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Homework Help: OP-Amp connected with Transistor whose Base and Collector are short

  1. Jun 4, 2013 #1
    1. The problem statement, all variables and given/known data

    Problem number 6a in the picture.

    2. Relevant equations
    Ic = βIb

    Ic = αIe

    Ic = Is*e^(Vbe/nVt)


    Vo = Acl*Vi

    Acl = 1 + (Rf/Ri)

    3. The attempt at a solution

    I tried to solve but just couldn't understand the behavior the transistor.
  2. jcsd
  3. Jun 4, 2013 #2

    rude man

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    First, find Vo1.

    Then find Vo.

    Note that Q1 and Q2 behave like p-n junction diodes.
  4. Jun 4, 2013 #3
    I tried that.
    Now Q1 must be forward biased so that the constant current Iref flows through Q1 and the voltage drop across Q1 is 0.7 Volts that is

    Vbe1 =.7 Volts

    Vo1 = Vi*(R4/R4+R3) - .7

    Now what will be the behavior of transistor Q2? The Voltage across it depends on Vi

    Vbe2 = .7 - Vi*(R4/R4+R3)

    So how can i take diode 2 as forward or reverse biased without knowing Vi?

    If the diode 2 is forward biased then where does the current go at NODE Vo1. Does it go into the OP-Amp1?

    Taking the Diode 2 reverse biased the circuit is open and Vo2 becomes 0 Volts

    PS: I don't have a camera right now or i could have shown you the full attempt.
  5. Jun 4, 2013 #4

    rude man

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    You can't!
    What current? Vo1 sinks currents from both diodes. Obviously, all net diode current has to go into op amp 1's output.

    Q1 current is always = I_ref.
    That is correct.


    Hint: You got Vo1 right.

    This circuit is a Rube Goldberg comparator. The output Vo tells you whether the input is above or below a certain threshold voltage. What is the correspondence between input voltage and output voltage? Vo is one voltage when Vin is < threshold and another when Vin > threshold.
    Last edited by a moderator: Jun 4, 2013
  6. Jun 5, 2013 #5
    Thanks for help. Now tell me is this the answer

    1) When Vi < .7*(R4 + R3)/R4, Vo = 0 Volts

    2) When Vi > .7*(R4 + R3)/R4, Vo = R3*Is*e^((.7 - Vi*(R4/R4+R3))/nVt)
  7. Jun 5, 2013 #6

    rude man

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    'Fraid not.

    First, what is Vo1 as a function of Vi?

    Second question, what happens if Vo1 is < -0.7V or > -0.7V?

    BTW your expression for Vo must be compared to the saturation output voltage. Your op amp can't go more positive than a certain voltage, say +12V for a 741 op amp running on +/- 15V supplies. This applies to terms like your Vo = R3*Is*e^((.7 - Vi*(R4/R4+R3))/nVt) which might be correct if your op amp had no saturation limit.
  8. Jun 5, 2013 #7
    Vo1 = Vi*(R4/R4+R3) - .7

    1) Vo1> -.7 that is Vi > 0, Vo = 0 Volts

    2) Vo1 < -.7, Vi < 0

    In this case the diode would be forward biased and second op-amp would behave as a negative comparator and then op amp will saturate at positive saturation voltage that is 12 volts.
    Is this the right answer?
  9. Jun 5, 2013 #8

    rude man

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    Absolutely correct. Good work!
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