Electronic Devices - Bipolar Junction Transistors - Discussion
Discussion Forum : Bipolar Junction Transistors - General Questions (Q.No. 3)
3.
Refer to this figure. If VCE = 0.2 V, IC(sat) is:
Discussion:
7 comments Page 1 of 1.
Soniya said:
4 years ago
What if Vce is not given? Please explain.
Austhene Nicolas said:
6 years ago
Use Kvl,
Vcc - Ic(Rc) - Vce = 0.
10 - Ic(4.7Kohm) - 0.2V = 0.
10 - 0.2v = Ic(4.7Kohm).
9.8V = Ic(4.7Kohm).
(9.8V/4.7Kohm) = (Ic(4. 7Kohm)/4.7Kohm).
Ic = 2.085mA.
Vcc - Ic(Rc) - Vce = 0.
10 - Ic(4.7Kohm) - 0.2V = 0.
10 - 0.2v = Ic(4.7Kohm).
9.8V = Ic(4.7Kohm).
(9.8V/4.7Kohm) = (Ic(4. 7Kohm)/4.7Kohm).
Ic = 2.085mA.
Rock star said:
8 years ago
Ic=vcc-vce/Rc.
Arnav Yadav said:
8 years ago
From where this 0.2v is coming?
Nvn said:
8 years ago
Please explain me clearly.
Tan KY said:
8 years ago
According to me,
10V - Ic(4.7k ohm) - 0.2V = 0,
Ic(4.7k ohm) = 10V - 0.2V,
Ic = (10V - 0.2V) / 4.7k ohm,
Ic = 2.085mA.
Please correct me if my method is wrong. I am still learning this.
10V - Ic(4.7k ohm) - 0.2V = 0,
Ic(4.7k ohm) = 10V - 0.2V,
Ic = (10V - 0.2V) / 4.7k ohm,
Ic = 2.085mA.
Please correct me if my method is wrong. I am still learning this.
SWATHI NAIR said:
1 decade ago
Apply kvl in o/p side
10= 4.7*10^3*Ic + 0.2
Ic=2.085mA
10= 4.7*10^3*Ic + 0.2
Ic=2.085mA
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