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Transistor circuit

  1. Aug 12, 2016 #1
    1. The problem statement, all variables and given/known data

    IMG_20160812_234158.png
    Hello, guys! I just wanted to make sure I do this right.
    2. Relevant equations
    Forget about the rest of the circuit and focus only on Q2 and Q4. How to calculate R3?

    3. The attempt at a solution
    The way I do it:
    R3 is basicaly the emitter resistance which purpose is to regulate Hfe. So in most cases the emitter is held at 1V at current of 1mA. We find R3 from the formula R3=1V/1mA = 1000ohms. When Q4 is connected to the emitter of Q2, the emitter will fall to 0,6V and the current through R3 will reduce, leaving room for the base current of Q4 I3=0,6/1000= 0,0006A=0,6A. I3 reduced from 0,001A to 0,0006A. Now we can find what the base current of Q 4 will be from IB=IE-I3=1mA-0,6mA=0,4mA.
     
  2. jcsd
  3. Aug 12, 2016 #2

    tech99

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    Gold Member

    I think R2 should equal R3 so that there is equal and opposite drive into the push-pull pair Q3 and Q4.
     
  4. Aug 12, 2016 #3
    But the calculations for R3 are done in such a way like I did, isn't that right?
     
  5. Aug 12, 2016 #4

    NascentOxygen

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    Staff: Mentor

    In order to follow your explanation, I think we need each mention of emitter to be identified as of Q2, Q3 or Q4.
     
  6. Aug 13, 2016 #5
    R3 is more a pull-down resistor. In this type of a circuit we do not care about Hfe much.
    If you have time try watch this lecture
     
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