Electronic Devices - Bipolar Junction Transistors - Discussion
Discussion Forum : Bipolar Junction Transistors - General Questions (Q.No. 29)
29.
Refer to this figure. Determine the minimum value of VIN from the following that will saturate this transistor.
Discussion:
7 comments Page 1 of 1.
Ragul said:
3 weeks ago
Thanks all for explaining the answer.
Susan Thapa said:
4 years ago
Vcc = IcRc+ vce.
Ic = (vcc-vce)/RC.
Voltage drop for vce at saturation is 0.2
Ic = 2.085mA.
Ib=Ic/B= 10.425 * 10^-6A.
Now,
Vin=IbRb+Vbe.
The Voltage drop for vbe at saturation is 0.7.
Vin = 10.425x10^-6 * 1.2 * 10^6 + 0.7.
Vin = 12.325V.
Thus, the given option is correct or incorrect?
Ic = (vcc-vce)/RC.
Voltage drop for vce at saturation is 0.2
Ic = 2.085mA.
Ib=Ic/B= 10.425 * 10^-6A.
Now,
Vin=IbRb+Vbe.
The Voltage drop for vbe at saturation is 0.7.
Vin = 10.425x10^-6 * 1.2 * 10^6 + 0.7.
Vin = 12.325V.
Thus, the given option is correct or incorrect?
Susan Thapa said:
4 years ago
Vcc = IcRc+ vce.
Ic = (vcc-vce)/RC.
Voltage drop for vce at saturation is 0.2
Ic = 2.085mA.
Ib=Ic/B= 10.425 * 10^-6A.
Now,
Vin=IbRb+Vbe.
The Voltage drop for vbe at saturation is 0.7.
Vin = 10.425x10^-6 * 1.2 * 10^6 + 0.7.
Vin = 12.325V.
Thus, the given option is correct or incorrect?
Ic = (vcc-vce)/RC.
Voltage drop for vce at saturation is 0.2
Ic = 2.085mA.
Ib=Ic/B= 10.425 * 10^-6A.
Now,
Vin=IbRb+Vbe.
The Voltage drop for vbe at saturation is 0.7.
Vin = 10.425x10^-6 * 1.2 * 10^6 + 0.7.
Vin = 12.325V.
Thus, the given option is correct or incorrect?
Susan Thapa said:
4 years ago
The answer should be option B.
Susan Thapa said:
4 years ago
The answer should be option B.
Bhargav.g said:
1 decade ago
For saturation Vce = 0.2.
And we know Vce = Vcc-IcRc => Ic(sat) = 2.085mA.
Now Ib = Ic/beta => Ib = 10.425 microA.
But Ib=(Vin-Vbe)/Rb => Ib = (Vin-0.7)/1.2 Mohms.
Solve the above two you'll get Ib =13.21.
And we know Vce = Vcc-IcRc => Ic(sat) = 2.085mA.
Now Ib = Ic/beta => Ib = 10.425 microA.
But Ib=(Vin-Vbe)/Rb => Ib = (Vin-0.7)/1.2 Mohms.
Solve the above two you'll get Ib =13.21.
Prakash Koju said:
1 decade ago
Consider, Vce=0.2V and Vbe=0.7V
Ic=(Vcc-Vce)/Rc
Ib=Ic/Bdc
Now,
Using KVL,
Vin=IbRb+Vbe=13.21V
Ic=(Vcc-Vce)/Rc
Ib=Ic/Bdc
Now,
Using KVL,
Vin=IbRb+Vbe=13.21V
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