Transistor question

Discussion in 'Homework Help' started by chaosdestro0, May 1, 2011.

  1. chaosdestro0

    Thread Starter New Member

    Apr 30, 2011
    10
    0
    I am having difficulty understanding how you would find the voltage across R2.
    Let's say I had this circuit, with Vout as 4.4v and a voltage drop across the diode at 2v. From the mark scheme, it says do 9-2-4.4. I am not sure why this does this, can anyone explain?
    [​IMG]
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Do you know II Kirchhoff's law (KVL)??
    For your circuit KVL must hold
    [​IMG]

    Vcc = Vout + VR + Vd = 9V
    VR = Vcc - Vd - Vout
     
    • 1.PNG
      1.PNG
      File size:
      7.4 KB
      Views:
      43
    chaosdestro0 likes this.
  3. chaosdestro0

    Thread Starter New Member

    Apr 30, 2011
    10
    0
    Wait, so why does the voltage from the v out of the transistor not add to the 9v?
    I have done work on it but without additional sources.
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Well, I don't understand you question.
    Can you specify your question.
    Your circuit has "external 9V additional sources".
     
  5. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,386
    1,605
    Because they are of opposite polarity. Look at the arrows drawn for you.

    If you put two 1.5 volt batteries in series, same polarity, they add, you see 3V across both.

    If you put two 1.5 volt batteries in series, but opposite polarity, they subtract, and you see about 0V across both.
     
  6. jegues

    Well-Known Member

    Sep 13, 2010
    735
    43
    If you are confused about the arrows, it basically indicates how one would attach a voltmeter to the circuit.

    See the figure attached.
     
    • KVL.JPG
      KVL.JPG
      File size:
      41.6 KB
      Views:
      16
  7. chaosdestro0

    Thread Starter New Member

    Apr 30, 2011
    10
    0
    Now I see it, Cheers :).
     
Loading...