BJT as switch...

Discussion in 'General Electronics Chat' started by RRITESH KAKKAR, Sep 9, 2013.

  1. RRITESH KAKKAR

    Thread Starter Senior Member

    Jun 29, 2010
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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..?
     
  2. ScottWang

    Moderator

    Aug 23, 2012
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    1. 9V 2A
    Assuming the V_LED=3V, I_LED=20mA, 9V can be in series with two LEDs and a 180Ω resistor.
    250 LEDs divided by 2 = 125 string, I_Total = 20mA x 125 = 2.5A

    2. BC337 which is capable of driving 800mA load.
    The BC337 should be using less than Ic=800mA/3=267mA is better.
    So it seems to using a logical MOSFET is a better choice.

    3. 20K variable R (POT)
    When you using a pot to adjust the value of resistance, you should in series with a 1K resistor to avoid if the pot=0Ω, it could be damaged the bjt.

    So, how about the Vf/I of LED?
     
  3. RRITESH KAKKAR

    Thread Starter Senior Member

    Jun 29, 2010
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    I want to avoid series connection and driving oval LED at ~4mA not 20mA..
     
  4. ScottWang

    Moderator

    Aug 23, 2012
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    Then how is the Vf for LED?
     
  5. RRITESH KAKKAR

    Thread Starter Senior Member

    Jun 29, 2010
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    Min should be ~3V ..
     
  6. ScottWang

    Moderator

    Aug 23, 2012
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    LED:3V/4mA

    250 LEDs divided by 2 = 125 string, I_Total = 4mA x 125 = 500mA
    each BC337 for LED string : 125 / 2 = 62.5.
    I_Q1 = 4mA x 62 = 248 mA, one BC337 for 62 string,
    I_Q2 = 4mA x 63 = 252 mA, another one BC337 for 63 string.

    V_R = 9V - (V_LED*2) = 9V - 6V = 3V
    I_R = 3V/4mA = 750Ω
    I_Rb = Ic / 10 = 4mA / 10 = 0.4mA

    Rb=(5V - Vce)/0.4mA = 4.3V/0.4mA = 10.75K
    so Rb can be 10K+750Ω or VR10K+1K
    And you should using two BC337 is better.

    [​IMG]
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    A Vf of 3V for a red LED? Those are usually closer to 2V. Where is that number coming from?

    Let's assume that your transistor is hard on and has negligible voltage across it. Now let's assume 2V across the diode array. That leaves you with 7V across the resistor and, with a 110Ω resistor, you should have somewhere around 64mA (probably closer to 60mA is more likely). If you have 250 LEDs in the array and, assuming that they share evenly (a VERY bad assumption, BTW), then you would have only about 0.25mA per LED, which would be quite dim.

    Your power dissipation in the resistor should only be about 0.5W, which should make a 2W resistor quite warm, but not "hot", depending on what you consider "hot", of course.

    Do NOT put LEDs in parallel such that they share the same current limiting resistor. This is asking for trouble. The reasons have been discused repeatedly in these forums, so do a search about LEDs in parallel. Another search term you might include would be "thermal runaway".
     
  8. RRITESH KAKKAR

    Thread Starter Senior Member

    Jun 29, 2010
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    Hi,

    thanks for calculation I h ave lot of 110ohm 2watt Resistor and 10k pre set so, can't we use it by changing some circuit to avoid 750ohms?
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    Put 7 of your 110Ω resistors in series to get 770Ω.

    Hey, you said you had a lot of them. :rolleyes:
     
    absf likes this.
  10. ScottWang

    Moderator

    Aug 23, 2012
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    1. If you dont' care about the size and space, yes, you can do as WBahn said.
    2. If you using 750Ω, then you just need 1/4w can afford it, so the size is very small, but if you using 110Ω/2W*7 in series to replace 750Ω/0.25W, then the space of resistor will be enlarge over 21 times.
    3. Another way is using boost circuit to increasing the voltage as 9V to 48 or some more, it will reducing the resistor of quantity.
    Examp as : 9V -- 3V LED x 2, resistor x 1
    48V can be using LED x 15pcs,
    Resistor = 250 LED / 15 LED = 16.7 = 17pcs.
    If you using 9V then you have to using 125 pcs of resistor, if you using 48V then you only need 17 pcs limiting resistor.
     
  11. RRITESH KAKKAR

    Thread Starter Senior Member

    Jun 29, 2010
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    The project is to make this type of led sign board.
    To avoid lot of heat/power diss. i will be using 5V 7805 9V to 5V and i think 1N4007 can be used to reduce .7 voltage to make less Diss across collector Resistor.
    so, if i will do this and make all parallel will it be fine.
    There is no such deep issue of making all exactly bright so, if there are some less/more brightness it will work..
     
  12. ScottWang

    Moderator

    Aug 23, 2012
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    You are using so many LEDs, if you using 7805 then the heat that you can't avoid it, if you using LM2576-5V then it will be better.
     
  13. RRITESH KAKKAR

    Thread Starter Senior Member

    Jun 29, 2010
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    I can use several 7805 regulator and i have found the all LEd 1500 are talking 400mA and giving good brightness at 9V the all LEd are solder so, i don't to make new connection and wire them all.
    Please tell and way to make it work.
     
  14. RRITESH KAKKAR

    Thread Starter Senior Member

    Jun 29, 2010
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    I want to know connecting series Resistance at collector or at Vcc will effect something or not??
    [​IMG]
    I have want to connecting Series Resistance to led at collector and at directly to vcc will affect anythin
     
  15. ronv

    AAC Fanatic!

    Nov 12, 2008
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    Are you trying to adjust the brightness with the pot?
    How many LEDs do you have for each transistor?
     
  16. RRITESH KAKKAR

    Thread Starter Senior Member

    Jun 29, 2010
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    The pic is from Google any way,
    Hi again,


    Today i have taken Vf for Led for white it is 3V and for Red is ~2V. and it is working well in 4mA to more than 80mA.
    so, my question is if i connect 2 Led in series and with 110Ohm V cc of 5V.
    so, 5-4=1v across 110ohm collector resistance of 2watt. p=V x i 2=1xi
    i=~2Amp so, if i make series of 2 Led so, how many parallel combination should be there??



    Hi again,


    Today i have taken Vf for Led for white it is 3V and for Red is ~2V. and it is working well in 4mA to more than 80mA.
    so, my question is if i connect 2 Led in series and with 110Ohm V cc of 5V.
    so, 5-4=1v across 110ohm collector resistance of 2watt. p=V x i 2=1xi
    i=~2Amp so, if i make series of 2 Led so, how many parallel combination should be there??
     
  17. ScottWang

    Moderator

    Aug 23, 2012
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    The below are the difference between 3V LED and 2V LED.
    3V LED,Imax = 20mA, using 16mA is better, high light.
    2V LED,Imax = 10mA, using 8mA is better, ordinary light.

    Do you have the datasheets for your 3V and 2V LED?

    If your calculation is right then you have to combined in parallel with 125 series.
     
  18. RRITESH KAKKAR

    Thread Starter Senior Member

    Jun 29, 2010
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    Yes i have measured for both white and red led...
    125 led with two in series ??

    I have want to connecting Series Resistance to led at collector and at directly to vcc will affect anything??
    [​IMG]
    I have want to connecting Series Resistance to led at collector and at directly to vcc will affect anything??
    [​IMG]
     
  19. ScottWang

    Moderator

    Aug 23, 2012
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    Go to 6# take a look the circuit.
     
  20. RRITESH KAKKAR

    Thread Starter Senior Member

    Jun 29, 2010
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    For every seies of two led i have to add 750ohm resistance rather than 110ohm very difficult.
    can we connect with one resistor
     
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