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Very simple question about transistors

  1. Nov 13, 2013 #1
    About the stages of a transistor.. Active,Saturation and Cutoff. Active its obviously on and cutoff off. May be a dumb question but is the transistor ON on Saturation mode?

    Also my textbooks says that when the BE junction is forward bias, Vbe= 0.7V. But at the same time in another problem it says that the cutoff voltage in the BE junction of an NPN transistor is Vbe=0.5V.

    Which leads to my question, What is the cutoff voltage of the BE junction in an NPN transistor, 0.7V or 0.5V? Is the 0.7 value given as cushion flor the transitor to be fully ON? I'm kind of confused about this whole thing so any help would be appreciated.

    And lastly, what is the difference of current behavior in Active with Saturation mode? Where is the current the largest?
    Last edited: Nov 13, 2013
  2. jcsd
  3. Nov 13, 2013 #2
    I've attached the datasheet for a popular transistor, the 2N3904. Note the values for VCE(sat) for Ic of 10mA and 50mA and VBE(sat) for Ib of 1mA and 5mA.

    Frankly an Ib of even 1mA is quite high for this transistor. Normally Ib runs in the tens of uA. I've measured VBE with normal Ib at room temperature and typically got around 0.65 V and it varies with temperature about -2mV/C. As you can see, VCE(sat) is lower than VBE(sat).

    The VBE(cutoff) can be anything down to the reverse breakdown voltage. As you will note it's not specified.

    The difference between the active region and the saturated region is that when a transistor is in the active region, a small change in Ib causes a much larger but proportional change in Ic. When a transistor is in saturation, a small change in Ib will not cause a perceptible change in Ic.

    Attached Files:

  4. Nov 13, 2013 #3
    I see, thanks a lot, I'll take a look at the file.
  5. Nov 13, 2013 #4
    To add to skeptic's answer, saturation mode simply means that the transistor is fully on.

    A transistor is somewhat like a relay if you only use cutoff mode and saturation mode.

    In the active region (triode mode) it's more like an amplifier because small changes in the input become large changes in the output.

    Vbe can vary due to manufacturing, temperature, and base current. 0.7V is sort of like the golden standard for doing calculations because it's mostly right. Unless you're told otherwise, assume that Vbe is 0.7V.
  6. Nov 14, 2013 #5
  7. Nov 14, 2013 #6
    Better yet, once you get a circuit working in the active region, measure it for yourself. Most engineers assume that Vbe is 0.7V and never measure it.
  8. Nov 14, 2013 #7
    I have a last question guys, and thanks for the input so far.

    Can the BJT transistor act as a voltage amplifier?
    Or does it just amplify current and power?
  9. Nov 14, 2013 #8


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    Science Advisor
    Homework Helper

    That depends how you interpret the question. The simplest way to understand the behavior of a transistor in the active region, ignoring everything else in the circuit, is that the collector current is a constant multiple of the base current.

    But if you are designing a complete working circuit, the objective is often to produce a voltage gain and the actual currents flowing are not so important. The basic idea is that if the current flows through a resistor in series with the transistor, the changes in current will produce changes in voltage across the resistor.

    Google for how to design a common-emitter amplifier stage, for the details.
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