# 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

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2. ### Ron H AAC Fanatic!

Apr 14, 2005
7,049
673
How do you know that Ic1=500mA?