Electronic Devices - Transistor Bias Circuits - Discussion
Discussion Forum : Transistor Bias Circuits - General Questions (Q.No. 27)
27.
Refer to this figure. Calculate the current I2.
32 mA
3.2 mA
168 
A
A320 A
ADiscussion:
5 comments Page 1 of 1.
Eoj Shores said:
2 years ago
Formula:
I2 = Vb÷R2
Solve for Vb:
Vb = Vbe + Ve.
But;
Vbe = 0.7V.
Ve = IeRe (Where Ie=β*Ib + Ib).
Ve = (β*Ib + Ib)(Re).
Ve = 2.475V.
Thus:
Vb = 0.7V + 2.475V.
Vb = 3.175V.
Therefore:
I2 = Vb/R2 = 3.175V÷10kΩ ≈ 320µA (Ans.)
I2 = Vb÷R2
Solve for Vb:
Vb = Vbe + Ve.
But;
Vbe = 0.7V.
Ve = IeRe (Where Ie=β*Ib + Ib).
Ve = (β*Ib + Ib)(Re).
Ve = 2.475V.
Thus:
Vb = 0.7V + 2.475V.
Vb = 3.175V.
Therefore:
I2 = Vb/R2 = 3.175V÷10kΩ ≈ 320µA (Ans.)
Anonimous said:
7 years ago
IC = IE = VE/RE.
IC = β * IB.
VB = VE+VBE.
I2 = VB/R2.
IC = β * IB.
VB = VE+VBE.
I2 = VB/R2.
Onodera said:
9 years ago
RE must be 500kiloohms because VE=IE * RE will give you: 5mA * 500ohms = 2.5mV.
(1)
Jaikishore.r said:
1 decade ago
Analyse the data
Given Ib=(50*10^-6)A
Calculate ic using beta value i.e ic=beta*ib=5mA
Ic=Ie therefore ve=ie*re=2.5v
vbe=vb-ve
vb=vbe+ve=0.7+2.5=3.2
Then i2=vb/r2=3.2/10000=320microamps
Given Ib=(50*10^-6)A
Calculate ic using beta value i.e ic=beta*ib=5mA
Ic=Ie therefore ve=ie*re=2.5v
vbe=vb-ve
vb=vbe+ve=0.7+2.5=3.2
Then i2=vb/r2=3.2/10000=320microamps
Jaikishore.r said:
1 decade ago
Analyse the data
Given Ib=(50*10^-6)A
Calculate ic using beta value i.e ic=beta*ib=5mA
Ic=Ie therefore ve=ie*re=2.5v
vbe=vb-ve
vb=vbe+ve=0.7+2.5=3.2
Then i2=vb/r2=3.2/10000=320microamps
Given Ib=(50*10^-6)A
Calculate ic using beta value i.e ic=beta*ib=5mA
Ic=Ie therefore ve=ie*re=2.5v
vbe=vb-ve
vb=vbe+ve=0.7+2.5=3.2
Then i2=vb/r2=3.2/10000=320microamps
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