l'analisi in continua di un transistore usando il suo modello ideale

1. Lezione sul transistor Prof. Silvano Natalizi, novembre 2009

4.  dc I B V CE Diodo ideale

5.  dc I B V CE V BE = 0.7 V

6. Ic = β x Ib

17. V CC R C R B V BB 12 V 1 k  12 V Mental short I C = 12 V 1 k 

18. 0 2 4 6 8 10 12 14 16 18 2 4 6 8 10 12 14 V CE in Volts I C in mA = 12 mA This is the saturation current. 20  A 0  A 100  A 80  A 60  A 40  A I C = 12 V 1 k 

19. V CC R C R B V BB 12 V 1 k  12 V Mental open

20. 0 2 4 6 8 10 12 14 16 18 2 4 6 8 10 12 14 V CE in Volts I C in mA V CE(cutoff) = V CC 20  A 0  A 100  A 80  A 60  A 40  A

21. 0 2 4 6 8 10 12 14 16 18 2 4 6 8 10 12 14 V CE in Volts I C in mA A circuit can operate at any point on the load line. 20  A 0  A 100  A 80  A 60  A 40  A

22. V CC R C R B = 283 k  V BB 12 V 1 k  12 V The operating point is determined by the base current. I B = 12 V - 0.7 V 283 k  = 40  A I B = V BB - V BE R B

23. 0 2 4 6 8 10 12 14 16 18 2 4 6 8 10 12 14 V CE in Volts I C in mA The operating point is called the quiescent point. Q This Q point is in the linear region. 20  A 0  A 100  A 80  A 60  A 40  A

24. 0 2 4 6 8 10 12 14 16 18 2 4 6 8 10 12 14 V CE in Volts I C in mA Saturation and cutoff sono punti di lavoro non lineari. Questi punti Q sono usati in switching applications. 20  A 0  A 100  A 80  A 60  A 40  A

26. V CC R C R E V BB 5 V 1 k  15 V 1 k  2.2 k  I E = V BB - V BE R E = 1.95 mA V C = 15 V - (1.95 mA)(1 k  ) = 13.1 V Emitter bias: V CE = 13.1 V - 4.3 V = 8.8 V I C  I E