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
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
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