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BJT as switch...

Discussion in 'General Electronics Chat' started by koolguy, Sep 8, 2013.

  1. koolguy

    koolguy Active Member

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    Hi,

    I am using BC337 which is capable of driving 800mA load, i am using 9V 2A power.
    The Load is LED connected to collector of BJT when i connect 110ohm 2Wattt Resistance it get Hot slowly and the brightness is also low. i need more bright LED and 20K variable R at the base from Vcc of 5V.
    so, i need to know all LED of Red colour are connected in Parallel with each other there are ~250 LED

    please tell how to increase the intensity of light from Led without changing the circuit of LED..?
     

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  2. Ian Rogers

    Ian Rogers Super Moderator Most Helpful Member

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    250 LED's on one 110 resistor!!!!! Surely not..... I would also redraw your circuit as you WILL get negative remarks...


    Each LED can vary from @1mA to @30mA.... Assume 20mA per LED... 250 * .020A = 6.25A

    You are going to need about 8 more transistors... But that will be driving 31 LED's..

    31*20mA = 625mA....
     
    Last edited: Sep 9, 2013
  3. koolguy

    koolguy Active Member

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    When i make two sets of LED with two resistance they get hot more than single..
     
  4. dave

    Dave New Member

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  5. Ian Rogers

    Ian Rogers Super Moderator Most Helpful Member

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    Have you heard of "Ohms law".... You are trying to get blood out of a stone....
     
  6. koolguy

    koolguy Active Member

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    OK, should i connect 15 led per 110watt resistor of 2watt..
     
  7. koolguy

    koolguy Active Member

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    Actually there are 1500LED so, i am confused with it and Pic is used to flash Led so, R should be connected at collector side before LED or to Vcc directly to LED?
    I am controlling sinking with BJT not sourcing.
     
  8. koolguy

    koolguy Active Member

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    I am using oval led which work fine at ~2mA. and BJT is not getting hot only resistance.
     
  9. Ian Rogers

    Ian Rogers Super Moderator Most Helpful Member

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    15 * .02A = 300mA ...

    You will need about 2 watt (with heat sink) 15Ω resistor..... You'll need to generate about 371mA to run 15 LED's at about 20mA each..... You'll need a heat sink on the BC337 aswell..
     
  10. koolguy

    koolguy Active Member

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    Why are you taking 15ohm ? and 20mA b'coz led is working fine in few ~4mA
     
  11. Ian Rogers

    Ian Rogers Super Moderator Most Helpful Member

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    Right!!!! As you really struggle with electricity, power and current...

    A bog standard LED can run as high as 30mA....... It can also run as low as 1mA.... What would be the point of adjusting the brightness if you didn't want the full brightness...

    Look at YOUR circuit..... You have parallel LED's on top of a resistor on top of a transistor.... Using the transistor as a variable resistor, you want to vary the current from 15mA to 371mA!!!

    Therefor if you use a 15 ohm resistor you can control the voltage across the resistor by varying the voltage across the transistor.... The way YOU had it the resistor was doing all the work and the transistor wasn't.... I would have thought it obvious why the resistor was getting hot... but you don't understand how to transfer power....

    Sorry for the rant.... but you don't listen.....
     
  12. koolguy

    koolguy Active Member

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    So, what is your suggestion?
     
  13. Ian Rogers

    Ian Rogers Super Moderator Most Helpful Member

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    Well...... That say's it all..... What do YOU think is my suggestion.... Try virtually all my posts...
     
  14. rumpfy

    rumpfy Active Member

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    Ritesh,
    You cannot single LED in parallel.
    Each LED will require about 1.6 volt and depending on the LED, will handle up to around 20 mAmp. You need to get the data sheet for the types you are using.
    Assuming each LED is 1.6 volt, you can connect 6 LED's in series and connect them directly to your 9 volt supply. In this configuration, the LED current will be around the same value as the current rating for each LED. If the current is too high, then connect 7 LED's in series. You need to test this with a meter.
    The transistor BC337 is rated at a collector current of 500 mAmp at a junction temperature of 150 degree C.
    The base current has to be 50 milliamp for this collector current. At 50 mA, the base emitter voltage is 1.2 max. The base drive resistor will be 150 ohm for your 9 volt supply.
    this base resistor will allow the collector to carry 500 mA. The maximum collector emitter voltage is then 0.7 volt.
    If each group of 6/7 LED's carries 20 mAmp, the you can connect 25 sets of LEDS in parallel on your 9 volt supply.
    Each group of LEDS in this arrangement will have different levels of brightness due to differences in the forward voltage of each LED. If you want to improve this, you need to stabilise the current in each group of LED's by arranging say only 4 LED's in each group, and using a series resistor to give say 20 mA in each group of LED's. That would need a resistor of say 130 ohm for each group of 4 LED using 20 mA led's.
    hope this helps.
     
  15. koolguy

    koolguy Active Member

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    I don't want to make series connection, as making parallel connection it is easy to solder all in one adding series will complicate it.
     
  16. koolguy

    koolguy Active Member

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    OK, adding two 2watt 110ohm Resistor in // will do something will it dissipate less heat?
     
  17. Ian Rogers

    Ian Rogers Super Moderator Most Helpful Member

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    You can.... I do it with 20 LED's in one project and 8 LED's in another..... Its perfectly ok..

    Ritesh...... You will be able to place 4 resistors in parallel... to increase your current... that's the same as 1 27 ohm 2 watt resistor..
     
  18. koolguy

    koolguy Active Member

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    Ok Now i have added 10watt 56Ohm Resistor little bit getting hot..
     
  19. Ian Rogers

    Ian Rogers Super Moderator Most Helpful Member

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    Where have you added it???
     
  20. koolguy

    koolguy Active Member

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    at collector removed that 110ohm and 2watt
     
  21. Ian Rogers

    Ian Rogers Super Moderator Most Helpful Member

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    [MODNOTE] wrong!!![/MODNOTE]

    Whoops sorry 56 ohm not 560 ohm.... I'll take a another look.
     

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