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Vbe, Darlington Pair 3v?(?)

Discussion in 'General Electronics Discussion' started by The New Guy, Nov 21, 2020.

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  1. The New Guy

    The New Guy

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    Jun 6, 2017
    Hey guys,

    I am just a hobbyist, and this is the first time I have looked into using a Darlington NPN transistor, a TIP142.
    I want to use it as an on/off switch (no amplification). I want to use this switch to turn a high power infrared LED (100w), at 3.5 amps, completely on and completely off. I furthermore want to pulse the base, held at a Thevenin voltage (voltage divider bias) with square waves derived from a 555, using capacitive coupling.
    Everywhere I look on the web, every article says that the Vbe drop for a silicon Darlington is approximately 1.4v (2x.7v), and the Vce(sat) drop is approximately .1 to .2 volts + 1 Vbe(.7), for a rough total of .8 volts.
    However, in the attached datasheet, I see a Vbe(on) of 3V, A Vbe(sat) of 3.5v, and a Vce(sat) of 2-3V.
    Can anyone help me to understand this? Is this just a result of doping the silicon?
    All I know is that I want to use the Darlington transistor as a switch at ELF frequencies.
    All of your help is appreciated.
     

    Attached Files:

  2. Bluejets

    Bluejets

    5,029
    1,050
    Oct 5, 2014
    If you want to pulse the base on and off, why hold it at any voltage other than ground?
    Why use a darlington if it is driven by a 555?
    Perhaps think of using an IRLZ44 mosfet.
     
  3. Harald Kapp

    Harald Kapp Moderator Moderator

    11,702
    2,717
    Nov 17, 2011
    A MOSFET as suggested by @Bluejets is definitely the better solution.
    But to answer your question, look at how a Darlington pair is constructed (from the datasheet):
    upload_2020-11-21_14-4-46.png
    To drive the base of the rightmost NPN transistor, the base voltage needs to be ~0.7 V (give or take a few mV, doesn't matter for the consideration). This is the voltage at the emitter of the left NPN transistor. Assuming a saturation voltage Vce(sat) = 0.1 V, the collector of the left NPB transistor will be at ~0.8 V (VC = VE + VCE(sat). As both collectors are connected, this will also be VCE of the right NPN transistor and thus VCE of the darlington pair.

    While a Darlington pair give high gain and is simple to use, this high VCE(sat) is a ral drawback in power hungry applications. Either use a MOSFET, as suggested, or use 2 discrete NPN transistors. Connect the collector of the left NPN to positive supply via a series resistor to limit the base current into the right NPNB transistor. In this way VCE(sat)of the right transistor will be around 0.1 V only and power dissipation will be greatly reduced.
    upload_2020-11-21_14-11-51.png
     
  4. bertus

    bertus Moderator

    2,060
    777
    Nov 8, 2019
    Hello,

    @Harald Kapp , I think you have shorted R1 in the diagram.

    Bertus
     
  5. The New Guy

    The New Guy

    8
    0
    Jun 6, 2017
    Thank you! That totally changed my approach.
     
  6. The New Guy

    The New Guy

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    0
    Jun 6, 2017
    Thank you for your help.
     
  7. Harald Kapp

    Harald Kapp Moderator Moderator

    11,702
    2,717
    Nov 17, 2011
    Yes indeed, thanks for noticing. Corrected version:
    upload_2020-11-21_16-56-17.png
     
  8. Audioguru

    Audioguru

    3,394
    733
    Sep 24, 2016
    Transistors have a range of specs. Some have minimum specs, some have mid and some have maximum specs.
    The maximum Vbe for the TIP140 is 3V at room temperature when the collector current is 10A (!) and the transistor has minimum specs. The graph in the datasheet shows 2V for one with mid specs and shows a mid spec one with a Vbe of 1.4V at 1A. The graph shows the Vbe dropping as the temperature rises.

    Why feed its base voltage when its base needs a certain current??
    Design the circuit for a TIP140 that has the worst specs for base voltage and current then ALL TIP140 darlingtons will work perfectly.
     
  9. The New Guy

    The New Guy

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    0
    Jun 6, 2017
    Thank you!
     
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