Question about BJT voltage divider biasing method

Discussion in 'General Electronics Chat' started by simpsonss, Sep 26, 2010.

  1. simpsonss

    Thread Starter Active Member

    Jul 8, 2008
    173
    0
    hi all,

    Sorry for mistake at the title. it is a base biasing but not a voltage divider biasing.

    Basically i'm trying to understand the operation of a BJT (2N3904). I had tried on plotting out the Ic versus Vce curve. What i get from the plot is that when i fixed Ib= 0.02mA Ic is almost constant when Vce is more than 0.6V. ok this make sense.

    Then, i go for base biasing method. Rc i put a 3Kohm and Rb i put a 3.9Kohm resistor. While for Vdd i supply in a 5Vdc. When i tried to measure Id, Ic and Vce i get this results. Ic=1.7mA and Ib=1.10mA and Vce = 0.02V. So it means my BJT is not biased(am i right?:confused:). How can i calculate the resistor value to bias 2N3904?

    thank you.
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Your BJT is in saturation region.
    You need Rb>>200*Rc to be in active region.
    For example we wont Ic = 1mA and Vce= 2.5V
    Rc = 2.5V/1mA = 2.5K = 2.2K
    And form datasheet we get "beta" (Hfe) Hfe = Ic/Ib = 230 (current gain)
    http://www.fairchildsemi.com/ds/2N/2N3904.pdf
    Figure 1 "Typical Pulsed Current Gainvs Collector Current"
    So Ib = 1mA/230 = 4.3uA
    Rb = (5V - 0.6V)/4.3uA = 1MΩ

    And this type of BJT biasing is good only for saturation.
    Becaues in active region hfe is Ic depending and virtually every single BJT will be have different current gain

    http://forum.allaboutcircuits.com/showthread.php?p=234782#post234782
     
  3. simpsonss

    Thread Starter Active Member

    Jul 8, 2008
    173
    0
    So do u mean that base biasing method is good only for saturation region but no for active region? i cant get what u mean because i thought that normally we bias a transistor to an active region so that it operate properly. Can u explain a bit more about your statement?

    ---------------------------------------------------------------------------------------------

    Another question is that if my BJT is in saturation region, i'm not able to calculate the Ic value by using this formula.

    Ic = β ((Vcc -Vbe)/Rb)
    because if Ic/Ib , β = 1.5 which is not the value stated in the datasheet as u mentioned.

    thanks again for the replies.
     
    Last edited: Sep 27, 2010
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Well of-course you can use "base biasing" in your circuit.
    But this is not a "good" circuit.
    Becaues the beta of a BJT vary with collector current and temperatures.
    In addition every single BJT will have different current gain.
    So you'll have to choose RB resistor individually each time when you change the BJT.
    So in practice we use this method:
    http://1.bp.blogspot.com/_sqidiQSxJp0/S_v5LyUsimI/AAAAAAAAAAo/SZGUomHMrkM/s1600/Imagen1.jpg

    Becaues in saturation Ic = β * Ic don't hold anymore.
    Collector current cannot be greater than:
    Vcc/Rc = 5V/3K = 1.6mA.
    And if base current is equal
    Ib = (5V - 0.65V)/3.9K = 1.1mA
    So if β=100
    Ic must be large then:
    Ic = 100* 1.1mA = 110mA
    Voltage drop on Rc resistor
    VRc = 110mA * 3K = 330V
    So this voltage is impassible in this circuit as VRc_max = 5V
    So BJT can not force this amount of current trough the load (ohms law and Kirchhoff law must hold ).
    The only think that BJT can do in this situation is to full open.
    And when transistor is full on (saturated) then he act just like a short circuit (switch)
    http://forum.allaboutcircuits.com/showthread.php?p=165135#post165135
     
    Last edited: Sep 27, 2010
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