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Voltage -Divider Biased transistor circuit

Discussion in 'Electronic Design' started by [email protected], Mar 19, 2007.

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  1. Guest

    i am trying to find resistors values R1 and R2 in a voltage divider
    biased transistor circuit.
    the following info is given: Vcc=+24 volts, Rc=680

    if anyone knows how to calculate this please let me know.
  2. Jim Thompson

    Jim Thompson Guest

    Yes. I do ;-)

    ...Jim Thompson
  3. There are many ways to calculate the resistor values,
    depending on the quality of the approximations used.

    Lets start with the simplest.

    Simplifying assumption #1: Assume the current gain of the
    transistor is infinite (no base current).

    Simplifying assumption #2: Assume the base to emitter drop
    is a typical value, like .6 volts.

    Simplifying assumption #3: Assume you want the collector
    bias voltage to be half way between the positive supply, and
    the voltage you would get if you replaces the transistor
    with a short circuit. That short circuit voltage would be
    24*47/(47+680)=1.55, s the half way point between that and
    +24 would be 12.77 volts. Since we are doing
    approximations, lets call that 13 volts.

    First we calculate what the emitter voltage is when the
    collector voltage is at this bias point of 13 volts. This
    bias point puts (24-13)/680=.016 amps. Bases on S.A.#1,
    this same current is passing through the 47 ohm emitter
    resistor, so its drop is .016*47=.75 volts.

    Adding the nominal base to emitter drop from S.A.#2, we find
    the base voltage is .75+.6=1.35 volts.

    So R1 and R2 have to divide the 24 volt supply to produce
    1.35 volts base voltage. So (if R1 is the resistor to the
    +24 volt supply), 24*R2/(R1+R2)=1.35 Solving for R1 in
    terms of R2, that comes out R1=16.68*R2

    Unfortunately, there are an infinite number of resistor
    pairs that will divide 24 volts down to 1.35 volts, so we
    need some additional piece of info to pick a pair.

    To do that , we have to give up a simplifying assumption.
    We have to estimate what the actual base current will be,
    because S.A.#1 can't be right. So we can slightly improve
    that assumption by guessing a current gain. So, I am going
    to change that assumption to assuming that the current gain
    is 100. If you have a transistor in mind, you can check the
    data sheet for a better value.

    So now we have to have a divider that produces about 1.35
    volts while delivering .016A/100=.00016A to the base.

    We can calculate the values with this exact current included
    in the formula, but since it is an educated guess that might
    be off by a considerable factor, either way, the
    approximation often used is to just make the current passing
    through the divider something like 10 times this estimated
    base current, so that the base current distorts the divider
    only a little. And when you get to picking actual values,
    err on the side that produces slightly more base voltage to
    compensate for the droop caused by the small base current.

    So we have the above ratio that R1=16.68*R2 and the total
    divider current 24/(R1+R2)=.0016A.

    Solving these two equations, I get:

    Picking the nearest 5% values (the E24 set available at:
    that will produce a slightly higher voltage, I would use:

    A check of 24*R2/(R1+R2)=1.57 volts, without any base
    current loading it down. The guessed .0016A of base current
    will drop this slightly to 1.57-.00016/(1/R1+1/R2)=1.43
    volts. So I may have over compensated by picking 910 for
    R2, instead of 820. My choices would produce an emitter
    voltage of about 1.43-.6=.83 volts. This represents an
    emitter current of about .83/47=.0177ma. and based on our
    guess of a current gain of 100, 99% of that will be
    collector current, so the drop across the collector resistor
    will be 680*.99*.0177=11.9 volts, down from 24 for a
    collector bias point of 12.1, where we were shooting for 13.
    Perhaps good enough, perhaps not.

    When you get tired of such approximations, you move up to a
    more accurate set of simplifying assumptions and
    approximations and solve more complicated equations that
    better model the details.
  4. Just a first fly over. Well, perhaps good enough for E100.

    I was hoping you would do a more detailed version.
  5. Guest

  6. quandong nut

    quandong nut Guest

    Not likely. I don't believe they (students) learn when others do their
    assignments, and I'm all for them actually learning their way through the
    education system.

    Mind you, I reckon you did a bloody fine job for him/her.
  7. Guest

    You still can learn a lot even if someone figures out the problem for
    you because sooner or later that same type of problem can come up and
    then you will know how to do it. Besides some teachers just dont know
    how to explain things good.
  8. Guest

    I seen these formulas on a web site for calculating R1 and R2

    R1 = (Vcc - Vb) / (10 x ib)
    R2 = ( Vb) / (9 x ib)
  9. Rich Grise

    Rich Grise Guest

    How can you say this with a straight face? When someone else does your
    homework for you, what you learn is to ask someone else to do your
    homework for you.
    Apparently your English teacher is an example of one.

  10. That will give you the voltage across the emitter resistor. Add about 0.1
    volts to that and that's where the base voltage should be. That's your
    starting point.
  11. YD

    YD Guest

    Uh, germanium transistor?

    - YD.
  12. What part of "starting point" did you fail to understand?
  13. That shows the same generality that the bias divider should
    pass about 10 times the base current that I used. I think
    this pair of equations can be pulled out of the two I gave you.

    Now, a question for you to think about. Since having the
    divider pass so much more current than is actually needed to
    bias the base, this really lowers the amplifier input
    impedance and wastes some supply power. Why do you think a
    current factor of 10 x ib is more common, than, say, 2 or 3?
  14. John  Larkin

    John Larkin Guest

    Um, 0.6 would be a better guess.

  15. Add about 0.1
    Maybe a Germanium Schottky ;-)
  16. John  Larkin

    John Larkin Guest

    Jfet near Idss?

    Somebody recently introduced a sort of floating-gate mosfet where gate
    threshold voltage can be factory programmed to be above zero
    (enhancement region) or below (depletion) or very close to zero (which
    I don't have a name for.)

  17. There are many ways to calculate the resistor values,
    Thank you, John, for a great explanation of a basic electronic concept that I
    never did learn.

    And to the critics: not all who ask such questions are teens who want help
    with homework. Some are 50-something hobbyists who need to understand similar
    things. Best to err on the side of "too much knowledge" (if there *is* such a
    thing in America...)

    Thanks again, John, for not doing the oh-so-common selfish "knowledge is
    power" USENET thing...

  18. joseph2k

    joseph2k Guest

    Well, the reply is from a really nice guy. Just the same, the question
    belongs in sci.electronics.basics. BTW i absolutely detest Google for
    listing this ng as "popular"; the spam and OT posts went up 10 fold
  19. Phil Hobbs

    Phil Hobbs Guest

    Used to be called "depletion-enhancement". Of course, that was before
    you could buy depletion enhancers on the Internet.


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