NPN and PNP BJT questions

Discussion in 'Homework Help' started by uofmx12, Oct 18, 2011.

  1. uofmx12

    Thread Starter Member

    Mar 8, 2011
    Here are two diagrams. I have some basic questions dealing with them.
    1) In 9.2, what is the reason to connect the base of Q2?
    2) In 9.3, If Q2 is conducting, which mode of operation is Q1?

    Question not related to those:
    Is alpha and beta formulas the same for a NPN as they are for a PNP BJT?
  2. crutschow


    Mar 14, 2008
    1) Not sure I understand your question. If the base of Q2 is not connected then the transistor is off and the circuit is not functional.

    2) You can determine that if you calculate the base-emitter voltage of both transistors under those conditions. Remember that the base-emitter diode junction (as shown by the arrow direction) must be forward biased for the transistor to be on.

    alpha and beta formulas are the same for an NPN and a PNP. It's just the operating voltage polarities that change.
  3. BJT_user


    Oct 9, 2011
    Okay uofmx12, let's see what we can tell here.

    First, 9.2 states it is a Class-A BJT follower with Emitter-Current Bias. The key word in that statement is class-A. What we know about class A is that the active element(s) remains conducting at all times so it works inside its most linear range. IE, the average DC voltage at its output (before any capacitor coupling) is typically around 1/2 Vcc. For this to happen, the output transistor has to be biased so that it conducts roughly equal to the load to maintain approximately 1/2 Vcc. In circuit 9.2, that is accomplished by biasing transistor Q2 so it is constantly on and allowing transistor Q1 to handle signal amplification.

    In 9.3 it states it is a Class-B Complimentary BJT Output Stage. This is also known a push-pull amplifier. What we know about a push-pull amplifier is that it has two active elements, each reacting to opposite halves of the signal phase. The mean DC voltage of the output is at or around 0 or ground for this stage, which makes it more efficient since it is not conducting 100% of the time. Since each element, or transistor (Q1/Q2) functions inversely to the other, then while one is conducting, the other is off.

    I really hope this helps.