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Discrete element logic circuits

These circuits are made with diodes or bipolar transistors. Let us take as an example the realization of the 3 basic logical operators.

Realization of the AND:

Realization of the AND

Operation:
Let us consider that the 1 corresponds to the voltage Vcc (5V for example), and the 0 to the mass: GND (ground) = 0V.

   

  • if A = B = 0 then the 2 diodes are forward biased => They are conducting => in Y, we have 0.6V or approximately 0V => Y = 0.
  • if A = 1, B = 0 then D1 is reverse biased => D1 is blocked. On the other hand, D2 is forward biased => D2 is conductive and imposes in Y: 0.6V (approximately 0V) => Y = 0.
  • if A = 0, B = 1 then symmetrically to the previous case, we have D2 blocked and D1 open => Y = 0.
  • if A = B = 1 then the 2 diodes are reverse biased => in Y we have about Vcc => Y = 1.

This operation is indeed that of an ET!

Realization of the OR:

Realization of the OR

Operation: 1 <=> Vcc (5V for example)
0 <=> GND (0V)

   

  • if A = B = 0 then the 2 diodes are blocked => point Y is linked to GND => Y = 0.
  • if A = 1, B = 0 then D2 is blocked but D1 is forward biased => D1 is conductive => in Y we have Vcc (Vcc – 0.6 in true) => Y = 1.
  • if A = 0, B = 1 then in the same way we have D1 blocked and D2 open => Vcc on Y => Y = 1.
  • if A = B = 1 then the 2 diodes are forward biased => they are both on => Vcc on Y => Y = 1.

This operation is indeed that of an OR!

Realization of NO:

Realization of NOT

Operation:
1 <=> Vcc (5V for example)
0 <=> GND (0V)

  • if A = 1 then the transistor T is saturated => Y = Vce_sat = 0.2V = 0 approximately (GND) => Y = 0.
  • if A = 0 then the transistor T is blocked => Y = Vcc => Y = 1.

This operation is indeed that of a NO!

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