The 8085 is an 8-bit general purpose microprocessor that can address 64K Byte of memory. It has 40 pins and uses +5V for power. It can run at a maximum. - презентация

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2 The 8085 is an 8-bit general purpose microprocessor that can address 64K Byte of memory. It has 40 pins and uses +5V for power. It can run at a maximum frequency of 3 MHz. The pins on the chip can be grouped into 6 groups: Address Bus. Data Bus. Control and Status Signals. Power supply and frequency. Externally Initiated Signals. Serial I/O ports. BLOCK SCHEMATIC

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4 The address bus has 8 signal lines A8 – A15 which are unidirectional. The other 8 address bits are multiplexed (time shared) with the 8 data bits. So, the bits AD0 – AD7 are bi-directional and serve as A0 – A7 and D0 – D7 at the same time. During the execution of the instruction, these lines carry the address bits during the early part, then during the late parts of the execution, they carry the 8 data bits. In order to separate the address from the data, we can use a latch to save the value before the function of the bits changes.

5 Arithmetic Logic Unit: The ALU performs the actual numerical and logic operation such as add, subtract, AND, OR, etc. Uses data from memory and from Accumulator to perform arithmetic. Always stores result of operation in Accumulator. Accumalator: The accumulator is an 8-bit register that is a part of arithmetic/logic unit (ALU). This register is used to store 8-bit data and to perform arithmetic and logical operations. The result of an operation is stored in the accumulator. The accumulator is also identified as register A.

6 Registers: The 8085/8080A-programming model includes six registers, one accumulator, and one flag register, as shown in Figure. In addition, it has two 16-bit registers: the stack pointer and the program counter. They are described briefly as follows. The 8085/8080A has six general-purpose registers to store 8-bit data; these are identified as B,C,D,E,H, and L as shown in the figure. They can be combined as register pairs - BC, DE, and HL - to perform some 16-bit operations. The programmer can use these registers to store or copy data into the registers by using data copy instructions.

7 The ALU includes five flip-flops, which are set or reset after an operation according to data conditions of the result in the accumulator and other registers. They are called Zero(Z), Carry (CY), Sign (S), Parity (P), and Auxiliary Carry (AC) flags; they are listed in the Table and their bit positions in the flag register are shown in the Figure below. The most commonly used flags are Zero, Carry, and Sign. The microprocessor uses these flags to test data conditions. For example, after an addition of two numbers, if the sum in the accumulator id larger than eight bits, the flip-flop uses to indicate a carry -- called the Carry flag (CY) -- is set to one. When an arithmetic operation results in zero, the flip-flop called the Zero(Z) flag is set to one. The first Figure shows an 8-bit register, called the flag register, adjacent to the accumulator. However, it is not used as a register; five bit positions out of eight are used to store the outputs of the five flip- flops. The flags are stored in the 8-bit register so that the programmer can examine these flags (data conditions) by accessing the register through an instruction.

8 Program Counter(PC): This 16-bit register deals with sequencing the execution of instructions. This register is a memory pointer. Memory locations have 16-bit addresses, and that is why this is a 16-bit register. The microprocessor uses this register to sequence the execution of the instructions. The function of the program counter is to point to the memory address from which the next byte is to be fetched. When a byte (machine code) is being fetched, the program counter is incremented by one to point to the next memory location Stack pointer(SP): The stack pointer is also a 16-bit register used as a memory pointer. It points to a memory location in R/W memory, called the stack. The beginning of the stack is defined by loading 16-bit address in the stack pointer. The stack concept is explained in the chapter "Stack and Subroutines." Instruction Register/Decoder: Temporary store for the current instruction of a program. Latest instruction sent here from memory prior to execution. Decoder then takes instruction and decodes or interprets the instruction. Decoded instruction.Then passed to the next stage.8

9 There are 4 main control and status signals. These are: ALE: Address Latch Enable. This signal is a pulse that become 1 when the AD0 – AD7 lines have an address on them. It becomes 0 after that. This signal can be used to enable a latch to save the address bits from the AD lines. RD: Read. Active low. WR: Write. Active low. IO/M: This signal specifies whether the operation is a memory operation (IO/M=0) or an I/O operation (IO/M=1). S1 and S0 : Status signals to specify the kind of operation being performed.Usually un-used in small systems.

10 There are 3 important pins in the frequency control group. X0 and X1 are the inputs from the crystal or clock generating circuit. The frequency is internally divided by 2. So, to run the microprocessor at 3 MHz, a clock running at 6 MHz should be connected to the X0 and X1 pins. CLK (OUT): An output clock pin to drive the clock of the rest of the system. We will discuss the rest of the control signals as we get to them.

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19 To understand how the microprocessor operates and uses these different signals, we should study the process of communication between the microprocessor and memory during a memory read or write operation. Lets look at timing and the data flow of an instruction fetch operation.

20 Lets assume that we are trying to fetch the instruction at memory location That means that the program counter is now set to that value. The following is the sequence of operations: The program counter places the address value on the address bus and the controller issues a RD signal. The memorys address decoder gets the value and determines which memory location is being accessed. The value in the memory location is placed on the data bus. The value on the data bus is read into the instruction decoder inside the microprocessor. After decoding the instruction, the control unit issues the proper control signals to perform the operation.

21 Now, lets look at the exact timing of this sequence of events as that is extremely important. (figure 3.3) At T1, the high order 8 address bits (20H) are placed on the address lines A8 – A15 and the low order bits are placed on AD7–AD0. The ALE signal goes high to indicate that AD0 – AD8 are carrying an address. At exactly the same time, the IO/M signal goes low to indicate a memory operation. At the beginning of the T2 cycle, the low order 8 address bits are removed from AD7– AD0 and the controller sends the Read (RD) signal to the memory. The signal remains low (active) for two clock periods to allow for slow devices. During T2, memory places the data from the memory location on the lines AD7– AD0. During T3 the RD signal is Disabled (goes high). This turns off the output Tri-state buffers in the memory. That makes the AD7– AD0 lines go to high impedence mode.

22 From the above discussion, we can define terms that will become handy later on: T- State: One subdivision of an operation. A T-state lasts for one clock period. An instructions execution length is usually measured in a number of T-states. (clock cycles). Machine Cycle: The time required to complete one operation of accessing memory, I/O, or acknowledging an external request. This cycle may consist of 3 to 6 T-states. Instruction Cycle: The time required to complete the execution of an instruction. In the 8085, an instruction cycle may consist of 1 to 6 machine cycles.

23 The 8085 executes several types of instructions with each requiring a different number of operations of different types. However, the operations can be grouped into a small set. The three main types are: Memory Read and Write. I/O Read and Write. Request Acknowledge. These can be further divided into various operations (machine cycles ).

24 The first step of executing any instruction is the Opcode fetch cycle. In this cycle, the microprocessor brings in the instructions Opcode from memory. To differentiate this machine cycle from the very similar memory read cycle, the control & status signals are set as follows: IO/M=0, s0 and s1 are both 1. This machine cycle has four T-states. The 8085 uses the first 3 T-states to fetch the opcode. T4 is used to decode and execute it. It is also possible for an instruction to have 6 T-states in an opcode fetch machine cycle.

25 The memory read machine cycle is exactly the same as the opcode fetch except: It only has 3 T-states The s0 signal is set to 0 instead.

26 To understand the memory read machine cycle, lets study the execution of the following instruction: MVI A, 32 In memory, this instruction looks like: The first byte 3EH represents the opcode for loading a byte into the accumulator (MVI A), the second byte is the data to be loaded. The 8085 needs to read these two bytes from memory before it can execute the instruction. Therefore, it will need at least two machine cycles. The first machine cycle is the opcode fetch discussed earlier. The second machine cycle is the Memory Read Cycle. Figure 3.10 page H 2001H 3E 32

27 Machine cycles and instruction length, do not have a direct relationship. To illustrate lets look at the machine cycles needed to execute the following instruction. STA 2065H This is a 3-byte instruction requiring 4 machine cycles and 13 T-states. The machine code will be stored in memory as shown to the right This instruction requires the following 4 machine cycles: Opcode fetch to fetch the opcode (32H) from location 2010H, decode it and determine that 2 more bytes are needed (4 T-states). Memory read to read the low order byte of the address (65H) (3 T-states). Memory read to read the high order byte of the address (20H) (3 T-states). A memory write to write the contents of the accumulator into the memory location. 2010H 2011H 2012H 32H 65H 20H

28 In a memory write operation: The 8085 places the address (2065H) on the address bus Identifies the operation as a memory write (IO/M=0, s1=0, s0=1). Places the contents of the accumulator on the data bus and asserts the signal WR. During the last T-state, the contents of the data bus are saved into the memory location.

29 There needs to be a lot of interaction between the microprocessor and the memory for the exchange of information during program execution. Memory has its requirements on control signals and their timing. The microprocessor has its requirements as well. The interfacing operation is simply the matching of these requirements.

30 The process of interfacing the above two chips is the same. However, the ROM does not have a WR signal. Address Lines Date Lines CS RD Output Buffer ROM Address Lines Data Lines CS RD Output Buffer RAM WR Input Buffer Data Lines

31 From the above description, it becomes obvious that the AD7– AD0 lines are serving a dual purpose and that they need to be demultiplexed to get all the information. The high order bits of the address remain on the bus for three clock periods. However, the low order bits remain for only one clock period and they would be lost if they are not saved externally. Also, notice that the low order bits of the address disappear when they are needed most. To make sure we have the entire address for the full three clock cycles, we will use an external latch to save the value of AD7– AD0 when it is carrying the address bits. We use the ALE signal to enable this latch.

32 Given that ALE operates as a pulse during T1, we will be able to latch the address. Then when ALE goes low, the address is saved and the AD7– AD0 lines can be used for their purpose as the bi-directional data lines. A15-A8 Latch AD7-AD0 D 7 - D 0 A 7 - A ALE

33 The result of address decoding is the identification of a register for a given address. A large part of the address bus is usually connected directly to the address inputs of the memory chip. This portion is decoded internally within the chip. What concerns us is the other part that must be decoded externally to select the chip. This can be done either using logic gates or a decoder.

34 Putting all of the concepts together, we get: A15-A8 Latch AD7-AD0 D 7 - D 0 A 7 - A ALE IO/M RD WR 1K Byte Memory Chip WR RD CS A 9 - A 0 A 15 - A 10 Chip Selection Circuit

35 Accessing memory can be summarized into the following three steps: Select the chip. Identify the memory register. Enable the appropriate buffer. Translating this to microprocessor domain: The microprocessor places a 16-bit address on the address bus. Part of the address bus will select the chip and the other part will go through the address decoder to select the register. The signals IO/M and RD combined indicate that a memory read operation is in progress. The MEMR signal can be used to enable the RD line on the memory chip.

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