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"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
 
J

John Popelish

Jan 1, 1970
0
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.

By total impedance, do you mean the magnitude of impedance, without
regard to its real/imaginary ratio?
Is it possible to mathematically extract a Bode plot from an inverse
plot?

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.
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.

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

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
 
J

John Popelish

Jan 1, 1970
0
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?

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.
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.

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.
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.

What is turning it?
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.

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.
 
John said:
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.

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

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.
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.

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
 
S

Scott Stephens

Jan 1, 1970
0
John Popelish wrote:

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

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

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.
My hypothesis is R and L will decrease linearly together with zero
intercept.

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!
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.

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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