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8085 Microprocessor Architecture

The document provides information on the architecture of the 8085 microprocessor. It discusses the different busses used - the address bus, data bus and control bus. It describes the internal architecture of the 8085 including registers like the accumulator, flags, program counter and stack pointer. It also discusses the different types of operations like memory read/write, I/O read/write. Memory organization and addressing are explained along with the concept of memory mapping.

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Introduces the chapter on microprocessor architecture. Explains programming and data manipulation specifics.

Describes the 8085 architecture focusing on its three buses: address bus (16-bit, unidirectional) for memory locations, data bus (8-bit, bi-directional) for data transfer, and control bus.

Classifies microprocessor operations: microprocessor-initiated (memory reads/writes) and peripheral-initiated. Treats I/O devices as memory locations.

Details the steps for reading a memory location, including address placement and activation of memory read control signal.

Describes the internal data operations of 8085 including registers, accumulator, and architecture needed for these operations.

Explains the flag register showing the microprocessor status, including sign, zero, auxiliary carry, parity, and carry flags.

Details the roles of the Program Counter (PC) for instruction sequencing and Stack Pointer for memory addressing.

Lists operations initiated by external devices like reset, interrupt, ready, and hold, along with their functions.

Explains main memory types (RAM, ROM) and accessing memory through address and data buses.

Describes tri-state buffers with logic states used in memory operations, explaining their function and behavior.

Illustrates memory elements similar to D latches and their control signals (WR, RD) in memory architecture.

Describes the construction of a memory register using latches, inputs, and outputs, showing grouped memory registers.

Outlines how to design a memory chip, including control signals for reading and writing data, and architecture overview.

Discusses memory measurement (length and width) and how a microprocessor utilizes address lines to access memory.

Explains chip selection inputs and how to manage multiple memory chips using address lines for unique identification.

Illustrates the memory map showing address ranges for different memory chips, comparing it to a post office box analogy.

Discusses how the 8085 uses address lines to establish address ranges across multiple chips for memory operations.

Describes how combinations of address lines define chip selection and address ranges for enabling/disabling memory chips.

Categorizes address lines into high-order (chip selection) and low-order (location selection) for memory design.

Discusses how to use multiple memory chips to create a byte-wide memory system, demonstrating data line organization.

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CHAPTER 2 MICROPROCESSOR ARCHITECTURE
MICROPROCESSOR ARCHITECTURE • The microprocessor can be programmed to perform functions on given data by writing specific instructions into its memory. • The microprocessor reads one instruction at a time, matches it with its instruction set, and performs the data manipulation specified. • The result is either stored back into memory or displayed on an output device.
THE 8085 ARCHITECTURE • The 8085 uses three separate busses to perform its operations • The address bus. • The data bus. • The control bus.
THE ADDRESS BUS • 16 bits wide (A0 A1…A15) • Therefore, the 8085 can access locations with numbers from 0 to 65,536. Or, the 8085 can access a total of 64K addresses. • “Unidirectional”. • Information flows out of the microprocessor and into the memory or peripherals. • When the 8085 wants to access a peripheral or a memory location, it places the 16-bit address on the address bus and then sends the appropriate control signals.
THE DATA BUS • 8 bits wide (D0 D1…D7) • “Bi-directional”. • Information flows both ways between the microprocessor and memory or I/O. • The 8085 uses the data bus to transfer the binary information. • Since the data bus has 8-bits only, then the 8085 can manipulate data 8 bits at-a-time only.
THE CONTROL BUS • There is no real control bus. Instead, the control bus is made up of a number of single bit control signals. • (Refer to Figure 3.1: 8085 Bus Structure)
OPERATION TYPES IN A MICROPROCESSOR • All of the operations of the microprocessor can be classified into one of three types: - Microprocessor-Initiated Operations - Internal Operations - Peripheral-Initiated Operations
MICROPROCESSOR-INITIATED OPERATIONS • These are operations that the microprocessor itself starts. • These are usually one of 4 operations: • Memory Read • Memory Write • I/O Read (Get data from an input device) • I/O write (Send data to an output device)
MICROPROCESSOR-INITIATED OPERATIONS • It is important to note that the microprocessor treats memory and I/O devices the same way. • Input and output devices simply look like memory locations to the microprocessor. • For example, the keyboard may look like memory address A3F2H. To get what key is being pressed, the microprocessor simply reads the data at location A3F2H. • The communication process between the microprocessor and peripheral devices consist of the following three steps: • Identify the address. • Transfer the binary information. • Provide the right timing signals.
THE READ OPERATION • To read the contents of a memory location, the following steps take place: • The microprocessor places the 16-bit address of the memory location on the address bus. • The microprocessor activates a control signal called “memory read” which enables the memory chip. • The memory decodes the address and identifies the right location. • The memory places the contents on the data bus. • The microprocessor reads the value of the data bus after a certain amount of time.
INTERNAL DATA OPERATIONS • The 8085 can perform a number of internal operations. Such as: storing data, Arithmetic & Logic operations, Testing for condition, etc. • To perform these operations, the microprocessor needs an internal architecture similar to the following: Accumulator Flags B C D E H L Program Counter Stack Pointer DataAddress 816
• Registers • Six general purpose registers used to store 8-bit data during a program execution • The registers are identified as B, C, D, E, H, and L • They can be combined as register pairs: BC, DE, and HL to perform 16-bit operations • H & L can be used as a data pointer (holds memory address) • Accumulator • 8-bit register that is part of the ALU • Used to store 8-bit data and in performing 8-bit arithmetic and logical operations, and in storing the results of operations
THE INTERNAL ARCHITECTURE • Flags • 8 bit register –shows the status of the microprocessor before/after an operation • S (sign flag), Z (zero flag), AC (auxiliary carry flag), P (parity flag) & CY (carry flag) • The bits of flag register are affected by the arithmetic & logic operations. • S-sign flag • The sign flag is set if bit D7 of the accumulator is set after an arithmetic or logic operation. D7 D6 D5 D4 D3 D2 D1 D0 S Z XX AC XX P XX CY
THE INTERNAL ARCHITECTURE • Z-zero flag • Set if the result of the ALU operation is 0. Otherwise is reset. This flag is affected by operations on the accumulator as well as other registers. (DCR B). • AC-Auxiliary Carry • This flag is set when a carry is generated from bit D3 and passed to D4 . This flag is used only internally for BCD operations. • P-Parity flag • After an ALU operation, if the result has an even # of 1s, the p-flag is set. Otherwise it is cleared. So, the flag can be used to indicate even parity. • CY-carry flag • This flag is set when a carry is generated from bit D7 after an unsigned operation.
THE INTERNAL ARCHITECTURE • The ProgramCounter(PC) • This is a register that is used to control the sequencing of the execution of instructions. • This register always holds the address of the next instruction. • Since it holds an address, it must be 16 bits wide. • Stackpointer • The stack pointer is also a 16-bit register that is used to point into memory. • The memory this register points to is a special area called the stack. • The stack is an area of memory used to hold data that will be retreived soon.
EXTERNALLY-INITIATED OPERATIONS • External devices can initiate (start) one of the 4 following operations: • Reset • All operations are stopped and the program counter is reset to 0000. • Interrupt • The microprocessor’s operations are interrupted and the microprocessor executes what is called a “service routine”. • This routine “handles” the interrupt, (perform the necessary operations). Then the microprocessor returns to its previous operations and continues.
EXTERNALLY-INITIATED OPERATIONS • Ready • The 8085 has a pin called RDY. This pin is used by external devices to stop the 8085 until they catch up. • As long as the RDY pin is low, the 8085 will be in a wait state. • Hold • The 8085 has a pin called HOLD. This pin is used by external devices to gain control of the busses. • When the HOLD signal is activated by an external device, the 8085 stops executing instructions and stops using the busses. • This would allow external devices to control the information on the busses. Example DMA.
THE DESIGN AND OPERATION OF MEMORY • Memory in a microprocessor system is where information (data and instructions) is kept. It can be classified into two main types: • Main memory (RAM and ROM) • Storage memory (Disks , CD ROMs, etc.) • The simple view of RAM is that it is made up of registers that are made up of flip-flops (or memory elements). • The number of flip-flops in a “memory register” determines the size of the memory word. • ROM on the other hand uses diodes instead of the flip-flops to permanently hold the information.
ACCESSING INFORMATION IN MEMORY • For the microprocessor to access (Read or Write) information in memory (RAM or ROM), it needs to do the following: • Select the right memory chip (using part of the address bus). • Identify the memory location (using the rest of the address bus). • Access the data (using the data bus).
TRI-STATE BUFFERS • An important circuit element that is used extensively in memory. • This buffer is a logic circuit that has three states: • Logic 0, logic1, and high impedance. • When this circuit is in high impedance mode it looks as if it is disconnected from the output completely. The Output is Low The Output is High High Impedance
THE TRI-STATE BUFFER • This circuit has two inputs and one output. • The first input behaves like the normal input for the circuit. • The second input is an “enable”. • If it is set high, the output follows the proper circuit behavior. • If it is set low, the output looks like a wire connected to nothing. Input Output Enable Input Output Enable OR
THE BASIC MEMORY ELEMENT • The basic memory element is similar to a D latch. • This latch has an input where the data comes in. It has an enable input and an output on which data comes out. QD EN Data Input Data Output Enable
THE BASIC MEMORY ELEMENT • However, this is not safe. • Data is always present on the input and the output is always set to the contents of the latch. • To avoid this, tri-state buffers are added at the input and output of the latch. QD EN Data Input Data Output Enable WR RD
THE BASIC MEMORY ELEMENT • The WR signal controls the input buffer. • The bar over WR means that this is an active low signal. • So, if WR is 0 the input data reaches the latch input. • If WR is 1 the input of the latch looks like a wire connected to nothing. • The RD signal controls the output in a similar manner.
A MEMORY “REGISTER” • If we take four of these latches and connect them together, we would have a 4-bit memory register WR RD EN Q D EN Q D EN Q D EN Q D EN I0 I1 I2 I3 O0 O1 O2 O3
A GROUP OF MEMORY REGISTERS • Expanding on this scheme to add more memory registers we get the diagram to the right. D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D0 D1 D2 D3 D0 D1 D2 D3 oooo oooo WR RD
A GROUP OF MEMORY REGISTERS • If we represent each memory location (Register) as a block we get the following Input Buffers Output Buffers Memory Reg. 0 Memory Reg. 1 Memory Reg. 2 Memory Reg. 3 I0 I1 I2 I3 O0 O1 O2 O3 WR EN0 EN1 EN2 EN3 RD
THE DESIGN OF A MEMORY CHIP • Using the RD and WR controls we can determine the direction of flow either into or out of memory. Then using the appropriate Enable input we enable an individual memory register. • What we have just designed is a memory with 4 locations and each location has 4 elements (bits). This memory would be called 4 X 4 [Number of location X number of bits per location].
THE ENABLE INPUTS • How do we produce these enable line? • Since we can never have more than one of these enables active at the same time, we can have them encoded to reduce the number of lines coming into the chip. • These encoded lines are the address lines for memory.
THE DESIGN OF A MEMORY CHIP • So, the previous diagram would now look like the following: Input Buffers Output Buffers Memory Reg. 0 Memory Reg. 1 Memory Reg. 2 Memory Reg. 3 I0 I1 I2 I3 O0 O1 O2 O3 WR RD A d d r e s s D e c o d e r A1 A0
THE DESIGN OF A MEMORY CHIP • Since we have tri-state buffers on both the inputs and outputs of the flip flops, we can actually use one set of pins only. • The chip would now look like this: Input Buffers Output Buffers Memory Reg. 0 Memory Reg. 1 Memory Reg. 2 Memory Reg. 3 WR RD A d d r e s s D e c o d e r A1 A0 D0 D1 D2 D3 D0 D1 D2 D3 A1 A0 RD WR
THE STEPS OF WRITING INTO MEMORY • What happens when the programmer issues the STA instruction? • The microprocessor would turn on the WR control (WR = 0) and turn off the RD control (RD = 1). • The address is applied to the address decoder which generates a single Enable signal to turn on only one of the memory registers. • The data is then applied on the data lines and it is stored into the enabled register.
DIMENSIONS OF MEMORY • Memory is usually measured by two numbers: its length and its width (Length X Width). • The length is the total number of locations. • The width is the number of bits in each location. • The length (total number of locations) is a function of the number of address lines. # of memory locations = 2( # of address lines) • So, a memory chip with 10 address lines would have 210 = 1024 locations (1K) • Looking at it from the other side, a memory chip with 4K locations would need Log2 4096=12 address lines
THE 8085 AND MEMORY • The 8085 has 16 address lines. That means it can address 216 = 64K memory locations. • Then it will need 1 memory chip with 64 k locations, or 2 chips with 32 K in each, or 4 with 16 K each or 16 of the 4 K chips, etc. • how would we use these address lines to control the multiple chips?
CHIP SELECT • Usually, each memory chip has a CS (Chip Select) input. The chip will only work if an active signal is applied on that input. • To allow the use of multiple chips in the make up of memory, we need to use a number of the address lines for the purpose of “chip selection”. • These address lines are decoded to generate the 2n necessary CS inputs for the memory chips to be used.
CHIP SELECTION EXAMPLE • Assume that we need to build a memory system made up of 4 of the 4 X 4 memory chips we designed earlier. • We will need to use 2 inputs and a decoder to identify which chip will be used at what time. • The resulting design would now look like the one on the following slide.
CHIP SELECTION EXAMPLE CS RD WR A0 A1 CS RD WR A0 A1 CS RD WR A0 A1 CS RD WR A0 A1 2 X4 DecoderA3 A2 A1 A0 RD WR D1 D0
MEMORY MAP AND ADDRESSES • The memory map is a picture representation of the address range and shows where the different memory chips are located within the address range.0000 FFFF AddressRange RAM 1 RAM 2 RAM 3 RAM 4 EPROM 0000 3FFF 4400 5FFF 6000 8FFF 9000 A3FF A400 F7FF Address Range of EPROM Chip Address Range of 1st RAM Chip Address Range of 2nd RAM Chip Address Range of 3rd RAM Chip Address Range of 4th RAM Chip
ADDRESS RANGE OF A MEMORY CHIP • The address range of a particular chip is the list of all addresses that are mapped to the chip. • An example for the address range and its relationship to the memory chips would be the Post Office Boxes in the post office. • Each box has its unique number that is assigned sequentially. (memory locations) • The boxes are grouped into groups. (memory chips) • The first box in a group has the number immediately after the last box in the previous group.
ADDRESS RANGE OF A MEMORY CHIP • The above example can be modified slightly to make it closer to our discussion on memory. • Let’s say that this post office has only 1000 boxes. • Let’s also say that these are grouped into 10 groups of 100 boxes each. Boxes 0000 to 0099 are in group 0, boxes 0100 to 0199 are in group 1 and so on. • We can look at the box number as if it is made up of two pieces: • The group number and the box’s index within the group. • So, box number 436 is the 36th box in the 4th group. The upper digit of the box number identifies the group and the lower two digits identify the box within the group.
THE 8085 AND ADDRESS RANGES • The 8085 has 16 address lines. So, it can address a total of 64K memory locations. • If we use memory chips with 1K locations each, then we will need 64 such chips. • The 1K memory chip needs 10 address lines to uniquely identify the 1K locations. (log2 1024 = 10) • That leaves 6 address lines which is the exact number needed for selecting between the 64 different chips (log2 64 = 6).
THE 8085 AND ADDRESS RANGES • Now, we can break up the 16-bit address of the 8085 into two pieces: A15 A14 A13 A12 A11 A10 A9 A8 A7 A6 A5 A4 A3 A2 A1 A0 • Depending on the combination on the address lines A15 - A10 , the address range of the specified chip is determined. Location Selection within the ChipChip Selection
CHIP SELECT EXAMPLE • A chip that uses the combination A15 - A10 = 001000 would have addresses that range from 2000H to 23FFH. • Keep in mind that the 10 address lines on the chip gives a range of 00 0000 0000 to 11 1111 1111 or 000H to 3FFH for each of the chips. • The memory chip in this example would require the following circuit on its chip select input: CS A10 A11 A12 A13 A14 A15
CHIP SELECT EXAMPLE • If we change the above combination to the following: • Now the chip would have addresses ranging from: 2400 to 27FF. • Changing the combination of the address bits connected to the chip select changes the address range for the memory chip. CS A10 A11 A12 A13 A14 A15
CHIP SELECT EXAMPLE • To illustrate this with a picture: • in the first case, the memory chip occupies the piece of the memory map identified as before. • In the second case, it occupies the piece identified as after. 0000 2000 23FF FFFF 0000 2400 27FF FFFF Before After
HIGH-ORDER VS. LOW-ORDER ADDRESS LINES • The address lines from a microprocessor can be classified into two types: • High-Order • Used for memory chip selection • Low-Order • Used for location selection within a memory chip. • This classification is highly dependent on the memory system design.
DATA LINES • All of the above discussion has been regarding memory length. Lets look at memory width. • We said that the width is the number of bits in each memory word. • We have been assuming so far that our memory chips have the right width. • What if they don’t? • It is very common to find memory chips that have only 4 bits per location. How would you design a byte wide memory system using these chips? • We use two chips for the same address range. One chip will supply 4 of the data bits per address and the other chip supply the other 4 data bits for the same address.
DATA LINES CS A0 … A9 CS CS D0 … D3 D4 … D7

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8085 Microprocessor Architecture

  • 1.
  • 2.
    MICROPROCESSOR ARCHITECTURE • The microprocessorcan be programmed to perform functions on given data by writing specific instructions into its memory. • The microprocessor reads one instruction at a time, matches it with its instruction set, and performs the data manipulation specified. • The result is either stored back into memory or displayed on an output device.
  • 3.
    THE 8085 ARCHITECTURE •The 8085 uses three separate busses to perform its operations • The address bus. • The data bus. • The control bus.
  • 4.
    THE ADDRESS BUS •16 bits wide (A0 A1…A15) • Therefore, the 8085 can access locations with numbers from 0 to 65,536. Or, the 8085 can access a total of 64K addresses. • “Unidirectional”. • Information flows out of the microprocessor and into the memory or peripherals. • When the 8085 wants to access a peripheral or a memory location, it places the 16-bit address on the address bus and then sends the appropriate control signals.
  • 5.
    THE DATA BUS •8 bits wide (D0 D1…D7) • “Bi-directional”. • Information flows both ways between the microprocessor and memory or I/O. • The 8085 uses the data bus to transfer the binary information. • Since the data bus has 8-bits only, then the 8085 can manipulate data 8 bits at-a-time only.
  • 6.
    THE CONTROL BUS •There is no real control bus. Instead, the control bus is made up of a number of single bit control signals. • (Refer to Figure 3.1: 8085 Bus Structure)
  • 7.
    OPERATION TYPES INA MICROPROCESSOR • All of the operations of the microprocessor can be classified into one of three types: - Microprocessor-Initiated Operations - Internal Operations - Peripheral-Initiated Operations
  • 8.
    MICROPROCESSOR-INITIATED OPERATIONS • These areoperations that the microprocessor itself starts. • These are usually one of 4 operations: • Memory Read • Memory Write • I/O Read (Get data from an input device) • I/O write (Send data to an output device)
  • 9.
    MICROPROCESSOR-INITIATED OPERATIONS • It isimportant to note that the microprocessor treats memory and I/O devices the same way. • Input and output devices simply look like memory locations to the microprocessor. • For example, the keyboard may look like memory address A3F2H. To get what key is being pressed, the microprocessor simply reads the data at location A3F2H. • The communication process between the microprocessor and peripheral devices consist of the following three steps: • Identify the address. • Transfer the binary information. • Provide the right timing signals.
  • 10.
    THE READ OPERATION •To read the contents of a memory location, the following steps take place: • The microprocessor places the 16-bit address of the memory location on the address bus. • The microprocessor activates a control signal called “memory read” which enables the memory chip. • The memory decodes the address and identifies the right location. • The memory places the contents on the data bus. • The microprocessor reads the value of the data bus after a certain amount of time.
  • 11.
    INTERNAL DATA OPERATIONS •The 8085 can perform a number of internal operations. Such as: storing data, Arithmetic & Logic operations, Testing for condition, etc. • To perform these operations, the microprocessor needs an internal architecture similar to the following: Accumulator Flags B C D E H L Program Counter Stack Pointer DataAddress 816
  • 12.
    • Registers • Sixgeneral purpose registers used to store 8-bit data during a program execution • The registers are identified as B, C, D, E, H, and L • They can be combined as register pairs: BC, DE, and HL to perform 16-bit operations • H & L can be used as a data pointer (holds memory address) • Accumulator • 8-bit register that is part of the ALU • Used to store 8-bit data and in performing 8-bit arithmetic and logical operations, and in storing the results of operations
  • 13.
    THE INTERNAL ARCHITECTURE •Flags • 8 bit register –shows the status of the microprocessor before/after an operation • S (sign flag), Z (zero flag), AC (auxiliary carry flag), P (parity flag) & CY (carry flag) • The bits of flag register are affected by the arithmetic & logic operations. • S-sign flag • The sign flag is set if bit D7 of the accumulator is set after an arithmetic or logic operation. D7 D6 D5 D4 D3 D2 D1 D0 S Z XX AC XX P XX CY
  • 14.
    THE INTERNAL ARCHITECTURE •Z-zero flag • Set if the result of the ALU operation is 0. Otherwise is reset. This flag is affected by operations on the accumulator as well as other registers. (DCR B). • AC-Auxiliary Carry • This flag is set when a carry is generated from bit D3 and passed to D4 . This flag is used only internally for BCD operations. • P-Parity flag • After an ALU operation, if the result has an even # of 1s, the p-flag is set. Otherwise it is cleared. So, the flag can be used to indicate even parity. • CY-carry flag • This flag is set when a carry is generated from bit D7 after an unsigned operation.
  • 15.
    THE INTERNAL ARCHITECTURE •The ProgramCounter(PC) • This is a register that is used to control the sequencing of the execution of instructions. • This register always holds the address of the next instruction. • Since it holds an address, it must be 16 bits wide. • Stackpointer • The stack pointer is also a 16-bit register that is used to point into memory. • The memory this register points to is a special area called the stack. • The stack is an area of memory used to hold data that will be retreived soon.
  • 16.
    EXTERNALLY-INITIATED OPERATIONS • External devicescan initiate (start) one of the 4 following operations: • Reset • All operations are stopped and the program counter is reset to 0000. • Interrupt • The microprocessor’s operations are interrupted and the microprocessor executes what is called a “service routine”. • This routine “handles” the interrupt, (perform the necessary operations). Then the microprocessor returns to its previous operations and continues.
  • 17.
    EXTERNALLY-INITIATED OPERATIONS • Ready • The8085 has a pin called RDY. This pin is used by external devices to stop the 8085 until they catch up. • As long as the RDY pin is low, the 8085 will be in a wait state. • Hold • The 8085 has a pin called HOLD. This pin is used by external devices to gain control of the busses. • When the HOLD signal is activated by an external device, the 8085 stops executing instructions and stops using the busses. • This would allow external devices to control the information on the busses. Example DMA.
  • 18.
    THE DESIGN ANDOPERATION OF MEMORY • Memory in a microprocessor system is where information (data and instructions) is kept. It can be classified into two main types: • Main memory (RAM and ROM) • Storage memory (Disks , CD ROMs, etc.) • The simple view of RAM is that it is made up of registers that are made up of flip-flops (or memory elements). • The number of flip-flops in a “memory register” determines the size of the memory word. • ROM on the other hand uses diodes instead of the flip-flops to permanently hold the information.
  • 19.
    ACCESSING INFORMATION IN MEMORY •For the microprocessor to access (Read or Write) information in memory (RAM or ROM), it needs to do the following: • Select the right memory chip (using part of the address bus). • Identify the memory location (using the rest of the address bus). • Access the data (using the data bus).
  • 20.
    TRI-STATE BUFFERS • Animportant circuit element that is used extensively in memory. • This buffer is a logic circuit that has three states: • Logic 0, logic1, and high impedance. • When this circuit is in high impedance mode it looks as if it is disconnected from the output completely. The Output is Low The Output is High High Impedance
  • 21.
    THE TRI-STATE BUFFER •This circuit has two inputs and one output. • The first input behaves like the normal input for the circuit. • The second input is an “enable”. • If it is set high, the output follows the proper circuit behavior. • If it is set low, the output looks like a wire connected to nothing. Input Output Enable Input Output Enable OR
  • 22.
    THE BASIC MEMORYELEMENT • The basic memory element is similar to a D latch. • This latch has an input where the data comes in. It has an enable input and an output on which data comes out. QD EN Data Input Data Output Enable
  • 23.
    THE BASIC MEMORYELEMENT • However, this is not safe. • Data is always present on the input and the output is always set to the contents of the latch. • To avoid this, tri-state buffers are added at the input and output of the latch. QD EN Data Input Data Output Enable WR RD
  • 24.
    THE BASIC MEMORYELEMENT • The WR signal controls the input buffer. • The bar over WR means that this is an active low signal. • So, if WR is 0 the input data reaches the latch input. • If WR is 1 the input of the latch looks like a wire connected to nothing. • The RD signal controls the output in a similar manner.
  • 25.
    A MEMORY “REGISTER” •If we take four of these latches and connect them together, we would have a 4-bit memory register WR RD EN Q D EN Q D EN Q D EN Q D EN I0 I1 I2 I3 O0 O1 O2 O3
  • 26.
    A GROUP OFMEMORY REGISTERS • Expanding on this scheme to add more memory registers we get the diagram to the right. D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D EN Q D0 D1 D2 D3 D0 D1 D2 D3 oooo oooo WR RD
  • 27.
    A GROUP OFMEMORY REGISTERS • If we represent each memory location (Register) as a block we get the following Input Buffers Output Buffers Memory Reg. 0 Memory Reg. 1 Memory Reg. 2 Memory Reg. 3 I0 I1 I2 I3 O0 O1 O2 O3 WR EN0 EN1 EN2 EN3 RD
  • 28.
    THE DESIGN OFA MEMORY CHIP • Using the RD and WR controls we can determine the direction of flow either into or out of memory. Then using the appropriate Enable input we enable an individual memory register. • What we have just designed is a memory with 4 locations and each location has 4 elements (bits). This memory would be called 4 X 4 [Number of location X number of bits per location].
  • 29.
    THE ENABLE INPUTS •How do we produce these enable line? • Since we can never have more than one of these enables active at the same time, we can have them encoded to reduce the number of lines coming into the chip. • These encoded lines are the address lines for memory.
  • 30.
    THE DESIGN OFA MEMORY CHIP • So, the previous diagram would now look like the following: Input Buffers Output Buffers Memory Reg. 0 Memory Reg. 1 Memory Reg. 2 Memory Reg. 3 I0 I1 I2 I3 O0 O1 O2 O3 WR RD A d d r e s s D e c o d e r A1 A0
  • 31.
    THE DESIGN OFA MEMORY CHIP • Since we have tri-state buffers on both the inputs and outputs of the flip flops, we can actually use one set of pins only. • The chip would now look like this: Input Buffers Output Buffers Memory Reg. 0 Memory Reg. 1 Memory Reg. 2 Memory Reg. 3 WR RD A d d r e s s D e c o d e r A1 A0 D0 D1 D2 D3 D0 D1 D2 D3 A1 A0 RD WR
  • 32.
    THE STEPS OFWRITING INTO MEMORY • What happens when the programmer issues the STA instruction? • The microprocessor would turn on the WR control (WR = 0) and turn off the RD control (RD = 1). • The address is applied to the address decoder which generates a single Enable signal to turn on only one of the memory registers. • The data is then applied on the data lines and it is stored into the enabled register.
  • 33.
    DIMENSIONS OF MEMORY •Memory is usually measured by two numbers: its length and its width (Length X Width). • The length is the total number of locations. • The width is the number of bits in each location. • The length (total number of locations) is a function of the number of address lines. # of memory locations = 2( # of address lines) • So, a memory chip with 10 address lines would have 210 = 1024 locations (1K) • Looking at it from the other side, a memory chip with 4K locations would need Log2 4096=12 address lines
  • 34.
    THE 8085 ANDMEMORY • The 8085 has 16 address lines. That means it can address 216 = 64K memory locations. • Then it will need 1 memory chip with 64 k locations, or 2 chips with 32 K in each, or 4 with 16 K each or 16 of the 4 K chips, etc. • how would we use these address lines to control the multiple chips?
  • 35.
    CHIP SELECT • Usually,each memory chip has a CS (Chip Select) input. The chip will only work if an active signal is applied on that input. • To allow the use of multiple chips in the make up of memory, we need to use a number of the address lines for the purpose of “chip selection”. • These address lines are decoded to generate the 2n necessary CS inputs for the memory chips to be used.
  • 36.
    CHIP SELECTION EXAMPLE •Assume that we need to build a memory system made up of 4 of the 4 X 4 memory chips we designed earlier. • We will need to use 2 inputs and a decoder to identify which chip will be used at what time. • The resulting design would now look like the one on the following slide.
  • 37.
    CHIP SELECTION EXAMPLE CS RDWR A0 A1 CS RD WR A0 A1 CS RD WR A0 A1 CS RD WR A0 A1 2 X4 DecoderA3 A2 A1 A0 RD WR D1 D0
  • 38.
    MEMORY MAP ANDADDRESSES • The memory map is a picture representation of the address range and shows where the different memory chips are located within the address range.0000 FFFF AddressRange RAM 1 RAM 2 RAM 3 RAM 4 EPROM 0000 3FFF 4400 5FFF 6000 8FFF 9000 A3FF A400 F7FF Address Range of EPROM Chip Address Range of 1st RAM Chip Address Range of 2nd RAM Chip Address Range of 3rd RAM Chip Address Range of 4th RAM Chip
  • 39.
    ADDRESS RANGE OFA MEMORY CHIP • The address range of a particular chip is the list of all addresses that are mapped to the chip. • An example for the address range and its relationship to the memory chips would be the Post Office Boxes in the post office. • Each box has its unique number that is assigned sequentially. (memory locations) • The boxes are grouped into groups. (memory chips) • The first box in a group has the number immediately after the last box in the previous group.
  • 40.
    ADDRESS RANGE OFA MEMORY CHIP • The above example can be modified slightly to make it closer to our discussion on memory. • Let’s say that this post office has only 1000 boxes. • Let’s also say that these are grouped into 10 groups of 100 boxes each. Boxes 0000 to 0099 are in group 0, boxes 0100 to 0199 are in group 1 and so on. • We can look at the box number as if it is made up of two pieces: • The group number and the box’s index within the group. • So, box number 436 is the 36th box in the 4th group. The upper digit of the box number identifies the group and the lower two digits identify the box within the group.
  • 41.
    THE 8085 ANDADDRESS RANGES • The 8085 has 16 address lines. So, it can address a total of 64K memory locations. • If we use memory chips with 1K locations each, then we will need 64 such chips. • The 1K memory chip needs 10 address lines to uniquely identify the 1K locations. (log2 1024 = 10) • That leaves 6 address lines which is the exact number needed for selecting between the 64 different chips (log2 64 = 6).
  • 42.
    THE 8085 ANDADDRESS RANGES • Now, we can break up the 16-bit address of the 8085 into two pieces: A15 A14 A13 A12 A11 A10 A9 A8 A7 A6 A5 A4 A3 A2 A1 A0 • Depending on the combination on the address lines A15 - A10 , the address range of the specified chip is determined. Location Selection within the ChipChip Selection
  • 43.
    CHIP SELECT EXAMPLE •A chip that uses the combination A15 - A10 = 001000 would have addresses that range from 2000H to 23FFH. • Keep in mind that the 10 address lines on the chip gives a range of 00 0000 0000 to 11 1111 1111 or 000H to 3FFH for each of the chips. • The memory chip in this example would require the following circuit on its chip select input: CS A10 A11 A12 A13 A14 A15
  • 44.
    CHIP SELECT EXAMPLE •If we change the above combination to the following: • Now the chip would have addresses ranging from: 2400 to 27FF. • Changing the combination of the address bits connected to the chip select changes the address range for the memory chip. CS A10 A11 A12 A13 A14 A15
  • 45.
    CHIP SELECT EXAMPLE •To illustrate this with a picture: • in the first case, the memory chip occupies the piece of the memory map identified as before. • In the second case, it occupies the piece identified as after. 0000 2000 23FF FFFF 0000 2400 27FF FFFF Before After
  • 46.
    HIGH-ORDER VS. LOW-ORDER ADDRESSLINES • The address lines from a microprocessor can be classified into two types: • High-Order • Used for memory chip selection • Low-Order • Used for location selection within a memory chip. • This classification is highly dependent on the memory system design.
  • 47.
    DATA LINES • Allof the above discussion has been regarding memory length. Lets look at memory width. • We said that the width is the number of bits in each memory word. • We have been assuming so far that our memory chips have the right width. • What if they don’t? • It is very common to find memory chips that have only 4 bits per location. How would you design a byte wide memory system using these chips? • We use two chips for the same address range. One chip will supply 4 of the data bits per address and the other chip supply the other 4 data bits for the same address.
  • 48.