IO AND MEMORY INTERFACE
ARCHITECTURE OF 8085
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Question
[CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
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For a common bus for eight registers of 16 bits each requires how many mux, size of mux?
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32 multiplexers, 16 input lines
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8 multiplexers, 8 input lines
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16 multiplexers, 32 input lines
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16 multiplexers, 8 input lines
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Explanation:
Detailed explanation-1: -If there are 8 registers of 16 bits each, then we need 16 multiplexers, each of which is an 8 : 1 multiplexer to create the 16 bit data bus.
Detailed explanation-2: -Multiplexer is a combinational circuit that has maximum of 2n data inputs, ‘n’ selection lines and single output line. One of these data inputs will be connected to the output based on the values of selection lines. Since there are ‘n’ selection lines, there will be 2n possible combinations of zeros and ones.
Detailed explanation-3: -Common bus transfers 8-bits at a time. Eight 4:1 Muxes are required to design this system, 4:1 MUX since there are 4 registers.
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