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How transistors amplify AC signals

  1. Sep 6, 2016 #1
    AC-Wave.png
    How are transistors able to amplify full AC signals? What I mean is that a transistor will conduct only if the base emitter junction is forward biased and when you apply high enough treshold voltage to turn it on. Like 0,6v. Then looking at the graph above, from 0 to 120 and from 120 to 0v the signal will be amplified perfectly. But what about when the polarity is reversed? How can transistors amplify that since they will be reverse biased?
     
  2. jcsd
  3. Sep 6, 2016 #2

    Drakkith

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    Staff: Mentor

    I suppose you could always rectify the signal so the transistor always sees the correct polarity.
     
  4. Sep 6, 2016 #3

    LvW

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    At first, I assume you mean 120mV rather than 120V, OK? (Or do you intend to amplify a 120V sinus?).
    My recommendation: Try to become familiar with the terms "BJT biasing", "DC operational point", "quiescent current", "coupling capacitors".
    If you know and understand the meaning of these terms you will see how ac signals are amplified.
     
  5. Sep 6, 2016 #4
    I think I get it. We first bias the transistor to set the Q point somewhere half Vcc and then we apply the AC signal. The confusion came from the assumption that the only voltage we apply to the base is the AC one while now as I recall upon old knowledges, we never just leave the base waiting for the AC signal, we use DC voltage to set the Q point through the base current so that Vc is half Vcc. But you will say here "Haven't you heard of Common emitter amplifier?" Well, I came here from differential amplifiers where basicaly no base bias is shown.
    And no, don't take the graph litteraly, I used it just as an example of AC sinusoidal wave.
     
  6. Sep 6, 2016 #5

    LvW

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    Yes - basically correct. All DC levels are to be decoupled from the ac signals (input and output) using suitable coupling capacitors.
    Regarding the diff. amplifier:
    Because this type usually must be able to amplify also DC signals, the base DCpotential is kept at 0V (no capacitive coupling required) - however, the emitter is biased by a suitable negative DC voltage of app. -0.7V. Hence, again we have app. Vbe=0.7V.
     
  7. Sep 6, 2016 #6
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