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Q. Implementation of a Simple Arithmetic?
So, by now we have concerned how logic and arithmetic micro-operations can be applied individually. If we combine these 2 circuits along with shifting logic then we may have a possible simple structure of ALU. Basically ALU is a combinational circuit whose inputs are contents of specific registers. ALU performs desired micro-operation as decided by control signals on input and places results in an output or destination register. Whole operation of ALU can be performed in a single clock pulse as it's a combinational circuit. Shift operation can be performed in a separate unit however sometimes it can be made as a part of overall ALU. The following diagram gives a simple structure of one stage of an ALU.
One stage of ALU with shift capability
Please note that in this diagram we have given reference to two previous figures for arithmetic and logic circuits. This phase of ALU has two data inputs; the ith bits of registers to be manipulated. But the (i - 1)th or (i+1)th bit is also fed for case of shift micro-operation of only one register. There are 4 selection lines that determine what micro-operation (arithmetic, logic or shift) on the input. The Fi is resultant bit after desired micro-operation. Let's see how value of Fi changes on the foundation of four select inputs. This is displayed in Figure below:
Please note that in Figure below arithmetic micro-operations have both S3 and S2 bits as zero. Input Ci is significant for only arithmetic micro-operations. For logic micro-operations S3, S2 values are 01. Values 10 and 11 cause shift micro-operations.
For this shift micro-operation S1 and S0 values and Ci values don't play any role.
Figure: Micro-operations performed by a Sample ALU
i want to know complete detail of 8086 microprocessor such as memory segment ,interface with ram rom ect
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