Electronic Devices - Transistor Bias Circuits - Discussion
Discussion Forum : Transistor Bias Circuits - General Questions (Q.No. 3)
3.
Refer to the given figure. The most probable cause of trouble, if any, from these voltage measurements is
Discussion:
4 comments Page 1 of 1.
Wasp said:
6 years ago
6.56V should be 0.656V.
By Thevenins:
Vth =Vcc*(R2/R1+R2) = 2.44V
Ib = Vcc/(R1+R2) = 0.298mA
Rth = R1//R2 = 5973.51ohms.
Vb = Vth-Ib * Rth = 0.656V.
The base-emitter junction is open because of reverse bias condition in BE junction.
0.656V<0.7V.
By Thevenins:
Vth =Vcc*(R2/R1+R2) = 2.44V
Ib = Vcc/(R1+R2) = 0.298mA
Rth = R1//R2 = 5973.51ohms.
Vb = Vth-Ib * Rth = 0.656V.
The base-emitter junction is open because of reverse bias condition in BE junction.
0.656V<0.7V.
Anna Baaba Jacobs said:
7 years ago
Since the base emitter junction is reverse biased instead of forward bias, the voltage will be zero meaning the circuit is opened.
Muralidhar Padavala said:
9 years ago
In the figure voltmeter across Re is ZERO means there is no voltage or current in emitter resistor this happens only when the base-emitter junction is reverse biased.
Yakub said:
9 years ago
Not understand this, someone explains it clearly.
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