Jony ,In answering your question, and more...

Discussion in 'General Electronics Chat' started by hobbyist, Feb 3, 2010.

  1. hobbyist

    Thread Starter Distinguished Member

    Aug 10, 2008

    for posing that question, to me in the homework section, thread about transistor amp design.

    Question, you asked was why I use such low base resistor values.
    My answer is, in all my course books, they say to make RB1 around 10-20 times RE.

    My assumption from prototype experimenting, was to keep the base current from upsetting the voltage divider.

    A higher resistor for RB1 also means RB2 goes higher, and if RB2 goes to high then base current loads the divider.
    From experimenting, I know this to be true...

    I took a different approach tonight, and designed a low Zout CC stage, and chose a collector current from the data sheet, of 10mA.
    For a 2n3904, IC @ 10mA and VCE @ 1v. gives a min. beta of 100.

    Not considering VCE @ 1v. mine is around 17v.

    But just going by beta min. 100 @ IC 10mA.
    I biased the stage as such.


    VBE 0.7v.
    IC = 10mA.
    RE = 7.5 ohms.
    VE = 75mV.
    VB = 775mV.
    Beta min = 100
    IB = (10ma / 100)
    IRB1 = (10 x IB) :Where RB1 is base to ground resistor.
    Rb1 calculated to be 750 ohms.
    Rb2 calc. = 18K ohms


    I breadboard prototyped it:
    Result voltage VE,
    VE = 60 - 70 mV.

    Now the test:

    replaced the transistor with another transistor 5 times
    and got the same round of voltage on the output.


    By designing for a specific IC using the data sheet,
    using the min. beta, then calculate resulting IB
    and ultimately current for the divider.

    This gives a much higher Zin and saves on power usage as well...

    This has been a good learning experiance.


  2. howartthou

    Active Member

    Apr 18, 2009
    Yeah, Jony is a wizard.

    I know this comment doesn't help this thread specifically, but Jony has helped me so many times, he is a real asset to these forums...thanks from me too Jony.
  3. Audioguru

    New Member

    Dec 20, 2007
    The output impedance is much lower than 7.5 ohms.
    If the hFE of the transistor is 100 and the base resistor is 750 ohms then the output impedance of the emitter-follower is 750/100= 7.5 ohms and the 7.5 ohm resistor is in parallel with it so the output impedance of the circuit is 3.75 ohms if the hFE of the transistor is 100.

    So you don't need an emitter resistor with a value so low that the amplifier will have a very low output swing.