex
PRINTINT 0CDBH HL=nnnn Print decimal number
DELAY 0CE1H none Delay ~0.5s

Steps to enable the LCD module and print a letter or text are as follows.

1. Initialize the module.

2. Write the ASCII code or text to LCD.

3. Set position, if need another location to be printed.

4. Clear display if needed.

37
Character code for LCD module is the same as ASCII code. For example letter 'A', the code
is 41H, 'B' is 42H. For numeric, letter '1', code is 31H.

For scientific or mathematics symbols, different manufacturers may provide different code.
We may test it by writing the code to the LCD and check it then.

38
Let us try print text “Hello worlds” on the first line and letter 'A' on the second line.

0001 1800 .ORG 1800H

0002 1800
0003 1800 INITLCD .EQU 0B64H
0004 1800 PRINTCHAR .EQU 0CCEH
0005 1800 PRINTTEXT .EQU 0CC8H
0006 1800 GOTOXY .EQU 0CC2H
0007 1800
0008 1800 CD 64 0B MAIN CALL INITLCD
0009 1803
0010 1803 21 16 18 LD HL,HELLO
0011 1806 CD C8 0C CALL PRINTTEXT
0012 1809
0013 1809 26 01 LD H,1
0014 180B 2E 00 LD L,0 ; X=0, Y=1
0015 180D CD C2 0C CALL GOTOXY
0016 1810
0017 1810 2E 41 LD L,'A'
0018 1812 CD CE 0C CALL PRINTCHAR
0019 1815
0020 1815 FF RST 38H
0021 1816
0022 1816 48 65 6C 6C 6F 20 HELLO .BYTE "Hello worlds",0
0022 181C 77 6F 72 6C 64 73 00
0023 1823
0024 1823 .END
0025 1823
tasm: Number of errors = 0

Main code begins with initialize the LCD module, then print text “hello worlds” and print
letter 'A' on LCD screen.

Procedure

1. Turn the kit's power off, then plug the LCD module to the 16-pin header.

2. Turn power on, adjust R18 until the first line becomes black.

39
3. Enter the hex code from 1800 to 1822.

4. Press PC and GO, Did you see any text on LCD screen?

Exercise

1. The message “hello worlds” is called ASCII string. At the end, it has terminator, 0. If you
want to show your name, how to do that?

2. Try shifting letter 'A' to center of 2nd line. How to do that?

__________________________________________________________________________

We see that the LCD module has more space for displaying many meaningful values.
Suppose we want to display the memory content 4-byte on the LCD. How to do that?

0001 1800 .ORG 1800H

0002 1800
0003 1800
0004 1800 INITLCD .EQU 0B64H
0005 1800 PRINTCHAR .EQU 0CCEH
0006 1800 PRINTTEXT .EQU 0CC8H
0007 1800 GOTOXY .EQU 0CC2H
0008 1800 CLEAR .EQU 0B06H
0009 1800 PRINTHEX .EQU 0CD4H
0010 1800
0011 1800 PRINTINT .EQU 0CDBH
0012 1800
0013 1800 DELAY .EQU 0CE1H
0014 1800
0015 1800
0016 1800
0017 1800 CD 64 0B MAIN CALL INITLCD
0018 1803
0019 1803 21 00 18 LD HL,1800H
0020 1806 7C LD A,H
0021 1807 E5 PUSH HL
0022 1808 CD D4 0C CALL PRINTHEX
0023 180B D1 POP DE
0024 180C
0025 180C 7B LD A,E
0026 180D CD D4 0C CALL PRINTHEX
0027 1810 2E 3A LD L,':'
0028 1812 CD CE 0C CALL PRINTCHAR 40
0029 1815
0030 1815 06 04 LD B,4
0031 1817 1A LOOP LD A,(DE)
0032 1818 CD D4 0C CALL PRINTHEX
0033 181B 2E 20 LD L,' '
0034 181D CD CE 0C CALL PRINTCHAR
0035 1820
0036 1820 13 INC DE
0037 1821 10 F4 DJNZ LOOP
0038 1823
0039 1823 FF RST 38H
0040 1824
0041 1824 .END
0042 1824
tasm: Number of errors = 0

We use PRINTHEX subroutine for printing one byte in HEX number. The value to be
printed is loaded into register A. For symbol, we use PRINTCHAR subroutine that prints
the symbol from ASCII code directly.

The printout on the LCD will be 1800: CD 64 0B 21.

Procedure

1. Enter the hex code from 1800 to 1823.

2. Press PC and GO, did you see the printout as described above?

Exercise

1. Modify the program to display the 2nd line for the next four bytes. Show the printout
result. Compare the printout with the memory contents by monitor key ADDR and key +.

The LCD driver also provides subroutine that prints unsigned 16-bit integer in decimal
number, PRINTINT. Let us test it.

0001 1800 .ORG 1800H

0002 1800
0003 1800 INITLCD .EQU 0B64H
0004 1800 GOTOXY .EQU 0CC2H
0005 1800 PRINTINT .EQU 0CDBH
0006 1800 DELAY .EQU 0CE1H
0007 1800
0008 1800 CD 64 0B MAIN CALL INITLCD
0009 1803
0010 1803 21 00 00 LD HL,0
0011 1806
0012 1806 E5 LOOP PUSH HL 41
0013 1807 CD DB 0C CALL PRINTINT
0014 180A CD E1 0C CALL DELAY
0015 180D
0016 180D 21 00 00 LD HL,0
0017 1810 CD C2 0C CALL GOTOXY
0018 1813
0019 1813 E1 POP HL
0020 1814 23 INC HL
0021 1815
0022 1815 C3 06 18 JP LOOP
0023 1818
0024 1818 .END
0025 1818
tasm: Number of errors = 0

The HL register is loaded with initial value of 0000 and printed it to LCD in decimal
representation. The actual incrementing by INC HL is binary counting up. Delay provides
approx. 0.5s delay, so we can see the incrementing clearly. To preserve the HL contents,
instruction PUSH HL and POP HL are used to save HL contents to stack memory.

Procedure

1. Enter the hex code from 1800 to 1817.

2. Press PC and GO, did you see the printout of number counting?

Summary

The LCD module is suitable for applications that need more details readout. The Z80 kit
provides both hardware interface and drivers subroutine for experimenting. The drivers is
composed of hardware level subroutine. To use the LCD we must first initialize it then we
can write the ASCII letter or string to the LCD.

42
LAB 12 SERIAL COMMUNICATION

The Z80 kit has RS232C communication port. We can test sending/receiving data between
two computers using the RS232C. The serial data is asynchronous format. The speed is
2400 bit/s, 8-data bit, no parity and one stop bit. The RS232C is designed for connecting
between two computers or between computer and data communication equipment (DCE)
such as MODEM.

The kit has two keys for upload and download Intel hex file. The Intel hex file is generated
from assembler or compiler program. We will learn how serial data is sent over the serial
port. Since the kit has no UART chip, serial data streams are generated by software control.
The monitor subroutines provide two basic functions, i.e. 1) COUT for sending a byte and
2) CIN for receiving a byte.

Connecting the kit to VT100 terminal (computer to computer)

VT100 terminal can be built using PC running communication program, like teraterm. The
RS232C cable is cross cable.

RS232 cross cable

Z80 Kit
VT100 Terminal

List of subroutines for serial communication.

Subroutine Address Parameters Function

CIN 08B9H A=byte Receive byte from 2400 bit/s terminal
COUT 0861H A=byte Send byte to 2400 bit/s terminal
PSTRING 0852H IX=pointer Print ASCII string to terminal
OUT2X 0875H A=byte Write hex to terminal
NEWLINE 088CH none Send CR LF to terminal
Let us test the code that prints “hello worlds” to terminal.

0001 1800 .ORG 1800H

0002 1800
0003 1800 PSTRING .EQU 0852H
0004 1800 CIN .EQU 08B9H
0005 1800 COUT .EQU 0861H
0006 1800 OUT2X .EQU 0875H
0007 1800 NEWLINE .EQU 088CH
0008 1800 DELAY .EQU 0CE1H
0009 1800
0010 1800
0011 1800 CD 10 18 START CALL INIT ; SET UART MARK LEVEL
0012 1803
0013 1803 CD 8C 08 MAIN CALL NEWLINE
0014 1806 DD 21 19 18 LD IX,HELLO
0015 180A CD 52 08 CALL PSTRING
0016 180D C3 03 18 JP MAIN
0017 1810
0018 1810 3E FF INIT LD A,0FFH
0019 1812 CD 61 08 CALL COUT
0020 1815 CD E1 0C CALL DELAY
0021 1818 C9 RET
0022 1819
0023 1819 48 65 6C 6C 6F 20 HELLO .BYTE "Hello worlds",0
0023 181F 77 6F 72 6C 64 73 00
0024 1826
0025 1826 .END
0026 1826
tasm: Number of errors = 0

The kit shares TXD pin with on-board speaker, thus to send data correctly, we must set mark
level before sending the serial data. This can be done with subroutine INIT. Main code
sends NEWLINE and print text using PSTRING.

Procedure

1. Enter the hex code from 1800 to 1825.

2. Connect the kit to terminal COM port using RS232C cross cable.

3. Run teraterm with 2400 bit/s, 8-data bit, no parity and one stop bit.

4. Run the program with key RESET, PC and GO. We will get the message on terminal
screen.

44
Exercise

1. Modify the program to send your message like, Hello friends.

Now we will try print the memory contents with OUT2X subroutine.

0001 1800 .ORG 1800H

0002 1800
0003 1800 PSTRING .EQU 0852H
0004 1800 CIN .EQU 08B9H
0005 1800 COUT .EQU 0861H
0006 1800 OUT2X .EQU 0875H
0007 1800 NEWLINE .EQU 088CH
0008 1800 ONESPACE .EQU 0897H
0009 1800 DELAY .EQU 0CE1H
0010 1800
0011 1800
0012 1800 CD 2C 18 START CALL INIT ; SET UART MARK LEVEL
0013 1803 21 00 18 LD HL,1800H
0014 1806
0015 1806 06 04 LD B,4
0016 1808
0017 1808 C5 LOOP PUSH BC
0018 1809 CD 10 18 CALL HEXDUMP
0019 180C C1 POP BC
0020 180D 10 F9 DJNZ LOOP
0021 180F
0022 180F FF RST 38H
0023 1810
0024 1810
0025 1810 45
0026 1810 CD 8C 08 HEXDUMP CALL NEWLINE
0027 1813 7C LD A,H
0028 1814 CD 75 08 CALL OUT2X
0029 1817 7D LD A,L
0030 1818 CD 75 08 CALL OUT2X
0031 181B CD 97 08 CALL ONESPACE
0032 181E
0033 181E 0E 10 LD C,16
0034 1820
0035 1820 CD 97 08 HEXDUMP2 CALL ONESPACE
0036 1823 7E LD A,(HL)
0037 1824 CD 75 08 CALL OUT2X
0038 1827 23 INC HL
0039 1828
0040 1828 0D DEC C
0041 1829 20 F5 JR NZ,HEXDUMP2
0042 182B C9 RET
0043 182C
0044 182C
0045 182C
0046 182C 3E FF INIT LD A,0FFH
0047 182E CD 61 08 CALL COUT
0048 1831 CD E1 0C CALL DELAY
0049 1834 C9 RET
0050 1835
0051 1835 .END
0052 1835
tasm: Number of errors = 0

The program begins with writing ADDRESS, 1800 then prints one space, followed with 16
bytes memory contents in hex number using OUT2X. Register C is inner loop counter for
16 bytes. The outer loop counter is register B at line 0015. The printout on terminal screen
would look like this.

Procedure

1. Enter the hex code from 1800 to 1835.

2. Run the program with key RESET, PC and GO. Did you get similar screen like above?46
Exercise

1. Modify the program to print more, says 16 lines.

The kit can receive serial data as well. We will see how to type message on VT10 terminal
and display it on Z80 Kit's LCD.

RS232C

Hello Z80 kit

Press enter

Z80 kit LCD

VT100 terminal

0001 1800 .ORG 1800H

0002 1800
0003 1800 INITLCD .EQU 0B64H
0004 1800 PRINTCHAR .EQU 0CCEH
0005 1800 GOTOXY .EQU 0CC2H
0006 1800 CLEAR .EQU 0B06H
0007 1800
0008 1800 CIN .EQU 08B9H
0009 1800 COUT .EQU 0861H
0010 1800 CR .EQU 0DH
0011 1800 ESC .EQU 1BH
0012 1800
0013 1800
0014 1800 CD 64 0B MAIN CALL INITLCD
0015 1803
0016 1803 CD B9 08 LOOP CALL CIN
0017 1806 FE 0D CP CR
0018 1808 20 08 JR NZ,SKIP1
0019 180A 21 00 01 LD HL,0100H
0020 180D CD C2 0C CALL GOTOXY
0021 1810 18 0D JR SKIP3
0022 1812
0023 1812 FE 1B SKIP1 CP ESC
0024 1814 20 05 JR NZ,SKIP2
0025 1816 CD 06 0B CALL CLEAR
0026 1819 18 04 JR SKIP3
0027 181B
0028 181B 6F SKIP2 LD L,A
0029 181C CD CE 0C CALL PRINTCHAR
0030 181F 47
0031 181F 18 E2 SKIP3 JR LOOP
0032 1821
0033 1821 .END
0034 1821
tasm: Number of errors = 0

Main code reads a character received from terminal and check if it is CR or ESC. For CR it
will set the new line, for ESC, it will clear the LCD display. The message that typed on the
terminal will be displayed on LCD concurrently.

Procedure

1. We will need 16x2 LCD, 2400 bit/s VT100 terminal and RS232C cable for this
experiment.

2. Enter hex code from 1800 to 1820.

3. Press PC and GO, then type a message on VT100 terminal. The message will show on the
Kit's LCD.

4. To enter newline, press key Enter or clear LCD screen with key ESC.

Summary

Sending or receiving digital data between computers mostly uses serial communication. The
Z80 kit has software control UART for asynchronous communication. Signal level of the
serial port is RS232C. The low level driver subroutines for serial port are COUT and CIN.
COUT is for sending 8-bit data in register A and CIN for receiving data to register A. Upper
level subroutines e.g., PSTRING will print ASCII string, OUT2X will print register in HEX
number.

We can use PC running terminal program to emulate VT100 terminal. The terminal uses
ASCII code for data exchange. Printable ASCII letters are from 20H to 7FH. Control codes
are from 00 to 1FH. The example of control codes are CR and LF for entering the new line.

The kit also has text file downloading key. The Intel hex file generated from assembler or c
compile programs is ASCII text file. For binary file transaction, the upper level software
will provide the protocol for byte sequence, acknowledgment, and error checking. The
popular are XMODEM, YMODEM.

48
49
 APPENDIX A

MONITOR SUBROUTINES

Subroutine Address Parameters Function

INITLCD 0B64H none Initialize LCD
PRINTCHAR 0CCEH L=ASCII Print letter on LCD
PRINTTEXT 0CC8H HL=pointer Print text on LCD
GOTOXY 0CC2H H=X, L=Y Set cursor position
CLEAR 0B06H none Clear LCD screen
PRINTHEX 0CD4H A=nn Print hex on LCD
PRINTINT 0CDBH HL=nnnn Print decimal on LCD
SCAN1 0624H IX=buffer Scan display one cycle
SCAN 05FEH IX=buffer Scan until key pressed
HEX7SEG 0678H A=nn Convert A to 7-segment
TONE1K 05DEH HL=period Produce tone 1kHz
TONE2K 05E2H HL=period Produce tone 2kHz
CIN 08B9H A=byte Receive byte from 2400 bit/s terminal
COUT 0861H A=byte Send byte to 2400 bit/s terminal
PSTRING 0852H IX=pointer Print ASCII string to terminal
OUT2X 0875H A=byte Write hex to terminal
NEWLINE 088CH none Send CR LF to terminal
DELAY 0CE1H none Delay ~0.5s

50
 APPENDIX B

Examples:

1. Letter 'A', the ASCII code is 41H.

2. Letter '1', the ASCII code is 31H.
3. CR code is 0DH, LF is 0AH, ESC is 1BH.

Hexadecimal Representation Example: FF = 1111 1111, 0D= 0000 1101

Decimal Binary Hexadecimal
0 0000 0
1 0001 1
2 0010 2
3 0011 3
4 0100 4
5 0101 5
6 0110 6
7 0111 7
8 1000 8
9 1001 9
10 1010 A
11 1011 B
12 1100 C
13 1101 D
14 1110 E
15 1111 F

51
 REDUCED Z80 INSTRUCTION SET REDUCED Z80 INSTRUCTION SET

Edited by Phil Townshend 2003 FLAG * = affected Edited by Phil Townshend 2003 FLAG * = affected
DEF's ? = unknown DEF's ? = unknown
Conditional jumps show 2 values: e.g. "RET Z = 1/3" (cycles if condition not met)/(cycles if condition met) x = no change Conditional jumps show 2 values: e.g. "RET Z = 1/3" (cycles if condition not met)/(cycles if condition met) x = no change

INSTR. HEX CODE CYC FREG - Flags register INSTR. HEX CODE CYC FREG - Flags register INSTR. HEX CODE CYC FREG - Flags register INSTR. HEX CODE CYC FREG - Flags register
b7 b6 b5 b4 b3 b2 b1 b0 b7 b6 b5 b4 b3 b2 b1 b0 b7 b6 b5 b4 b3 b2 b1 b0 b7 b6 b5 b4 b3 b2 b1 b0
S Z - H - - N C S Z - H - - N C S Z - H - - N C S Z - H - - N C
ADC A,C 89 1 * * * 0 * HALT 76 1 x x x x x ADC A,C 89 1 * * * 0 * HALT 76 1 x x x x x
ADC A,n CE nn 2 * * * 0 * IN A,(nn) DB nn 3 x x x x x ADC A,n CE nn 2 * * * 0 * IN A,(nn) DB nn 3 x x x x x
ADD A,(HL) 86 2 * * * 0 * INC (HL) 34 3 * * * 0 x ADD A,(HL) 86 2 * * * 0 * INC (HL) 34 3 * * * 0 x
ADD A,A 87 1 * * * 0 * INC A 3C 1 * * * 0 x ADD A,A 87 1 * * * 0 * INC A 3C 1 * * * 0 x
ADD A,B 80 1 * * * 0 * INC B 04 1 * * * 0 x ADD A,B 80 1 * * * 0 * INC B 04 1 * * * 0 x
ADD A,C 81 1 * * * 0 * INC BC 03 1 x x x x x ADD A,C 81 1 * * * 0 * INC BC 03 1 x x x x x
ADD A,D 82 1 * * * 0 * INC C 0C 1 * * * 0 x ADD A,D 82 1 * * * 0 * INC C 0C 1 * * * 0 x
ADD A,E 83 1 * * * 0 * INC D 14 1 * * * 0 x ADD A,E 83 1 * * * 0 * INC D 14 1 * * * 0 x
ADD A,H 84 1 * * * 0 * INC DE 13 1 x x x x x ADD A,H 84 1 * * * 0 * INC DE 13 1 x x x x x
ADD A,L 85 1 * * * 0 * INC E 1C 1 * * * 0 x ADD A,L 85 1 * * * 0 * INC E 1C 1 * * * 0 x
ADD A,n C6 nn 2 * * * 0 * INC H 24 1 * * * 0 x ADD A,n C6 nn 2 * * * 0 * INC H 24 1 * * * 0 x
ADD HL,BC 09 3 x x x 0 * INC HL 23 1 x x x x x ADD HL,BC 09 3 x x x 0 * INC HL 23 1 x x x x x
ADD HL,DE 19 3 x x x 0 * INC L 2C 1 * * * 0 x ADD HL,DE 19 3 x x x 0 * INC L 2C 1 * * * 0 x
ADD HL,HL 29 3 x x x 0 * INC SP 33 1 x x x x x ADD HL,HL 29 3 x x x 0 * INC SP 33 1 x x x x x
AND (HL) A6 2 * * 1 0 0 JP (HL) E9 1 x x x x x AND (HL) A6 2 * * 1 0 0 JP (HL) E9 1 x x x x x
AND A A7 1 * * 1 0 0 JP C,mn DA nn mm 3 x x x x x AND A A7 1 * * 1 0 0 JP C,mn DA nn mm 3 x x x x x
AND B A0 1 * * 1 0 0 JP mn C3 nn mm 3 x x x x x AND B A0 1 * * 1 0 0 JP mn C3 nn mm 3 x x x x x
AND C A1 1 * * 1 0 0 JP N,mn FA nn mm 3 x x x x x AND C A1 1 * * 1 0 0 JP N,mn FA nn mm 3 x x x x x
AND D A2 1 * * 1 0 0 JP NC,mn D2 nn mm 3 x x x x x AND D A2 1 * * 1 0 0 JP NC,mn D2 nn mm 3 x x x x x
AND E A3 1 * * 1 0 0 JP NZ mn C2 nn mm 3 x x x x x AND E A3 1 * * 1 0 0 JP NZ mn C2 nn mm 3 x x x x x
AND H A4 1 * * 1 0 0 JP Z,mn CA nn mm 3 x x x x x AND H A4 1 * * 1 0 0 JP Z,mn CA nn mm 3 x x x x x
AND L A5 1 * * 1 0 0 JR C d 38 dd 2/3 x x x x x AND L A5 1 * * 1 0 0 JR C d 38 dd 2/3 x x x x x
AND n E6 nn 2 * * 1 0 0 JR dd 18 dd 3 x x x x x AND n E6 nn 2 * * 1 0 0 JR dd 18 dd 3 x x x x x
CALL C,mn DC nn mm 3/5 x x x x x JR NC,d 30 dd 2/3 x x x x x CALL C,mn DC nn mm 3/5 x x x x x JR NC,d 30 dd 2/3 x x x x x
CALL mn CD nn mm 5 x x x x x JR NZ,d 20 dd 2/3 x x x x x CALL mn CD nn mm 5 x x x x x JR NZ,d 20 dd 2/3 x x x x x
CALL NC,mn D4 nn mm 3/5 x x x x x JR Z d 28 dd 2/3 x x x x x CALL NC,mn D4 nn mm 3/5 x x x x x JR Z d 28 dd 2/3 x x x x x
CALL NZ,mn C4 nn mm 3/5 x x x x x LD (BC),A 02 2 x x x x x CALL NZ,mn C4 nn mm 3/5 x x x x x LD (BC),A 02 2 x x x x x
CALL Z,nn CC nn mm 3/5 x x x x x LD (DE),A 12 2 x x x x x CALL Z,nn CC nn mm 3/5 x x x x x LD (DE),A 12 2 x x x x x
CCF 3F 1 x x x 0 * LD (HL),A 77 2 x x x x x CCF 3F 1 x x x 0 * LD (HL),A 77 2 x x x x x
CP (HL) BE 2 * * * 1 * LD (HL),B 70 2 x x x x x CP (HL) BE 2 * * * 1 * LD (HL),B 70 2 x x x x x
CP A BF 1 * * * 1 * LD (HL),C 71 2 x x x x x CP A BF 1 * * * 1 * LD (HL),C 71 2 x x x x x
CP B B8 1 * * * 1 * LD (HL),D 72 2 x x x x x CP B B8 1 * * * 1 * LD (HL),D 72 2 x x x x x
CP C B9 1 * * * 1 * LD (HL),E 73 2 x x x x x CP C B9 1 * * * 1 * LD (HL),E 73 2 x x x x x
CP D BA 1 * * * 1 * LD (HL),H 74 2 x x x x x CP D BA 1 * * * 1 * LD (HL),H 74 2 x x x x x
CP E BB 1 * * * 1 * LD (HL),L 75 2 x x x x x CP E BB 1 * * * 1 * LD (HL),L 75 2 x x x x x
CP H BC 1 * * * 1 * LD (HL),n 36 nn 3 x x x x x CP H BC 1 * * * 1 * LD (HL),n 36 nn 3 x x x x x
CP L BD 1 * * * 1 * LD (mn),A 32 nn mm 4 x x x x x CP L BD 1 * * * 1 * LD (mn),A 32 nn mm 4 x x x x x
CP n FE nn 2 * * * 1 * LD (mn),HL 22 nn mm 5 x x x x x CP n FE nn 2 * * * 1 * LD (mn),HL 22 nn mm 5 x x x x x
CPL 2F 1 x x 1 1 x LD A,(BC) 0A 2 x x x x x CPL 2F 1 x x 1 1 x LD A,(BC) 0A 2 x x x x x
DEC (HL) 35 3 * * * 1 x LD A,(DE) 1A 2 x x x x x DEC (HL) 35 3 * * * 1 x LD A,(DE) 1A 2 x x x x x
DEC A 3D 1 * * * 1 x LD A,(HL) 7E 2 x x x x x DEC A 3D 1 * * * 1 x LD A,(HL) 7E 2 x x x x x
DEC B 05 1 * * * 1 x LD A,(mn) 3A nn mm 4 x x x x x DEC B 05 1 * * * 1 x LD A,(mn) 3A nn mm 4 x x x x x
DEC BC 0B 1 x x x x x LD A,A 7F 1 x x x x x DEC BC 0B 1 x x x x x LD A,A 7F 1 x x x x x
DEC C 0D 1 * * * 1 x LD A,B 78 1 x x x x x DEC C 0D 1 * * * 1 x LD A,B 78 1 x x x x x
DEC D 15 1 * * * 1 x LD A,C 79 1 x x x x x DEC D 15 1 * * * 1 x LD A,C 79 1 x x x x x
DEC DE 1B 1 x x x x x LD A,D 7A 1 x x x x x DEC DE 1B 1 x x x x x LD A,D 7A 1 x x x x x
DEC E 1D 1 * * * 1 x LD A,E 7B 1 x x x x x DEC E 1D 1 * * * 1 x LD A,E 7B 1 x x x x x
DEC H 25 1 * * * 1 x LD A,H 7C 1 x x x x x DEC H 25 1 * * * 1 x LD A,H 7C 1 x x x x x
DEC HL 2B 1 x x x x x LD A,L 7D 1 x x x x x DEC HL 2B 1 x x x x x LD A,L 7D 1 x x x x x
DEC L 2D 1 * * * 1 x LD A,n 3E nn 2 x x x x x DEC L 2D 1 * * * 1 x LD A,n 3E nn 2 x x x x x
DEC SP 3B 1 x x x x x LD B,(HL) 46 2 x x x x x DEC SP 3B 1 x x x x x LD B,(HL) 46 2 x x x x x
DJNZ,d 10 dd 2/3 x x x x x LD B,A 47 1 x x x x x DJNZ,d 10 dd 2/3 x x x x x LD B,A 47 1 x x x x x
EX AF,AF' 08 1 x x x x x LD B,B 40 1 x x x x x EX AF,AF' 08 1 x x x x x LD B,B 40 1 x x x x x
EX DE,HL EB 1 x x x x x LD B,C 41 1 x x x x x EX DE,HL EB 1 x x x x x LD B,C 41 1 x x x x x
EXX D9 1 - (2) x x x x x LD B,D 42 1 x x x x x EXX D9 1 - (2) x x x x x LD B,D 42 1 x x x x x
 REDUCED Z80 INSTRUCTION SET REDUCED Z80 INSTRUCTION SET

FLAG * = affected FLAG * = affected

DEF's ? = unknown DEF's ? = unknown
Conditional jumps show 2 values: e.g. "RET Z = 1/3" (cycles if condition not met/condition met) x = no change Conditional jumps show 2 values: e.g. "RET Z = 1/3" (cycles if condition not met/condition met) x = no change

INSTR. HEX CODE CYC FREG - Flags register INSTR. HEX CODE CYC FREG - Flags register INSTR. HEX CODE CYC FREG - Flags register INSTR. HEX CODE CYC FREG - Flags register
b7 b6 b5 b4 b3 b2 b1 b0 b7 b6 b5 b4 b3 b2 b1 b0 b7 b6 b5 b4 b3 b2 b1 b0 b7 b6 b5 b4 b3 b2 b1 b0
S Z - H - - N C S Z - H - - N C S Z - H - - N C S Z - H - - N C
LD B,E 43 1 x x x x x NEG ED 44 2 * * * 1 * LD B,E 43 1 x x x x x NEG ED 44 2 * * * 1 *
LD B,H 44 1 x x x x x OR (HL) B6 2 * * 0 0 0 LD B,H 44 1 x x x x x OR (HL) B6 2 * * 0 0 0
LD B,L 45 1 x x x x x OR A B7 1 * * 0 0 0 LD B,L 45 1 x x x x x OR A B7 1 * * 0 0 0
LD B,n 06 nn 2 x x x x x OR B B0 1 * * 0 0 0 LD B,n 06 nn 2 x x x x x OR B B0 1 * * 0 0 0
LD BC.mn 01 nn mm 3 x x x x x OR C B1 1 * * 0 0 0 LD BC.mn 01 nn mm 3 x x x x x OR C B1 1 * * 0 0 0
LD C,(HL) 4E 2 x x x x x OR D B2 1 * * 0 0 0 LD C,(HL) 4E 2 x x x x x OR D B2 1 * * 0 0 0
LD C,A 4F 1 x x x x x OR E B3 1 * * 0 0 0 LD C,A 4F 1 x x x x x OR E B3 1 * * 0 0 0
LD C,B 48 1 x x x x x OR H B4 1 * * 0 0 0 LD C,B 48 1 x x x x x OR H B4 1 * * 0 0 0
LD C,C 49 1 x x x x x OR L B5 1 * * 0 0 0 LD C,C 49 1 x x x x x OR L B5 1 * * 0 0 0
LD C,D 4A 1 x x x x x OR n F6 nn 2 * * 0 0 0 LD C,D 4A 1 x x x x x OR n F6 nn 2 * * 0 0 0
LD C,E 4B 1 x x x x x OUT (n),A D3 nn 3 x x x x x LD C,E 4B 1 x x x x x OUT (n),A D3 nn 3 x x x x x
LD C,H 4C 1 x x x x x POP AF F1 3 x x x x x LD C,H 4C 1 x x x x x POP AF F1 3 x x x x x
LD C,L 4D 1 x x x x x POP BC C1 3 x x x x x LD C,L 4D 1 x x x x x POP BC C1 3 x x x x x
LD C,n 0E nn 2 x x x x x POP DE D1 3 x x x x x LD C,n 0E nn 2 x x x x x POP DE D1 3 x x x x x
LD D,(HL) 56 2 x x x x x POP HL E1 3 x x x x x LD D,(HL) 56 2 x x x x x POP HL E1 3 x x x x x
LD D,A 57 1 x x x x x PUSH AF F5 3 x x x x x LD D,A 57 1 x x x x x PUSH AF F5 3 x x x x x
LD D,B 50 1 x x x x x PUSH BC C5 3 x x x x x LD D,B 50 1 x x x x x PUSH BC C5 3 x x x x x
LD D,C 51 1 x x x x x PUSH DE D5 3 x x x x x LD D,C 51 1 x x x x x PUSH DE D5 3 x x x x x
LD D,D 52 1 x x x x x PUSH HL E5 3 x x x x x LD D,D 52 1 x x x x x PUSH HL E5 3 x x x x x
LD D,E 53 1 x x x x x RET C9 3 x x x x x LD D,E 53 1 x x x x x RET C9 3 x x x x x
LD D,H 54 1 x x x x x RET C D8 1/3 x x x x x LD D,H 54 1 x x x x x RET C D8 1/3 x x x x x
LD D,L 55 1 x x x x x RET NC D0 1/3 x x x x x LD D,L 55 1 x x x x x RET NC D0 1/3 x x x x x
LD D,n 16 nn 2 x x x x x RET NZ C0 1/3 x x x x x LD D,n 16 nn 2 x x x x x RET NZ C0 1/3 x x x x x
LD DE,mn 11 nn mm 3 x x x x x RET Z C8 1/3 x x x x x LD DE,mn 11 nn mm 3 x x x x x RET Z C8 1/3 x x x x x
LD E,(HL) 5E 2 x x x x x RLA 17 1 x x 0 0 * LD E,(HL) 5E 2 x x x x x RLA 17 1 x x 0 0 *
LD E,A 5F 1 x x x x x RLCA 07 1 x x 0 0 * LD E,A 5F 1 x x x x x RLCA 07 1 x x 0 0 *
LD E,B 58 1 x x x x x RRA 1F 1 x x 0 0 * LD E,B 58 1 x x x x x RRA 1F 1 x x 0 0 *
LD E,C 59 1 x x x x x RRCA 0F 1 x x 0 0 * LD E,C 59 1 x x x x x RRCA 0F 1 x x 0 0 *
LD E,D 5A 1 x x x x x SBC A,C 99 1 * * * 1 * LD E,D 5A 1 x x x x x SBC A,C 99 1 * * * 1 *
LD E,E 5B 1 x x x x x SBC A,n DE nn 2 * * * 1 * LD E,E 5B 1 x x x x x SBC A,n DE nn 2 * * * 1 *
LD E,H 5C 1 x x x x x SCF 37 1 x x 0 0 1 LD E,H 5C 1 x x x x x SCF 37 1 x x 0 0 *
LD E,L 5D 1 x x x x x SUB A,(HL) 96 2 * * * 1 * LD E,L 5D 1 x x x x x SUB A,(HL) 96 2 * * * 1 *
LD E,n 1E nn 2 x x x x x SUB A,B 90 1 * * * 1 * LD E,n 1E nn 2 x x x x x SUB A,B 90 1 * * * 1 *
LD H,(HL) 66 2 x x x x x SUB A,C 91 1 * * * 1 * LD H,(HL) 66 2 x x x x x SUB A,C 91 1 * * * 1 *
LD H,A 67 1 x x x x x SUB A,D 92 1 * * * 1 * LD H,A 67 1 x x x x x SUB A,D 92 1 * * * 1 *
LD H,B 60 1 x x x x x SUB A,E 93 1 * * * 1 * LD H,B 60 1 x x x x x SUB A,E 93 1 * * * 1 *
LD H,C 61 1 x x x x x SUB A,H 94 1 * * * 1 * LD H,C 61 1 x x x x x SUB A,H 94 1 * * * 1 *
LD H,D 62 1 x x x x x SUB A,L 95 1 * * * 1 * LD H,D 62 1 x x x x x SUB A,L 95 1 * * * 1 *
LD H,E 63 1 x x x x x SUB n D6 nn 2 * * * 1 * LD H,E 63 1 x x x x x SUB n D6 nn 2 * * * 1 *
LD H,H 64 1 x x x x x XOR (HL) AE 2 * * 0 0 0 LD H,H 64 1 x x x x x XOR (HL) AE 2 * * 0 0 0
LD H,L 65 1 x x x x x XOR B A8 1 * * 0 0 0 LD H,L 65 1 x x x x x XOR B A8 1 * * 0 0 0
LD H,n 26 nn 2 x x x x x XOR C A9 1 * * 0 0 0 LD H,n 26 nn 2 x x x x x XOR C A9 1 * * 0 0 0
LD HL,(mn) 2A nn mm 5 x x x x x XOR D AA 1 * * 0 0 0 LD HL,(mn) 2A nn mm 5 x x x x x XOR D AA 1 * * 0 0 0
LD HL,mn 21 nn mm 3 x x x x x XOR E AB 1 * * 0 0 0 LD HL,mn 21 nn mm 3 x x x x x XOR E AB 1 * * 0 0 0
LD L,(HL) 6E 2 x x x x x XOR H AC 1 * * 0 0 0 LD L,(HL) 6E 2 x x x x x XOR H AC 1 * * 0 0 0
LD L,A 6F 1 x x x x x XOR L AD 1 * * 0 0 0 LD L,A 6F 1 x x x x x XOR L AD 1 * * 0 0 0
LD L,B 68 1 x x x x x XOR n EE nn 2 * * 0 0 0 LD L,B 68 1 x x x x x XOR n EE nn 2 * * 0 0 0
LD L,C 69 1 x x x x x LD L,C 69 1 x x x x x
LD L,D 6A 1 x x x x x LD L,D 6A 1 x x x x x
LD L,E 6B 1 x x x x x LD L,E 6B 1 x x x x x
LD L,H 6C 1 x x x x x LD L,H 6C 1 x x x x x
LD L,L 6D 1 x x x x x LD L,L 6D 1 x x x x x
LD L,n 2E nn 2 x x x x x LD L,n 2E nn 2 x x x x x
NOP 00 1 x x x x x Reproduced by P. Townshend 2003 NOP 00 1 x x x x x Reproduced by P. Townshend 2003
NOTE
NOTE

53
NOTE