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 1 of 3.
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)
JOE said:
9 years ago
Everyone who has posted thus far is incorrect.
If you assume that Vc = 0.2V then Vce = 0.2 This means that the circuit is already in saturation mode and BIc = Ib does not apply.
The BIc = Ib equation only applies to transistors in forward active mode.
If you assume that Vc = 0.2V then Vce = 0.2 This means that the circuit is already in saturation mode and BIc = Ib does not apply.
The BIc = Ib equation only applies to transistors in forward active mode.
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.
KingLevis said:
7 years ago
Vce at sat is btn 0.2 to 0.3 for silicon diode transistor.
From Ic sat=(Vcc-Vce)/Rc.
(10-0.2)/4700ohms.
Ic=2.085mA.
But Ib=Ic/Bdc.
2.085/200.
Ib=10.425micro-ampere.
Assumptions assume Vce=0.2V.
From Ic sat=(Vcc-Vce)/Rc.
(10-0.2)/4700ohms.
Ic=2.085mA.
But Ib=Ic/Bdc.
2.085/200.
Ib=10.425micro-ampere.
Assumptions assume Vce=0.2V.
(1)
ENHLE. said:
6 years ago
VCC = RC.IC(sat) + VCE(sat).
IC(sat) = VCC-VCE(sat)/RC.
IC(sat) = (10-0.2)/4700.
IC(sat) = 2.085 micro Amps.
IC(sat) = Bdc.IB(min).
IC(sat) = (200)(2.085 micro Amps).
IC(sat) = 10.43 mcro amps.
IC(sat) = VCC-VCE(sat)/RC.
IC(sat) = (10-0.2)/4700.
IC(sat) = 2.085 micro Amps.
IC(sat) = Bdc.IB(min).
IC(sat) = (200)(2.085 micro Amps).
IC(sat) = 10.43 mcro amps.
Hariom Gupta said:
1 decade ago
Apply kirchoff's voltage law in the output side
10=4.7*10^3*Ic+0.2
on solving Ic=2.085 ma.
For common emmiter configuration we know Ic=b*Ib
where b=200
So, Ib = 2.085/200 = 10.425 uA .
10=4.7*10^3*Ic+0.2
on solving Ic=2.085 ma.
For common emmiter configuration we know Ic=b*Ib
where b=200
So, Ib = 2.085/200 = 10.425 uA .
Hapakhoa Riad said:
1 decade ago
Vc/Rc = 10/4700 ohms = 0.0021276595744681 A.
Betta = Ic/Ib.
Ib = Ic/Betta = 2.1276595744681/200 = 1.063829787234043 e-5 = 10.63829787234043 micro A (D).
Betta = Ic/Ib.
Ib = Ic/Betta = 2.1276595744681/200 = 1.063829787234043 e-5 = 10.63829787234043 micro A (D).
Eric Orellana Romero said:
1 decade ago
About of first answer, the equation ic=B*ib is not valid for saturation zone in a bjt transistor. In this zone, bjt behavior is not lineal.
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.
Isha said:
9 years ago
Use VCE = 0.2V for the saturated region and apply KCL in the outer loop and get Ic and use beta to get Ib.
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