Originally Posted by Bob S
Hi Sudar-
I will try to answer your questions. First, study the figure
http://www.physicsforums.com/attachm...0&d=1258770808
Power supply V1 is 12 volts. The base bias voltage is 1/3 of 12 volts, or 4 volts. So the emitter voltage is 0.6 volts less, or 3.4 volts. So the emitter current is 3.4 volts over 500 ohms. or 6.8 milliamps. The collector current is also about 6.8 milliamps, so the voltage drop across R2 is about 3.4 volts. So the collector voltage is 12V - 3.4V = 8.6V. Power supply V2 is ac coupled to the base. It has a 1 volt (peak) 500 Hz sinewave on it, and on the base. In this configuration, the transistor is an emitter follower, so all three sine waves (base, emitter, collector) are about 1 volt peak. I hope this helps.
Bob S
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1) i cant understand the basics
i want to know what do u mean by voltage between emitter and base and voltage between collector and base
is it the barrier voltage ???
i.e the barrier potential at the juctions EB an CB???
the depletion region is responsible for the voltage drop in diode
in forward bias also there exists voltage drop according to my book
but in forward bias the depletion region would have vanished,then how will there be voltage drop.
2)also i cant understand voltage divider bias
i have posted the circuit diagram
i have many doubts there about how the voltage drop between R1 and R2 gives forward bias to the base emitter
also which current's voltage is getting dropped there
in my diagram its showing collector current
but collector current itself arises from the emitter
also u have not biased the emitter till now then how did u get Ic
forward bias provided by the voltage drop of Ic
see u have not biased it and u r about to bias the emitter only from voltage drop of Ic
then where did u get that Ic from
Ic comes only after biasing EB
plz BOB S answer first question 1 and then 2
if u cant understand my question plz mention it i will try to ask it in a better way
