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ALU, Data-path & Control Unit

Short Notes

This topic covers the internal structure of the Processor.

Components

  • ALU (Arithmetic Logic Unit): Performs calculations and logical operations.
  • Datapath: The pathway (buses, registers) where data flows between CPU components.
  • Control Unit (CU): Directs the operation of the processor. It tells the datapath what to do.

Control Unit Types

  • Hardwired: Uses logic gates, flip-flops. Fixed, fast, but difficult to modify.
  • Microprogrammed: Stores control signals in a "Control Memory" (ROM) as microinstructions. Slower but flexible.

Key Theories & Formulas

1. Control Memory (ROM) Size

Size = (Number of microinstructions) \(\times\) (Length of microinstruction).

  • Horizontal Microprogramming: Many control bits (unencoded). High parallelism, large memory.
  • Vertical Microprogramming: Encoded bits (require decoders). Smaller memory, slower.

2. Control Signal Encoding

If a group of \(N\) control signals are mutually exclusive (never active together), they can be encoded using \(log_2(N+1)\) bits.

Example Problems

Problem: A processor has 64 distinct control signals. If they are all active together (worst case), how many bits are needed for horizontal microprogramming?

  • Result: 64 bits.
  • If they are grouped into 8 mutually exclusive groups of 7 signals each:
  • Each group needs \(log_2(7+1) = 3\) bits.
  • Total = \(8 \times 3 = 24\) bits. (Vertical/Encoded)

Hardest GATE Questions

Topic: Micro-instruction Sequencing Tricky Question (GATE 2005/2016): Calculate the size of control memory for a processor with 128 micro-instructions, each requiring 20 control signals and a 7-bit branch address. Control signals are grouped into 4 groups of size 5, 4, 6, and 5 respectively (mutually exclusive within groups).

  • Analysis:
  • Group 1 (5): \(log_2(5+1) \uparrow = 3\) bits.
  • Group 2 (4): \(log_2(4+1) \uparrow = 3\) bits.
  • Group 3 (6): \(log_2(6+1) \uparrow = 3\) bits.
  • Group 4 (5): \(log_2(5+1) \uparrow = 3\) bits.
  • Control Signal bits = \(3+3+3+3 = 12\) bits.
  • Branch bits = \(7\) bits.
  • Total bits per word = \(12 + 7 = 19\).
  • Control Memory Size = \(128 \times 19\) bits.
  • The "Trap": Forgetting the \(+1\) in \(log_2(N+1)\). The \(+1\) represents the "No-op" state where none of the \(N\) signals are active

References