npn base resistor

Discussion in 'General Electronics Chat' started by duxbuz, May 21, 2014.

  1. duxbuz

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    Feb 23, 2014
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  2. pwdixon

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    Oct 11, 2012
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    The resistor limits the current into the base and prevents overvoltage on the base. Without the resistor driving the base would probably blow the device due to overvoltage unless you were using a high impedance drive to the base.
     
  3. duxbuz

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    Feb 23, 2014
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    Thanks

    So the transistor would get damaged? That would lead to the microcontroller then getting damaged?

    In that order?
     
  4. samuel.whiskers

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    Mar 17, 2014
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    I would think it depended on the transistors power rating.... Generally I'd think the microcontroller/IC would go first.... I've killed a 555 similarly (and they are pretty robust), that had a beefy power transistor connected to it...
     
  5. pwdixon

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    Oct 11, 2012
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    Chips are pretty robust these days and often limit their output current so the processor or perhaps even both might survive but you don't want to design that way with everything heating up and running close to death.
     
  6. duxbuz

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    Feb 23, 2014
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    I have been looking at some postings and I am trying to gain a better understanding of the issues.

    Does anyone have pointers to tutorials that might help me get a better understanding?




    I saw some comments from a thread on another forum here:
    http://electronics.stackexchange.co...the-diode-and-capacitor-in-this-motor-circuit

    but when I read through the posts its quite hard to understand.

     
  7. Jony130

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    Feb 17, 2009
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    Digital output "gives" 5V at the output. Transistor base-emitter needs max 0.8V or so.
    If base voltage is force higher then this 0.8V the huge amount of current will start to flow. And this current will start generates the heat inside transistor. This heat in transistor will increase the current even further. And this increase in current will again increase the heat in transistor and this will lead to a destructive result (transistor will burn). Also this heat is also grow inside microcontroller, so the microcontroller can also be damage.
    So the prevent this disaster to happen we add RB resistor. This resistor will limited the base current and drop the voltage ( From 5V to 0.8V) also according to the Ohm's law.

    Which part you don't understand?
     
  8. alfacliff

    Well-Known Member

    Dec 13, 2013
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    use the base resistor and put in another transistor following the first one to handle more current. this will help isolate the chip from the load better. spikes might blow one transistor, but probably will not get through two.
     
  9. duxbuz

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    Feb 23, 2014
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    Makes more sense now thanks

    Its all the terms I am unfamiliar with, but I read a bit before so things are slowly making more sense. I mean a minute ago I didn't really understand sink and source... thanks google got it now.

    All the simple things when you don't know much, just add up and its almost like reading Chinese, for a non-Chinese speaker.

    Where is that 750mV from? Is that a standard base emitter saturation voltage?
     
  10. Alec_t

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    Yes. Driving more current through the b-e junction will increase the voltage only slightly.
     
  11. Jony130

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  12. duxbuz

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    Feb 23, 2014
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    If the I/O pin has max current 40mA

    Would I use a resistor to the base of the NPN with a value between 15 - 22 Ohms?
     
  13. alfacliff

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    Dec 13, 2013
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    a 22 ohm resistor in series with 5 volts applied would pass 190 ma or so to drop 4.3 volts. a little high? use a 220 to 1000 ohm resistor.
     
  14. Jony130

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    No, why such low values?
     
  15. duxbuz

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    Feb 23, 2014
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    I used a blooming online calculator and its still wrong!
     
  16. Jony130

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    Can you post the link? And what value you have put into this calculator?
     
  17. duxbuz

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    Feb 23, 2014
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    Ok I was doing it wrong. Trying to 'will' the voltage down as if it was an unknown quantity.... when I already had a voltage value.... tut
     
  18. duxbuz

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    Feb 23, 2014
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    I have a voltage: 5V, but I am trying to get the voltage to drop to fit this 0.75V

    I was using the calculator in an insane manner. The problem being... it made sense to me. :(

    So now I put in the correct values...
    5v
    20mA
    results = 250 Ohms

    but I want the voltage dropping to a value, so what do I use.... voltage divider calculator?

    These questions might be criminal... please bare with me
     
  19. duxbuz

    Thread Starter Member

    Feb 23, 2014
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    I might have it.

    I do what I was doing, and the subtract results from 5V?

    so work out a resistance with 20mA that will give me 4.25V

    leaving 0.75V?
     
  20. Jony130

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    Why you want specific voltage drop? As I said earlier the Vbe voltage in not so much important here. The base and collector current is important here.
    For example if your load connected between collector and VCC consume 60 mA of current. The base current need to be higher then
    Ib > 60mA/20 = 3mA. So

    Rb < (5V - 0.7V)/3mA = 1.4k so you use 1K resistor.
     
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