How to bias a transistor using emitter bias method?

Discussion in 'Homework Help' started by simpsonss, Oct 7, 2010.

  1. simpsonss

    Thread Starter Active Member

    Jul 8, 2008
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    hi,

    I would like to learn about the emitter bias biasing. From my reference book, the circuit is as below.

    untitled3.JPG

    But i'm having problem in identifying the 3 resistors value. How am i gonna put in the value? what value to choose?

    thank you.
     
  2. simpsonss

    Thread Starter Active Member

    Jul 8, 2008
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    anyone to help me on this?
     
  3. Ghar

    Active Member

    Mar 8, 2010
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    To pick components you actually need some targets in mind.

    Gain?
    Signal swing?
    Output / input resistance?
     
  4. simpsonss

    Thread Starter Active Member

    Jul 8, 2008
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    ok.if i would like to have a 10 times in gain.
    signal swing means what? the Amplitude?
    Another question is why i need to know the output or input resistance?

    I'm a newbie in design BJT biasing.
     
  5. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    I think I found example that have the same circuit. The key to the whole thing is to have transistor in forward-active mode, that is Vce>Vbe(on).

    Pull up the data sheet for that transistor and look up the Vbe(on). Then put voltmeters into the circuit to measure Vbe and Vce. Start playing with resistors until you get Vce > Vbe while Vbe=Vbe(on).

    Also, the book says that the rule of thumb is that amplification factor is about 50 to 300. So gain of 10 might be too low to experiment with.
     
  6. simpsonss

    Thread Starter Active Member

    Jul 8, 2008
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    how can i know whether my transistor is properly bias or not? By referring to Ic value?

    thank you.
     
  7. simpsonss

    Thread Starter Active Member

    Jul 8, 2008
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    From my datasheet, i can see Vbe(sat) but no Vbe(on). So is this the same?

    thank you.
     
  8. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    I am going to take the coward's way out. Assume Vbe(on) is between 0.6 and 0.7 volts. Now start playing with resistors until you get Vbe equal to 0.7 volts, and Vce greater than 0.7 volts.
     
  9. Wendy

    Moderator

    Mar 24, 2008
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    Hint: base emitter to -5VDC resistance is β (transistor beta) X R3. There is no Base to Collecter resistance. You should now be able to establish the voltage on the base, and work it from there.

    2nd Hint: Ic + Ib = Ie
     
  10. simpsonss

    Thread Starter Active Member

    Jul 8, 2008
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    your R3 means Rb or Rc on my circuit?

    thanks.
     
  11. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Does not make any sense.
     
  12. simpsonss

    Thread Starter Active Member

    Jul 8, 2008
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    can i said that when Vce is equal to Vcc means my transistor is in cutoff mode?

    So for currently i vary my Rc value. And i get Vbe=0.66 and Vce = 2.45V. So could i say that i'm working in the correct result?

    thanks.
     
  13. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Cutoff mode:
    Vbe < 0
    Vbc < 0

    You need to meet both requirements listed above.
     
  14. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    That looks good. You are in forward-active mode, which is where you have nice linear relationship Ic=β(Ib) for current gain.
     
  15. simpsonss

    Thread Starter Active Member

    Jul 8, 2008
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    How do i know whether it is in active mode? depends on the value for my Vce? What if my Vce = Vcc? Does it still in active mode?

    thanks.
     
  16. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    I can give you the definition from the book.
    Forward-active mode:
    Vbe > 0
    Vbc < 0

    You need to meet both of these conditions. Stick the voltmeters on that circuit and see what you got. You already said that Vbe=0.66 volts, so first condition is met. What do you have for Vbc?
     
  17. simpsonss

    Thread Starter Active Member

    Jul 8, 2008
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    about 2.21V for Vbc.
     
  18. Wendy

    Moderator

    Mar 24, 2008
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    Perhaps because you have dismissed it without studying it? I will explain in my next post.

    In a common collector amplifier (which relates) the resistance as measure from base to ground (or in this case, V2 (-5VDC)) is the formula I stated.
     
    Last edited: Oct 8, 2010
  19. Wendy

    Moderator

    Mar 24, 2008
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    [​IMG]

    I meant the component designation on your schematic, which is not labeled Re, Rc, or Rb, but is labeled R1, R2, and R3 (though I understood what you meant). Component designations are your friend, always.

    If the β of the transistor is 100, you will have 100KΩ from the base of the transistor and through R3. The battery is in parallel with this resistance, but the resistance is real and can be used to determine the voltage you will see at the base and across R2.

    The currents in this circuit (true for almost all transistor circuits) is

    IR3 = IR1 + IR2, or Ie = Ib + Ic

    In general for these designs, Ic ≈ Ie

    You provided the schematic, so I am working with that.

    One last Hint, the collector current and voltage is almost entirely dependent on the emitter current, unless the transistor saturates.

    .
     
    Last edited: Oct 8, 2010
  20. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    For voltage gain 10V/V
    Rc/Re = 10
    So for Rc = 1KΩ ---> Re = 100Ω
    Vce = 0.5 * V1+V2 = 5V
    So
    Ic = 5V/1.1K = 4.5mA

    Rb = ( V2 - Vbe - Ve)/Ib

    Ve = 4.5mA * 100Ω = 0.45V
    Vbe = 0.7V
    Ib = Ic/β = 4.5mA/160 = 28μA

    Rb = (5V - 0.7V - 0.45V)/ 28μA = 3.85V/28uA = 137K = 140KΩ

    Read this post
    http://forum.allaboutcircuits.com/showthread.php?p=153702#post153702
    And remember Rin>> R_source and Rout<<Rload
     
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