NPN Transistor in Cut-Off & Saturation Function.

Discussion in 'Homework Help' started by SanRath, May 17, 2011.

  1. SanRath

    SanRath Thread Starter New Member

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    Oct 28, 2010
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    Dear All.

    This is Sachin Rathod from Mumbai, India.

    Can you explain me the basic idea of NPN Bipolar Junction Transistor in Cut-Off & Saturation Mode?

    Waiting to have a help from you very soon.

    Thank You,
    Sachin Rathod.
  2. jegues

    jegues Active Member

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    There is plenty information on these topics available on the web.

    For example a quick wiki gives the following,

    http://en.wikipedia.org/wiki/Bipolar_junction_transistor

    Anyways I'll give you a quick idea of what I understand of the two modes of operation.

    Saturation mode and cut off mode are commonly used amongst switching applications (e.g., logic circuits)

    As the name implies, in the cutoff mode no current flows because both junctions are reverse biased.

    In saturation mode, both junctions are forward biased.

    In active mode, the collector current Ic is independent of the collector base voltage Vcb, this situation extends for Vcb going negative to approximately -0.4V.

    Below this value of Vcb, the CBJ begins to conduct sufficiently (i.e. it is said to saturate) that the transistor leaves active mode and enter the saturation mode of operation, where Ic decreases.

    This change in Vcb (or conversely, Vbc) and decrease in Ic causes the ratio for the DC current gain, β to decrease.

    The \beta_{forced} of a saturated transistor can be tweaked to any desired value lower than β by adjusting Vbc.

    \beta_{forced} = \frac{i_{c}}{i_{b}}\right|_{saturation} \leq \beta

    To determine wheter the BJT is in saturation one of asks himself either of the two following questions.

    Is the CBJ forward biased by more than 0.4V?

    Is the ratio of \frac{i_{c}}{i_{b}} lower than β?

    The collector-emitter voltage of a saturated transistor is simply the difference between the forward-bias voltages of the EBJ and CBJ,

    V_{CEsat} = V_{BE} - V_{BC}

    This is typically in the range of 0.1V to 0.3V.

    I hope this helps!
  3. Jony130

    Jony130 Senior Member

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    When the BJT is in Cut-Off mode, no current is flow through BJT.
    Simply the BJT is in OFF state.
    The saturation is opposite to Cut-off.
    In saturation increase of the base current (or voltage) causes no further change in collector current (or voltage).
    In saturation the BJT is in full open (ON), conduct the max. allowed current by the load.
    And If NPN transistor is full ON (saturated) then he act just like a short (a switch). Short collector to GND. Or in general short collector to emitter.
    And in real circuit the collector to emitter voltage in saturation will not gona be equal 0V. But will be equal Vce(sat) = 0.2V typical.

    [​IMG]



    PS. jegues witch book prefer such a "strange" definition of saturation ?
    I never check Vcb voltage to see if BJT is in saturation region.
    Last edited: May 17, 2011
    anhnha likes this.
  4. jegues

    jegues Active Member

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    My definition comes out of the Sedra/Smith text.

    Usually I'd start off the DC analysis with the assumption the transistors in active mode, and verified this by looking at the voltage across the CBJ in order to reassure myself the transistor isn't in saturation.
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