Vc vs Vin of NPN

Discussion in 'Homework Help' started by TsAmE, Oct 4, 2010.

  1. TsAmE

    Thread Starter Member

    Apr 19, 2010
    Calculate the collector voltage Vc as Vin increases from 0-5V and hence plot the transfer characteristic Vc vs Vin (consider many points where Vc is changing rapidly). Calculate Vin when Vc = 2V. An NPN with a gain of B = 100 is used.


    When Vin = 0V and 1V Vc = 0.

    Vin = 2V: Ib = 2 - 0.7 / 10 x 10^3
    = 1.3 x 10^-4

    Ic = IbB
    = 1.3 x 10^-4 x 1000
    = 0.013

    Vc = 5 - 0.013 x 1 x 10^3
    = -8V

    I dont understand why I got a negative answer.
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  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    In principle once you have enough base current to drive the collector voltage to 0V (actually it would be to Vce_sat) it can't be driven less than 0V - a condition called "saturation".

    So for Vc to be zero then Ic would be 5mA. To just drive Ic to 5mA would require Ib=5mA/β=50uA.

    To get Ib=50uA with Vbe assumed as 0.7V would require


    An input voltage greater than this would lead to the saturation condition.

    This isn't exactly what will happen in practice - but will suffice for the sake of this simple problem statement.
    Last edited: Oct 4, 2010
  3. Markd77

    Senior Member

    Sep 7, 2009
    You are getting a negative answer because you have calculated a collector current that is greater than is possible to pass through the 1K resistor with the given voltage (5mA).
  4. TsAmE

    Thread Starter Member

    Apr 19, 2010
    Howcome Vin = 1.2V would lead to saturation? Since the rule that the transistor will turn on if Vin - 0.7 >= 0.7V isnt fufilled? (with 1.2 - 0.7 = 0.5V)
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
    To open small signal BJT you need Vbe grater then 0.5V.

    Your circuit will be in saturation if Ib is grater then Ib > ( Vcc / β*Rc ) = 5V/100KΩ = 50μA
    This base current will cause voltage drop on RB resistor equal to :
    Vb = Ib * RB = 0.5V
    So for Vin grater then Vin > ( Vcc*Rb / β*Rc ) + Vbe BJT will be in saturation.

    Of course we assume that β is constant, witch is not true .
  6. Georacer


    Nov 25, 2009
    Saturation isn't a state before the transistor turn-on (or active region as it is called). It is actually achieved if you turn on your transistor "too-much".
    In increasing order of ib (and Vbe consequently) the transistor goes from cut-off, to active, to saturation. In the last state it cannot increase any more the current of the collector and therefore the relation Ic=β*Ib isn't valid anymore.