A = B = 0: diodes D1 and D2 conduct. Consequently, the current flowing through the 2 kOhms resistor passes entirely through these 2 diodes => Ib (T1) = 0 => T1 blocked => Ib (T2) = 0 => T2 blocked => V (Y) = Vcc = > Y = 1.

In fact, if at least one of the 2 inputs is at 0, then the diode corresponding to it will be conducting which will block T1, then T2 and Y will be equal to 1.

A = B = 1: D1 and D2 are blocked => the current flowing through the 2 kOhms resistor goes on the basis of T1 => T1 led => we have a current based on T2 => T2 led (saturated) => V (Y) = 0.2 V = 0 approximately => Y = 0.

We find the operation of a NAND: Y = /(A.B)!

By logic gate (<=> by realization of an operator): – consumption: 35 mW, important in electronics! – propagation time: 100 ns, slow in electronics!

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