Vol IV - Chap 3 - The NOT Gate - Transistor Q1

Discussion in 'General Electronics Chat' started by Florian Mainguy, Apr 30, 2015.

  1. Florian Mainguy

    Thread Starter New Member

    Apr 30, 2015
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    0
    Hello everyone,

    I'm a bit confused about the behaviour of the transistor Q1 in the NOT circuit showed in one of the textbooks. It says that it behaves as a back-to-back pair of diodes. I always try to understand at a low level so I would like to know what's going on beetwen the PN junctions.

    So it behaves as a back-to-back pair of diodes. Why there isn't a current flow beetwen Emitter and Collector ? Is it because:

    - The collector is linked to the Base of a NPN transistor Q2. If Q2 was a PNP transistor, there would be a flow.

    - Q1 is a different transistor. The P doped area is bigger, so that there can't be any flow beetwen Emitter and Collector.

    I hope I have been clear.
    Thanks.
     
  2. MrChips

    Moderator

    Oct 2, 2009
    12,447
    3,363
    As pointed out in the text, Q1 is not used as a transistor but as a back-to-back pair of diodes.
    Pay attention to the current through R1. The current can only go through diode A or diode B.
    When the input is at logic-HIGH, the current through R1 goes to the base-emitter junction of Q1.
    When the input is at logic-LOW, the current is steered away from the base of Q1.
     
  3. Florian Mainguy

    Thread Starter New Member

    Apr 30, 2015
    2
    0
    Actually I understand the circuit once I consider the transistor Q1 as a back-to-back pair of diodes, but what I don't understand is why we consider the transistor as a back-to-back pair of diodes.

    When the input is at logic-low, there is a current going through the base-emitter of Q1. And I thought that as soon there is a base-emitter current, there is a much bigger emitter-collector current. But if I'm right, this is not possible here because the collector of Q1 is connected to the P Base of Q2.
    So no emitter-collector current in Q1.
    So the transistor = back-to-back pair of diodes.

    If I'm right, of course.
     
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