Cut off or Saturation of double resistor bas

Discussion in 'Homework Help' started by Stelyios Vasilopoulos, Jun 8, 2016.

  1. Stelyios Vasilopoulos

    Thread Starter New Member

    May 26, 2016
    7
    0
    Hi everyone, and thank you for accepting me onto this forum.

    Now I am really curios, I sat a paper today and it involved some elementary electrical calculations. I just wanted to know if for this (i know I can't change my answer), if I was right, or following the right procedure. Essentially what threw me out was that there was a resistor connected to the base with one to group and one to 5V.

    The thing is if I use voltage divider and nodal analysis at the junction where the two resistors meet at Vb, we have Ve as 0, as it is connected to ground. Threw me out a bit!

    I have attached a picture of the question and my working.

    Thanks all

    IMG_1314_785x901.jpg
     
  2. Stelyios Vasilopoulos

    Thread Starter New Member

    May 26, 2016
    7
    0
    Just realized I posted in wrong thread. Sorry people!
     
  3. kubeek

    AAC Fanatic!

    Sep 20, 2005
    4,670
    804
    You should use thevenin theorem to get the equivalent of the 2x1k resistors as seen from the base of the transistor. It should get you to Ib=(2.5V-0.7V)/500ohm=3.6mA.
     
  4. Stelyios Vasilopoulos

    Thread Starter New Member

    May 26, 2016
    7
    0

    Can I ask where you got to 2.5V?

    Is that because you assume 2.5 is heading to the collector?
     
  5. kubeek

    AAC Fanatic!

    Sep 20, 2005
    4,670
    804
  6. Stelyios Vasilopoulos

    Thread Starter New Member

    May 26, 2016
    7
    0
  7. kubeek

    AAC Fanatic!

    Sep 20, 2005
    4,670
    804
    Yes that is correct.
     
  8. Stelyios Vasilopoulos

    Thread Starter New Member

    May 26, 2016
    7
    0

    Okay since we have Ve=0V

    Ib=3.6mA or 0.0036A
    Ic=0.0036A * 100=0.36A
    ie=0.3636A

    Do we thus say Vbc=2.5V and thus in Saturation mode?
     
  9. dl324

    Distinguished Member

    Mar 30, 2015
    3,242
    619
    Not quite.

    You started out analyzing the circuit correctly in your original post. The base is at approx. 0.7V, current through the top 1K is about 4.3mA. Since this saturates the transistor, you can't use the nominal beta to calculate collector current and have to recalculate. Vce=0V so Ic=5V/200Ω=25mA. Current in the bottom 1K would be 0.7V/1K=0.7mA, Ib=4.3mA-0.7mA=3.6mA.

    The nominal beta doesn't apply when the transistor is operating in saturation mode. Besides, beta varies with current and C-E voltage and is more appropriate to use when the transistor is biased in active mode and Ic is near the current that beta was specified for.

    Ib=3.6mA
    Ic=25mA
    Ie=28.6mA
    Vbc=0.7V

    If you're going to use electron current flow, the direction assigned for Ie is wrong.

    EDIT: Corrected Vbc.
     
    Last edited: Jun 8, 2016
  10. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,387
    1,605
    Your calculation of ib is correct. The be junction essentially forces the voltage across the lower 1k resistor.

    Calculating ic and then Vc quickly shows the transistor is no longer in the active region.
     
Loading...