Controling the ALU

• This material is from section 4.4 and B.5 the first part
• Recall, control of the ALU consists of four control lines
  • 2 bit operation.
  • 1 bit carray in.
  • 1 bit A_invert
  • 1 bit B_invert
• From the data at the end of last class we can construct the following table:

  • Function	ALU Control Lines
    	A_invert	B_invert	Carry In	Operation
    AND	0	0	X	00
    OR	0	0	X	01
    Add	0	0	0	10
    Subtract	0	1	1	10
    Set Less Than	0	1	1	11
    NOR	1	1	X	00

  • An X means we don't care.
  • Note that B_invert == Carry In for every case where it matters.
  • So we could just use B_invert to drive Carry In.
  • 
  • And externally, the ALU would appear (source code)
• In 4.4 they begin to constuct a simple MIPS machine
  • This is the one we have been looking at.
  • In this exercise we look at
    • LW, SW (I type)
    • BEQ (I type)
    • ADD, SUB, AND, OR and SLT (R types)
  • There are other instructions but these will do for now.
• How will we control the ALU?
  • LW, SW and others need to compute the effective address, so the ALU must add.
  • BEQ needs to subtract and determine if the result is 0.
  • Others
    • ADD, funct: 20₁₆
    • SUB, funct: 22₁₆
    • AND, funct: 24₁₆
    • OR, funct: 25₁₆
    • SLT, funct: 2a₁₆
  • For the R type the funct code indicates what to do.
  • For the I type the Opcode must do this.
  • Note that for I type, we must either add (LW, SW) or subtract (BEQ)
• We would like to build a unit to control the ALU
  • The input would be
    • 2 bit indicator of the op code (ALUOp).
      • 00, the ALU should add (LW, SW, ...)
      • 01, the ALU should subtract (SLT, BEQ, ...)
      • 10, use funct to determine operation.
    • The 6 bit funct field.
      • Directly from the instruction.
  • The output would be the 4 bit control code for the alu.
• Consider the following truth table.

  Operation	ALUOp	FUNCT	Control
  I-type ADD	00	XXXXXX	0010
  I-type SUB	01	XXXXXX	0110
  ADD	10	100000	0010
  SUB	10	100010	0110
  AND	10	100100	0000
  NOR	10	100111	1100
  OR	10	100101	0001
  SLT	10	101010	0111

• Given this, we can compute the boolean expression for each of the control bits.

        C3 only occurs on one line:	 NOR
  
  	 C3 = A1A0F5F4F3F2F1F0
  
        C0 is a bit more complex
  	 Only when A1A0
  	 Two Funct codes     100101
  	                     101010
  
  	      In Both Cases F5F4
  	        then F3F2F1F0
  	         or  F3F2F1F0
  
             C0=A1A0F5F4(F3F2F1F0+ F3F2F1F0)
  
  
        Bits C1 and C2 are more complex 
           The funct does not matter for ALUOp of 00 or 01, which is selectd by A1
  	 The funct des matter with A1A0  so
  
  	    C1 =A1 + A1A0funct
  
              Three Operations remain:
  	         100000
  		 100010
  		 101010
  
  	        F5 is always on
  		F4, F2 and F0 are always off
  
  		What remains is (F3F1+ F3F1 + F3F1)
  		   This can be simplified to F3 + F3F1
  
  		Or F5F4F2F0(F3 + F3F1)
  
            C1= A1 + A1A0F5F4F2F0(F3 + F3F1)
  
            C2 
  	       Only when the ALUOP is 01 
  	       or the ALUOP is 10 and the function code is    100010
  	                                                      100111
  					                      101010
  	           F5F4F1
  		      ( F3F2F0
  		      + F3F2F0
  		      + F3F2F0)
          
            C2 = A1A0 + A1A0F5F4F1(F3+F3F2F0)
  	      
      

  • This is just a little complex, but
  • 
  • 
• And the code is here