difference in measured and designed value

Discussion in 'General Electronics Chat' started by venkyvxl, Feb 25, 2010.

  1. venkyvxl

    Thread Starter New Member

    Feb 24, 2010
    1
    0
    As per the schematic i attached and according to my calculations which shown below , where the known parameters are Ie = 500mA Vcc= 24V

    For For Q1 :


    Vcc = IcRc + Vce + IeRe + IeRm where Rm = resistance of multimeer
    So,
    24 = Ic * 3.9 + Vce + 500mA * 38 + 500mA * 0.98
    Since Ic= Ie,

    24 = 500ma * 3.9 + Vce + 500mA * 38 + 500mA * 0.98

    ð Vce = 24 – 1.95 – 19 - 0.49
    Vce = 2.56 V


    From Datasheet, β = 115 when Vce = 2v and Ic = 500mA

    Ib = Ic / β = 500mA / 115 = 4.34 mA.

    So , Vce = 2.56v and Ib = 4.34mA.

    For For Optocoupler :

    Ie = 4.34 mA,

    Since Ic ≈ Ie ,

    So Ic = 4.34mA
    From Datasheet, Vce = 0.75v , Diode forward current If = 0.76mA and β = 1000
    So , Ib = Ic / β =4.34 mA / 1000,

    Ib = 4.34uA.

    Now the Question is as I measured Vce across Q1 , it is 0.03V and by design it is 2.56V. So can anyone please explain why there is such a differecne
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    How do you know that Ic1=500mA?
     
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