Input impedance of emitter follower v ch5 bipolar

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Input Impedance of Emitter Follower = V CH5 Bipolar Amplifiers 258 head2right The input impedance of emitter follower is exactly the same as that of CE stage with emitter degeneration. This is not surprising because the input impedance of CE with emitter degeneration does not depend on the collector resistance. E X X R r i v ) 1 ( β π + + = A
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Emitter Follower as Buffer CH5 Bipolar Amplifiers 259 head2right Since the emitter follower increases the load resistance to a much higher value, it is suited as a buffer between a CE stage and a heavy load resistance to alleviate the problem of gain degradation.
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Output Impedance of Emitter Follower CH5 Bipolar Amplifiers 260 head2right Emitter follower lowers the source impedance by a factor of β +1 improved driving capability. E m s out R g R R || 1 1 + + = β
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Emitter Follower with Early Effect O E v R r R A 1 || = CH5 Bipolar Amplifiers 261 head2right Since r O is in parallel with R E , its effect can be easily incorporated into voltage gain and input and output impedance equations. ( )( ) O E m s out O E in m S O E r R g R R r R r R g r R || || 1 1 || 1 1 || + + = + + = + + + β β β π
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Current Gain CH5 Bipolar Amplifiers 262 head2right There is a current gain of ( β +1) from base to emitter. head2right Effectively speaking, the load resistance is multiplied by ( β +1) as seen from the base.
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Emitter Follower with Biasing CH5 Bipolar Amplifiers 263 head2right A biasing technique similar to that of CE stage can be used for the emitter follower. head2right Also, V b can be close to V cc because the collector is also at V cc.
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Supply-Independent Biasing CH5 Bipolar Amplifiers 264 head2right By putting a constant current source at the emitter, the bias current, V BE , and I B R B are fixed regardless of the supply value.
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Summary of Amplifier Topologies CH5 Bipolar Amplifiers 265 head2right The three amplifier topologies studied so far have different properties and are used on different occasions. head2right CE and CB have voltage gain with magnitude greater than one, while follower’s voltage gain is at most one.
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Amplifier Example I CH5 Bipolar Amplifiers 266 head2right The keys in solving this problem are recognizing the AC ground between R 1 and R 2 , and Thevenin transformation of the input network. S E m S C in out R R R R g R R R R v v + + + + - = 1 1 1 2 1 1 || || β
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Amplifier Example II CH5 Bipolar Amplifiers 267 head2right Again, AC ground/short and Thevenin transformation are needed to transform the complex circuit into a simple stage with emitter degeneration. S m S C in out R R R R g R R R v v + + + + - = 1 1 2 1 1 1 || β
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Amplifier Example III r R r R + + = CH5 Bipolar Amplifiers 268 head2right The key for solving this problem is first identifying R eq , which is the impedance seen at the emitter of Q 2 in parallel with the infinite output impedance of an ideal current source. Second, use the equations for degenerated CE stage with R E replaced by R eq . 2 1 1 2 1 1 1 1 1 m m C v in g R g R A + + + - = β π π
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Amplifier Example IV CH5 Bipolar Amplifiers 269 head2right The key for solving this problem is recognizing that C B at frequency of interest shorts out R 2 and provide a ground for R 1 .
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