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Basically , Demultiplexing is breaking of multiplexed signal .Recall that A/D0 -A/D15 and A16/S3-A19/S6 are the multiplexed signals in 8086.To do so, three demultiplexing latches are used .ALE (Address Enable Latch) is used for strobe Demultiplexing.8086 is 16bit data lines and 20 bit address line microprocessor.BY the Demultiplexing ,we Get A0-A19 separate Address lines and D0-D15 Data lines .
Ajmal Shahbaz
Wiki User
∙ 12y ago
Wiki User
∙ 14y agoIn an 8085, the low order address and the data bus share the same pins. During the first clock cycle of a memory access, the data lines contain the low order address. External hardware is expected to latch that address on the trailing edge of ALE, Address Latch Enable. During the next clock cycle, the data lines either float (for read, S1=1) or change to data value (for write, S1=0).
Wiki User
∙ 14y agoTo demultiplex the data and addess lines on a 8085, you connect a latch to AD0-AD7 and strobe it with ALE. At the falling edge of ALE, AD0-AD7 is guaranteed to represent A0-A7. Following that, AD0-AD7 becomes D0-D7.
Wiki User
∙ 14y agoThe address and data lines in an 8085 are demultiplexed because they are multiplexed to start with. It is necessary to separate them as far as external (outside the CPU chip) logic is concerned.
Note that the address and data bus started separately. They are multiplexed in order to reduce the pin count on the chip. Whether multiplexing has any net value due to the added cost of demultiplexing is debateable, but that is the way it is.
Wiki User
∙ 14y agoIn the 8085, the data bus is shared with the low order address bus. During ALE, which is about the first half T state, starting about one sixth clock after the falling edge of clock, the data bus presents the low address bus. At the falling edge of ALE, about one half clock cycle later, external logic is expected to strobe that value. One half clock cycle later, during the second T state, the data bus becomes the data bus. This saves 7 pins on the chip. (8 pins, minus one for ALE)
Wiki User
∙ 12y agoThere are 16 address lines (8 shared by the data bus), and 8 data lines in the 8085.
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Why all the 16 address lines are not act as a data lines in 8085 microprocessor?
the 8085 microprocessor is a 8-bit microprocessor and these are bidirectional but the address lines are unidirectional.these address lines are used to address the location of the instruction in memory .these data lines are used to transfer data between processor and peripheral devices. when the address of the instruction will be recognized by the address lines the data will be send to the processor therefore the 16 address lines are not act as a data lines in 8085
How many data lines in 8086?
There are eight datalines, D0 through D7, in the 8085 microprocessor. They are shared, or multiplexed with the eight low order address lines, A0 through A7, and are called AD0 through AD7 on the pinout drawing.
Draw a schemetic to demultiplex bus ad0-ad7 using any octal latch in 8085 microprocessor?
The 8085 microprocessor is used IC 74LS373 to latch the address of 8085. Basically latch is consists of 8 flip flops. Generally we used D-flip flops (Delay).The clock of these flip flops are connected together and available as a output pin called enable.Working : The address will appear on AD0 AD7 lines. The ALE will go high and forcingEnable = 1. This will make latch enable and ready to work. Before address disappears ALE = 0. This will make latch disable. AD0 - AD7 will now be used as data bus.Hence, AD0 - AD7 (low order) address bus of the 8085 microprocessor is multiplexed (time-shared) with the data bus. The buses need to be demultiplexed.
How are the data bus and address bus are demultiplexed?
The data and address buses are multiplexed in order to save pin count on the chip. In the first clock cycle of a read or write cycle, the address is emitted on the address/data bus. The ALE signal is used to strobe the address, after which the address/data bus becomes the data bus. External logic is expected to strobe the address at the trailing edge of ALE. ALE is generated directly by the 8085, and by the 8086/8088 in minimum mode. In maximum mode in the 8086/8088, ALE is generated by the 8288 Bus Controller.
How many datalines are supported by Intel 8085?
8 data lines
Explain the need to demultiplex the bus AD7-AD0 in 8085 microprocessor?
The AD0-AD7 lines in an 8085 are multiplexed to reduce the pin count of the IC. Several added features were added to the 8085 from the 8080 design, and Intel did not want to require a larger package.
What is bus in 8085?
there are mainly 3 buses are there in 8085. They are: Address bus :-Used to carry address Data bus :- Used to carry data Control bus :-Used to carry signals such as control and timing signals
How is demultiplexing done in 8085 microprocessor?
it is nothing
Why does microprocessor 8086 has 16 data bus and 20 address bus why aren't they equal?
The 8086 has 16 data bus lines and 20 address bus lines because that is how Intel designed it. They wanted a processor that was more powerful than the 8085, which has an 8 bit data bus and a 16 bit data bus, so they increased both bus sizes accordingly.
How many Address line of 8085 microprocessor?
There are 20 address lines and 16 data lines in the 8086 microprocessor. The low order 16 address lines are multiplexed with the data lines. Some of the high order address lines are multiplexed with status lines.
Which is more usefull a multiplexer or a demultiplexer?
One is useful without the other. Data that is multiplexed at one end of a communication channel must be demultiplexed at the other end.One is useful without the other. Data that is multiplexed at one end of a communication channel must be demultiplexed at the other end.One is useful without the other. Data that is multiplexed at one end of a communication channel must be demultiplexed at the other end.One is useful without the other. Data that is multiplexed at one end of a communication channel must be demultiplexed at the other end.
Why are the data and address lines demultiplexed?
Address multiplexing allows you to use fewer pins on the processor, and thus fewer bus lines. So instead of having some bus lines for the address, and some more for the data, you put the address on the data line, it gets read, then you put the data on the same lines, and it gets read and stored at the previously read address. For the 8085, it allowed the design to add one pin, but cut 8, for a net gain (loss?) of 7 pins (reduced physical/manufacturing complexity at the cost of increased logical/programming complexity). Computers have different devices operating at different speeds. As such, there are often multiple devices competing for the bus at the same time. To allow transactions to occur in parallel instead of "taking a ticket", the system needs to be able to hold data when it becomes available but the bus is busy, until the bus gets freed. It holds that data in a buffer. I hope that was clear enough. If not, feel free to ask for clarification on anything you didn't understand
