Collector-emitter feedback bias Dc calculations help

Discussion in 'Homework Help' started by Ronscott1, Nov 10, 2009.

  1. Ronscott1

    Ronscott1 Thread Starter New Member

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    I was given the problem for homework to design the values of Rc, Rb, and Re given the following specifications:

    Vcc=9V
    Vce at midpoint
    Ic=1 mA
    βdc=300

    Using the guidelines for VDB from Albert Malvinos "Electronic Principles" I came up with:
    Ve=0.1*Vcc
    Ve=(0.1)(9V)
    Ve=0.9V
    Re=Ve/Ie Since Ie≈Ic
    Re=0.9V/1mA
    Re=900Ω
    Rc=4Re
    Rc=4(900Ω)
    Rc=3.6kΩ
    Rb= βdc(((Vcc-Vbe)/Ie)-Rc-Re)
    Rb=300(((9V-0.7V)/1mA)-3.6kΩ-900Ω)
    Rb=1.14MΩ

    But when I plugged these numbers back into the equation
    Ie=(Vcc-Vbe)/Rc+Re+Rb/βdc, I got Ie=1.8mA.

    Is there something that I am doing wrong? How should I go about calculating these resistor values? Also are there any specific guidelines that I should follow?

    Best,
    Ron

    Attached Files:

  2. hobbyist

    hobbyist Well-Known Member

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    Are they the guidelines your suppose to use, or are they your choice to use?
  3. ELECTRONERD

    ELECTRONERD Senior Member

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    Your schematic diagram is rather strange, since Rb usually connects above Rc to the Vcc source. But, as you may have realized, they form a voltage divider when connected to the collector. A useful equation is that I_B=\frac{I_C}{hFE}

    Austin
  4. Ronscott1

    Ronscott1 Thread Starter New Member

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    Hobbyist-
    Those are the guidelines I chose to follow. There is just about no information in my textbook about this type of biasing, let alone there is not much information available on the web. I am not really sure what guidelines to use. I just could not find any except for voltage divider bias in my text. Do you know of any?

    Austin- Ideally what should I try to bias the base around?
  5. ELECTRONERD

    ELECTRONERD Senior Member

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    Ron, a good resource is our current AAC e-book. Just go under "Volume III Semiconductors" and scroll down to the BJT biasing calculations. When beginning to bias transistors, Ohm's Law is virtually the only technique necessary, besides understanding linear region, saturation, and gain. Using the formula I suggested, I_B=\frac{1mA}{300}=3uA. That's a very small Ib current. By seeing the formula alone, you can see that Ic, Ib, and β are all related.

    Austin
  6. hobbyist

    hobbyist Well-Known Member

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    It looked like in your equation, for RB value, you were using the full 9V. instead of the drop across RC.

    I may be looking at your equation wrong. Though too.

    Is imidpoint at collector with respect to emitter or at the collector with respect to ground? That's my main question.
  7. t_n_k

    t_n_k Senior Member

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    Hi Ron,

    It's those wretched parentheses ....

    Ie=\frac{(Vcc-Vbe)}{(Rc+Re+\frac{Rb}{(1+\beta)})}

    Ie=\frac{(9-0.7)}{(3600+900+\frac{1.14e6}{(301)})}

    Ie=\frac{(8.3)}{(8287.4)}

    Ie=1mA
  8. Ronscott1

    Ronscott1 Thread Starter New Member

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    Thanks so much for your help guys. Hobbyist - I am pretty sure that midpoint is with respect to ground.
    Best,
    Ron
  9. Audioguru

    Audioguru New Member

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    No.
    That is the worst way to bias a transistor since the wide range of beta, the different base-emitter voltage and different temperature all determine its amount of needed base current.

    When a single base resistor connects to the collector then it provides AC and DC negative feedback.
    If the beta or temperature is high then the transistor tries to conduct more which reduces its collector voltage that reduces the base current. Then the transistor does not turn on too much. If the beta or temperature is low then the opposite happens and the transistor does not turn off too much.
  10. hobbyist

    hobbyist Well-Known Member

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    See post #7.

    he confirms your answer of 1mA.
  11. ELECTRONERD

    ELECTRONERD Senior Member

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    Good Audioguru, very good. But I've usually seen a resistor at the base that connects directly to Vcc. In fact, this is the first time I've seen that method. Next they usually add a voltage divider which helps with temperature and the low voltage and current output.

    Austin
  12. hobbyist

    hobbyist Well-Known Member

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    This one is known as collector feedback or better known as "voltage feedback"

    Just like the emitter resistor is used for neg. feedback that's called "series current feedback" method,
    and the one in this thread is "shunt voltage feedback" ,as well as series current, with the emitter resistor added.

    If there was a base to ground resistor, it would be a very stable circuit, with a base voltage established with external resistors.
    But as it is this configuration depends on the base current alone, for it's base voltage, which is not what you want, because the base parameters can differ greatly.
  13. Ronscott1

    Ronscott1 Thread Starter New Member

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    hobbyist-
    So what you are saying is that if I add a resistor from base to ground using collector-emitter feedback bias it would result would be a much more stable bias? What would this do to my previous calculations though? In other words, how would this alter the formulas I used this bias?
  14. hobbyist

    hobbyist Well-Known Member

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    volt feedback.jpg

    I never designed this feedback type.
    But this is what I think it works at,
    someone more knowledgeable with this config. can show you better how to do this and how to answer your question.

    So this is just a guess at how I would do this.

    parameters:
    VCC = 9V.
    IC = 1mA.
    DC B~= 70
    VC = VCC /2
    ---------------------------

    Design:
    VC = 4.5V. @ IC 1mA.
    RC has to be 4.5K.

    arbitrarily chose VB to be around 2V. (anything less than VC for it to work as a transistor ,collector base reversed bias)

    since B min. = 70 for this transistor 2N3904. @ 1mA.
    then IB = IC / B = around 14uA.
    Made divider current to be around 10 x IB at around 140uA. (ID)

    Now VB / ID = (2v. / 140uA) = around 13K. for R3.

    Now the top of R2 is at 4.5V, so {(VC - VB) / ID}= R2 = around 18K.

    With VB @ 2V. then VE will be 0.7v. less or around 1.3V.

    (VE / IC) = RE = (1.3V. / 1mA) = 1.3K for R4.

    To answer part of your question,

    We want base current (IB) to be the RESULT of base voltage (VB)

    NOT the base current to be the cause of base voltage.

    SF = (RB / RE) around 6 for this close approximation. (where RB = R2 // R3) which with this feedback theres more involved than just parrallelling the 2 base resistors, input impedance is actually much lower , but that's more advanced, circuit analysis, No need to goi nto right now...


    So Now VB is established with outside resistors. voltage divider network.
    Last edited: Nov 12, 2009
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