Wienbridge Question Help

Discussion in 'Homework Help' started by blah2222, Dec 8, 2011.

  1. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
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    Hello all,

    I have tried to work at this problem for a while but I am having some trouble with it. Here is the given multi-part question and I will discuss what I have done so far.

    Design a Wienbridge Oscillator using the circuit configuration shown. Assume that the voltage supply to the op-amp is +/- 10 V and that the diode turn-on voltage, Vd,on is 0.7 V. The oscillation frequency required is 10 kHz and the output amplitude is 5 V. Select R1 = 5 kΩ, R = 10 kΩ and set hte gain of the circuit at the start of oscillations to be 2.2.

    [​IMG]

    a) Determine the value of C and R2.
    b) Write an expression for the output voltage given the conditions shown in the figure.
    c) Determine the value of R3 that provides the output amplitude of 5 V.
    d) Accurately draw two cycles (periods) of the output waveform vs. time. Carefully labelling the drawing.


    My solutions:

    a)

    f_{0} = 10 kHz -> w0 = \frac{1}{RC} = 2*pi*f_{0} -> C = 1/(2*pi*f_{0}*R) = 1.59 nF

    Gain = 1 + \frac{R_{2}}{R_{1}} = 2.2 -> R_{2} = 1.2*R_{1} = 6k Not sure though...

    b) KCL @ V- ?

    \frac{V_{o} - V_{d,on} - V_{-}}{R_{3}} + \frac{V_{o} - V_{-}}{R_{2}} = \frac{V_{-}}{R1}

    V_{-} = \frac{V_{o}}{3} Solve for Vo... Not sure if this is correct as it doesn't take into account the other diode direction...

    V_{o} = \frac{(30M)V_{d,on}}{(14k)R_{3} + 60M}

    c) Not sure...

    d) 5 Vpp sinewave with a frequency of 10 kHz...

    Wondering if anyone can check my solutions and help me out. Thanks!
     
    Last edited: Dec 8, 2011
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I'd be surprised the circuit could get into oscillation with an initial gain of 2.2. Would one not need a gain of at least 3x ...??
     
  3. t_n_k

    AAC Fanatic!

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    Say with an initial gain of 3.3 it would most likely get going OK.
     
  4. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
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    haha well this is actually a test question that I got back and did not answer correctly. My prof did not post solutions...
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I see.

    Well you seem to have calculated the 'correct' capacitance value.

    You'll find R2 based on a 'sensible' linear region gain - I'm thinking not 2.2X...??

    To find the value of R3 a simple approach would be:

    The voltage at the amp -ve input will be 5/3 volts. So the current in the 5k to ground will be 1/3 mA.

    The voltage drop across R will be (5-5/3) V. The current in R2 will be I_R2=(5-5/3)/R2.

    So the current in R3 will 1/3mA - I_R2

    The voltage across R3 will be ~(5-5/3-0.7). This will enable you to find R3 since you also know the current.
     
    blah2222 likes this.
  6. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
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    I used the G = 2.2 = 1 + R2/R1 because it said that upon startup so I figured that the diodes would not be on, thus cutting off R3, so it would be a simple gain circuit between R1 and R2. Also, what is the significance of the current through R, if there is a capacitor in series with the other R?

    It is exactly the same as this example I found in my Blalock textbook, with the exception that 'their' R2 = 0 in my question.

    [​IMG]

    Thank you again!
     
  7. t_n_k

    AAC Fanatic!

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    Does the text not give a worked example from which you can model your answer?

    Yes - the diode pair + R3 branch only (rather progressively) comes into play as the output approaches either the positive or negative peak value. So the initial 'linear' gain at low output amplitude (<0.7V) will be (R1+R2)/R1.
     
    Last edited: Dec 9, 2011
  8. t_n_k

    AAC Fanatic!

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    If your work was marked as incorrect what about your classmates? I'd like to be there when your prof explains how the circuit works with a gain less than 3x.
     
  9. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
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    The 2.2 gain is just to start of with and build up the gain to 3, then the diode keeps it from going under or over.
     
  10. t_n_k

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    You are missing the point.

    If the gain is less than 3x in the low output (linear) region then it will decrease as the signal level increases - not increase. If you are notionally adding R3 in parallel with R2 then the stage gain is dynamically varying downwards - not upwards. If it starts out at 2.2 x then it will decrease with increasing output signal excursion. That's the very rationale of including the D1/D2 diode array + R3 circuit branch.
     
  11. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
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    Hmm, that seems reasonable, but I'm wondering why our prof would give us this sort of question if it were unsolvable (unrealistic)... I'll email him again.

    Thanks
     
  12. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    It's not unreasonable to allow for a typo to have occurred in the posting of the problem. It's most likely a simple issue of that nature.
     
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