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For normal operation it is said to remember:

  1. Emmitter Base junction is always forward biased
  2. Collector Base junction is always reverse biased

Can someone explain to me what this means? My view is that the Emitter is +ve wrt to Base and Collector is ove wrt to Base. I am new to this topic, please explain?

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1 Answer 1

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If you consider an NPN BJT to be two diodes in series opposition with their anodes connected together, then the collector is the cathode of one diode, the emitter is the cathode of the other, and the base is the junction of the two anodes.

For an NPN BJT, then, it means that the base must be positive with respect to the emitter and the collector must be positive with respect to the base.

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  • \$\begingroup\$ Thank you sir, your reason makes complete sense. I really appreciate your help. \$\endgroup\$
    – Saavin
    Feb 11, 2016 at 1:13
  • \$\begingroup\$ @EM_Feilds: will the base be negative with respect to the emitter and the collector be negative with respect to the base for PNP transistor (ie: Emmitter Base junction is always negative biased, Collector Base junction is always forward biased)? \$\endgroup\$
    – Saavin
    Feb 11, 2016 at 2:23
  • \$\begingroup\$ @Saavin: For a PNP the diode polarities are reversed, so the cathodes will be connected together for the base, one of the anodes will be the emitter,and the other anode will be the collector. That way, when the base is more negative than the emitter that diode will be forward biased, and when the collector is more negative than the base, that diode will be reverse biased. \$\endgroup\$
    – EM Fields
    Feb 11, 2016 at 2:38
  • \$\begingroup\$ In fact for a transistor, it is not strictly required that the C-B junction be reverser biased -- a small amount (< 300 mV) of forward bias has a negligible effect on the performance. In a switching application, where the transistor is driven strongly, the bias is often so much that the C-E voltage is only a few 100 mV -- meaning that the C-B forward bias is nearly as much as the B-E bias. \$\endgroup\$
    – jp314
    Feb 11, 2016 at 4:05

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