.1v saturated on darlington?

Discussion in 'General Electronics Chat' started by Mrdouble, May 12, 2015.

  1. Mrdouble

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    Aug 13, 2012
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    I have a simple circuit to experiment with a darlington transistor MJE183.
    I am reading Vce of .1v and I only have Ib of 2ma. When I remove base current the expected vcc Is present. I trying to use it to drive a small 3v motor. I swapped that transistor out (thinking inductive kick may have killed it) but new one did same thing. Any ideas?

    Thanks In advance.
    Micheal.
     
  2. blocco a spirale

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    Jun 18, 2008
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    Sounds like it's working ok, what Ib do you think you need ?
     
  3. Jony130

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    Impossible. For Darlington Vce(sat) voltage cannot be smaller than Vbe.
    And your BJT is ??
     
  4. Papabravo

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    Are you sure about that?
     
  5. Jony130

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    Yes I'm sure. For darlington transistor Vce(sat) cannot be smaller than Vbe.
     
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  6. wayneh

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    Even I know that's not true. You may be confusing it if it's in an emitter-follower configuration. Been there, done that.

    Nevermind, I misunderstood what we were talking about.
     
    Last edited: May 12, 2015
  7. Papabravo

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    Any saturated transistor, including a darlington, can have Vce(sat) very close to 0.1 to 0.2V with Vbe at 0.7V. Try it in simulation or on a breadboard with a forced beta of 10.
     
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  8. crutschow

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    No, Jony130 is correct.
    Look at the Darlington configuration. When fully on, the top transistor will saturate which then is the same as connecting the collector of the bottom transistor to its base. Thus the bottom transistor Vce can not be less than its Vbe.
    You should follow you own suggestion.
    Below is a simulation. As you can see the fully-on Vce is nearly .77V.

    Darlington.gif
     
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  9. Jony130

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    I do not have to do any simulations to know for sure.
    For me the "classic" Darlington connection look like this:

    [​IMG]
    And for this circuit Vce(sat) cannot be larger then Vbe. You do not see why is that ?

    We have completely different situation for this circuit:
    12.PNG
     
  10. joeyd999

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    Jun 6, 2011
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    Wikipedia:

     
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  11. WBahn

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    Mar 31, 2012
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    For the classic Darlington Vcesat has to be no less than Vbe + Vcesat for the input transistors and the output transistor can't saturate (it's the input transistor that saturates).

    You can break the collector connections, as Jony130 shows above to let the output transistor saturate, but that requires a four-terminal device.

    I've been trying to find the datasheet for the MJE183 and have had no luck. Anyone else able to find it?
     
  12. blocco a spirale

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    I think the MJE183 must have been discontinued for some considerable time.

    Reading through the TSs original post, I still can't figure out what the question is.
     
  13. ian field

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    Oct 27, 2012
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    That was my understanding too.

    If the output transistor were to saturate down to 0.4V - that is also the input transistor's collector. Ignore for the minute, the VCEsat of the input transistor, its emitter needs to achieve at least 0.7V to produce said saturation of the output transistor.
     
  14. Mrdouble

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    Aug 13, 2012
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    My question, is Vce of .1v possible? Not that I've read. Thank you for all your responses.
     
  15. WBahn

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    It's actually not quite that simple.

    The floor on Vcesat for a Darlington is based on the premise that the base current of the output transistor is provided predominantly by the collector current of the input transistor. However, it is possible to overdrive the input transistor so heavily that there is little to no collector current in the input transistor and all of the emitter current is due to the base current. At that point you are effectively replacing the input transistor with a diode, in which it is obvious that the output transistor CAN saturate.

    How much current is the Darlington supplying to the motor? If it is anything less than, say, 50 mA to 200 mA then you might well be driving it that hard with 2 mA of base current.
     
  16. ian field

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    Short the base/emitter - that should switch the transistor off so it doesn't pass any current.

    If the collecter is fed by a voltage through a current limiting resistor, the collector voltage should rise - if not, the transistor is shorted.
     
  17. crutschow

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    Mar 14, 2008
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    And Aristotle said it was obvious that a heavy object would fall faster than a light object.

    They both sound plausible but there's a hitch in those deductions.
    For the Darlington, when you overdrive the input transistor the excess base current goes through the now forward-biased base-collector junction, not the base-emitter, which keeps the output transistor from saturating.

    Below is a simulation of this. You can see that the collector current of Q1 is negative (current out of the collector) until Ibq1 equals Ibq2. It's only after Ibq1 is less the Ibq2 that Q1 actually starts to amplify its base current and act like a Darlington with gain higher than Q2 by itself..
    Note that Ibq2 is nearly constant as the base current of Q1 varies over a wide range.
    And Vc never goes below 768mV.

    So no matter what you do to a Darlington's input current, the Vce minimum will never be lower that about one base-emitter drop.

    As to Aristotle's deduction, that was disproved several centuries ago, as Aristotle hadn't realized that the inertia of an object goes up by the same factor as its attraction by gravity. ;)

    Darlington.gif
     
    Last edited: May 13, 2015
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  18. Mrdouble

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    Aug 13, 2012
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  19. Mrdouble

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    Aug 13, 2012
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    I am absolutely so sorry if this is not a darlington I notice it doesn't exactly say. Just thought it was based on current capacity
     
  20. DickCappels

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    Aug 21, 2008
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    Still, this is an interesting thread and I for one was surprised at crutschow's observation.
     
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