1. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Transistor de Junção Bipolar

2. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Transistor Três regiões dopadas: emissor, base, e coletor Duas junções pn: base-emissor e base- coletor NPN ou PNP Silício o germânio

3. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. COLETOR (dopagem média) BASE (dopagem leve) EMISSOR (dopagem pesada) N P N Imagem representando o transistor de junção bipolar

4. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Transistor polarizado Diodo polarizado diretamente no emissor Diodo polarizado reversamente no coletor

5. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Transistor currents A é corrente do coletor proporcional a corrente de base, este ganho é β dc ou h FE O ganho de corrente é geralemente entre 100 e 300

6. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sentido convencional Sentido do elétron ICIC IBIB IEIE ICIC IBIB IEIE I E = I C + I B I C  I E I B << I C  dc = ICIC IEIE  dc = ICIC IBIB

7. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. V CE V CC RCRC RBRB V BB V BE The base circuit is usually analyzed with the same approximation used for diodes. I B = V BB - V BE RBRB

8. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. V CC RCRC RBRB V BB V BE = 0.7 V I B = V BB - V BE RBRB I B = 5 V - 0.7 V 100 k  5 V 100 k  = 43  A

9. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. V CC RCRC RBRB V BB 5 V 100 k  I B = 43  A  dc = 100 I C =  dc I B I C = 100 x 43  A = 4.3 mA

10. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. V CC RCRC RBRB V BB 5 V 100 k  I B = 43  A I C = 4.3 mA 1 k  12 V V R C = I C x R C V R C = 4.3 mA x 1 k  = 4.3 V

11. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. V CC RCRC RBRB V BB 5 V 100 k  I B = 43  A I C = 4.3 mA 1 k  12 V V CE = V CC - V R C V CE V CE = 12 V - 4.3 V = 7.7 V

12. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.