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Transistor-saturation-input characteristics

  1. Jan 16, 2012 #1
    So I want to discuss a few things about transistors in saturation region.

    First the saturation part.

    Lets say the saturation occurs at 0.2 V and transistor cannot go below that voltage. We say that it is bottomed and its now in saturation region.

    How do charges go from base to collector? Does transistor effect still occur?
    Let me clarify what I mean.

    I know when in linear polarization, charges, get injected into base-collector depletion region and get swept across it because of large electric field. (without much discussion, lets say that this the effect)

    But when we have saturation region, base-collector is directly polarized, and charges don't have that depletion region to get swept across.

    Do they then, pass over to collector normally like in diodes?

    I generalized this pnp and npn. You may answer like so, too.

    Now for the second part, which is closely related.

    http://pokit.org/get/c4446cd88893f30ed96ced2ec01d0b14.jpg [Broken]

    I was told that this is input characteristic of a BJT. I believe these curves represent the diode characteristic of base-emitter.

    I want to discuss the difference between 2 curves, when transistor is in linear and saturation region.

    My explanation of the curves is:

    When transistor is in saturation region, base only has depletion region from emitter-base junction. And indeed, this part really acts like a diode, and will have that characteristic. Peaks out at 0.7 V

    But when transistor is in linear region, we have that depletion region from collector-base too. So base is being "squished" between 2 depletion regions.

    Because of this, we will need a larger voltage drop, to achieve the same current as in saturation region case. In a nutshell, base "offers" smaller path for current to go through and ergo we need more voltage to get more current?

    Am I thinking right here?
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 16, 2012 #2
    Yes, you are very correct. In forward active the BC jn acts like controlled source. In saturation the transistor is just two back to back diode.
  4. Jan 16, 2012 #3
    Thank you, makes sense. Any thoughts on the second part?
  5. Jan 16, 2012 #4
    Base - emitter junction is essentially pn junction. Thus it will not change its current until and unless its properties are changed (by collector). Usually emitter is heavily doped n-type. Thus approximately all Ibe is carried by electrons. Now electrons in base is minority carriers. And its concentration gradient (this quantity is proportional to the base emitter current) depends on how close is the CB depletion region. Actually we deliberately keep base width small such that collector can easily control base carriers. Otherwise it wont work.
  6. Jan 16, 2012 #5
    Yes, I know that. But the curves, is the explanation right? You do see the difference between first and second curve?
  7. Jan 16, 2012 #6
    If base offers smaller path for current flow and you consider drift transport, would not the current increase due to lower resistance of smaller base?

    Diffusive transport cannot be thought of as a simple intuitive rule. Try doing that for a pn junction.

    The point here is the bjt diffusive transport has a parameter minority carrier. In linear they have to vary very steep as base is small. In saturation they can relax and even store extra charge due to large base width. The greater the slope of the minority carrier the more "diffusion pressure" the more current.
  8. Jan 16, 2012 #7
    Ok thanks, I see I will have to dig a little more deep into this base and minority carrier thing.

    Thank you very much, kind sir, you have been very helpful.
  9. Jan 16, 2012 #8
    You're welcome. :smile:
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