Head2right r 1 appears in parallel with r c and the

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head2right R 1 appears in parallel with R C and the circuit simplifies to a simple CB stage. m S C v g R R R A 1 || 1 + =
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Amplifier Example V CH5 Bipolar Amplifiers 270 head2right The key for solving this problem is recognizing the equivalent base resistance of Q 1 is the parallel connection of R E and the impedance seen at the emitter of Q 2 . 1 2 1 || 1 1 1 1 m E m B in g R g R R + + + + = β β
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Amplifier Example VI O E out R R R R R r R R r R R v v + + + = 1 2 || 1 || || || || CH5 Bipolar Amplifiers 271 head2right The key in solving this problem is recognizing a DC supply is actually an AC ground and using Thevenin transformation to simplify the circuit into an emitter follower. S S m O E in g + 1 1 2 1 β O E m S out r R R g R R R || || || 1 1 || 2 1 + + = β
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Amplifier Example VII ( ) 3 2 2 1 1 1 1 1 1 1 m B C out m B E in g R R R g R R r R + + + = + + + + + = β β β π CH5 Bipolar Amplifiers 272 head2right Impedances seen at the emitter of Q 1 and Q 2 can be lumped with R C and R E, respectively, to form the equivalent emitter and collector impedances. 1 2 1 3 2 1 1 1 1 1 m m B m B C v g g R g R R A + + + + + + - = β β
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