Write Short notes on 2 input TTL NAND gate
1 Answer

The circuit diagram of a 2 input TTL NAND gate is as follows:

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  • A two input TTL NAND is shown above. A and B are two inputs while Y is the output.

  • Operation of the gate:

    a) A and B both low: both B-E junctions of Q1 are forward biased. Hence D1 and D2 will conduct to force the voltage at point C to 0.7V. This voltage is insufficient to forward bias B-E junction of Q2. Hence Q2 remains OFF. Therefore its collector voltage rises to $V_{CC}$. As Q3 is operating in emitter follower mode, output Y will be pulled up to high voltage Y= 1

    b) Either A or B low: If any one input is connected to ground with other left open or connected to $V_{CC}$ the corresponding diode (D1 or D2) will conduct. This will pull down voltage at C o 0.7V. This voltage is insufficient to turn on Q2 so it remains OFF. So collector voltage of Q2 will be equal to VCC. This voltage acts as base voltage for Q3. As Q3 acts as an emitter follower, output Y will be pulled to $V_{CC}$. Y= 1

    c) A and B both high: If both A and B are connected to then both diodes D1 and D2 will be reverse biased and do not conduct. Therefore D3 is forward biased and base current is supplied to transistor Q2 via R1 and D3. As Q2 conducts, the voltage at X will drop down and Q3 will be OFF, whereas voltage at Z will increase to turn ON Q4. As Q4 goes into saturation, the output voltage Y will be pulled down to low. Y = 0

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