Example 5.12 in Sedra/Smith book

Discussion in 'General Electronics Chat' started by kyrieli, Dec 25, 2004.

  1. kyrieli

    Thread Starter New Member

    Dec 25, 2004
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    Dear friends,
    I am having trouble understanding example 5.12 pg 435 (5th edition) in the Microelectronic Circuits book of Sedra and Smith. This same circuit appears in previous editions as well. It shows two transistors, one npn and one pnp. They are connected at the base which is biased through a +5V supply and a 10k resistor. The collector of the npn (which is on the top part of the circuit) is connected directly to +5V, while the collector of the pnp is connected directly to -5V. Their emitters are connected together to ground through a 1k resistor.

    The auhtors say that "by examining the circuit we conclude that the two transistors cannot be simultaneously conducting." Can anyone explain to me why this is so? Thank you very much in advance.
    Elias
     
  2. dragan733

    Senior Member

    Dec 12, 2004
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    It is obvious. The NPN transistor is conducted on positive voltages, the PNP transistor is conducted on negative voltages. If their bases are connected together, then only one transistor can be conducted, in this case the NPN transistor.
     
  3. Dave

    Retired Moderator

    Nov 17, 2003
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    It is because in a pnp transistor, the emitter voltage should be higher than the base voltage for current to flow through the emitter, i.e. the emitter-base voltage Veb has to be positive for a pnp transistor to operate in its active region. If you look at the example you have stated, the emitter voltage is 3.9V and the base voltage is 4.6V, Veb = 3.9 - 4.6 = -0.7, therefore the pnp transistor is off.

    On the other hand for the npn transistor in this circuit, the base-emitter voltage Vbe should be positive for the npn transistor to be active, in this example the base voltage is 4.6V and the emitter voltage is 3.9V, Vbe = 4.6 - 3.9 = 0.7, therefore the npn transistor is operating in its active region.

    Hope that helps.
     
  4. nanobyte

    Senior Member

    May 26, 2004
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    Dave, you got me confused. In the post previous to your dragan733 said that PNP transistors were conducting on negative voltages. Then you said that the emitter-base voltage (Veb) had to be positive for the PNP transistor to operate in its active region. Are these two different things? Forgive me if my question sounds stupid, but I am a beginner in electronics and I am a slow learner.
     
  5. Dave

    Retired Moderator

    Nov 17, 2003
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    In the context of the question, I'm not sure what exactly dragan733 is saying, maybe he could clarify what he means.

    What I have said is relating to the current-voltage characteristics of bipolar transistors where we consider two voltages, one connected between the base and emitter and one connected between the base and collector. Remember BJT are current controlled devices, however current is conversely controlled by voltages, and for npn and pnp transistors the polarity of the appled voltages Veb and Vbe respectively are what define the transistors in the active region.

    If you have a copy of Sedra and Smith (5th Edition), or can get to see one in a library, have a look on page 393 for a more detailed explaination. Earlier editions will have the same explaination under the title current-voltage characteristics of bipolar transistors.

    I hope I'm making a little more sense, if you still don't understand anything, for example, why Veb and Vbe must be positive, post back and I will try to explain in more detail. These concepts are fundamental elements in understanding BJT and electronics in general.
     
  6. nanobyte

    Senior Member

    May 26, 2004
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    Hi Dave. Could you go in more detail for me please?
     
  7. dragan733

    Senior Member

    Dec 12, 2004
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    When I said negative voltages I thought the voltage of the base of the NPN transistor in reference to the emitter of the NPN transistor. In the example: Vb=3,9V<Ve=4,6V and this voltage is negative in refference to the emitter of the NPN transistor. When one mesures the voltage of the emitter of the NPN transistor in reference to the base of the NPN transistor this voltage is positive. In the praxis the emitter of the NPN transistor is connected to positive voltage, and the NPN transistor always conducts on voltages on its base lower from the voltage on its emitter, then this voltage of the base is negative in refference to its emitter.
     
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