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 1 of 2.
Denn said:
3 years ago
I don't get how VBE is 0.7V? It isn't given in the question, right?
(4)
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)
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)
Rachel said:
7 years ago
I Don't understand this problem. Please explain anyone clearly.
(1)
Sakthi said:
8 years ago
How Vce? will anyone explain me clearly?
(1)
Rahul said:
8 years ago
Can anyone solve the question in detail?
(1)
Manoj said:
9 years ago
Apply kvl to base emitter loop.
-Vb + (Ib * 5k) + Vbe = 0. Vbe = 0.7v
Thus, Ib= 8.6*10^-4A.
B=Ic/Ib.
Thus, Ic = 0.0215A.
Appliy KVL to collector to base loop.
-20+Ic*470 +Vcb - Ib * 5k + 5 =0.
Thus, Vcb = 9.195v
Thus, Vbc = -9.195~ -9.2v.
-Vb + (Ib * 5k) + Vbe = 0. Vbe = 0.7v
Thus, Ib= 8.6*10^-4A.
B=Ic/Ib.
Thus, Ic = 0.0215A.
Appliy KVL to collector to base loop.
-20+Ic*470 +Vcb - Ib * 5k + 5 =0.
Thus, Vcb = 9.195v
Thus, Vbc = -9.195~ -9.2v.
Sabar said:
9 years ago
How Vce is 9.895?
Shivam said:
9 years ago
The equation is Vce = Vcb + Vbe.
Vbe = 0.7V is forward bias voltage of pn junction diode which is formed by base and emitter.
And Vce = 9.895 V as we calculating.
So, Vcb = Vce - Vbe = 9.195V~ = 9.2V.
So, Vbc = -Vcb = -9.2v is correct option.
Only the sign of magnitute of voltage measured changed.
Vbe = 0.7V is forward bias voltage of pn junction diode which is formed by base and emitter.
And Vce = 9.895 V as we calculating.
So, Vcb = Vce - Vbe = 9.195V~ = 9.2V.
So, Vbc = -Vcb = -9.2v is correct option.
Only the sign of magnitute of voltage measured changed.
(1)
Isha said:
9 years ago
For active region, Vbe = 0.7v get ib by using kcl in input loop ib = (Vbb - Vbe) / Rb and ic = β Ib and use Kcl in outer loop and get Vce and then Vbc = Vbe - Vce = -9.2.
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