Biasing a transistor

Discussion in 'General Electronics Chat' started by Abhinavrajan, Sep 1, 2016.

  1. Abhinavrajan

    Thread Starter Member

    Aug 7, 2016
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    Please have a look at the below image.

    I am unclear about the working of that NPN transistor with it base biased.
    I understood what biasing is.
    But, please help me to understand how the output waveform is generated when that type of input (assume the input is 20mV) is given.
    Please help in simple english.
    What is the value of Vc and Vcc in the output waveform?
     
  2. #12

    Expert

    Nov 30, 2010
    16,247
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    Vcc is 10 volts as the label says. Vc depends on the gain of the transistor. That is why this is generally considered to be a poor design.
     
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  3. ci139

    Member

    Jul 11, 2016
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    ________your Rb sets Ub - - - your Uinp disturbs or changes the Ub ± d(Ub)
    _________your Ub sets Ib - - - your Ub ± d(Ub) sets Ib ± d(Ib) , note that the relation d(Ib) = f( d(Ub) ) depend on input impedances
    ______your Ib sets Ic.max - - - your Ib ± d(Ib) sets Ic ± ß·d(Ib)
    your Rc and Ic.max set Uc - - - your Rc and Ic ± d(Ic) set Uc ± d(Uc) = U - ( Ic ± ß·d(Ib) )·Rc , !!! while such applies . . .
     
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  4. DickCappels

    Moderator

    Aug 21, 2008
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    Current flows from Vcc through RB into the base of the transistor.

    That base current causes current to flow in the collector.

    A voltage appears across RL -that is the collector current times the resistance of RL.

    The value of RB can be adjusted so that 1/2 of the Vcc is across RL.

    That means that the collector will also be at about 1/2 Vcc.

    If a small signal is applied to the input through Cc, that voltage will cause an increase or decrease of the base current.

    On the positive peaks the base current will increase, which causes an increase in collector current, which increases the voltage drop across LR which means that the collector voltage be get lower.

    On the negative peaks the base current will decrease and the collector voltage will increase.
     
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  5. Abhinavrajan

    Thread Starter Member

    Aug 7, 2016
    77
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    Thank you all! Clear now! :)
     
  6. BR-549

    Well-Known Member

    Sep 22, 2013
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    Think of a seesaw, with the fulcrum in the center. The input is the left side, and the output is the right side. If we move the left side up and down 1 foot, the right hand side will move 1 foot also, but in the opposite direction. The right hand movement is said to be inverted.

    Now move the fulcrum towards the left side, off center. Now not only is the movement on the right side inverted, but the output movement of the right side is much greater than the input movement on the left side.

    Make any sense?

    Edit...the fulcrum is the bias.
     
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  7. Abhinavrajan

    Thread Starter Member

    Aug 7, 2016
    77
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    WOW.. Really cool analogy. Thanks a lot. Really got an idea.
     
  8. BR-549

    Well-Known Member

    Sep 22, 2013
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    It's far from perfect. It's hard to find an analogy for electricity.

    In the seesaw example......the fulcrum controls the gain. And there is a wide range of fulcrum points.

    But in a transistor this range is much smaller. And the gain is built in. The bias just adjust the operating point around the small linear range.

    I was just trying to show the idea of proportion, and how a small wiggle can be turned into a large wiggle.

    The bias keeps the proportion linear, or not, if you want to.
     
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  9. ian field

    Distinguished Member

    Oct 27, 2012
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    Poor design - but *VERY* common.

    The bias resistor calculations aren't trivial, I usually decide what collector resistor to use and use a 1 or 2.2M pot in place of the base resistor.

    For general audio use; adjust for 1/2 Vcc at the collector.

    Then all you do is measure the pot and fit the nearest fixed resistor.
     
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  10. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Note, as mentioned above, that this circuit (called "dangle-biasing") is the absolute worst way to design a linear amplifier. If you disconnect the upper end of Rb from Vcc and connect it directly to the collector of Q1, you have a circuit with much more predictable and stable gain, and lower distortion. The biasing equations are a bit more complicated, but it is worth it.

    ak
     
  11. ian field

    Distinguished Member

    Oct 27, 2012
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    Obviously I didn't look at the original image properly.

    From various books and magazines published in the very early days, it seems to have taken some 15 years for experimenters to figure out that taking the top of the base resistor to collector instead of Vcc, gave some degree of compensation - and that was in the days of very thermally sensitive germanium transistors!
     
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