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A more efficient way to multiplex registers to the ALU input

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• Previously discussed
  • We saw how to use multiplexors to switch the output of the registers to the input of the ALU:

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• Using tri-state-buffers to multiplex n outputs onto 1 input
  • Consider the problem of sending n output values to the same input.

    Suppose we just connect all the outputs to the input:

    This solution will not work because the outputs o0, o1, o2 and o3 are connected together (burn out).

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  • The next logical step is:

    Add a tri-state-buffer to each output (to make sure that exactly one output will be on at any one time).

    Result:

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  • In order to make sure that exactly one tri-state-buffer is turned on at any one time, we use a decoder to regular the settings of the tri-state-buffers:

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• Using tri-state-buffers to multiplex n register outputs onto the ALU input
  • Apply the above idea n times, and you have the following solution:

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  • Example Program: (Demo above code)                                                

    /home/cs355001/bin/cs355sim /home/cs355001/demo/circuits/reg-TSB-alu

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  • In this circuit demo, I have used 2 sets of tri-state-buffers to "multiplex" the values in the registers to both inputs of the ALU:

    /home/cs355001/bin/cs355sim /home/cs355001/demo/circuits/reg-TSB-alu2

    The "ALU" is simulated by a 4 bit adder for simplicity - you can see the addition (= sum) in the output of the circuit (at the bottom)

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• Why is this solution more efficient (less hardware)
  • Old solution:

    uses:

    n multiplexor to multiplex n register to 1 ALU input A multiplexor is a very complex ciruit.

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  • New solution:

    uses:

    1 decoder to multiplex n register to 1 ALU input A decoder is similar in complexity as a multiplexor. A tri-state-buffer is a very light weight circuit.