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Wien bridge amplitude calculation

  1. Dec 4, 2016 #1
    Hi all!

    How can I set up the amplitude of a wien bridge oscillator made like this?
    wien.jpg

    Let's suppose that the bottom resistances are (from top to bottom) R1, R2 and R3.

    We know that for oscillations to start and converge we need:

    R1 / R3 > 2 (let's suppose 2.2)
    (R1 || R2) / R3 < 2 (let's suppose 1.8)

    But how can I managed the amplitude?
     
  2. jcsd
  3. Dec 4, 2016 #2

    Svein

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    The two diodes manage the amplitude, so I expect that the amplitude will be ≈0.7V. If you want a greater amplitude, add an amplifier after the oscillator (it is a good idea to do so anyhow).
     
  4. Dec 4, 2016 #3

    tech99

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    You need a manual gain control, so that the diodes are only just limiting. Then you will obtain a good sine wave.
     
  5. Dec 4, 2016 #4

    LvW

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    In case you have the correct I=f(V) characteristic of the diodes you can exctly find the oscillation amplitude.
    However, is this really your problem? Or do you rather want to DESIGN the circuit for a certain (desired) amplitude?
     
  6. Dec 6, 2016 #5
    Actually, I got at least 2 V peak-to-zero.

    Yes my aim is to understand how to manage the amplitude to either design or calculate the amplitude given this type of circuit, without using any second stage amplifier.

    I'm doing several ltspice simulations and I've found out that if I change the resistors values, but keeping their ratios like before, the amplitude changes. But I cannot empirically understand the law.
     
  7. Dec 6, 2016 #6

    LvW

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    Different resistor values give different currents (even in case of constant ratios).
    And different currents allow different voltage drops across the diodes.
    Remember: Diodes are NON-LINEAR elements.
     
  8. Dec 7, 2016 #7
    Yes I know but still cannot understand how to start the analysis. Because with normal opamp analysis I get an equation where Vo is simplified. I found this explanation, with this circuit, which is almost the same as mine, except for the capacitor instead of the resistor.

    Using that formula I found the value peak-to-zero, instead of peak-to-peak as said there, but I correctly get about 1.2 Vp, which stay fixed (whatever the resitances) as far as the ratio is constant, and the resistor in series with diodes has the same resistance of the one to the ground. BUT if I change that resistance the amplitude changes, so in a general treatise I must consider it.

    And again, how can I get the Vo amplitude as function of those three resistance and diodes forward voltage drop? I just don't get it.
     
  9. Dec 7, 2016 #8

    LvW

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    Ok - I will give you some hints. Hower, I think that the calculation is somewhat involved - and, finally, it is a graphical solution.
    The following applies to a simplified version with the diodes directly across the feedback resistor. However, if you understand the principle, you can apply it to your configuration because the key of the calculation is the required (final) value for the diode resistance Rd. This can be found easily for my (simplified) circit - and without too much effort for your circuit also.

    1.) Steady-state oscillation case: Gain Acl=3=(1+Rp/R1) >>>> Rp=2R1 with Rp=R2||Rd (Rd=static diode resistance=Vd/Id)
    2.) Because R1 and R2 are given, we can solve for the final value of Rd. That means:The required Rd has a known value.
    3.) We know Vd=(2/3)Vout (Vout is the wanted output voltage) >>>> (2/3)Vout=Vd=Rd*Id >>>>Vout=(3/2)Rd*Id.
    Note that you cannot simply set Vd=(0.6...0.7)V because we dont know Id (Vd can be as low as 0.2V)
    4.) For finding Vout, we need the current Id. However, Id depends on Vd=(2/3)Vout based on the known exponenetial relationship.
    5.) Therefore: Vout=(3/2)Rd*Id=(3/2)Rd*Io{exp[(2/3)Vout/Vt] -1} .
    6.) This equation cannot be solved for Vout. Hence, we need a graphical solution and find the point where the two curves Id=f(Vd) cross each other.
     
    Last edited: Dec 7, 2016
  10. Dec 7, 2016 #9

    Baluncore

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    One difference from reality might be the idealised models used by your LTspice circuit.
    If you could rename a copy of your “wein_bridge.asc” file to “wein_bridge.asc.txt”, you could attach it to your next post. We could then run it and see how real it is.
     
  11. Dec 10, 2016 #10
    Sorry for the delay in the answer.

    Thank you very much indeed! One of my errors was to use a costant voltage drop of 0.7 across the diode. One question though: you have supposed that Vd is the voltage drop between Vo and the opposite node (let's say Vx), equal to the voltage drop across the diode. Did you make an approximation right? I mean, the real Vd, should have been calculated between Vo and the point before Rd, which is different from Vo - Vx. Am I correct?

    Yes I had used the costant voltage drop model like this:

    .model D D(Ron=1n Roff=1G Vfwd=0.7)

    EDIT:
    Anyway I get unrealistically big Id if I had to find Rd knowing Vo... From your equations:

    Vo = 3/2 * Rd * Is * (e2/3 * Vo/Vt - 1)

    or

    Rd = 2/3 * Vo / (Is * (e2/3 * Vo/Vt - 1) )

    Which for Vt = 25 mV and Vo anywhere from 1 or greater gives a HUGE current, and consequently Rd = 0 about. So the approximation I think you made cannot hold, can it?
     
    Last edited: Dec 10, 2016
  12. Dec 10, 2016 #11

    jim hardy

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    I have the same trouble with Wein bridges. They need something to keep amplitude stable , usually a voltage dependent component like an incandescent lamp.


    IMHO the approach is to realize they're a feedback system and it's operating right on the cusp of (in)stability.

    To your question about ciphering amplitude,
    i'd bite the bullet and admit the math will run away from me
    then take a low-tech look at the circuit.

    How does feedback work ?
    A fraction of output gets applied to both of the opamp's inputs,
    and that fraction must be same for both inputs . There's the key....

    A quick glance at the upper voltage divider circuit for +input suggests to me that fraction is 1/3 at 1 hzEDIT oops, 1khz... (I didn't calculate , so correct me if i'm wrong i just want to demonstrate an approach here.)

    weinbridge.jpg

    Now take a look at lower feedback voltage divider:

    If 1/3Vo is to appear across that 10K resistor,
    then in order to drop 2/3Vo , the two resistor & two diode part of that divider must appear to be 20K.
    So, what resistance in parallel with 22.1K would make 20K ? Answer = 210K .

    Therefore, the combination of those diodes and the 100K must look like 210 K.

    How does that 2/3 Vo divide across the 210K of resistor and diodes ?
    100/210 = 48% across the resistor, leaving 52% across the diodes. Again it's nonlinear but we're estimating here so just call it a 50/50 split.
    That means ~1/3 Vo appears across the resistor and ~1/3Vo appears across the diodes.

    1/3 Vo across the diodes?
    That tells me Vo at peak (when negative feedback turns the output around) should be in the neighborhood of 3X a diode drop ,
    if i guess a diode drop is in the range of 0.5 to 0.7 volts around its knee
    i get 1.5 to 2.1 volts for Vo at peak.

    Does that 'poor man's' approach skin the cat ?

    Was your 2 volt number measured or from a simulation ?




    old jim


    EDIT OOPS!
    I just noticed your capacitor is 1 nf not 1uf so my frequency is off by three decades. 1khz not 1hz.

    Sorry about that. I had been working from the smaller schematic . My bad.

    Approach doesn't change though.
    .
     
    Last edited: Dec 10, 2016
  13. Dec 10, 2016 #12

    Baluncore

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    I am surprised it is possible for members to follow the discussion without a schematic labelled with component references R1, R2 … C1, C2 … etc. I see little point in adding to the confusing.
    @ Frank-95. Why do you not post an LTspice oscillator model with clear component references ?
     
  14. Dec 11, 2016 #13
    I get around 1.75, 1.78 by running the circuit via LTSpice, using a real diode model (2 V using an idealised one)

    Here the code:

     
  15. Dec 11, 2016 #14

    LvW

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    To Jim Hardy: It is no problem to follow your approach - and, of course, I agree, with one exception:

    "I guess a diode drop is in the range of 0.5 to 0.7 volts around its knee".

    That means, you are assuming that the diode will be "open" with a voltage-current relation around the "knee".
    I think, this assumption is rather "vague" and not justified sufficiently..
    When the value of the static diode resistance (Rd=Vd/Id) should be app. 100 kohms (and I agree to this rough assumption), the corresponding forward voltage may be lower (why not 0.1 or 0.2 volts?). However, this operating point will strongly depend on the diode type used.
     
  16. Dec 11, 2016 #15

    LvW

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    I must admit that I do not understand your question.
    I did not make any assumption. Vd is the voltage across the diode - that`s all.
    In your specific case, there is another resistor (100kohms) in series with the diode (which, as I have mentioned, does not appear in my calculation).
    As I have mentioned, the value of Rd is the "key value" for the final result - and Jim Hardy has given in his contribution a rough - perhaps sufficient - guess for the Rd value (100...110 kOhms). The remaining task is to find the corresponding voltage Vd for the particular diode type.
     
  17. Dec 11, 2016 #16

    jim hardy

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    One might take the equation I=I0eqv/kt , solve for I at various v's, find at what I does v/I approximate 110KΩ .

    If Windows still supported Qbasic i'd have done it.


    It'd be interesting to put zener reference diodes into that simulator .
     
  18. Dec 11, 2016 #17

    LvW

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    I think, the problem is a suitable and realistic selection (guess?) for the current Io.
     
  19. Dec 11, 2016 #18

    Baluncore

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    The voltage rating of the light globe used in the original HP oscillators decided to a great extent the output amplitude. The resistor ratios are now selected to support reliable oscillation. The diodes then reduce the gain as the amplitude increases. Since the diodes have a typical drop of about 0.65 volts, that is the order of magnitude you can expect for oscillator amplitude. Variation of production component values makes it difficult to build reliable WBOs with higher gains.

    By using several diodes in series you can multiply the signal amplitude.
    Or, instead of silicon diodes, use a couple of parallel back-to-back LEDs to get about 5Vpp.
    A couple of 6.2V Zenners, reversed in series, will give about 22Vpp.
     
  20. Dec 12, 2016 #19

    LvW

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    I rather think that this "typical drop" is not relevant in this case. We cannot automatically assume that the diode voltage will be in this range.
    In contrary - this would be a very bad design (with a rather large TDH for the oscillating signal).
    The designers goal must be to keep the diodes influence as small as possible.
    As we have estimated earlier, the diodes static resistance will be in the order of 100kOhms.
    A SPICE simulation for a typical diode (1N4148) gives a value of Vd=0.385V for Rd=100kOhms
    (and Vd=0.34V for the type BA315).
     
  21. Dec 12, 2016 #20

    Baluncore

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    I was talking orders of magnitude.

    Running the LTspice spoiler code shows that;
    With R5=100k, peak voltage across 1N4148 diodes is 333mV. The output voltage is 2.203Vpp.
    The 3'rd harmonic is 36 dB below the fundamental.

    Change R5 to 180k and the 1N4148 voltage rises to 397mV, Output = 10.04Vpp.
    The 3'rd harmonic is 48 dB below the fundamental.

    Change R5 to 200k and the 1N4148 voltage rises to 451mV, Output = 33.5Vpp.
    The 3'rd harmonic is 58 dB below the fundamental.

    But we have reached coffin corner, where component and thermal variation will make huge differences to stability and amplitude.
    The critical point is when R5 = 1/((1/20k)-(1/22k1)) = 210k4762.
    How silly can it get?

    Change R5 to 205k and the 1N4148 voltage rises to 483 mV, Output = 67.4Vpp.
    The 3'rd harmonic is 66 dB below the fundamental.

    Change R5 to 210k and the 1N4148 voltage rises to 588 mV, Output = 676.Vpp.
    The 3'rd harmonic is more than 71 dB below the fundamental, below the FFT noise floor.

    The distortion is least when the diode voltage is greatest.
     
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