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Experiência 6: Transistor como Chave

Experiência 6: Transistor como Chave. PSI-318 Laboratório de Eletrônica I. Circuito transistor pnp. I C. R C. I B. R B. V EC. -V CC. V A. V EB. Modelo do transistor pnp. região ativa e saturado (cargas na base). p n = p n0 e V EB /V T.  BF. I B

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Experiência 6: Transistor como Chave

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  1. Experiência 6:Transistor como Chave PSI-318 Laboratório de Eletrônica I

  2. Circuito transistor pnp IC RC IB RB VEC -VCC VA VEB PSI-318: Transistor como Chave

  3. Modelo do transistor pnp • região ativa e saturado (cargas na base) pn = pn0 eVEB/VT BF IB VEB VECsat B C E emissor base coletor BF pn = pn0 eVCB/VT A PSI-318: Transistor como Chave

  4. Curva IC X VCE (parâmetro:IB) saturação IB5 IB4 IB3 IB2 IB1 VCE IC ICsat ICQ VCEsat VCQ VCC corte PSI-318: Transistor como Chave

  5. Equações do Modelo • qF(t) = qF(0) - IB1(1 - e-t/) IB1 < 0 • ic(t) = ic(0) - IB1 (1 - e-t/) • = - IB1 (1 - e-t/) • |IB1| > (VCC/RC)(1/) • ic(t1) = - VCC /RC = ICsat ICsat < 0 PSI-318: Transistor como Chave

  6. ligamento • F =  / n = ICsat / IB • n = IB1 /(ICsat /) • t1 = BFln[1/ (1 - ICsat /IB1)] • t1 = BFln[1/ (1 - 1/n)] • ts = BFln[(1 - 0.1/n)/ (1 - 0.9/n)] PSI-318: Transistor como Chave

  7. desligamento • ic(t) = -VCC/RC +  (IB2 +VCC/RC)(1 - e-t/) • m = -IB2 /(ICsat /) • tq = BFln[(1 + 0.9/m)/ (1 + 0.1/m)] • tA = Aln[(IB2 - IB1)/ (IB2- ICsat/)] = Aln[(1 + n/m)/ (1 + 1/m)] PSI-318: Transistor como Chave

  8. Formas de onda de corrente V2 V1 ta ts tA td 0.1 ICsat 0.9 ICsat tliga tdesliga IB2 IB1 PSI-318: Transistor como Chave

  9. ligamento IB1 VCC/RC  IB1 ICsat PSI-318: Transistor como Chave

  10. desligamento IB2 IB1  IB2 ICsat PSI-318: Transistor como Chave

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