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Experiment with NPN transistor BC547

Discussion in 'General Electronics Discussion' started by Djsarkar, Nov 1, 2020.

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  1. Djsarkar

    Djsarkar

    46
    4
    Jul 27, 2020
    Hi
    I am trying to switch on LED using transistor. I have a one BC547 transistor, potentiometer 470 ohm resistor and LED.

    When I adjust pot LED doesn't turn ON What's wrong with connection?

    bc457.jpg
     
  2. bertus

    bertus Moderator

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    Nov 8, 2019
    Hello,

    Can you post a schematic, in stead of the breadboard?
    Schematics are the universal electronic language.

    Bertus
     
    Djsarkar likes this.
  3. Djsarkar

    Djsarkar

    46
    4
    Jul 27, 2020
    okay here is my schematics diagram

    sc.jpg
     
  4. bertus

    bertus Moderator

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    Nov 8, 2019
    Hello,

    I hope you did not turn the potmeter fully to the +5 Volts.
    If you did so, the transistor might not have survived the action and got burned out.
    The base current may not exceed 20 mA.

    Bertus
     

    Attached Files:

  5. Djsarkar

    Djsarkar

    46
    4
    Jul 27, 2020
    Hi Bertus
    What's wrong in my connection ?
    How to verify transistor weather its okay or burnt ?
     
  6. bertus

    bertus Moderator

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    Nov 8, 2019
  7. bertus

    bertus Moderator

    1,282
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    Nov 8, 2019
    Hello,

    If you want to make better looking schematics, you can copy paste symbols from this sheet into your schematic drawing:
    Electrical_symbols_library.png

    Your schematic could look like this:

    DJS_led.png


    Bertus
     
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  8. Djsarkar

    Djsarkar

    46
    4
    Jul 27, 2020
    Hi
    I have attached test result. Does led will damage if I set pot to full 5 volt?
     

    Attached Files:

  9. bertus

    bertus Moderator

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    Nov 8, 2019
    Hello,

    No, the led is protected by the 470 Ohms current limiting resistor.
    The transistor can be dameged by a to large base current.
    When you put a resistor of 1K between the top of the potentiometer and the +5 Volts, the transistor is protected from a to high base current.
    DJS_led base protected.png
    Bertus
     
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  10. Djsarkar

    Djsarkar

    46
    4
    Jul 27, 2020
    Hi
    I want to perform experiment to see how the transistor work as switch. I have attached a circuit diagram
     

    Attached Files:

    Cannonball likes this.
  11. bertus

    bertus Moderator

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    Nov 8, 2019
    Hello,

    That should work fine.
    The base current is limited by the 1K resistor.
    The led is protected by the 470 Ohms resistor.

    Bertus
     
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  12. kpatz

    kpatz

    318
    84
    Feb 24, 2014
    Your "transistor as a switch" circuit should work fine. Try it out!

    One characteristic of transistors you should know about is that they have what's called "gain". Which means a small current flowing from base to emitter can control a much larger current from collector to emitter. The BC547 has approx 100-800x current gain, so you could use a much higher resistor on the base and the transistor will still turn on fully. For example, a 100K resistor would work, though 10K is more commonly used to ensure the transistor is fully "on" even if you're switching larger currents. For your little LED, a higher base resistor will work no problem.

    With your circuit using a pot, you can see how the transistor responds over its "linear region" as well. This is the region in between "fully off" and "fully on". When the B-E voltage reaches 0.55V or so the junction starts to conduct and the transistor starts to turn "on". At around 0.7V it will reach saturation and be fully "on". This is a narrow range near the bottom travel of your pot, where it transitions between fully off, partially on, and fully on.

    Since a transistor can control a much larger current using a smaller current, it works as an amplifier, like those in radios, cell phones, etc., and it can also be used to allow a small voltage/current to switch on/off a larger current.
     
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  13. Cannonball

    Cannonball

    191
    53
    May 6, 2017
    You may instead do away with the pot and use a 10k resistor from collector to the switch and from the switch to the base. That should work as well.
     
  14. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    This will work but will not be efficient: When the transistor turns "on", Vce will fall towards 0 V thus reducing Vbe (via Rcb from collector to base). As the transistor needs Vbe = 0.6 V ... 0.7 V to turn on, Vce cannot fall below this value.
    Better: Connect the resistor to V+ as shown by Bertus in post #9. Here the transistor can go into full saturation where Vce = 0.1 V (in the ballpark of), thus reducing power dissipation across the transistor and supplying 0.5 V ... 0.6 V more to the load.

    In the case of a simple LED as load this may not have a noticeable influence at all as the load is small. But if you were to drive a high load through e.g. a power transistor, every millivolt counts. At a load current of e.g. 1 A (! the BC547 will not be able to drive this current!), a difference of 0.5 V in Vce results in 0.5 W power dissipation that would have to be taken care of.
    This is one reason why Darlington transistors are not well suited to drive high current loads despite their high gain.
     
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  15. ratstar

    ratstar

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    13
    Aug 20, 2018
    One more thing to add, if you dont put a resistor with the potentiometre, youll actually steal all the current from the collector as well, because itll short to the base.

    And it definitely seems like a situation where the transistor would be fried.

    But if the 2 5v entry points, where separate isolations, this current theivery actually doesnt happen, but u probably would have fried anyway.
     
  16. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Huh? Makes no sense at all, please explain.
     
  17. ratstar

    ratstar

    308
    13
    Aug 20, 2018
    if you have 2 isolated positives, (must go with) 2 isolated negatives. the current will add up on the lines, and they only short circuit within themselves.
    To get an isolated pos+neg, you need a capacitor, a battery, or a transformer.
    1 isolation, will only ever attract to itself, and 2 isolations can literally pass straight through each other (in a subtraction) and pop out the other sides full strength again.

    If youve never known that your whole life, it may be a bit of a shock to you...
     
  18. Harald Kapp

    Harald Kapp Moderator Moderator

    10,770
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    Nov 17, 2011
    I dare think I know a tad more about electronics than you do :rolleyes:
     
    ratstar likes this.
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