Electronic Devices - Bipolar Junction Transistors - Discussion
Discussion Forum : Bipolar Junction Transistors - General Questions (Q.No. 6)
6.
Refer to this figure. The value of VBC is:
Discussion:
16 comments Page 2 of 2.
Rahul said:
8 years ago
Can anyone solve the question in detail?
(1)
Sakthi said:
8 years ago
How Vce? will anyone explain me clearly?
(1)
Rachel said:
7 years ago
I Don't understand this problem. Please explain anyone clearly.
(1)
Shashi said:
6 years ago
We know Vce = Vbe+Vbc so we need Vbc = Vbe-Vce.
Using KCL in input loop we get Ib = Vbb-Vbe/Rb = 5-0.7/5*mA = 4.3/5 mA = 0.86mA.
Where Ic = b*Ib = 25*0.86mA = 0.0215A.
Using KCL outer loop we nee the voltage of Vce =?
Using V = IR form,
Vce = Vcc-Ic*Rc.
Vce = 20-0.0215*470 = 9.895.
Vbc = Vbe-Vce = 0.7-9.895 = -9.2V.
Using KCL in input loop we get Ib = Vbb-Vbe/Rb = 5-0.7/5*mA = 4.3/5 mA = 0.86mA.
Where Ic = b*Ib = 25*0.86mA = 0.0215A.
Using KCL outer loop we nee the voltage of Vce =?
Using V = IR form,
Vce = Vcc-Ic*Rc.
Vce = 20-0.0215*470 = 9.895.
Vbc = Vbe-Vce = 0.7-9.895 = -9.2V.
(7)
Tapioed said:
4 years ago
The formula VBC = VBE - VCE originated from VCE = VCB + VBE, and VCB = -VBC
So substituting, VCE = -VBC + VBE, rearranging the formula we get, VBC = VBE - VCE.
So substituting, VCE = -VBC + VBE, rearranging the formula we get, VBC = VBE - VCE.
(2)
Denn said:
3 years ago
I don't get how VBE is 0.7V? It isn't given in the question, right?
(4)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers