Microprocessor

INTRODUCTION TO MICRO COMPUTER AND MICROPROCESSOR

Introduction

A Microprocessor is a multipurpose programmable logic device which reads the binary instructions from a storage device called Memory accepts binary data as input and process data according to the instructions and gives the results as output. So, you can understand the Microprocessor as a programmable digital device, which can be used for both data processing and control applications. In view of a computer student, it is the CPU of a Computer or heart of the computer. A computer which is built around a microprocessor is called a microcomputer. A microcomputer system consists of a CPU (microprocessor), memories (primary and secondary) and I/O devices as shown in the block diagram in Fig 1. The memory and I/O devices are linked by data and address (control) buses. The CPU communicates with only one peripheral at a time by enabling the peripheral by the control signal. For example to send data to the output device, the CPU places the device address on the address bus, data on the data bus and enables the output device. The other peripherals that are not enabled remain in high impedance state called tri-state.

Fig.1 Block diagram of a Microcomputer

The Intel 8085 Microprocessor:Intel 8085A is a single chip 8-bit N-channel microprocessor which works at +5V DC power supply. It is a 40 pin IC available as a DIP (Dual Inline Package) chip. 8085A can operate with a 3MHZ single phase clock and 8085A-2 version can operate at a maximum frequency of 5MHZ. This 8085 is an enhanced version of its predecessor the 8080A. Its instruction set is upward compatible with that of the 8080A. 8085A has an on-chip clock generator with external crystal, LC or RC network. This 8085 microprocessor is built with nearly 6200 transistors. The enhanced version of 8080 is the Intel 8085AH. It is an N channel depletion load, silicon gate (HMOS) 8-bit processor. Here 3MHZ, 5MHZ and 6MHZ selections are available. It has 20% lower power consumption than 8085A for 3MHZ and 5MHZ. Its instruction set is 100% software compatible with the 8085A. It is also 100% compatible with 8085A.

Central Processing Unit (CPU)

The Central Processing Unit of any microcomputer is the microprocessor. Hence microprocessor is also known as the heart of the computer. The CPU performs the various activities in response to a set of instructions called a program. Programs are stored in the memory. The CPU reads in data control signals (instructions) through the input ports and executes one instruction at a time. So, generally speaking, a microprocessor is nothing but the CPU .The Intel 8085 CPU is an 8-bit device with a clock speed of 3 - 5 MHZ. It has 80 basic instructions and 246 op-codes. Its clock cycle is 320 ns. The time for the clock cycle of Intel 8085 is 200 ns. The block diagram of 8085 microprocessor is shown in Fig 2. The 8085 CPU consists of three major sections, They are:(i) Arithmetic and logic unit (ALU)(ii) Registers(iii) Timing and Control unit.

Arithmetic and logic unit (ALU)

The ALU performs all the arithmetic and logical operations like addition, subtraction, complementing, logical AND, logical OR, logical Exclusive OR, incrementing and decrementing, rotate, shift and clear. An ALU is made of many logic gates and adders etc.The arithmetic and logic unit consists of the following units (a) Accumulator (A).(b) Temporary register.(c) Flag register.

(a) Accumulator (A):It is an 8-bit register which is treated as a special function register. Most of the arithmetic and logic operations are performed using this accumulator. All the I/O data transfers between 8085 and I/O devices are performed via accumulator. One of the operands for arithmetic operations in ALU is from the accumulator. After performing the arithmetic operations the result is stored back in accumulator. It is from the accumulator only, the data is sent out to an output device. Similarly, the data from an input device is read only through the accumulator. The data in the accumulator alone can be rotated or shifted. No other register can be used for these operations. Certain instructions like DAA are performed using only accumulator. So, many times the Accumulator register is treated as a default register.

Figure 2 .The Block Diagram of 8085 Microprocessor

(b) Temporary register: This is an 8-bit register which is not accessible to the user. This register is used by the microprocessor to load the second operand during arithmetic/logical operations in ALU. The final result is stored in the Accumulator and the flags are set or reset according to the result of the operation. For example when MVI M, 17H instruction is fetched, IR register will receive the opcode for MVI M and the Temporary register will receive 17H.In arithmetic and logical operations, that involve two operands, the accumulator provides one operand. The other is provided by the temporary register. For example in ADD C instruction C register contents are moved to the Temp. Register and the addition of A and Temp. Register contents is performed by the ALU.(c).Flag register: The flag register is an 8- bit register which generally reflect data conditions in the accumulator with certain exceptions. Hence this flag register is also known as Status register. Though this flag register is an eight bit register, it contains only 5 flag bits and the remaining three bits are undefined as shown in Fig.3.3 In the Flag register each flag bit is a Flip-Flop. i.e., the bit may be either in the flip state or flop state

Figure 3. Flag RegisterS - Sign Flag After execution of an arithmetic and logic operation, if bit D7 of the result (Normally in the Accumulator) is 1, the sign flag is set. This Flag is used with signed numbers.For example in a given byte, if D7 is 1, the number is treated as a negative number. Else (if it is zero), it is viewed as a positive.In arithmetic operations with signed numbers bit D7is reserved for indicating the sign and the remaining seven bits are used to denote the magnitude of the number.Z - Zero FlagThis Flag is set (made 1) if the result after any arithmetic operation is zero, and the flag is reset (made 0) if the result is not zero. So, this flag is set or reset based on the results in the accumulator as well as in the other registers.AC Auxiliary carry Flag In this arithmetic operation, when a carry is generated by and passed on to bit 4 , the AC flag is set. This flag is used internally for BCD arithmetic and is not available for the programmer to change the sequence of a program with a jump instruction. But the Z and CY flags can be used for this purpose.P-Parity FlagIf the result after an arithmetic and logical operation has an even number of 1s, this parity flag is set to 1 otherwise (if number of 1s is odd) the flag is reset (made0).For example the data byte 10111101 has even parity and the data byte 10011011 has odd parity. So P bit=0.CY-Carry flagAfter an arithmetic operation, like addition, subtraction if there exists a carry or barrow, this flag CY is set to 1 else it is reset (made0)Example : Let us consider the addition of two binary numbers 11011001 and 11101101 and check the Flag register. D7 D0 1 1 0 1 1 0 0 1

1 1 1 1 0 1 1 0 1 1 1 0 0 0 1 1 0

In the result, the sum is not zero, So Z-Flag is reset (Z=0).There is a carry from the third bit to fourth bit. So AC Flag is set (AC=1).The D7 bit=1, so, the sign Flag is set (S=1).In the result, the no. of 1s is even. So, parity is even (P=1).After addition, there is a carry. So carry Flag is set (CY=1).

The Flag register contents after addition are shown below: D7 D0

Register OrganizationThe 8085 microprocessor has different types of registers. It includes six , 8 bit registers (B, C, D. E, H and L), one 8-bit Accumulator and two 16-bit registers (SP and PC). Also there are two 8-bit temporary registers W and Z. Among these registers W and Z are not accessible to the user, They are used by the processor for internal, intermediate operations. The remaining registers are accessible to the user. The organization of 8085 registers is shown in Fig. 3.4 The various registers of 8085 are classified into three types. They are (i).Temporary registers.(ii).General purpose registers(iii).Special purpose registers.

(i) Temporary registers: The Temporary registers are temporary data registers, W register and Z register. All are 8-bit registers. The temporary data register is associated with the ALU operations. One of the operand is stored in this register. This is not accessible to user. Similarly W and Z are also temporary registers used to hold 8-bit data during execution of certain instructions. As these registers are internally used by the CPU, they are not accessible to the user.The W and Z registers are used by the processor during CALL instruction. When a CALL instruction is encountered in any program, the current Program counter (PC) contents are pushed on to the stack and the given address is loaded on to PC. The given address is temporarily stored in W and Z registers and placed on the bus for the fetch cycle. Thus the program control is transferred to the address given in the instruction. Another example is, during the execution of XCHG instruction, the contents of H-L pair are exchanged with D-E pair. At the time of exchange W and Z registers are used for temporary storage of data.

(ii) General purpose registers: B, C, D, E, H and L are six, 8-bit general purpose registers to store data. These registers can be used as separate 8-bit registers and also can be paired as 16-bit registers to store the address of a memory location. But they must be paired as B-C; D-E and H-L register pairs only as shown below.

Figure 4. Register organization

When used as pair, for example B-C, the higher order byte moves to the first register (B) and the low order byte moves to the second register (C). The H-L pair also functions as a data pointer or memory pointerFor Ex: LXI H, 8500 H.This will load immediately the address of memory location (8500H) in to H-L pair .Now the H-L pair points to the location 8500 H.

(iii) Special purpose registers: The Special purpose registers, as their name indicates, are used for some specific purpose. The Special purpose registers are Accumulator (A), Flag Register, Instruction Register (IR), Program Counter (PC) and Stack Pointer (SP).Accumulator (Register A): It is an 8-bit tri-state register. It is mainly used for arithmetic, logic, load and store operations. It is also used in I/O operations. In most of operations, the result is stored in Accumulator after execution.Flag Register: It is an 8-bit register, which consists of only five flags. Each flag bit is a flip flop that indicates either a set or reset state. The five flags are Sign, Zero, Auxiliary carry, Parity and Carry as shown below

Here X means undefined.

Sign Flag: The sign flag is set to1 if the most significant bit of the result of an arithmetic or logic operations is 1. Else it is reset (0).

Zero Flag: The Zero status flag is set to 1 if the result of an arithmetic or logic operation is Zero For non-Zero result it is reset to 0.G

Auxiliary carry Flag: This flag is set if there is a carry from 3rd bit to 4th bit during BCD operations (carry from lower nibble to higher nibble). This flag is not accessible to the user.

Parity Flag: Parity is defined by the number of 1s present in a binary number stored in A register. After any arithmetic or logical operation, if the result has an even number of 1s it is called even parity and the Parity Flag is set to 1. Otherwise. i.e. If there is odd number of 1s in the result, it is called Odd Parity and the Parity flag is set 0.

Program Counter (PC): It is a 16-bit special purpose register, which stores the address of the next instruction to be fetched or executed. The execution of a program is initiated by loading the PC by the address of the first instruction of the program. Once the first instruction is executed, the PC is automatically incremented to point to the next instruction unless a jump to some specific address occurs. This process is repeated till the last instruction of the program. In case of JUMP or CALL instructions, current address is stored in the Program Counter. The processor then fetches the next instruction from the new address specified by the JUMP or CALL instruction. In conditional JUMP and conditional CALL instructions, if the condition is not satisfied, the processor increments the Program Counter by three so that it points the instruction followed by the conditional JUMP or CALL instruction. Otherwise the processor fetches the next instruction from the new address specified by JUMP or CALL instruction.

Stack Pointer (SP): It is a 16-bit special purpose register which always stores the address of top of the Stack. i.e. it always points to top of the Stack. Stack is a part of the memory location used to store the data temporarily. A stack works on Last in First out (LIFO) basis. As the Stack pointer always points to the top of the Stack, only top of the Stack of the memory can be accessed. When a Write operation (PUSH) takes place, the contents of the stack pointer is decremented by two so that the SP points to the new location. Similarly when the Read operation (POP) occurs, the Stack pointer is incremented by two to point to the next data on top of the Stack.The Stack Pointer is initialized by load register pair immediate instruction.

Ex: LXI SP, 8530 HHere 8530 H is the 16 bit address of the top of Stack location.

The remaining blocks of 8085 microprocessor block diagram

Instruction Register and Decoder: The instruction register and the decoder are also part of the ALU. When an instruction is fetched from memory, it is loaded in the instruction register. The Decoder decodes the instruction and develops the sequence of events to follow. The instruction register is a 8 bit special register, but it is not a programmable and is not accessible to the user. The instruction decoder decodes the instruction at a binary level and sends the appropriate signals to the control unit.

Increment/ Decrement Address Latch: This is a 16 bit special register not accessible to the user. This register is used by the CPU to increment/ decrement the contents of the Stack Pointer (SP) and increment program counter (PC) during instruction execution. During first T - state of op code fetch machine cycle (T1) the microprocessor increments the PC register contents to point to the next location. This increment operation takes place on increment/ decrement register address latch. The 16 bit address that is sent out through AD0-AD7 and A8-A15 are latched into this register. The address bus AD0-AD7 continues to be available on the bus after T1 state from this latch.

Address Buffer: This is an 8-bit unidirectional buffer. It is used to drive external higher order address bus. It is also used to tri-state the higher order address (A8-A15) bus under certain conditions like reset, hold, and halt and also when address lines are not in use.

Address/Data Buffer: This is an 8-bit bi-directional buffer. It is used to drive multiplexed address/data bus. It means low order address bus (A7-A0) and data bus (D7-D0). It is also used to tri-state the multiplexed address/data bus under certain conditions like reset, hold, and halt and also when A/D bus lines are not in use.The address and data bus buffers are used to drive external address and data buses respectively. Due to these buffers the address and data buffers can be tri-stated when they are not in use. In actual practice, in a microprocessor the driving capacity of the address pins after the internal buffering may not be adequate. So, there will be external buffer chips also available.

Serial I/O control:This control provides two lines SOD (Serial Out Data) and SID (serial In Data) for serial communication. These lines are used during serial data transmission over long distance where data is transmitted and received bit by bit. The Serial Output Data (SOD) pin is used to send data out serially and serial Input Data (SID) pin is used to receive data serially by the 8085 microprocessor.

Interrupt control:This is an important block related to interrupts. This block is linked to the CPU through the 8-bit internal data bus. This interrupt control has five interrupt signals. They are TRAP, RST 7.5, RST 6.5, RST 5.5 and INTR. The control block will take care of enabling and disabling of these interrupts etc

8085 Interrupts: The 8085 microprocessor has five interrupts. They are TRAP, RST 7.5, RST 6.5, RST5.5 and INTR. Among all these interrupts TRAP has the highest priority and INTR (Interrupt Request) has the lowest priority. The TRAP is also a non maskable interrupt. The numbers succeeding the RST (7.5, 6.5, and 5.5) are related to the call locations. The various interrupts, their locations in the order of highest to lowest priority are given in Table 3.1. Here RST means RESTART. Among these interrupts INTR is the only non-vector interrupt whereas the other interrupts are vectored interrupts.

TRAP: It is a non maskable interrupt with highest priority. It means that whenever the pin is activated, the 8085 will always get interrupted even if the 8085 is in DI (Disable Interrupt) state. Trap input is both edge and level sensitive. So, the microprocessor is interrupted when the input is both edge and level sensitive. So, the microprocessor is interrupted when the input pulse goes from low to high or when it remains high .When interrupted, the microprocessor loads the program counter with 0024H.

RST 7.5: It is an edge sensitive pin. Internal to 8085 there is a flip-flop connected to RST 7.5 interrupt pin . This flip flop is set 1, when a positive going edge occurs on RST 7.5 input. RST 7.5 interrupt has a higher priority than RST 6.5, RST 5.5 and INTR. This RST 7.5 is a maskable interrupt known as MI. This interrupt is enabled under program control with two instructions EI (Enable Interrupt) and SIM (Set Interrupt Mask)

RST 6.5and RST 5.5: These interrupts are level sensitive, it means the triggering level should be on until the microprocessor completes the execution of the current instruction. If the microprocessor is not able to respond to the requests immediately, they should be stored or held by external hardware. These two interrupts are also maskable interrupts. RST 6.5 and RST 5.5 have higher priority than INTR interrupt. The condition of these interrupts can be known using RIM (Read Interrupt Mask) instruction and the condition of the masking interrupt can be set and reset using SIM instruction (Set Interrupt Mask).

INTR: It is only non-vectored interrupt in 8085 microprocessor. This interrupt has the lowest priority among all the interrupts. This is also a maskable interrupt and can be disabled using the instruction DI (Disable Interrupt). The mask on INTR can be removed by executing EI (Enable Interrupt) instruction. When EI instruction is executed, the flip flop associated with this is set and the mask is removed. This is a non-vectored interrupt because when the remaining interrupts are initialized, they are automatically transferred (vectored) to specific locations on memory page 00H without any external hardware. They do not require the signal. The necessary hardware is already implemented inside the 8085. But coming to INTR, interrupt, it executes interrupt acknowledge machine cycle. During this cycle, the device that has interrupted this microprocessor will provide the operation code. The signal works as a signal during acknowledge machine cycle. During this time, the microprocessor loads the code into instruction register from I/O device. Based on the code, the remaining operation is executed by the processor.

S. NoInterruptsCall locations

1TRAP (Highest priority)0024H

2RST 7.5(7.5 x 8)H = 003CH

3RST 6.5(6.5 x 8)H = 0034H

4RST 5.5(5.5 x 8)H = 002CH

5INTR (least priority)No location

Table 1 . Various Interrupts ,Call locations in order of highest to lowest priorityTiming and control Unit This unit of the microprocessor issues necessary timing and control signals for the execution of instructions. It generates three types of signals namely status, control and timing signals required for the operation of memory and I/O devices. This unit with the help of these signals controls the entire operation of the microprocessor and the peripherals. The signals associated with this unit are two control signals. and three status signals IO/, S1 and S0 to identify the nature of the operation and one special signal ALE which indicates the starting of the operation. These signals are explained below in detail.

-Read (active low): This is a Read control signal. This signal indicates that the selected I/O or memory device is to be read and data are available on the data bus.

- Write (active low): This is a Write control signal. This signal indicates that the data on the data bus are to be written into a selected memory or I/O device.Table 2. Status signals of 8085

S. No

IO/S1S0Status

1

0

0

0Memory Write

2

0

1

0Memory Read

3

1

0

1I/O Write

4

110I/O Read

5

1/0

1

1Opcode fetch

6

1

1

1

InterruptAcknowledge

7

*

0

0

Halt

8*X

X

Hold

9*X

X

RESET

IO/ : This is a status signal used to differentiate between I/O and memory operations. When this signal is high, it indicates an I/O operation, when it is low it denotes a memory operation. This signal is combined with Read () and Write to generate necessary I/O and memory control signals.

S1 and S0: These signals are also status signals like IO/, used to identify various operations. The complete operation of the microprocessor can be understood by these three signals. The various operations and the associated status signals are shown in Table 3.2 .

ALE (Address Latch Enable): This is a positive going pulse generated every time the 8085 begins an operation, It indicates that the bits on AD7 - AD0 are address bits. This signal is used primarily to latch the low-order address from the multiplexed bus and generate a set of eight address lines A7 A0.

Address, Data and Control Buses: Intel 8085 has 16-bit unidirectional address bus which carries the address of memories and peripheral devices. A bus is nothing but a group of electrical lines used to transmit the information as electrical signals. So, this 16-bit parallel address bus carries address from microprocessor to memories/peripherals. Hence it is Uni-directional(because the converse is not possible). The width of the parallel bus determines how much memory that a microprocessor can address.The 8085 microprocessor with 16-bit address bus can address a maximum of 216= 65536=64 KB of memory locations. The size of the address bus is independent of the size of the microprocessor. In INTEL 8085 microprocessor, the 8 most significant bits of the address are transmitted by the high order address bus A8-A15. But the 8 least significant bits of the addresses are transmitted by Address/Data bus or A/D bus. i.e. the lower order address lines are multiplexed with the data bus. So, the A/D bus operates in a time shared mode. i.e. the data and address are sent on the same lines but at different instants of time. A0-A7 will always have the address during the first T state (T1) of the machine cycle. To demultipl exe the A/D bus the pin ALE is used. When ALE=1 (high) the A/D bus acts as a lower order address bus else it acts as Data bus.The Data bus is a bidirectional bus which is used to send data to and from the microprocessor. This is also a parallel bus. The size of the data bus determines the size of the microprocessor. The 8085 microprocessor has 8-bit data bus and hence it is called an 8-bit microprocessor. This refers to the width of the data bus but not the address bus. Similarly8086 is a 16-bit microprocessor and its data bus width is 16 bits INTEL 8085 has Address/ Data bus namely AD7-AD0. i.e. at some instances it acts as a 8- bit address bus and at other instances it works as a 8-bit data bus. INTEL used this time multiplexing technique to save the pins. Generally the size of the internal general purpose registers matches the size of the data bus. Thus, the INTELs 8-bit general purpose registers matches with its 8-bit data bus. The size of the data bus matches the size of the internal registers, so that all the bits on the bus can at one time come into or go out of any of the registers. The control bus of 8085 is a uni-directional bus because the microprocessor alone sends control signals to memories or peripheral devices. The size of the control bus depends upon the specific microprocessor. Typical control signals are Read or Write signals. It means whether the microprocessor operation is a read or writes and whether it is memory or I/O operation. In addition to this it includes state signals, and address latch enables. A microprocessor may also have certain additional control signals and such as interrupt signals, acknowledgement signals and hold signals. But they are not considered as part of control bus even through they take part in control of microprocessor based systems. The above three buses that interface the CPU to the system components are combinedly known as the System bus.

Pin configuration

The pin diagram of 8085 microprocessor is shown in Fig 5. From the figure it is clear that it is 40 pin DIP chip. The various pins of 8085 microprocessor can be grouped in the following categories Power Supply and Clock pins Data bus and Address bus Control and Status signals Interrupt signals DMA signals Serial I/O signals

The description of various pins is given below.

Power supply and clock pins: Vcc: +5V power supply

Vss: Ground reference.

X1 and X2: A Crystal (or RC, LC Network) is connected at these two pins. The internal clock generator divides oscillator frequency by 2, therefore to operate a system at 3MHZ, the crystal of the tuned circuit should have a frequency of 6MHZ.

CLK (OUT): This signal is used as a system clock for other devices. Its frequency is half the oscillator frequency

Data bus and Address bus: AD0-AD7: These lines are Address/Data lines, which are bidirectional with dual purpose. They are used as the low-order address bus as well as the data bus. During the first part of the machine cycle (T1), lower 8 bits of memory address or I/O address appear on the bus. During the remaining part of the machine cycle (T2,T3) these lines are used as a bi-directional data bus

. Figure 5. Pin Diagram of 8085 A

A8-A0: These are the upper half of the 16 bit address lines. These lines are exclusively used for the most significant 8 bits of the 16 bits of the16 bit address bus.

Control and Status Signals: ALE (Address Latch Enable): This is a positive going pulse generated every time the 8085 begins an operation. The ALE=High indicates that the bits on AD7-AD0 are address bits. This signal is mainly used to latch the low order address from the multiplexed bus and generate a separate set of eight address lines (A7-A0)

(Read): This is an active low read control pin. This signal indicates that the selected I/O or memory device is to be read and data are available on data bus.

(Write): This is an active low write control pin. It indicates that the data on the data on the data bus are to be are to be written into a selected memory or I/O location

IO/: This is a status signal used to differentiate between IO and memory operations. When it is high, it indicates an I/O operation and when it is low, it indicates a memory operation. This signal is combined with and signals to generate I/O and memory control signals.

S1 and S0: These are status signals and they indicate the type of machine cycle in progress during execution of an instruction.

READY (Input): Through this pin, the microprocessor will know whether peripheral device is ready or not for data transfer. If the device is not ready the processor waits. So, this pin helps to synchronize slow devices to the microprocessor.

Interrupt signals:

TRAP, RST 7.5, RST 6.5, RST5.5 and INTR: These are the interrupt signals which are externally initiated.

INTR (Interrupt Request): This is used as a general purpose interrupt. It has a lowest priority and it is the only non-vectored interrupt.

RST 7.5: It is a restart interrupt pin. It has higher priority than RST 6.5, RST5.5 and INTR. It is a maskable vectored interrupt.

RST 6.5 and RST5.5: These two are maskable vectored interrupt with higher priority than INTR.

TRAP: It is a non-maskable vectored interrupt. It has higher priority.

(Output): It is an active low interrupt acknowledge pin. This will acknowledge the receival of interrupt request to the peripheral device.

DMA Signals: Hold: This pin is used during the Direct Memory Access. A high on this pin indicates that, a peripheral like DMA controller is requesting the use of address and data buses.

HLDA (Output): A high on this p in acknowledges the hold request from peripheral.

: It is an active low signal. When the signal on this pin goes low, the system is in reset i.e. the program counter is set to zero, the address & data buses are tristated.

RESETOUT: This signal is used to Reset other devices in microprocessor system.

Serial input/ Output signals: SID: Serial input Data is a pin through which serial data are brought into the micro processor accumulator after the RIM instruction is executed.

SOD: Serial output Data pin is used by the microprocessor to output data serially to the external devices. Serial data is sent out of the microprocessor by executing SIM instruction. The most significant bit of accumulator should have the serial bit and D6 bit of the accumulator must be made high to enable the serial data transfer.

Timing Diagram :

The graphical representation of the time taken for the execution of each instruction by a microprocessor is known as timing diagram. The execution time is denoted by T-states. One T-state is equal to the time period of the internal clock signal of the microprocessorFor Ex: If the internal clock frequency of 8085 microprocessor is 3 MHZ, One T-state is equal to

= =0.333x10-6sec=333x10-9sec. (333 nano seconds nearly)

As far as execution of instructions is concerned, in 8085 microprocessor, each instruction is divided into two parts: The operation code (opcode) and the operand. The opcode tells us what the operation is and the operand is the necessary information required for the instruction. The operand may be either data or an address or other information required for the instruction.Each instruction is divided into machine cycles and each machine cycle is divided into clock cycles or T- statesThe first machine cycle is every instruction is the op-code fetch. During this time the opcode is fetched from memory and returned on the data bus to the microprocessor. The 8085 machine cycles are divided into following six types. They are1. Opcode fetch2. Memory read3. Memory write4. I/O read5. I/O write6. Interrupt acknowledge

Opcode fetch Machine cycle:

The first operation in every instruction is the opcode fetch. The opcode fetch cycle is called the M1 machine cycle and is usually for four T-states or clock cycles (certain instructions may also have 6T states in their opcode fetch machine cycle). During T1-T3 states the address is placed on the address bus and the opcode is returned on the data bus. The T4-state is used to decode and execute the opcode. The next machine cycles (M2, M3------) that follow depend upon what the instruction actually is.The timing diagram for execution of MVI A, 25 machine cycle is shown in Fig 6 as shown in the timing diagram, in T1 state, the 8085 places the contents of the program counter o n the address bus. The high order byte of the P.C (80) is placed on the A8-A15 lines. The low-order byte of the P.C(00) is placed on the AD0-AD7 line which stays only on only during T1. So, the microprocessor activates the ALE (Address Latch Enable) pin which is used to latch the low-order byte of the address in external latch before it vanishes. During T1 state, 8085 also sends status signals IO/, S1 and S0. The IO/ signal specifies whether the operation is read or write. In opcode fetch machine cycle status signals are IO/=0, S1 =1 and S0=0In T2 state, the lower order address disappears from AD0-AD7 lines and 8085 sends signal low to enable the addressed memory location. The memory device then places the contents of the addressed memory location on the data bus (AD0-AD7)During T3 state, the microprocessor loads the data from the data bus in its instruction register and raises to high which disables the memory deviceIn T4 state, the microprocessor decodes the opcode and based on the instruction it decides whether to T0 state T5 or to enter state T1of the next machine cycle (M2). All the one byte instructions which operate on 8-bit data like MOV A, B, ADDB, DCRC, RAL etc..are executed T4 state. One byte instructions which operate on 16-bit data are executed in T5 and T6 states. For example: INXH, SPHL, 5DCXH etc. Figure 6 Timing diagram for Opcode fetch Machine cycle

Memory Read cycle:

Memory read machine cycle is a machine cycle during which memory is read. For example, the instruction LDA 8900H which is a 3-byte instruction has three memory read cycles immediately after the opcode fetch cycle. The first two cycles are to get the memory address, in two 8-bit groups (the low-order part of the address and then the high-order address). The third read cycle is needed to read the data located at the address previously retrieved. This data is then loaded into accumulator.The timing diagram for memory read cycle [For Ex: LDA 2020H] is shown in Fig 7. As shown in the memory read timing diagram, after the opcode fetch cycle, the first two read cycles have the address going out over the address bus first for the low-order of the address (2001H) and then for the high-order of the address(2002H). In the third read cycle, the address of the instruction just read from memory (2020H) is sent back over the address bus in T1 and then data from that memory location is returned over the data bus in T2-T3. IO/ goes low at the beginning of the opcode fetch cycle and remain low during the next three cycles., on the other hand goes low each time data on the data bus is to be read into the microprocessor.

Figure 7. Timing diagram for Memory Read machine cycle

Memory Write cycle:

This memory write cycle is used when the microprocessor needs to send data out from accumulator or specific register and then write into the memory. As an example let us consider the instruction MOV M, A (50H). This instruction requires two machine cycles-an opcode fetch machine cycle followed by one write cycle. Because, after fetching the opcode, the instruction has to write the data in the accumulator out to memory at the address location in the H-L register. This operation requires 7-T states for opcode fetch and three T states for the memory write.

Figure 8 .Timing diagram for Memory Write machine cycle

The timing diagram of the instruction MOV A, M is shown in Fig .8. The opcode fetch cycle shows the address [2000H] going out over the address bus and the opcode for the MOV M, A (50H) returning over the data bus. During the write cycle (M2), the address that was stored in the H-L pair goes out from the microprocessor during T1 and data to be written from the accumulator goes out during T2-T3. The remaining signals like ALE, IO/, and have their usual meaning as shown in the timing diagram.

I/O Read cycle:

This I/O read cycle occurs when the microprocessor executes IN instruction and during the I/O read cycle, data is read in from an I/O device. In the case of IN PORT, there are three machine cycles. The opcode fetch cycle, a memory read cycle and an I/O read cycle. The three machine cycles combinedly taken 10-states.The Fig 9. shows the timing diagram of the instruction IN 80H, Here 80H is the port address of the device being read.The opcode fetch cycle shows the address of the instruction (2000H) going out over t he address bus and the opcode (DBH) for the IN instruction returning on the data bus. The memory read cycle (M2) displays the address of the second byte of the instruction (8001H) going out over the address bus and the port address (80H) returning on the data bus. During the I/O read cycle, the port address of the device being read is sent over the address bus and the lower 8-bits carry the same 8-bit port address and the data from the input device is returned on the data bus during T2-T3.

Figure 9. Timing diagram for I/O Read machine cycle

I/O write cycle:

During this I/O write cycle DATA is written into I/O device specified by the port address from the accumulator. The out port instruction has three machine cycles. The first one is opcode fetch cycle, second one is memory read to get the port address and the third cycle is an I/O write cycle. The OUT command writes the data stored in the accumulator over the data bus to the device whose port address was sent out over the address bus. The three machine cycle that forms the outport command has a total of ten T-states. The Fig 10 Shows the timing diagram for OUT 04H instruction. 04H is the address of the output deviceThe opcode fetch cycle sends the address of the instruction (2000H) over the address bus while the opcode (D3H) for the OUT instruction is returns on the data bus. The second machine cycle shows the address of the second byte of the instruction (2001H) going out over the address bus with the port address (84H) returning over the data bus.From the timing diagram it is clear that, In the third machine cycle the port address (84H) is sent out over both the upper and lower parts of the address bus, similar to I/O read cycle whenever the microprocessor addresses an I/O device, the port address of the device being read is sent out over both the lower and higher order parts of the address bus.It is also to be noted that the data in the accumulator is being written to the output device and goes out over the data bus during T2 -T3 states after the port address has been sent over the two parts of the address bus during T1 of the I/O write cycle

Figure 10 Timing diagram for I/O Write machine cycle

Interrupt Acknowledge

This Interrupt acknowledges machine cycle is a special machine cycle that is used in place of opcode fetch cycle in the RST (restart) instruction. It is same as an opcode fetch except that it 0sends out an signal instead of an signal and the status lines IO/, S0 and S1 are 111 instead 011. Another difference is, the interrupt acknowledge is six-T states whereas opcode fetch is only four T-states

Instruction cycle, Machine cycle, fetch and execute cycles Instruction cycle

An instruction is a command given to the microprocessor to perform a specific operation on the given data. Sequence of instructions written for a processor to perform a particular task is called a program. Program & data are stored in the memory. The microprocessor fetches one instruction from the memory at a time & executes it. It executes all the instructions of the program one by one to produce the final result. The necessary steps that a microprocessor carries out to fetch an instruction & necessary data from the memory & to execute it constitute an instruction cycle. In other words, an instruction cycle is defined as the time required completing the execution of an instruction. An instruction cycle consists of a fetch cycle and an execute cycle. The time required to fetch an opcode (fetch cycle) is a fixed slot of time while the time required to execute an instruction (execute cycle) is variable which depends on the type of instruction to be executed. Instruction cycle(IC) = Fetch cycle(FC) + Execute cycle(EC)

This is shown diagrammatically in the Fig.3.11

Figure 11 Instruction cycle

Machine cycle:

Machine cycle is defined as the time required for completing the operation of accessing either memory or I/O device. In the 8085, the machine cycle may consist of three to six T states. The T-state is defined as one sub-division of the operation performed in one clock period. These sub-divisions are internal states synchronized with the system clock. In every machine cycle the first operation is op-code fetch and the remaining will be read or write from memory or IO devices.

Fetch operation:

The first byte of an instruction is its op-code. An instruction may be more than one byte long. The other bytes are data or operand address. The program counter (PC) keeps the memory address of the next instruction to be executed. In the beginning of a fetch cycle the content of the program counter, which is the address of the memory location where op-code is available, is sent to the memory. The memory places the op-code on the data bus so as to transfer it to the microprocessor. The entire operation of fetching an op-code takes three clock cycles.

Execute operation:

The op-code fetched from the memory goes to the instruction register (IR). From the instruction register it goes to the decoder circuitry which decodes the instruction. After the instruction is decoded, execution begins. If the operand is in general purpose registers execution is immediately performed. The time taken for decoding and execution is one clock cycle. If an instruction contains data or operand and address which are still in the memory, the microprocessor has to perform some read operations to get the desired data. After receiving the data it performs execute operation. A read cycle is similar to a fetch cycle. In case of a read cycle the quantity received from the memory are data or operand address instead of an op-code. In some instructions write operation is performed. In write cycle data are sent from the microprocessor to the memory or an output device. Thus we see that in some cases an execute cycle may involve one or more read or write cycles or both.

Instruction set of 8085

An Instruction is a command given to the microprocessor to perform a given task on specified data. Each instruction has two parts one is the task to be performed called the operation code (op-code) and the second is the data to be operated on, known as operand. The operand or data can be specified in various ways.

Instruction and data formats:

The format of a typical instruction is composed of two parts: an operation code or op-code and an operand. Every instruction needs an op-code to specify what the operation of the instruction is and then an operand that gives the appropriate data needed for that particular operation code. According to the word or byte size the 8085 instructions are classified into three types. They are (a) One byte (single) instructions. (b)Two byte instructions. (c) Three byte instructions.

Onebyte instructions: An instruction with only opcode and do not require any dat or address is called a one byte instruction.

Ex: 1. MOV C, A Hex code = 4FH (one byte) 2. ADD B Hex code = 80H (one byte) 3. CMA Hex code = 2FH (one byte)

Twobyte instructions: A two byte instruction is one which contains an 8-bit op-code and 8-bit operand (Data).

Ex: 1. MVI A, 09 Hex code = 3E, 09 (two bytes) 2. ADD B, 07 Hex code = 80, 07 (two bytes) 3. SUB A, 05 Hex code = 97, 05 (two bytes)

Threebyte instructions: A three byte instruction contains an opcode plus a 16 bit address.

Ex: 1.LXI H, 8509 Hex code = 21, 09, 85 (Three bytes) 2 .LDA 8509 Hex code = 3A, 09, 85 (Three bytes) 3. JMP 9567 Hex code = C3, 67, 95 (Three bytes) 4. STA 3525 Hex code = 32, 35, 25 (Three bytes)

DATA FORMATS: The 8085 is an 8-bit microprocessor which process only binary numbers. But it is very difficult to understand these binary numbers by a common user. So, we have to code these binary numbers into different data formats. The commonly known data formats are ASCII, BCD, signed integers and unsigned integers. The ASCII code is a 7-bit alpha-numeric code that represents decimal numbers, English alphabets and certain special characters. The ASCII stands for American Standard code for Information Interchange

The term BCD stands for binary coded decimal, used for decimal numbers from 0-9.An 8-bit register can store two BCD numbers. A signed integer is either a positive or a negative number. In 8085 microprocessor the most significant bit is used for the sign. Here 0 denotespositive sign and 1 denotes the negative sign. An integer without a sign can be represented by all the 8-bits in a microprocessor register. So, the largest number that can be processed at one time is FFH. The numbers larger than 8-bits like 16, 24, 32 bits can be processed by dividing them in groups of 8-bits.

CLASSIFICATION OF INSTRUCTIONSAn instruction is a binary pattern designed inside a microprocessor to perform a specific function. The entire group of instructions, called the instruction set, determines what functions the microprocessor can perform. The 8085 microprocessor instruction set has 74 operation codes that result in 246 instructions. This instruction set includes all the 8080A instructions plus two additional instructions namely SIM and RIM.The instruction set of 8085 microprocessor is classified into five groups. They are:1. Data transfer (copy) group.2. Arithmetic group3. Logic group4. Branch control group5. Machine control and I/O group.

Data transfer (copy) instructions

The data transfer instructions are used to transfer data from one register to another register, from memory to register or register to memory but not from one memory location to another memory location. Actually this data transfer instruction copies the data from source to destination and the contents of the source are not altered. So, the data transfer instruction performs basically copy operation.Examples of data transfer instructions are MOV, MVI (Move Immediate), LXI (Load Immediate H-L Pair), LDA (Load Accumulator), STA (Store Accumulator), LHLD (Load H-L pair direct), SHLD (Store H-L pair direct), XCHG (Exchange the contents of H-L pair with D-E pair) etc

Ex: MVI A, 55H ; Move the data 55H into Accumulator MOV B, C ; Copies the contents of C register into B register IN 00H ; Read the Input port(00H is the port address) OUT 01H ; write data to an output port(01H is the port address) LXIH 8570H ; Load H-L pair by address 8570H.In the 8085 microprocessor, data transfer instructions do not affect any flags.

Arithmetic InstructionsThe arithmetic operations like addition, subtraction, increment and decrement are performed by the 8085 microprocessor using the following arithmetic instructions.ADD, ADI (Add Immediate), SUB (Subtract), SUI (Subtract Immediate), INR (Increment), DCR (Decrement) etcThe arithmetic operations Add and subtract are performed in relation to the contents of the accumulator. But, the increment or the decrement operations can be performed in any register.

Ex:ADD B, C ; Add the contents of B register to the B register contents ADI 08 ; Add the data 08 to the accumulator. SUB A, B ; Subtract the contents of B register from accumulator. SUI 05 ;Subtract immediate the 8-bit data from accumulator INR B ; Increment the B register contents by one bit DCR C ; Decrement the C register contents by one bit.

Arithmetic instructions modify all the flags according to the data conditions of the result. The INR and DCR instructions affect all flags except the carry flag.

Logical Group of Instructions: Since the microprocessor is a programmable logic chip, it can be perform all the logic functions of the hard-wired logic through its instruction set. The 8085 processor can perform the logic instructions like, AND, OR, NOT (Complement) and X-OR (Exclusive OR) etc The mnemonics of these instructions are given below.

ANA : Logically AND the contents of a register ANI : Logically AND immediate the 8-bit data. ORA : Logically OR the contents of a register. OR : Logically OR immediate the 8-bit data. XRA : Exclusive-OR the contents of a register. XRI : Immediate Exclusive-OR the 8-bit data CMA : Complement the accumulator

All the logic operations are performed in relation to the contents of the accumulator. The CMA instruction does not affect any flags. The executions of the logical instruction do not affect the contents of the operand register.

Branch InstructionsThese instructions are very important because they allow the microprocessor to change the sequence of a program either conditionally or unconditionally. The conditional branch instructions transfer the program to the specified label when certain condition is satisfied. The unconditional branch instructions transfer the program to the specified location unconditionally.We know that the microprocessor is a sequential machine. So, it executes machine codes from one memory location to the next. Branch instructions instruct the microprocessor to go to a different memory location and the processor continues executing machine codes from the new location. The address of the new locations either specified explicitly or provided by the microprocessor or some times by additional hardware. The Branch instructions are classified into three categories. They are

(a). Jump instructions(b). Call and return instructions(c). Restart instructions.Jump instructions specify memory locations explicitly and they are 3-byte instructions. These Jump instructions are of two types. They are , Unconditional Jump and Conditional Jump.

Unconditional Jump:This is similar to Unconditional Go to statement in BASIC. When this instruction is executed the 16-bit address available immediately in the instruction is loaded into the program counter , so that the next sequence of instruction execution starts from this location. This Unconditional Jump instruction enables the programmer to create continuous loops.JMP (16 bit address). So, this is a 3-byte instruction where the first byte is op-code and the second, third bytes specify memory address.For example, the instruction JMP 8500H, instructs the microprocessor to go to the memory location8500H unconditionally. Sometimes, the jump location is specified using a label also.

Conditional Jump:

This instruction allows the microprocessor to make decision depending on certain conditions indicated by flags. The 8085 processor Jump instruction is associated with four flags. Namely Carry flag (CY), Zero flag (Z), Sign flag (S) and Parity flag (P). The following instructions shown in Table 3.3 transfer the program sequence to the memory location specified under the given conditions.

S. NoInstructionDescription

1JC (16 bit Addr)Jump on carry (if CY=1)

2JNC (16 bit Addr)Jump on no carry (if CY=0)

3JZ (16 bit Addr)Jump on Zero (if Z=1)

4JNZ (16 bit Addr)Jump on no Zero (if Z=0)

5JP (16 bit Addr)Jump on plus (if D7=0; S=0)

6JM (16 bit Addr)Jump on minus (if D7=1; S=1)

7JPE (16 bit Addr)Jump on Even Parity (if P=1)

8JPO (16 bit Addr)Jump on Odd Parity (if P=0)

Table 3 various conditional jump instructions

To understand the instructions, let us consider the instruction JC (16 bit address). The meaning of this instruction is, the microprocessor is instructed to jump the specified 16 bit memory location if there exists a carry after the arithmetic operation else it will execute the next instruction in the sequence.

CALL and RETURN InstructionsThe microprocessor uses the two instructions CALL and RETURN to implement subroutines. Here CALL instruction calls a subroutine program which is not a part of the main program and the RET instruction at the end of the subroutine program to return the control to the main program.

Ex: CALL (16 bit memory address) RETRESET (RST) InstructionThe 8085 processor provides eight RST instructions to transfer the program control to a specific location on page 00H. These instructions are 1-byte instructions. The various RST instructions and their call locations are given in the following Table 3.4

S. NoMnemonicsHex codeCall location In Hex

1RST 0C70000

2RST1CF0008

3RST2D70010

4RST3DF0018

5RST4E70020

6RST5EF0028

7RST6F70030

8RST7FF0038

Table 4 Various RST instructions and their call locations

Machine control and I/O Instructions

There are six basic machine control instructions. They are EI (Enable Interrupt) DI (Disable Interrupt) NOP (No Operation) SIM (Set Interrupt Mask) RIM (Read Interrupt Mask) HLT (Halt)

EI (Enable Interrupt): This is a one byte instruction used to enable the interrupt. This instruction is used to enable the interrupts when the microprocessor is reset or the interrupt enable flag is reset after interrupt acknowledge. This instruction takes one machine cycle with four states. The op-code is FBH.

DI (Disable Interrupt): This is a one byte instruction which resets the interrupt enable flag to disable all the interrupts except TRAP. It takes one machine cycle with four states. The op-code is F3H.

NOP (No Operation): when this instruction is executed, the microprocessor performs nothing. Microprocessor spends four states doing nothing. It is a one byte instruction whose op-code is 00H.This instruction is normally used to generate very small time delays of the order of few micro seconds. This NOP instruction is also very useful when we are required to insert a few instructions in the main program additionally .

SIM (Set Interrupt Mask): This instruction masks the interrupt as desired. This is a dual purpose instruction. The first purpose is to set or reset the mask of the maskable interrupt. The second purpose is to send the data out through the SOD pin at pin number 4 of the microprocessor.

RIM (Read Interrupt Mask): This instruction copies the status of the interrupts into the accumulator. It is also used to read the serial data through the SID pin

HLT (Halt): After execution of this instruction the microprocessor goes into the halt state. The processor can be restarted by a valid interrupt or by applying a RESET signal. The microprocessor takes 5T states to implement the halt instruction.

I/O instructions: There are two important instructions to input the data into the microprocessors accumulator through the input port and output the data from the accumulator to the output port. They are

IN (port address) OUT (port address)

This port address is an 8-bit address. In both these instructions the default register is Accumulator.

Ex: (i) IN 01H. This instruction will copy the contents into the Accumulator through the port whose address is 01H. It takes three machine cycles and takes 10 states. The op-code is DBH. (ii)OUT 02H. This instruction sends the contents of Accumulator to the outport whose address is 02H. It is a two byte instruction which requires 10 states. The op-code for this instruction is D3H.

DETAILED INSTRUCTION SET DATA TRANSFER INSTRUCTIONS

ARITHMATIC INSTRUCTIONS

SOD Serial output Data .Bit D7 of accumulator is latched in to the SOD output line made available to serial peripheral if bit D6 =1 .SOE Serial output enable.If this bit =1 ,it enables the serial output.XXX Dont care conditionR7.5 Reset RST7.5.If this bit = 1 ,RST7.5 flip-flop is reset .This is an additional control to reset RST7.5MSE Mask set Enable.If this bit is high ,it enables the function of bits D2,D1 and D0.This is a master control over all the interrupt masking bits. M7.5 D2=0 ,RST 7.5 is enabled D2=1 RST7.5 is masked or disabledM6.5 D1=0 RST6.5 is enabled D1=1 RST 6.5 masked or disabled.M5.5 D0=0 RST5.5 is enabled D0=1 RST 5.5 is disabled or masked.

ADDRESSING MODES

Majority of the instructions of 8085 microprocessor requires an operand (either data or address) on which the intended operation can be performed. Some instructions may require only one operand and some other instructions require two operands for its instruction execution. The speed of execution mainly depends on the position of the operand in the instruction. The scheme involved in identifying the position of operands in an instruction is known as addressing mode.

There are five addressing modes 8085 processor.They are(i). Immediate addressing mode(ii). Direct addressing mode(iii). Register addressing mode.(iv). Register indirect addressing mode.(v). Implicit addressing mode.

(i). Immediate Addressing mode: The mode of addressing in which the operand is a part of the instruction itself is known as Immediate Addressing mode. If the immediate data is 8-bit, the instruction will be of two bytes. If the immediate data is 16 bit, the instruction is of 3 bytes.Ex: (1). ADI DATA ; Add immediate the data to the contents of the accumulator. (2).LXIH 8500H : Load immediate the H-L pair with the operand 8500H (3). MVI 08H ; Move the data 08 H immediately to the accumulator (4). SUI 05H ; Subtract immediately the data 05H from the accumulator

(ii) Direct Addressing mode: The mode of addressing in which the 16-bit address of the operand is directly available in the instruction itself is called Direct Addressing mode. i.e., the address of the operand (data) is available in the instruction itself. This is a 3-byte instruction.Ex: (1). LDA 9525H ; Load the contents of memory location into Accumulator. (2). STA 8000H ; Store the contents of the Accumulator in the location 8000H (3). IN 01H ; Read the data from port whose address is 01H.

(iii). Register addressing modes: The mode, in which the operand is in one of the general purpose registers, is known as the register addressing mode.Ex: (1). MOV A, B; Move the contents of B register to A register. (2). SUB D; Subtract the contents of D register from Accumulator. (3). ADD B, C; Add the contents of C register to the contents of B register.

(iv). Register indirect addressing modes: The 16-bit address location of the operand stored in a register pair (H-L) is given in the instruction. The address of the operand is given in an indirect way with the help of a register pair. Hence it is called Register indirect addressing modeEx: (1). LXIH 9570H ; Load immediate the H-L pair with the address of the location 9570H MOV A, M ; Move the contents of the memory location pointed by the H-L pair to accumulator

(v). Implicit Addressing mode: The mode of instruction which do not specify the operand in the instruction but it is implicated, is known as implicit addressing mode. i.e., the operand is automatically considered to be in the Accumulator.Ex: (1).CMA; complement the contents of Accumulator (2).CMC; Complement carry (3). RLC; Rotate Accumulator left by one bit (4). RRC; Rotate Accumulator right by one bit (5). STC; Set carry.

ASSEMBLY LANGUAGE PROGRAMMING EXAMPLES:

Addition Programs

Example 1: Addition of two 8-bit numbers whose sum is 8-bits.

Explanation: This assembly language program adds two 8-bit numbers stored in two memory locations .The sum of the two numbers is 8-bits only. The necessary algorithm and flow charts are given below.

ALGORITHM:Step1. : Initialize H-L pair with memory address XX00 (say: 9000).Step2. : Clear accumulator.Step3. : Add contents of memory location M to accumulator.Step4. : Increment memory pointer (i.e. XX01).Step5. : Add the contents of memory indicated by memory pointer to accumulator.Step6. : Store the contents of accumulator in 9002.Step7. : Halt

PROGRAM:

Address of the memory locationHex codeLabelMnemonicsComments

Op-codeOperand

800021,00,90LXIH, 9000Initialise memory pointer to point the first data location 9000.

80033EMVIA, 00Clear accumulator

800400

800586ADDA, MThe first number is added to accumulator [A] [A] + M

800623INXHIncrement the memory pointer to next location of the Data.

800786ADDA, MThe 2nd number is added to contents of accumulator

800832STA9002The contents of accumulator are stored in memory location 9002.

800902

800A90

800B76HLTStop the execution

Ex: Input: Ex: (i) 9000 29 H Ex :(ii) 9000 49 H 9001 16 H 9001 32 H

Result: Ex: (i) 9002 3F H Ex :( ii) 9002 7B Flow Chart Start

Initialise H-L Pair with XX00

Clear Accumulator(00) AAdd contents of M to Accumulator# Increment memory pointer(XX01) # Add contents of M to ACopy the contents of Reg. A to memory location XX02

Stop

Fig 3.18

Example 2: Addition of two 8-bit numbers whose sum is 16 bits.

Explanation: The first 8-bit number is stored in one memory location (say 8500) and the second 8-bit number is stored in the next location (8501).Add these two numbers and check for carry. Store the LSB of the sum in one memory location (8502) and the MSB (carry) in the other location(8503).ALGORITHM:Step1. : Initialize H-L pair with memory address X (say: 8500).Step2. : Clear accumulator.Step3. : Add contents of memory location M to accumulator.Step4. : Increment memory pointer (i.e. 8501).Step5. : Add the contents of memory indicated by memory pointer to accumulator.Step6. : Check for CarryStep 7 : Store the sum in 8502.Step8 : Store the Carry in 8503 locationStep 9 : Halt StartFlow Chart


Clear Accumulator(00) A

Add contents of M to Accumulator

# Increment memory pointer(XX01) & # Add contents of M to A

Store the sum in the location xx02

Is Carry exists ? Yes

Stores carry in the XX03 location

No

Store Zero in the XX03 location

Stop

PROGRAM:


Op-codeOperand

800021,00,85LXIH, 8500 HInitialise memory pointer to point the first data location 9000.

80033EMVIA,00Clear accumulator

800400

800586ADDA, MThe first number is added to accumulator [A] [A]+M

80060EMVIC,00Initial value of Carry is 0

800700



800A32JNCFWDIs Carry exists ? No,go to the label FWD

800B0E

800C80

800D0CINRCMake carry =1

800E32FWDSTA8502 HThe sum is stored in memory location 8502.

800F02

801085

801179MOVA,C

801232STA8503 HStore the carry at 8503 location

801303

801485

801576HLTStop the execution

Ex: Input: Ex : 8500 97 H RESULT: 8502 32 H 8501 98H 8503 -- 01 H

Example 3: Decimal addition of two 8-bit numbers whose sum is 16 bits.

Explanation: Decimal addition of two 8-bit numbers is same as that of two 8-bit numbers program. Except that the use of DAA instruction. The first 8-bit number is stored in one memory location (say 8500) and the second 8-bit number is stored in the next location(8501).Add these two numbers and use the DAA instruction to get the result in decimal. Also check for carry. Store the LSB of the sum in one memory location(8502) and the MSB (carry) in the other location(8503).

ALGORITHM:Step1. : Initialize H-L pair with memory address XXXX (say: 8500).Step2. : Clear Carry register C.Step3. : Move contents of memory location M to accumulator.Step4. : Increment memory pointer (i.e. 8501).Step5. : Add the contents of memory indicated by memory pointer to accumulator.Step6. : Apply the instruction DAA(Decimal adjust after addition)Step7: Check for CarryStep8: Store the sum in XX02.Step9: Store the Carry in XX03 locationStep10: Halt

StartFlow Chart


Clear Carry register(00) C

Add contents of M to Accumulator

# Increment memory pointer(XX01) & # Add contents of M to A

Apply DAA instruction to convert Hex data to decimal form

Store the sum in the location XX02

Is Carry Exists?Yes

Stores carry in the XX03 location

Store Zero in the XX03 location No

Stop

PROGRAM


Op-codeOperand

800021, 00,85LXIH, 8500 HInitialise memory pointer to point the first data location 9000.

80030EMVIC, 00Clear accumulator

800400

80057EMOVA, MThe first number is added to accumulator [A] [A]+M



800827DAA

8009D2JNCFWDIs Carry exists? No, go to the label FWD

0D

80

800C0CINRCMake carry =1

800D32FWDSTA8502 HThe contents of accumulator are stored in memory location 8502.

800E02

800F85

801079MOVA, CCarry is moved to accumulator

801132STA8503 HA Carry is stored in the location 8503

801203

801385


Ex: Input: Ex : 8500 67 D RESULT: 8502 52 D 8501 85 D 8503 01 (Carry)

Example 4: Addition of two 16-bit numbers whose sum is 16 bits or moreExplanation: First 16-bit number is stored in two consecutive locations (Ex 8500 &8501) because in each location we can store only one 8-bit number. Store the second 16-bit number in the next two consecutive locations (For Ex: 8502 &8503).Add the LSB of the first number to the LSB of the second number and the MSB of the first number to the MSB of the second number using the DAD instruction. Store the sum in the next two locations and the carry (if any) in the third locationALGORITHM: Step1: First 16 bit number is in locations 8500 & 8501 respectively Step2: Second 16-bit number is in locations 8502 & 8503 Step3: Add the two 16-bit numbers using DAD Instruction.Step4: Sum is stored in locations 8504 & 8505.Step5: Carry (if any) is stored in the location 8506. Step6: Halt

StartFlow Chart

Initialise H-L Pair with 8500

Load the first 16-bit number in to H-L Pair

Exchange this number in to D-E pair

Load the second 16-bit number in to H-L pair

Add the LSB of Ist number to the LSB of the second number & MSB of the second number to MSB of the second number using DAD instruction

Store the LSBs of the sum in 8504 & 8505 locations

Is Carry Set? Yes

Store1 in the 8506 location No

Store 0 in the location 8506

Stop

Fig 21

PROGRAM:ADDRESSHEX CODELABEL MNEMONICCOMMENTS

OPCODEOPERAND

80002A,00,85LHLD8500 HFirst 16-bit number in H-L pair

800100

800285

8003EBXCHG Exchange first number to D-E Pair

80042ALHLD8502 H

800502

800685

80070EMVI00MSB of the sum is initially 00

800800

800919DADDAdd two 16 bit numbers

800AD2JNC FWDIs Carry? If yes go to the next line .Else go to the 800E LOCATION

800B0E

800C80

800DOCINRCIncrement carry

800E22FWDSHLD8504 HStore the LSB of the Sum in 8504 & MSB in 8505 locations

800F04

801085

801179MOVA, CMSBs of the sum is in Accumulator

801232STA8506 HStore the MSB (Carry) of the result in 8506 location

801306

801485

801576HLTStop execution

Ex: INPUT: 8500- 12 H LSB of the Ist Number RESULT : 8504 - 25H LSB of the Sum 8501- 13 H MSB of the Ist Number 8505 25H MSB of the Sum 8502 -13 H LSB of the IInd Number 8506 -- 00 Carry . 8503 -12H MSB of the IInd numberSubtraction Programs:Example 5: Subtraction of two 8-bit numbers without borrows.Explanation: Its a simple program similar to addition of two 8- bit numbers, except that we use the instruction SUB instead of ADD. The first 8-bit number is stored in XX00 memory location and the second 8-bit number is stored in the XX01 location .Use the SUB instruction and store the result in the XX02 location.ALGORITHM:Step1. : Initialize H-L pair with the address of minuend.Step2. : Move the minuend into accumulatorStep3. : Increment H-L pair Step4. : Subtract the subtrahend in memory location M from the minuend.Step5. : Store the result in XX02.Step6. : Stop the execution

StartFlow Chart

Initialize H-L Pair with XX00

Move the contents of M to Accumulator

Increment memory pointer(XX01)

Subtract the subtrahend in memory location M from the minuend in A

Store the result in the location XX02

Stop

Figure 22

ADDRESSHEX CODELABELMNEMONICCOMMENTS

OPCODEOPERAND

800021LXIH, 8500Initialise H-L pair and get the First number in to 8500 location

800100

800285

80037EMOVA,M[A] [M]

800423INXH[M+1] [M]

800596SUBMA [A] [M]

800623INX HNext memory location

800777MOVM,AStore the result in the location 8502


PROGRAM:

INPUT: Ex : 8500- 59H Result: 8502 29H 8501- 30H

Example 6: Subtraction of two 8-bit Decimal numbers.

Explanation: In this program we cant use the DAA instruction after SUB or SBB instruction because it is decimal adjust after addition only. So, for decimal subtraction the number which is to be subtracted is converted to 10s complement and then DAA is applied.

ALGORITHM:Step1. : Initialise H-L pair with the address of second number (XX01).Step2. : Find its tens complementStep3. : Decrement the H-L pair for the first number (XX00)Step4. : Add the first number to the 10s complement of second number.Step5. : Store the result in XX02.Step6. : Stop the execution

Flow Chart

Start

Initialise H-L Pair with XX01 and get the second number

Find its 10s complement

Add first number to the 10s complement of the second number

Apply the DAA instruction to get decimal result

Store the result in the location XX02

Stop

Figure 23

PROGRAM:


OPCODEOPERAND

800021LXIH,8500Initialise H-L pair and get theSecond number in to 8501 location

800100

800285

80033EMVIA,99[A] 99

800499

800596SUBM9s complement of second number

80063CINRA10s complement of second number

80072BDCX HAddress of the first number

800886ADDMAdd first number to 10s complement of second number

800927DAA

800A32STA8502Store the result in the location 8502

800B02

800C85

800D76HLTStop the execution

Ex: Input: 8500 -76 D Result: 8502 - 41 D

8501- 35 D

Example 6: Subtraction of two 16 bit numbers.

Explanation: It is very similar to the addition of two 16-bit numers.Here we use SUB &SBB instructions to get the result .The first 16-bit number is stored in two consecutive locations and the second 16-bit number is stored in the next two consecutive locations.The lsbs are subtracted using SUB instruction and the MSBs aare subtracted using SBB instruction.The result is stored in different locations.

ALGORITHM:Step1. : Store the first number in the locations 8500 & 8501.Step2. : Store the second number in the locations 8502 &8503. Step4. : Subtract the second number from the first number with borrow.Step5. : Store the result in locations 8504 & 8505.Step6. : Store the borrow in location 8506Step 7: Stop the execution

StartFlow Chart

Initialize H-L Pair with XX00

Load the first 16-bit number in to H-L Pair A

Exchange this number in to D-E pair

Load the second 16-bit number in to H-L pair

Subtract LSB of 2 nd number from the LSB of the Ist number & MSB of the 2 nd number from MSB of the first number

Store the result in XX04 & XX05 locations

Is borrow Set ? Yes

Store 1 in the XX06 location No

Store 0 in the location XX06 No

Stop

Fig 24

PROGRAM:


OPCODEOPERAND

80002A, 00,85LHLD8500 HFirst 16-bit number in H-L pair

8003EBXCHG Exchange first number to D-E Pair

80042ALHLD8502 HGet the second 16-bit number in H-L pair

800502

800685

80077BMOVA, EGet the lower byte of the First number in to Accumulator

800895SUBLSubtract the lower byte of the second number

80096FMOV L, AStore the result in L- register

800AMOVA, DGet higher byte of the first number

800A9CSBB HSubtract higher byte of second number with borrow

800B67MOVH, A

800C22SHLD8504Store the result in memory locations with LSB in 8504 & MSB in 8505

800D04

80OE85

80OF76HLTStop execution

Ex: INPUT : 8500- FF H LSB of the Ist Number RESULT: 8504 - 11H LSB 8501 - FF H MSB of the Ist Number 8505 11 H MSB 8502 -EE H LSB of the IInd Number 8503 EE H MSB of the IInd number

Multiplication Programs

Example 7: Multiplication of two 8-bit numbers. Product is 16-bits.Explanation: The multiplication of two binary numbers is done by successive addition. When multiplicand is multiplied by 1 the product is equal to the multiplicand, but when it is multiplied by zero, the product is zero. So, each bit of the multiplier is taken one by one and checked whether it is 1 or 0 .If the bit of the multiplier is 1 the multiplicand is added to the product and the product is shifted to left by one bit. If the bit of the multiplier is 0 , the product is simply shifted left by one bit. This process is done for all the 8-bits of the multiplier.

ALGORITHM: Step 1 : Initialise H-L pair with the address of multiplicand.(say 8500)Step 2 : Exchange the H-L pair by D-E pair. so that multiplicand is in D-E pair.Step 3 : Load the multiplier in Accumulator.Step 4 : Shift the multiplier left by one bit.Step 5 : If there is carry add multiplicand to product.Step 6 : Decrement the count.Step 7 : If count 0; Go to step 4Step 8 : Store the product i.e. result in memory location.Step 9 : Stop the execution

Flow Chart

Start

Load the multiplier in to Accumulator & multiplicand in D-E pair

Initial value of product=00.Count=08

Shift product left by one bit &Shift multiplier left by one bit

Is carry exists from multiplier ?

No

Yes

Product = Product + Multiplicand

Count = count -1

Is count = 0 ?

No

Yes

Store result

Stop

Figure 25

PROGRAM:

ADDRESSHEX- CODELABELMNEMONICCOMMENTS

OPCODEOPERAND

80002A,00,85LHLDH, 8500Load the multiplicand in to H-L pair

8003EBXCHGExchange the multiplicand in to D-E pair

80043ALDA8502Multiplier in Accumulator

800502

800685

800721LXI H.0000Initial value in H-L pair is 00

800800

800900

800A0EMVIC,08Count =08

800B08

800C29LOOPDADHShift the partial product left by one bit.

800D17RALRotate multiplier left by one bit

800ED2JNCFWDIs Multiplier bit =1? No go to label FWD

800F12

801080

801119DADDProduct =Product +Multiplicand

80120DFWDDCRCCOUNT=COUNT-1

8013C2JNZLOOP

80140C

801580

801622SHLD8503Store the result in the locations 8503 & 8504

801703

801885


INPUT :AddressData

85008AH LSB of Multiplicand

850100 H MSB of Multiplicand

850252 H - Multiplier

Result:850334 H LSB of Product

85042C H MSB of Product

Division Programs

Example 7: Division of a 16- bit number by a 8-bit number.

Explanation: The division of a 16/8-bit number by a 8-bit number follows the successive subtraction method. The divisor is subtracted from the MSBs of the dividend .If a borrow occurs, the bit of the quotient is set to 1 else 0.For correct subtraction process the dividend is shifted left by one bit before each subtraction. The dividend and quotient are in a pair of register H-L.The vacancy arised due to shifting is occupied by the quotient .In the present example the dividend is a 16-bit number and the divisor is a 8-bit number. The dividend is in locations 8500 &8501.Similarly the divisor is in the location 8502.The quotient is stored at 8503 and the remainder is stored at 8504 locations.

ALGORTHM: STEP1. : Initialise H-L pair with address of dividend.STEP2. : Get the divisor from 8502 to register A & then to Reg.BSTEP3. : Make count C=08STEP4. : Shift dividend and divisor left by one bit STEP 5: Subtract divisor from dividend.STEP6. : If carry = 1 : goto step 8 else step7.STEP7. : Increment quotient register.STEP8. : Decrement count in CSTEP9. : If count not equal to zero go to step 4STEP10: Store the quotient in 8503 STEP11: Store the remainder in 8504 STEP12: Stop execution.

Flowchart

START

Initialise H-L pair with address of dividend

Copy the divisor in to Reg. A

Move the divisor in to Reg. B

Move count =08 in to C register

# Shift dividend left by one bit# Shift Quotient left by one bit+

Is MSBs of Dividend > Divisor ?

No

Yes

Quotient = Quotient +1

8 MSBs of dividend =8 MSBs of dividend -divisor

Count = Count-1

IsCount =0?

Stopstore th quotient in 8503 &remainder in 8504

No Yes Fig.26

PROGRAM: ADDR-ESSHEX CODELABELMNEMONICCOMMENTS

OPCODEOPERAND

800021LHLDH, 8500Initialize the H-L pair for dividend

800100

800285

80033ALDA8502 HLoad the divisor from location 8502 to accumulator

800402

800585

800647MOVB,AMove Divisor to Reg.B from A

80070EMVIC,08Count =08

800808

800929BACKDADHShift dividend and quotient left by one bit

800A7CMOVA,HMSB of dividend in to accumulator

800B90SUBBSubtract divisor from MSB bits of divisor

800CDAJCFWDIs MSB part of dividend > divisor ? No,goto label FWD

800D11

800E80

800F67MOVH,AMSB of the dividend in Reg.H

80102CINRLIncrement quotient

80110DFWDDCRCDecrement count

8012C2JNZBACKIf count is not zero jump to8009 location

801309

801480

801522SHLD8503HStore quotient in 8503 and remainder in 8504 locations

801603

801785

801876HLTStop execution

AddressData

850064 LSB of Dividend

850100 MSB of Dividend

850207 Divisor

85030E Quotient

850402 Remainder

Ex: Input & Result

Largest & Smallest numbers in an Array

Example 8: To find the largest number in a data array

Explanation: To find the largest number in a data array of N numbers (say)first the count is placed in memory location (8500H) and the data are stored in consecutive locations.(8501.onwards).The first number is copied to Accumulator and it is compared with the second number in the memory location. The larger of the two is stored in Accumulator. Now the third number in the memory location is again compared with the accumulator. And the largest number is kept in the accumulator. Using the count, this process is completed , until all the numbers are compared .Finally the accumulator stores the smallest number and this number is stored in the memory location.85XX.

ALGORTHM:

Step1: Store the count in the Memory location pointed by H-L register.Step2: Move the I st number of the data array in to accumulatorStep3: Compare this with the second number in Memory location.Step4: The larger in the two is placed in AccumulatorStep5: The number in Accumulator is compared with the next number in memory .Step 6: The larger number is stored in Accumulator.Step 7; The process is repeated until the count is zero.Step 8: Final result is stored in memory location.Step 9: Stop the execution

Flow Chart

START

Initialise H-L pair

Get the count in Register C

Get first number in Accumulator

IsNumber in Accumulator> Next number ?

Yes

Get Larger number inAccumulator No

No

Decrement Count

Is count = 0 ? No

Yes

store the result in the location 85XX

STOP

Fig 27

PROGRAM

ADDR-ESSHEX CODELABELMNEMONICCOMMENTS

OPCODEOPERAND

800021,00,85LXIH, 8500INITIALISE H-L PAIR

80037EMOVC,MCount in the C register

800423INXHFirst number in H-L pair

80054EMOVA,MMove first number in to Accumulator

80060DDCRCDecrement the count

800791LOOP1INXHGet the next number

8008BECMPMCompare the next number with previous number

8009D2JNCLOOP2Is next number >previous maximum?No,go to the loop2

800A0D

800B80

800C7EMOVA,MIf,yes move the large number in to Accumulator

800D0DLOOP2DCRCDecrement the count

800EC2JNZLOOP1If count not equal to zero,repeat

800F07

801180

801278

801332STA85XXStore the largest number in the location 85XX

8014XX

801585


Ex : Input : 8500- N(Say N=7 ) Result : 8508 - 7F 8501-05 8502-0A 8503-08 8504-14 8505 -7F 8506-25 8507-2D

Example 9 : To find the smallest number in a data array.

Explanation: To find the smallest number in a data array of N numbers (say)first the count is placed in memory location (8500H) and the data are stored in consecutive locations.(8501.onwards).The first number is copied to Accumulator and it is compared with the second number in the memory location.The smaller of the two is stored in Accumulator.Now the third number in the memory location is again compared with the accumulator.and the smallest number is kept in the accumulator.Using the count,this process is completed until all the numbers are compared .Finally the accumulator stores the smallest number and this number is stored in the memory location.85XX.

ALGORTHM :

Step1: Store the count in the Memory location pointed by H-L register.Step2: Move the I st number of the data array in to accumulatorStep3: Compare this with the second number in Memory location.Step4: The smaller in the two is placed in AccumulatorStep5: The number in Accumulator is compared with the next number in memory .Step 6: The smaller number is stored in Accumulator.Step 7; The process is repeated until the count is zero.Step 8: Final result is stored in memory location.Step 9: Stop the execution

Flow Chart

START

Initialise H-L pair

Get the count in Register C

Get first number in to Accumulator

IsNumber in Accumulator< Next number ?

Yes

No

Get smaller number inAccumulator

Decrement Count

Is count = 0 ?

No Yes

store the result at 85xx

STOP

Fig.28

PROGRAM

ADDR-ESSHEX CODELABELMNEMONICCOMMENTS

OPCODEOPERAND

800021LXIH, 8500Initialise the H-L pair.

800100

800285

80037EMOVC,MCount in the C register

800423INXHFirst number in H-L pair

80054EMOVA,MMove first number in to Accumulator

80060DDCRCDecrement the count

800791LOOP1INXHGet the next number

8008BECMPMCompare the next number with previous number

8009D2JCLOOP2Is next number

Microprocessor

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