Electronic Devices - Bipolar Junction Transistors - Discussion
Discussion Forum : Bipolar Junction Transistors - General Questions (Q.No. 20)
20.
Refer to this figure. The value of 
DC = 100 and VIN = 8 V. Determine IC(sat).
DC = 100 and VIN = 8 V. Determine IC(sat).
18 mA
7.92 mA
1.8 mA
8 
A
ADiscussion:
7 comments Page 1 of 1.
ITULU VALENTINE said:
3 years ago
@All.
Here, we have to consider the base-emitter voltage drop (Vbe)
It should be;
Ic = (Vcc-Vce-Vbe)/Rb.
Where Vbe=0.7 and Vce=0.2 for saturation to occur.
Ic = (20 - 0.2 - 0.7)/2500.
= 19.1/2500.
= 0.0000764,
= 764mA.
Here, we have to consider the base-emitter voltage drop (Vbe)
It should be;
Ic = (Vcc-Vce-Vbe)/Rb.
Where Vbe=0.7 and Vce=0.2 for saturation to occur.
Ic = (20 - 0.2 - 0.7)/2500.
= 19.1/2500.
= 0.0000764,
= 764mA.
SUJITH KS said:
9 years ago
While saturation (ON) the Vce value of the npn (si) transistor is 0.2 (fixed) , it's not the cutoff voltage.
Minitkumar said:
1 decade ago
For saturation Vce = 0.2 So,
Ic = (Vcc - Vce)/Rc.
= (20 - 0.2)/2500.
= 19.8/2500.
= 0.00792 A.
= 7.92 mA.
Ic = (Vcc - Vce)/Rc.
= (20 - 0.2)/2500.
= 19.8/2500.
= 0.00792 A.
= 7.92 mA.
Nitesh said:
1 decade ago
What is cut off voltage mean here?
Bhuvneshgupta said:
1 decade ago
Here 0.2 is the cutoff voltage.
Priya said:
1 decade ago
What is this 0.2 value ?
Naga said:
1 decade ago
ic=(vcc-0.2)/rc
= (20-0.2)/2500
=7.92ma
= (20-0.2)/2500
=7.92ma
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