From-Transistors-to-Gates

• Claude Shannon

  • His master's thesis in 1937, A Symbolic Analysis of Relay and Switching Circuits, is considered as "possibly the most important, and also the most famous, master's thesis of the century.”
  • He came up with the idea that electrical switches can be used to do Boolean logic
  

• Relay 继电器

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• Vacuum Tube 真空管

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• Transitor 晶体管

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Transistor

A transistor is a discrete electronic component that can behave like a switch 可视为开关

• Tiny, cheap, flexible and reliable
  • onducts when VGS is high (N Type transistor) (close)
  • Blocks when VGS is 0 (open)

CMOS Transistors

Complementary Metal-Oxide Semiconductor 互补金属氧化物半导体

• Two types: P-type (positive) and N-type (negative)
  • P-type
    • Open (insulating) if gate is “on” = 1 激活时(1)阻断
    • Closed (conducting) if gate is “off” = 0 休眠时(0)导通
  • N-type
    • Open if gate is “off” = 0 激活时(1)导通
    • Closed if gate is “on” = 1 休眠时(0)阻断

Logical Gates

Boolean functions are implemented in digital computer circuits called logic gates.

• A gate is an electronic device that produces a result based on two or more input values.
• In reality, gates consist of one to six transistors, but digital designers think of them as a single unit.
• Integrated circuits contain collections of gates suited to a particular purpose.

Inverter Gate (NOT) 非

• In = 0v -> P conduct N insulate -> Out is 1 (2.9 v)
• In = 1v -> P insulate N conduct -> Out is 0 (0 v)

AND/NAND Gate 与

AND consists of NAND and NOT gates.

NOR Gate 异或

OR Gate 或

OR Gate consists of NOR and NOT gates.

Universal Logical Gate

A universal gate is a gate which can implement any Boolean function without need to use any other gate type.

• The NAND and NOR gates are universal gates.
• Demorgans'Law
  • $\overline{X\ast Y}=\bar{X}+\bar{Y}$
  • $\overline{X+Y}=\bar{X}\ast \bar{Y}$

Prove NAND is a universal gate $\overline{XY}$

• $NOT:\bar{X}=\overline{XX}$
• $OR:X+Y=\overline{\stackrel{―}{X}}+\overline{\stackrel{―}{Y}}=\overline{\stackrel{―}{X}\ast \stackrel{―}{Y}}$
• $AND:XY=\overline{\stackrel{―}{XY}}$

Prove NOR is a universal gate $\overline{X+Y}$

• $NOT:\bar{X}=\overline{X+X}$
• $OR:X+Y=\overline{\stackrel{―}{X+Y}}$
• $AND:XY=\overline{\stackrel{―}{X}}\ast \overline{\stackrel{―}{Y}}=\overline{\stackrel{―}{X}+\stackrel{―}{Y}}$

Demorgans'Law 记忆技巧

• $\ast ->+$ 帽子不变,一分一和
  • $\overline{X\ast Y}=\bar{X}+\bar{Y}$
  • $\overline{X+Y}=\bar{X}\ast \bar{Y}$