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BJT as switch...

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

  1. audioguru

    audioguru Well-Known Member Most Helpful Member

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    You NEVER use a little pot for low values/high currents. Instead you use a fixed resistor calculated to do the job with its size big enough so it survives.
    With a +9V supply I calculated a series base resistor of 160 ohms and it will dissipate 0.4w. A half-watt resistor will be very hot so use a 1W resistor.

    A little pot is rated for about 0.5W for its entire resistive track, 0.25W for half of the track and 0.05W for one-tenth of its track. When the 20k pot is set to only 160 ohms then only 0.008 of its track (if it is possible to set it so low) must dissipate 0.4W which will destroy it.

    This thread is getting confusing because Ritesh will use a PNP transistor (with a +5V supply then +5V into the series base resistor turns it off and 0V turns it on) and other people are talking about NPN transistors (with a +9V supply then +5V or +9V into the series base resistor turns it on and 0V turns it off).
     
  2. ronv

    ronv Well-Known Member Most Helpful Member

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    Yes it is. As Ritesh has it drawn it will be pretty unstable.
     
  3. Ian Rogers

    Ian Rogers Super Moderator Most Helpful Member

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    Audio.... It has always been a NPN... the bc337 is a NPN transistor..
     
  4. dave

    Dave New Member

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  5. audioguru

    audioguru Well-Known Member Most Helpful Member

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    OOPs. You are correct.
    I hope Ritesh understands that for the transistor to saturate then its base current must be 1/10th the collector current for high currents and must be about 1/20th of the collector current for low currents. Then the PWM from his micro-controller might not have enough output current to drive the transistor at high currents.
     
  6. alec_t

    alec_t Well-Known Member Most Helpful Member

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    We've all been hoping that but the evidence so far isn't promising :banghead:
     
  7. koolguy

    koolguy Active Member

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    Hi again,

    I am using 12V transformer giving 18V and 6V giving 9.2V..
    i have connected 6 series of white Led with 2W 110 or 220ohm resistance working fine with 18V
    and Red led with 9V b'coz uC and series of 4 led working nice......
     
  8. Little Ghostman

    Little Ghostman Well-Known Member Most Helpful Member

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    i dont understand any of the above??? where do you get 9.2 from?
    RITESH why did you choose emitter follower configuration? i am interested in the design decision to go with this.
     
  9. Ian Rogers

    Ian Rogers Super Moderator Most Helpful Member

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    How are you measuring the voltage.... Chances are the peak voltage of a 12V transformer may be higher than 12V but I'm not sure its that high, 12V RMS is about 16.5V Pk... If you place a load on the transformer it will reduce quite a bit...

    I don't think we're singing from the same hymn book here...
     
  10. tvtech

    tvtech Well-Known Member Most Helpful Member

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    ^^^
    I don't think we're singing from the same hymn book here...

    I have not seen that little quote in Years. Thanks Ian for reminder :)

    Regards,
    tvtech
     
  11. audioguru

    audioguru Well-Known Member Most Helpful Member

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    The LED currents are very low so the 12V and 6V transformer currents are low enough to allow their voltages to rise to about 14V RMS and 7.8V RMS which when full-wave rectified and filtered produce 18VDC and 9.2VDC.

    Six 3V white LEDs need 18V and the current-limiting resistor has about 1V so the current with 110 ohms is only 9mA.
    Four 2V red LEDs need 8V and the 110 ohm resistor has 1.2V so the current is only 11mA.

    Ritesh shows his circuit in post #59 where the supply is very high at 24VDC and there is only one LED. The transistor is not an emitter-follower, it is a switching common-emitter transistor.
     
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  12. Little Ghostman

    Little Ghostman Well-Known Member Most Helpful Member

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    thank you mr AG it's that it much clearer for me, i thought he was working from the other diagram i somehow had missed his one. many thanks for the clear explanation's.
    LG
     
  13. koolguy

    koolguy Active Member

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    Hi again,

    You are right on load it is ~15.9V BRIDGES TYPE and 7.2V... CENTER TAP
    yes it is common collector, i forgot to make common emitter. should i change it??
    and now 110 ohm to 55ohm.
    The red led are 160 in one bc 5447 with 55ohm 2watt or 4watt they are // comb of 110ohm
     
  14. koolguy

    koolguy Active Member

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    In post #59 i snot my circuit it is searched in Google for examples..
     
  15. Ian Rogers

    Ian Rogers Super Moderator Most Helpful Member

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    Your specification changes like the wind...

    Your original circuit, the hand drawn one, was definitely a common emitter.... AND was a BC337... I am getting very confused.. :confused:.. see what you've done:arghh:
     
  16. Ian Rogers

    Ian Rogers Super Moderator Most Helpful Member

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    Gotta love MrT.....
     
  17. koolguy

    koolguy Active Member

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    Hi,
    update:
    I have removed that 55ohm and directly connected ~160 total red led making series of 4 so,
    there are 40 series led string.
    while removing the resistance the current from BC337 was ~10mA so, the led was giving better brightens...why this BJT is not giving more current? i am using 10K and pot is set nearly 1K or 700 ohms. the voltage across BJT is ~1.8V. the led are flashing.
    and common is collector is the problem ???
     
  18. audioguru

    audioguru Well-Known Member Most Helpful Member

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    A common collector transistor is an emitter-follower. It has no voltage gain. Its emitter voltage will be 0.7V or more less than its base voltage.
    If its base is +5V then its emitter is about +4.3V or less, even if its collector voltage is much higher. With a current of a few hundred mA then its emitter voltage is about 4V.

    You connected four 2V red LEDs in series as a string and connected forty strings in parallel. Without seeing your schematic I am guessing that the supply is +9V and the 700 ohm series base resistor (it is not needed with an emitter-follower) is also connected to +9V.

    If you want a total current of 400mA then the base current must be 40mA which is impossible with a series 700 ohm base resistor. For 40mA in 700 ohms then the voltage across the resistor must be 40mA x 700 ohms= 28V.
    If the base is connected directly to +9V then the emitter voltage will be about +8V and the LEDs will have a good amount of current.

    Look at the datasheets:
    A BC547 has a maximum allowed current of only 100mA and it conducts poorly above 50mA.
    A BC337 has a maximum allowed current of 500mA and it conducts poorly above 200mA.
     
  19. koolguy

    koolguy Active Member

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    OK, AG should i change to common emitter or something else?
    how to increase current? I have removed the series resistance connected to led to have more current.please tell
     
  20. audioguru

    audioguru Well-Known Member Most Helpful Member

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    An emitter-follower has a VBE voltage drop that causes the transistor to get hot and restricts the supply voltage to the signal voltage (5V from a micro-controller).
    A common-emitter transistor does not have these problems. But the base current for a bipolar transistor will be higher than you have available.

    A logic-level N-channel Mosfet as a common-source switch can provide much more current than a bipolar transistor and will not get hot. The gate current is nothing if it is not switched at a high frequency. The supply voltage for the LEDs can be anything.

    LEDs always need a series current-limiting resistor. Your transistor was turning on poorly so it was acting like a series resistor.
     
  21. tvtech

    tvtech Well-Known Member Most Helpful Member

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    Hi AG

    Does not help. Mr KAKKAR will never get it.
    I deal with this every day of my working life.

    If only somebody out there could sit next to me and witness daily what I deal with...

    And just see :wideyed:

    tvtech
     

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