Khaled A. Al-Utaibi
alutaibi@uoh.edu.sa
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Buffering and Latching
8086 Bus System
−Bus Timing
−Bus Write Cycle
−Read Cycle
−Ready and Wait States
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8086 Bus System Design
−Objective
−Requirements
−Procedure
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Before starting with 8086 bus system design, one
needs to understand the following concepts:
−(1) Fan-out limitations of logic gates
−(2) Buffering
−(3) Latching
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If the output of some gate A
is connected to the input of
another gate B, gate A is said
to be driving gate, while gate
B is said to be the load gate.
The fan-out of a gate is the
largest number of gate
inputs this gate can drive.
The fan-out for TTL is 10
loads.
The fan-out for CMOS is 8
loads
A
Driver
Gate
B
Load
Gate
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A buffer is a single input single output gate
that provides the necessary drive capability
which allows driving larger loads.
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The outputs of 2 logic gates can
not be directly connected
because this causes a short
circuit that results in huge
current flow may result in
damaging the circuit
The solution to this problem is
to use tri-state buffers.
A tri-state buffer is a special
gate with 3 states controlled by
an enable input (E) :
 Logic 0 (Low Voltage),
 Logic 1 (High Voltage), and
 Hi-Impedance (Open Circuit)
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74LS245 3-State Bidirectional Octal Buffers:
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74LS244 3-State Unidirectional Octal Buffers:
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The D-Latch is a memory device which can store
one bit of data for as long as the device is
powered.
The D-Latch has one data input (D), one control
input (C) and one data output (Q/Q’).
−When C = 0, the output Q keeps its value (No Change)
−When C = 1, the output Q follows the input D (Q = D)
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74LS374 3-State Octal D-Type Transparent
Latches :
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The 8086 has three buses
−(1) Address Bus: provides memory and I/O with the
memory address or the I/O port number
−(2) Data Bus: transfers data between the
microprocessor and the memory and I/O
−(3) Control Bus: provides control signals to the memory
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8086 access memory and I/O devices in periods
called bus cycles.
Each cycle equals 4 system-clocking periods (T
states).
If the clock is operated at 5 MHz, one 8086 bus
cycle is complete in 800 ns.
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The READY input causes wait states for slower memory and I/O
components.
A wait state (Tw) is an extra clocking period inserted between
T2 and T3.
The READY input is sampled (checked) at the end of T2 and in
the middle of Tw.
If READY is a logic 0 at the end of T2, then T3 is delayed and
Tw is inserted between T2 and T3
READY is next sampled at the middle of Tw to determine if the
next state is Tw or T3.
The READY signal is synchronized with clock using the 8284A
clock generator.
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Objective: to provide the system the three main
buses required to access memory and I/O devices
−(1) Address Bus
−(2) Data Bus
−(3) Control Bus
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Requirements: bus design must satisfy two main
requirements
−(1) Bus Buffering: the system must be buffered if the
number of devices interfaced to it is more than its fan-out
−(2) Bus Demultiplexing (Latching): All time multiplexed lines
of the processor must be first demultiplexed (latched) to
split address, data, status and control lines before interfacing
them to memory and I/O devices.
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Designing a bus system involves the following
steps:
−(1) Identify input, output and input/output pins of the
processor. Note that only output and input/output
pins are considered for designing the bus system (i.e.
input pins are not part of the bus system).
−(2) Identify time multiplexed pins of the processor.
These pins need to be demultiplexed.
−(3) Use latches to demultiplex time multiplexed pins.
−(4) Use buffers to buffer all data, address and control
lines to be connected to memory and I/O devices.
 Latched lines are already buffered.
 Input/output lines require bidirectional buffers
 Output lines require unidirectional buffers
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Design a fully buffered and de-multiplexed bus
system for a minimum mode 8086-based
microcomputer system.
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Inputs: these are not part of
bus system design
−VCC
−GND
−NMI
−INTR
−CLK
−RESET
−READY
−TEST’
−HOLD
−MN/MX’
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Outputs:
−A16/S3
−A17/S4
−A18/S5
−A19/S6
−RD’
−HOLDA (not required, why?)
−WR’
−M/IO’
−DT/R’
−DEN’
−ALE
−INTA’ (not required, why?)
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Inputs/Output:
−AD0-AD15
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Time Multiplexed Pins:
−AD0-AD15
−A16/S3-A19/S6
−BHE’/S7
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As shown in Step(2), the 8086 has 21 multiplexed
lines:
−AD0-AD15
−A16/S3-A19/S6
−BHE’/S7
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To latch (demultiplex) these lines using 74LS373
octal latches, we need 21/8 = 3 chips.
The input controls of the 74LS373 octal latches
should be connected as follows :
−Output Control (OE’) connected to GND (Why?)
−Latch Enable (G’) connected to ALE (Why?)
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The data lines D15-D0 should buffered using
bidirectional buffers (74LS245). Why?
−We have 16 data lines. So, we need 16/8 = 2 chips.
−The Buffer Enable (G’) should be connected to DEN’
(Why?)
−The Direction Control (DIR) should be connected to
DT/R’ (Why?)
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The Control Lines M/IO’, RD’ and WR’ should be
buffered using unidirectional buffers (74LS244).
Why?
−The Buffer Enable (G’) should be connected to GND
(Why?)
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Since address lines A19-A0 & BHE’ are latched
using 74LS373 octal latches, they are already
buffered.
The following control lines are will not be
buffered:
−DT/R’ & DEN’ because they are will not be connected
to memory or I/O devices.
−HOLDA because the direct memory access is not
supported by our system.
−INTA’ because interrupts are not supported by our
system.