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  1. #1
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    understanding bode plot characteristics

    Hello,


    i have a circuit where i try to understand how bode plot works..

    Bellow is the schematic, the opamp circuit has a Gain of 2,

    Click image for larger version. 

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    I am feeding a 200mhz sinewave into the circuit, bellow is the phase and gain .

    Phase
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    Gain
    Click image for larger version. 

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    For 200mhz i have a phase shift of -19,98degrees, and a gain of 5.6dB.

    When tring to compare the input waveform with the output form, i can see some phase shift between them

    On the plot i have the initial time (t) and a given point on the signal period (t1).

    VG1 (green) - 200MHZ 2Vp signal
    VF1 (blue) - output signal.

    Click image for larger version. 

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    For example i wanted to check the voltage at 1.8ns from the initial (t) for phase shift of -19,98deg on the output sine,how should i proceed?

    I know the voltage on a point in the phase shifted sine is

    U_point = U_peak * sin( 2PI * f * t + phi),

    But how can i find the voltage if the peak is diferent? Because i want to measure the voltage phase shift in respect with my input sine which has a smaller amplitude.

    My input sine has 1Vpeak, and the output sine has almost 2Vpeak?

    I am a beginner in this "bode plotting" , just wanted to know if my assumptions are somehow correct

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  2. #2
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    Re: understanding bode plot characteristics

    Each order of filter has asymptotic slope or -6 dB/2f or -20dB/decade

    You have several unknowns but appears to be around 4th order @ 200MHz

    Op AMps are 1st order integrators with a pole at GBW but after that becomes higher order. Also input RC has some pole you can compute.

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    A best design is easily achieved with good test specs™
    A better question deserves a better answer. ™
    ... so include all your acceptance criteria ( values, % tolerance) and assumptions in your question or any design.

    ... Tony Stewart EE since 1975
    - slightly north of Toronto



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    Re: understanding bode plot characteristics

    1. What's the 1 pf capacitor in parallel to the input voltage source good for?

    2. THS3202 is a current feedback OP, the open loop gain can be programmed by feedback resistor R2. 1K is too large to utilize the available THS3002 bandwidth.

    3. Bode plot is a small signal measurement and must be not measured with several volts magnitude. In simulation, it's obtained in AC rather than transient analysis.

    4. In case you want to derive phase and magnitude from time domain measurements, I don't see a problem to measure peak voltage and phase shift separately. Simulation tools and digital oscilloscopes can directly display these quantities.



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    Re: understanding bode plot characteristics

    What are your design goals?

    I just read datasheet and I agree Rf and RL load need to be reduced

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    A best design is easily achieved with good test specs™
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    ... so include all your acceptance criteria ( values, % tolerance) and assumptions in your question or any design.

    ... Tony Stewart EE since 1975
    - slightly north of Toronto



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    Re: understanding bode plot characteristics

    Hello,

    the goal for this was just to see a practical approach to help me understand bode plots, and how they are affect my circuit. So i started out just ploting a freq response from datasheet

    To be honest, 4 days ago i had no idea what the phase response in the bode plot means, i knew its referring to the phase of the voltage...

    I chose ths3202 because i have some parts in my lab where i can build a small circuit and test this in "Real life" ...

    I have changed the feedback resistos and the load to the values suggested in the datasheet. I will try latter on to plot everything again.

    Regarding the 1pf value at the non-invertin amp, in the datasheet it specified a input capacitance on the non-inverting pin of min 1pf

    But i have some questions before i continue

    asymptotic slope or -6 dB/2f or -20dB/decade
    - that means phase shift will also be -45degree at the -3db point?



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    Re: understanding bode plot characteristics

    It seems to me your question was very simple: how do I compare the phase of two sin waves with different amplitudes? If that was your question the answer is very simple: Compare any two similar points. Zero crossing to zero crossing or peak to peak etc

    Measure the distance between those points in time and calculate what that means in terms of phase.


    Very good exercise by the way. I remember drawing bode plots in some class and having no clue what they related to in real life.



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    Re: understanding bode plot characteristics

    that means phase shift will also be -45degree at the -3db point?
    Yes, for a first order (single pole) transfer function. The closed loop transfer function in post #1 isn't first order because it doesn't show -3 dB at 45° point.



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    Re: understanding bode plot characteristics

    Forgive my bad eyes and disregard my previous plots

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    A best design is easily achieved with good test specs™
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    ... so include all your acceptance criteria ( values, % tolerance) and assumptions in your question or any design.

    ... Tony Stewart EE since 1975
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  9. #9
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    Re: understanding bode plot characteristics

    I came back with some new result, after updating the resistor values, the plots changed
    This is my new circuit
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    After ploting the response i have the following curves.

    Click image for larger version. 

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    I am a bit confused here, the -3db point is much higher in frequency and much lower in phase than the -45 phase.

    Why is this?



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    Re: understanding bode plot characteristics

    As before, it's a higher order system, but now much more obviously, simply look at the gain peaking.



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    Re: understanding bode plot characteristics

    i have a circuit where i try to understand how bode plot works..
    The idea is really really very simple.

    Consider the circuit as a black box with an input and an output.

    The Bode plots (and the cousin, Nyquist plot) describe the amplitude and phase relations of the input and output: as a function of frequency.

    The Bode plots are always a pair: one says the gain vs log (f) and the other the phase as a function of log (f).

    This describes the transfer function of the black box.

    For any arbitrary input function, decompose the input into a set of sine functions (Fourier transformation) and calculate the gain and phase for that amplitude and frequency and combine back the result (another Fourier transformation) to get the output function.

    You need to get into the black box if you want to improve the transfer function.

    This is the basic principle.



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