"Inverse" Bode Plot

I define the inverse plot as the graph of the total impedance as a function of variable L or C, but not R, in a parallel "tank" circuit.

Is it possible to mathematically extract a Bode plot from an inverse plot?

I understand the Bode plot to be the graph of total impedance as a function of variable f, not L or C or R, in a paralle "tank" circuit.

Yours,

Doug Goncz Replikon Research Falls Church, VA 22044-0394 DGoncz at aol dot com email

Reply to
DGoncz
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You have my thanks for writing, John.

Yes, I mean |Z|, the magnitude of impedance w/ort phase.

Is it possible to use a numerical method to convert a graph from a list of capacitance values on the X axis and |Z| on the Y axis to a graph of f on the X axis and |Z| on the Y axis?

You see, I haven't mastered Daqarta yet... but I have a cap sub box of

11.111 ufd in increments of 0.001 ufd, and am building a sub box of 5-50 ufd. I find 1.0 VAC output from my self-excited induction generator (SEIG) with a 30 ufd cap and oscilloscope and DMM load, but 0.9 VAC output with 50 ufd. So I believe I can produce a meaningful graph.

Of course, one must know f, the frequency of measurement with varying C. My LCR meter uses 100 Hz. My scope tells my my SEIG is running near

60 Hz, and I can get a frequency meter in another DMM inexpensively.

The purpose of the experiment is to look at whether a rewind with much heavier wire at great expense would produce enough Q for the SEIG to reach 110 VAC output with an 8 watt load.

Doug

Reply to
DGoncz

By total impedance, do you mean the magnitude of impedance, without regard to its real/imaginary ratio?

A Bode plot is just magnitude (expressed as the logarithm) and phase versus log of frequency. You can express those two properties of any complex variable, as long as you keep track of what you are describing.

More often, it is a plot of log of gain and linear phase shift, with respect to log frequency, but you can substitute any complex variable for gain.

Reply to
John Popelish

I don't see how you change the independent variable from capacitance to frequency. You could make a whole bunch of graphs, one for each capacitance value with changing frequency.

If the SEIG is turning at a fixed speed, the frequency should also be fixed, allowing you to make a graph og output voltage versus capacitance, but I don't understand exactly what you are trying to accomplish.

What is turning it?

At best, I think your experiments will help yo find the optimum capacitor value for a given speed and load, but I don't see that it tells you much about what a different winding would accomplish.

Reply to
John Popelish

Frankly, neither do I, but I think I can do it in Mathcad if I use the solver. There must be a transform.

Burden's number 10-1134 motor has a 1/8 pipe fitting in the end bell opposite the shaft. I fitted two motors together with a short pipe nipple. The alignment was very good. I slipped a threaded rod through the hollow shafts and added some washers and cap nuts. One motor turns the other. The load is light and the speed near 400 rpm synchronous. I do not have a tach.

Well, if I can do this inverse plot I will have Q as a function of frequency. That peak value of Q tells me some ratios like L/R or some damn thing. With that, and a quick micrometer reading on a stator wire, I can work out so many thicker wires would have such an inductance and resistance. That will tell me if there's any hope of making R < sqrt(L/C) by reducing R.

My hypothesis is R and L will decrease linearly together with zero intercept. That means R will decrease more than sqrt(L) will decrease. However, since each value of L needs a value of C, I do not yet know whether R > sqrt(L/C) can be changed to R < sqrt(L/C) by a rewind and a new cap. I haven't worked that out. Unlike the inverse plot, this is mere algebra.

You see, with R > sqrt(L/C) there is no hope of SEIG function. With R < sqrt(L/C) there is some hope. Q must be high enough for SEIG function.

For proof of concept I need only 8 watts. To run a laptop, GPS, Pocket PC, radio, etc, I would need more.

Doug

Reply to
DGoncz

Ah, but that's the essence of Doug's query! Is there a transform that can replace the Laplacian *S* correspondence with *jOmega* with a transformed reactance?

I suspect there is, for a well-behaved 2nd order differential system. But I suspect that wouldn't do for Doug's naughty generator app.

There is, but how much method belies the madness is a devil in detail!

Awe, F'it Doug, make a binary C-box out of fine quality, over-sized motor caps, and switch'em in to fine-tune the system for the energy regime of interest, as you switch binary taps on your generator winding.

An empircal excersize that you can write a paper about fitting a curve to.

I'de be freaking over the saturation flux, but that shows my ignorance. G'D power electronics is horribly naughty. Play games with your F'ng computer, not simulations. Only G'D' thing their good for; aside from wasting time on their troubles avoiding a fem's troubles.

L is dependent on flux-density, that being determined mainly be the core and its saturation flux, current and speed.

That means R will decrease more than sqrt(L) will decrease.

But isn't saturation flux, or the decrease in inductance with increasing current (due to decreasing R) at issue? What is limiting your power and efficiency? The amount of iron you are willing to peddle around!

Follow the energy. You want to change force into volt-amps, by means of amp-meters in magnetized iron. Your motor core is the independent variable. Copper-loss and window-fact for the desire volt-amps the dependent variable.

Think of it as a transformer.

Scott

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Reply to
Scott Stephens

From a separate email, I think we need to know more about the details of the SEIG and what is intended to be accomplished. I suggest a new post with details about this project.

Paul E. Schoen

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Reply to
Paul E. Schoen

The new post will be titled "Inverting a Plot of SEIG Output to Produce a Bode Plot".

Doug

Reply to
DGoncz

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