Transistor Switch

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

  1. IamConfused

    Thread Starter New Member

    Oct 4, 2010
    I have built this circuit and measured Ic, Ib and VCE for 0, 5, 10, and 15 Vbb input voltages. Now, I have to calculate the values of each.

    For the base current:
    Ib = (Vbb - Vbe)/Rb

    I can't figure out how to calculate Ic
  2. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    Ic≈ Ie-Ib.
    Since Ib is negligable thus Ic≈Ie. (Ic is almost equal to Ie) Ie being in the μA range in most cases.

    Ib will tend to increase in low gain transistors, sometimes in order of mA

    Vbe will drop 0.6 to 0.7V typically for a silicone transistor when it is fully on

    by the way, IMHO, IamConfused is not good nick to begin with. After a couple of posts u might still be regarded is still confused.
    Last edited: Oct 4, 2010
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
    Your circuit will be in saturation if Ib is grater then Ib > [ (Vcc -V_led ) / β_min*Rc ] = (15V - 2V) / 50*1KΩ = 13V/50K = 260μA

    And this will occurs for Vbb greater the
    Vbb > 260μ * 10KΩ + Vbe = 2.6V + Vbe ≈ 3.2V
    So If Vbb is grater then 3.2V BJT will be in saturation region.
    And thats mean that Ic will be unchanged, and will be equal
    Ic_max = ( Vcc - Vled - Vce(sat) ) / Rc ≈ 12.9mA
    screen1988 and ignetiusjohnpaul like this.
  4. ignetiusjohnpaul

    New Member

    Mar 24, 2011
    since IB is equal to VBB-VBE/RB.THEN IC =BDCIB..........:)
  5. Adjuster

    Well-Known Member

    Dec 26, 2010
    Except that, as has already been pointed out by Jony130, the transistor is likely to be saturated.

    IC = βDCIB is only valid when the transistor is in the active region, away from saturation.
  6. SgtWookie


    Jul 17, 2007
    Well, in order for a 2N3904 transistor to be in saturation, from National Semiconductor's datasheet, Ib=Ic/10.
    Min hFE is specified as 40, but that's when Ic=10mA and Vce=1.0, which is not in saturation; it is in the linear region.

    In order to calculate the base resistor for most single transistors as saturated switches, you use β = 10.
    So, to calculate Rbase, you'd use:
    Rbase ~= (Vbb - Vbe) / (Ic/10)

    Generally, you want Vce to be below 0.2v; this is operation in the saturated region. If the transistor comes out of saturation, power dissipation in the transistor will increase dramatically.

    But, you first need to know the specifications of the LED for Vf @ current. A typical red LED might have a Vf of 2.1v @ 20mA current.

    Then you need to calculate the value for your LED current limiting resistor R1.
    R1 >= (Vsupply - (Vf_LED - Vce_sat)) / Desired_Current
    For example:
    R1 >= (15v - (2.1v - 0.1v)) / 20mA
    R1 >= (15 - 2) / 0.02
    R1 >= 13/.02
    R1 >= 650 Ohms.
    This is not a standard E24 value of resistance.
    A table of standard values is here:
    Bookmark that page.
    Scanning down the green E24 columns, you'll see that 680 Ohms is the closest standard E24 value.

    13v/680 Ohms = 19.1mA.

    Now you have the values to calculate Rbase (R2)
    Rbase <= (Vbb - Vbe) /(Ic/10)
    Rbase <= (5v - 0.7v) / (19.1mA/10)
    Rbase <= 4.3/.00191
    Rbase <= 2251.3 Ohms
    Looking back at the table of standard resistance values referenced earlier, you can see that 220 * 10 = 2,200 Ohms (2.2k) is the closest standard value of resistance <= 2251.3 Ohms.
  7. Audioguru

    New Member

    Dec 20, 2007
    The minimum hFE of a 2N3904 when Vce is 1.0V and Ic is 10mA is 100, not 40.
    The minimum hFE is 40 at a collector current of 0.1mA but its typical hFE (shown on a graph) is 230.

    Senior Member

    Jun 29, 2010


    It's sounds you have very good knowledge of biasing transistors for many applications, i have readied many books but don't get any good details of it, please tell any link for learning biasing of BJT.