BJT NPN pull-down Resistor

Discussion in 'Homework Help' started by wuz, Jul 8, 2011.

  1. wuz

    Thread Starter New Member

    Oct 4, 2010
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    Hi, I know that it's good practice to include a pull-down resistor R2 between the base - emitter junction to make sure that the transistor remains off when small leakage current is present, and understand that it has to be much larger than the base resistor R1, but I don't fully understand the effect of adding R2 to R1.
    When calculating the base current, are R1 & R2 in parallel? E.g. if Vcc = Vin = 5V, R1 = 1K, R2 = 10K.
    Is it correct that the input impedance = (1/R1+1/R2)^(-1) = 909, so Ib = 4.3V/909 = 4.73mA? So if we have a smaller R2, it's effectly reducing the base current?
    Thanks!
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    You are very close to the right answer.
    You can use the Thevenin's theorem to find Ib current.
    For example Vin = 5V and R1 = 1K, R2 = 10K we have
    Vth = R2/(R1+R2) * Vin = 4.5V
    And the base current is equal
    Ib = (Vth - Vbe) / R1||R2 = (4.5V - 0.6V) / 909Ω = 3.9V/909Ω ≈ 4.3mA
     
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  3. wuz

    Thread Starter New Member

    Oct 4, 2010
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    Hi Jony130,

    Thanks very much for your reply. I still don't quite understand how you got 4.3mA from the Thevenin's theorem, and what 2 points you are using for Vth (is it between b & e)? Could you please show me your Thevenin equivalent circuit?
     
  4. Jony130

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    Feb 17, 2009
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    Sure I can show you.
    We replace this part of the circuit ( in this gray rectangle)
    With his Thevenin's equivalent circuit

    Vth = Vin * R2/(R1+R2)


    Rth = R1||R2 = (R1*R2)/(R1+R2)
     
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  5. wuz

    Thread Starter New Member

    Oct 4, 2010
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    Thank you for your explanation. It makes perfect sense to me now! I can now go to sleep without feeling confused about the circuit :)
     
  6. Jony130

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    I'm very glad that you can sleep well.
     
  7. kmohanra

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    Feb 4, 2008
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    Hello together,

    Can you apply kirchoffs law in the same circuit and explain the base current concept from the Vin.

    Thanks
     
  8. kkaczor

    New Member

    Aug 4, 2013
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    Hi,

    Could you tell me why we cannot calculate the base current by simple kirchoff's law:
    1. Assume base-emitter voltage drop of 0.7V,
    2. The same voltage drop is on R2, so this means the current which flows through R2 is 0,07mA,
    3. The voltage drop on R1 is 5-0,7 = 4,3V. This gives us the current which flows through R1 of (5-4,3)/1k = 4.3mA

    Finally base current could be calculated as 4,3mA - 0,07mA = 4,23mA

    What I'm doing wrong?
     
  9. Jony130

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    Feb 17, 2009
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    For sure you can use kirchoff's law to find IB current. I don't see anything wrong in your solution.
     
  10. kkaczor

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    Aug 4, 2013
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    Okay, but maybe it would be stupid question - why the result values differ?
     
  11. #12

    Expert

    Nov 30, 2010
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    Post#2 did not account for the current through the 10k resistor subtracting from the 4.3ma. Your version is right, kk.
     
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  12. Jony130

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    Feb 17, 2009
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    I assume Vbe = 0.6V and you assume 0.7V
    So for Vbe = 0.6V we have

    IR2 = 0.6V/10KΩ = 60μA and IR1 = (5V - 0.6V)/1KΩ = 4.4mA
    So Ib = 4.4mA - 60μA = 4.34mA
     
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  13. kkaczor

    New Member

    Aug 4, 2013
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    I used to assume 0.6V, but I've measured several NPNs yesterday and it does not want to be other than 0.7V.

    Thanks guys for your explanations and help!
     
  14. Jony130

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    Feb 17, 2009
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    This is not true.


    Well its all depend on the Vin value and Rb value.

    I measure Vbe Vs Ib for BC337-40 for Vcc = 10V

    RB = 680kΩ....Vbe = 0.614V....Ib = 13.8µA

    RB = 470kΩ....Vbe = 0.616V....Ib = 20µA

    RB = 220kΩ....Vbe = 0.624V....Ib = 42.61µA

    RB = 100kΩ....Vbe = 0.639V....Ib = 93.61µA

    RB = 50kΩ......Vbe = 0.659V....Ib = 187µA

    RB = 10kΩ......Vbe = 0.719V....Ib = 928µA

    RB = 5kΩ........Vbe = 0.748V....Ib = 1.85mA

    RB = 2kΩ........Vbe = 0.787V....Ib = 4.6mA

    RB = 1kΩ........Vbe = 0.819V....Ib = 9.18mA

    RB = 500Ω......Vbe = 0.856V....Ib = 18.29mA

    RB = 200Ω......Vbe = 0.989V....Ib = 45mA
     
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  15. kkaczor

    New Member

    Aug 4, 2013
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    Vcc was equal to Vin, both 10V?
     
  16. screen1988

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    Mar 7, 2013
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    Could anyone help me explain it?
     
  17. kkaczor

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    Aug 4, 2013
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    When you drive transistor from microcontroller for example, when it is starting the state of output pins in unknown and there is a possibility a small current leakage. If so, a small current may drive your transistor.

    To prevent this, you can place a resistor between base and emitter to force current not drive the base, but go through pullup resistor instead.
     
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  18. screen1988

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    Mar 7, 2013
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    Thanks, kkaczor.
    I have just read that the pull-down resistor is very big (in this case it is much larger than R1). If so, in order for current flows this resistor to ground instead of flowing into base of transistor, we also need the pull-down resistance << input resistance of transistor?
    Is that right?
     
  19. kkaczor

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    Aug 4, 2013
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  20. ramancini8

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    Jul 18, 2012
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    The calculated 4.5V is actually 4.545V. One writer assumed Vbe=0.6V and you assumed 0.7V. Not theory, just quick, sloppy calculations.
     
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