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Example – see Example 21.2 from course text

Example – see Example 21.2 from course text Determine the quiescent output voltage of this circuit. Base current is small, so Emitter voltage V E = V B – V BE = 2.7 – 0.7 = 2.0 V Emitter current Since I B is small, collector current I C  I E = 2 mA

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Example – see Example 21.2 from course text

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  1. Example – see Example 21.2 from course text Determine thequiescent outputvoltage of thiscircuit

  2. Base current is small, so Emitter voltage VE = VB – VBE = 2.7 – 0.7 = 2.0 V Emitter current Since IB is small, collector currentIC  IE = 2 mA Output voltage = VCC – ICRC = 10 - 2 mA 2.2 k = 5.6 V

  3. A common-collector amplifier • unity gain • high input resistance • low output resistance • a very goodbuffer amplifier

  4. 21.7 Other Bipolar Transistor Applications • A phase splitter

  5. A voltage regulator

  6. A logical switch

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