Electronic Devices - Bipolar Junction Transistors - Discussion
Discussion Forum : Bipolar Junction Transistors - General Questions (Q.No. 1)
1.
Refer to this figure. Determine the minimum value of IB that will produce saturation.
0.25 mA
5.325 
A
A1.065 A
A10.425 A
ADiscussion:
25 comments Page 3 of 3.
S.M.Owais said:
5 years ago
Ice(sat)=Vcc-Vce(sat)/Rc => 10 - 0.2/4.7k => 2.1 mA.
Ib = Ice(sat)/Bdc=2.1x10^(-3)/200 => 10.425 microAmpere.
Ib = Ice(sat)/Bdc=2.1x10^(-3)/200 => 10.425 microAmpere.
Rahul said:
5 years ago
I don't understand this. Please explain in a simple way.
Keshar said:
5 years ago
Please give its complete explanation.
SRaj said:
4 years ago
We know that VCC= Ic Rc+ VCE
At saturation VCE=0.
10V = Ic x 4.7 x 10^3,
Ic = 10/4.7 x 10^3,
Ic = 1.127 mA.
Ic = B IB.
B = 200.
IB = Ic/B.
IB = 2.12mA/200.
IB = 10.6 micro Amps.
At saturation VCE=0.
10V = Ic x 4.7 x 10^3,
Ic = 10/4.7 x 10^3,
Ic = 1.127 mA.
Ic = B IB.
B = 200.
IB = Ic/B.
IB = 2.12mA/200.
IB = 10.6 micro Amps.
SRaj said:
4 years ago
AT SATURATION.
In an ideal condition, VCE =0 V.
bt in practical condition, VCE=0.2 V (usually we take VCE as 0.2V).
We know that ,
VCC=IC RC +VCE.
IC RC= VCC-VCE.
IC RC = 10-0.2 = 9.8.
IC = 9.8/RC
IC = 9.8/4.7X10^3 = 2.085 mA.
we know that;
IC = β IB ( beta=200).
IB=IC/β.
IB=2.085 mA/200=0.010425 mA =10.425 micro Amps.
In an ideal condition, VCE =0 V.
bt in practical condition, VCE=0.2 V (usually we take VCE as 0.2V).
We know that ,
VCC=IC RC +VCE.
IC RC= VCC-VCE.
IC RC = 10-0.2 = 9.8.
IC = 9.8/RC
IC = 9.8/4.7X10^3 = 2.085 mA.
we know that;
IC = β IB ( beta=200).
IB=IC/β.
IB=2.085 mA/200=0.010425 mA =10.425 micro Amps.
(8)
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