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Transistor mystery

Discussion in 'General Electronics Discussion' started by Ecce_lex, Nov 30, 2016.

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  1. Ecce_lex

    Ecce_lex

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    Nov 30, 2016
    Hello,

    I seek your guidance in trying to understand something related to transistors. Below is a picture of the circuit.

    I have two NPN transistors sharing the collector pin. One transistor receives constant 5v to is base, the other is connected to the switch (control voltage). A LED is connected to the Q2 emitter and ground.

    How it works: when 5V is applied to Q1, the LED turns off; when the switch is off, the LED turns on.

    I drew a red line where I think current flows but I am clueless what happens in case B in Q2. Would any of you be so kind as to explain ?

    The transistors I'm using are these: http://www.onsemi.com/pub_link/Collateral/BC546-D.PDF
    R=460ohm
     

    Attached Files:

  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,448
    2,809
    Jan 21, 2010
    That's a pretty bizarre circuit.

    Q2 acts as a pair of diodes with current flowing through the led via Q2's base resistor and the base emitter junction.

    When you choose the switch, Q1 turns on and current now flows through the base collector junction of Q2 and then through Q1. The voltage across Q1 is less than the forward voltage of the led, so the led turns off.

    I won't say that a transistor is never used like Q2 here, but I can't recall ever seeing it. A pair of diodes would perform the same task, and do it better.
     
    Last edited: Dec 1, 2016
  3. Ecce_lex

    Ecce_lex

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    Nov 30, 2016
    Hello *steve*,

    Thank you for the quick reply.

    I understand that in case A (switch OFF) current flows in Q2 from base to emitter, then through the LED. This makes sense.

    In case B (switch ON) current flows in Q1 from base to emitter. This also makes sense. In Q2 does current flow from base to collector? How about from base to emitter?
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    2,809
    Jan 21, 2010
    Q2 is a complete distraction. Short all three leads together and the circuit will operate essentially the same.

    In Q1, when you have a current flowing in the base emitter junction you set to the transistor to allow current to flow from collector to emitter. In this case the practical upshot is that you short out the led (it's slightly more complex than that, but not much).

    As I said earlier, Q2 is used in a weirdly bizarre way that I doubt anyone who knew what they were doing would do -- except maybe as a puzzle.
     
  5. AnalogKid

    AnalogKid

    2,425
    690
    Jun 10, 2015
    The circuit is not drawn in a traditional way, so it is a bit hard to understand at a glance. I've re-drawn it below. This is not a "normal" circuit, in that there is no source for collector current (no connection between the collectors and Vcc through a resistor, inductor, etc.).

    When SW1 is open, Q1 is off. The current path to the LED is through the Q2 base-emitter diode. Q2 is acting as a simple diode, and there is no collector current, transistor action, or current gain.

    When SW1 is closed, Q1's collector pulls Q2's collector to GND. This provides a bypass path for the Q2 base current, shunting it away from the LED. How? Because deep down inside, a NPN transistor is built as two diodes back-to-back. When a transistor is not properly biased, you can drive the base-emitter or base-collector junctions as not-very-good diodes. So there are two current paths from the Q2 base to GND: a) through the emitter and a LED with a 2 V forward voltage, or b) through the collector and a saturated transistor with a 0.1 V forward voltage.

    In a more standard circuit, there would be a resistor at point X. With this in place, Q2 acts a saturated emitter follower when SW1 is off. 99% of the LED current comes through resistor X and the Q2 collector. Q2 is saturated, so there is about 0.1 V from the collector to emitter. The lower end of the X resistor is now the LED power source. When SW1 closes, Q1 turns on and shorts this source to GND, again diverting the LED current through a lower voltage-drop path.

    ak
    TransistorMystery-1-c.gif
     

    Attached Files:

  6. Ecce_lex

    Ecce_lex

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    0
    Nov 30, 2016
    Imagine the puzzle when you *don't* know what you're doing :)

    I started with a circuit that had the LED OFF when no voltage was supplied. I wanted to build one where the LED would be ON when no voltage was supplied.

    I built this, it worked - somehow - and then I was unable to understand what I did. So yes, it may very well be bizarre.

    I shall persevere and may come back with more questions. Maybe I'll find a circuit simulator somewhere.

    Thank you for taking the time.

    Cheers,
     
  7. Ecce_lex

    Ecce_lex

    5
    0
    Nov 30, 2016
    AnalogKid: Thanks for the re-draw. So if I understand correctly all I did is that I created a path of least resistance through the saturated Q1 when flipping the switch on. The current will not flow through the LED but choose to "short" through Q1.

    The current through Q2 was sufficient to go against the forward-biasness and actually travel from base to collector I suppose.

    Indeed, not very elegant :)

    I shall try to put a resistance where you suggested.

    Thank you
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,448
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    Jan 21, 2010
    "The path of least resistance" does not apply to electronics.

    And the rest of what you say suggests you don't understand transistors.

    Perhaps you should try to learn the basics first?

    What are you actually trying to do? And I don't mean "turn on a led without power applied", I mean the bigger picture.
     
  9. Ecce_lex

    Ecce_lex

    5
    0
    Nov 30, 2016
    Was it when I said "I know nothing about this" when - in your immense wisdom - realised I knew nothing about electronics?

    Listen, buddy, I'll tell you what i wanna do - the bigger picture. I wanna share the things I know with others - like AnalogKid, who took the time to look at what I did, redraw and improve it. He actually made me learn things, even if on a very basic level. Very important for someone who is clueless like me.

    I'm also trying to be unlike you, on your high horse, bashing people cause you're better. All you did was stress the fact that you're the best, and I'm clueless. Zero knowledge shared.

    Get it?
     
    Last edited: Dec 1, 2016
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,448
    2,809
    Jan 21, 2010
    Hey, all I wanted to know is whether you wanted to learn electronics or you had some other problem you were trying to resolve.

    I hope you'll forgive me for giving you essentially the same answer that ak did (twice), and accept my profound apologies for being on a mobile device at the time where I could not redraw your circuit.

    I'm sorry if I interpreted your experimentation and reasonable choice of component values as meaning that you appeared to know a little more that the nothing you professed to know.

    My first piece of advice to you is that it is hard to read the emotional content of postings, and it is wise NOT to assume that people putting effort into answering your questions are doing it out of malice (because that would just be weird).

    My second piece of advice is that it's probably unwise for a person with only a handful of posts under their belt to unload on a moderator.

    So back to my questions...

    If you feel you'd like to give us a little more information then we may be able to offer more assistance or point you in the right direction to achieve some goal.

    If you read something into my posts that I didn't intentionally put there, send a message to any of the moderators. I'm sure any of us would be happy to remove your post, my response, and forget it ever happened.
     
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