Chapter 2 Microprocessor Bus Transfers. Big- and Little-Endian Ordering Bit-endian processor architecture –High-order-byte-first (H-O-B-F) map the highest-order.

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Presentation transcript:

Chapter 2 Microprocessor Bus Transfers

Big- and Little-Endian Ordering Bit-endian processor architecture –High-order-byte-first (H-O-B-F) map the highest-order byte of an internal register to the lowest memory byte-address The address of this data item is the address of its most significant byte EX: Motorola, Sparc RISC CPU

Little-endian processor architecture –Low-order-byte-first (L- O-B-F) Map the lowest-order byte of an internal register to the lowest memory byte-address The address of this data item is the address of its least significant byte EX: Intel 80x86 CPU

16- and 32-bit Microprocessors In generally, n-bit microprocessor is –N-bit internal register –N-bit ALU –N-bit external data bus –Exception: Intel 80386SX: 32-bit internal architecture and a 16-bit data bus –Smallest byte address of the addressed field

16-bit big-endian microprocessor 16-bit memory is connected to a 16-bit data bus It is made up of two byte sections, section 0 (even section, 2N) and section 1 (odd section, 2N+1) For byte-write operation, byte-enable signal may issue by the processor (EX: BHE#, section 0 can be triggered by the A0=0)

16-bit big-endian microprocessor The least significant byte (B0) is stored in the lowest memory location For the 16-bit word alignment operation, it only requires one bus cycle

32-bit big-engidn microprocessor

From 4N address to read a word data, it need multiplex in the CPU If read from 4N+2, it don’t need multiplex in the CPU

32-bit little-endian microprocessor

The figure can operate under both the big- and little- endian modes

2.2.5 operand and instruction alignment Operand (data) alignment: –For maximum performance, when data is stored in memory it must be aligned –2-byte  address A0 = 0  2N –4-byte  address A1, A0 = 0  4N –8-byte  address A2, A1, A0 = 0  8N Instruction alignment: –It is assumed that code is always loaded in memory in an aligned form

2.3.1 synchronous/asynchronous buses Synchronous bus operation –All events take place within a specified time period in synchronism with a system-wide clock –Both the processor master (the initiate the bus cycle) and memory or I/O slave (the respond to the request) are clocked by the same system clock. Address, data, and control signals are synchronous by the system clock –Clock skew: As distances increase, there are differences in clock arrival time at different points in the system

Synchronous design may yield a higher speed (don’t need to wake up; the CPU waits for device) and lower number of control lines (don’t need handshake control signals) –Imposes speed constraints on the external devices  since a bus cycle is made up of a number of clock cycles, its duration will be constrained by the speed of the slowest device connected to the bus –Synchronous bus implementation must be based on the worst case analysis Synchronous buses use a “wait protocol” to overcome this Device use “ready” input pin to tell that the addressed device did not have enough time to respond and therefore the bus master must insert a wait state

Asynchronous bus operation Bus master and bus slave have their own individual input clocks (operate at their own different speeds) Based on a handshake process –Ex: bus master issues a “strobe” signal to indicate the information on the bus line is “valid” –The addressed slave device will return a “data acknowledge” signal informing the master that it has responded to the request; i.e., it has either placed data on the data bus (read request) or latched the data off the data bus (write request)

2.3.2 Intel 80x8 bus transfers 30-bit address A31 – A2 Byte enable : BE1#, BE2#, BE3#, BE4# Little-Endian

80386 uses 4 bus clock cycle I/O mapped I/O Use M/IO# to indicate I/O or memory operation bus cycle begins with ADS# active, and end with “read” active