I have measured the DC resistance of a split-phase capacitor run
induction motor's two windings. They are 53 and 35 ohms. The motor is
impedance protected. Using the nameplate voltage and current, I have
calculated the total impedance at 60 Hz to be 110 ohms. This total
impedance is larger than the DC resistance, and so I have algebraically
subtracted the resistance from the total impedance to get the inductive
impedance, but I don't know if I did that step right:

http://users.aol.com/DGoncz/Publicat...orAnalysis.bmp

I am pretty sure about R1/R2 = X1/X2 although the winding *are*
different colors and could be different gages, but I am not sure about

1/(1/(R1+X1) + 1/(R2+X2)) = 110 ohms

I don't know if you can add a resistance and an inductive impedance
arithmetically this way. I have seen things like

R1 at angle 0 degrees + X1 at angle 90 degrees =
sqrt(R1^2 + X1^2)

I have invested hundreds of dollars into this motor/generator and while
I would like to avoid a rewind, these high resistances make a rewind
look inevitable. If I can get a good model, though, I may find a Q1
for some capacitance, and that would indicate, I think, that
self-excitation could commence.

What is not shown in MotorAnalyis.bmp is R1 in series with L1 and so
X1, and R2 in series with L2 and so X2, and R1/L1/X1 in parallel with
R2/L2/X2 and the capacitor C.

Yours,

Doug Goncz
Replikon Research
Falls Church, VA 22044-0394

On 1 Jun 2005 12:02:57 -0700, wrote:

I have measured the DC resistance of a split-phase capacitor run
induction motor's two windings. They are 53 and 35 ohms. The motor is
impedance protected. Using the nameplate voltage and current, I have
calculated the total impedance at 60 Hz to be 110 ohms. This total
impedance is larger than the DC resistance, and so I have algebraically
subtracted the resistance from the total impedance to get the inductive
impedance, but I don't know if I did that step right:

http://users.aol.com/DGoncz/Publicat...orAnalysis.bmp

I am pretty sure about R1/R2 = X1/X2 although the winding *are*
different colors and could be different gages, but I am not sure about

1/(1/(R1+X1) + 1/(R2+X2)) = 110 ohms

I don't know if you can add a resistance and an inductive impedance
arithmetically this way. I have seen things like

R1 at angle 0 degrees + X1 at angle 90 degrees =
sqrt(R1^2 + X1^2)

I have invested hundreds of dollars into this motor/generator and while
I would like to avoid a rewind, these high resistances make a rewind
look inevitable. If I can get a good model, though, I may find a Q1
for some capacitance, and that would indicate, I think, that
self-excitation could commence.

What is not shown in MotorAnalyis.bmp is R1 in series with L1 and so
X1, and R2 in series with L2 and so X2, and R1/L1/X1 in parallel with
R2/L2/X2 and the capacitor C.


Reactance is at a right angle to resistance so it is incorrect to add
them arithmetically. This is easy to accomodate in MathCAD since it
can do arithmetic with complex variables. Just multiply your X
terms by i which MathCAD understands to be the complex operator
sqrt(-1). You can direct MathCAD to use j instead of i since j is
more commonly used in electrical engineering. Using the j
convention, an R in series with an X has impedance R + j*X. In
polar notation, this would be a Z with an angle theta. Z is the
magnitude, volts/amps and theta is the phase angle between voltage and
current.

Sample MCad sheet available if you'd like. Email me -- and tell me
which version of MCad you're running.

Doug, you don't simply add reactance and resistance, since they are
orthogonal quanities (at right angles to one another)..

To compute Impedance, Z, simply use the same formula as you would use
for computing the hypotenuse of a right triangle (The Pythagorean
Thorem). In this case:

Z = SQRT(Resistance Squared + Reactance Squared).

A simple empirical method of determining the impedance is simply to
connect a variable resistor (potentiometer) in series with the
generator, apply a 60-Hz AC voltage (say 24-Volts), then adjust the
variable resistor until the AC voltage across it is equal to the AC
voltage across the generator windings. You can then measure the
resistance of the adjustable resistor with a simple ohm meter and its
value will be numerically equal to the impedance of the generator,
since the voltage drop across each will be equal.

E = IZ = IR, where the AC current passing though both the
generator and the variable resistor (when connected in series) are of
course equal.

Harry C.

Doug, as a follow on, it is important to note that the current draw of
a motor is typicallly stated for its maximum horsepower load, hence
tells you almost nothing about its impedance/reactance.

Harry C.

Harry C. wrote:

Z = SQRT(Resistance Squared + Reactance Squared).

Thanks, Harry, that was what I meant when I wrote

sqrt( R1^2 + X1^2)

You've confirmed that these are vector quantities.

I have a calibrator output on my 'scope that is a 60 Hz *square wave*
but I'd need a sine wave to use the potentiometer method, right?

Doug

Doug, since reactance varies with frequency you really need a 60-hz
sinewave source (something like a doorbell transformer or toy train
transformer should do nicely).

A pulse or square wave contains many higher frequency harmonics which
would confuse the measurement.

Hope this helps, Harry C.

I have ordered two 10-1134 motors and eleven 5 mfd capacitors from
Surplus Center. I have ordered an 11.1111 mfd 50 VDC cap sub box from
Electronic Parts. I intend to mate the hanger threads on the end bells
of the motors with a close nipple and install a threaded rod and cap
nuts to join the shafts. One motor will be the prime mover, and the
other will be the test generator.

I have written to Terry Given to see if he will recruit me into the
IEEE.

I have ordered an LCR meter and a less expensive cap sub box that
should be rated 200 VDC instead of 50 VDC.

Doug

I wrote:
I have ordered an LCR meter and a less expensive cap sub box that
should be rated 200 VDC instead of 50 VDC.

The LCR meter is on the way from a private ebay seller overseas by air
mail. The cap sub box has not been shipped yet. The motors are on their
way by Parcel Post.

Doug

wrote in sci.electronics.design and
rec.crafts.metalworking:
I have ordered an LCR meter and a less expensive cap sub box that
should be rated 200 VDC instead of 50 VDC.

Doug

The LCR meter has arrived from Hong Kong.

The inductance of the motor coils in parallel is 112 microhenries.

The resonant capacitance is (confirm?) 62 microfarads.

There are 36 poles on this split-capacitor motor.

The synchronous speed would be (confirm?) 400 rpm.

The motor runs at 225 (rated) rpm.

The motor runs with 175/400 = 44% slip.

I need to run the motor at 400+44% = 576 rpm.

I have to do that because of the impedance protection, right?

So the efficiency will indeed be low.

At 90 pedal rpm, with my existing cog, I will need a

x * 90 / 8 = 576;
x = 576 * 8 / 90 = 51 tooth cog, which is just what I have.

However, if I splice two motors together at 225 rpm, I will have to
recompute.

I recall that damping reduces the apparent frequency of an impulse
driven resonant system, and wonder if the substantial resistance of
this impedance protected motor will reduce the continuously driven
resonant frequency, or whether my recollection only applies to impulse
driven resonant systems.

Yours,

Doug Goncz
Replikon Research
Seven Corners, VA 22044-0394

(see previous thread in Google; it is likely expired in your reader.)

wrote:
The inductance of the motor coils in parallel is 112 microhenries.
No. 112 millihenries.

Doug

On 10 Jun 2005 15:00:48 -0700, wrote:



wrote in sci.electronics.design and
rec.crafts.metalworking:
I have ordered an LCR meter and a less expensive cap sub box that
should be rated 200 VDC instead of 50 VDC.

Doug

The LCR meter has arrived from Hong Kong.

The inductance of the motor coils in parallel is 112 microhenries.

The resonant capacitance is (confirm?) 62 microfarads.

There are 36 poles on this split-capacitor motor.

The synchronous speed would be (confirm?) 400 rpm.

The motor runs at 225 (rated) rpm.

The motor runs with 175/400 = 44% slip.

I need to run the motor at 400+44% = 576 rpm.

I have to do that because of the impedance protection, right?

So the efficiency will indeed be low.

At 90 pedal rpm, with my existing cog, I will need a

x * 90 / 8 = 576;
x = 576 * 8 / 90 = 51 tooth cog, which is just what I have.

However, if I splice two motors together at 225 rpm, I will have to
recompute.

I recall that damping reduces the apparent frequency of an impulse
driven resonant system, and wonder if the substantial resistance of
this impedance protected motor will reduce the continuously driven
resonant frequency, or whether my recollection only applies to impulse
driven resonant systems.

Yours,

Doug Goncz



I've picked up this thread late so I've probably missed important
bits. However the following comments may be useful.

If I've understood the post correctly you are aiming to use an
impedance protected 36 pole motor as a self excited induction
generator.

Self excited induction generators rely on the tiny residual
pattern of magnetisation of the rotor being reinforced by the current
flowing in the near resonant stator winding circuit. It has to be
operating close to resonance for the current build up to be large
enough to reinforce the rotor field pattern. It has to be on the
capacitative side of resonance to permit the phase angle of the stator
current to reinforce the rotor field pattern.

It is a positive feedback regenerative system and on a large
efficient motor the output can build up to far beyond its rated motor
power until limited by magnetic saturation. This effect is sometimes
used for regenerative braking of single and three phase motors and can
result in a spectacularly short stopping time.

With a care and control of speed, self excited induction
generator systems are possible but they're pretty touchy devices.
If you're unlucky with the the rotor iron they may not retain enough
initial magnetism to enable the output to build up (manufacturers
strive to reduce this because it degrades the efficiency when used as
a motor) Also it must use a reasonably efficient motor for the
magnetic feedback to exceed the system losses.

Efficiency is your major problem. An impedance protected motor
means a motor with deliberately large leakage inductance so that the
impedance of this inductance limits the current that flows when the
motor is stalled or overloaded. With limited stalled current the
starting torque (already poor because it is a capacitor run machine)
has to be boosted by the use of a high resistance rotor and this
results in your observed very high slip speed. Even if there were no
other losses of any kind the motor efficiency could not be any better
than the % synchronous speed - 56%. With other losses taken into
account the motor efficiency is probably no better than 40%.

With the uH to mH correction your sums are OK but this level
of efficiency is too low for a succesful induction generator.

Jim

I find the rotor has 48 conductive bars. I could have this wrong.

By substituting the rotor, could I increase the efficiency of this
motor? That might be much easier than rewinding. I have removed and
replaced rotors from shafts before. It seems any rotor with the right
number of bars and less than or equal to the linear size, with the
right or smaller bore, would do to make *some* improvement. I'm
guessing I'd want a rotor with fewer bars, but I don't know the
physics.

Can any reader show me the math relating 60Hz, 36 poles, 400 rpm, 48
bars, and 225 rpm?

I don't get it. I take 7200 / 32 = 225. To get it, I need to go to
Falls Church's library, the Mary Riley Stiles library. There are not
one but two editions of Audel's there.

Doug

I find the following encouraging note at:

http://www.ecmweb.com/mag/electric_m..._ac_induction/

"As seen in the Table above, smaller motors and lower-speed motors
typically have higher relative slip. However, high-slip large motors
and low-slip small motors are also available."

I need a low-slip small motor.

Doug

On 11 Jun 2005 10:26:54 -0700, wrote:

I find the rotor has 48 conductive bars. I could have this wrong.

By substituting the rotor, could I increase the efficiency of this
motor? That might be much easier than rewinding. I have removed and
replaced rotors from shafts before. It seems any rotor with the right
number of bars and less than or equal to the linear size, with the
right or smaller bore, would do to make *some* improvement. I'm
guessing I'd want a rotor with fewer bars, but I don't know the
physics.

Can any reader show me the math relating 60Hz, 36 poles, 400 rpm, 48
bars, and 225 rpm?

I don't get it. I take 7200 / 32 = 225. To get it, I need to go to
Falls Church's library, the Mary Riley Stiles library. There are not
one but two editions of Audel's there.

Doug

The precise number of bars in the rotor makes little difference to
motor efficiency and bears no particular relation to motor rated or
synchronous speed. The only thing that matters is the inductance to
resistance ratio of the effective shorted turn that surrounds the
rotor iron circuit i.e. the L/R time constant.

If there a lot of iron but not much copper/aluminium, the resistive
component is high and this is generally called a high resistance
rotor. This gives good starting torque but lousy efficiency because
of the extra power dissipated in the rotor. This results in full load
speed well below synchronous speed and it is this speed difference
that is responsible for the extra rotor dissipation.

Unless starting torque is a big problem, motors are
designed with the maximum possible amout of copper/aluminium in the
rotor. Only the minimum necesary amount iron is retained to enable it
to handle the stator induced flux density. This results in very low
slip frequency and high efficiency.

While fitting a low resistance rotor to your machine may
give some efficiency improvement it is unlikely to be enough to get it
to operate as useful induction generator. The main problem which will
still remain is the extremely high leakage inductance which is
inherent in the geometry of a small 36 pole stator.

Jim

I understand that leakage inductance is a defect in an otherwise
perfect transformer, and it is clear why toroidal transformers would
have low leakage inductance.

But "where" is the leakage inductance in a motor? Restated, in what way
is a motor like a transformer?

I've looked all over the web and haven't found this information. I did
find many documents referring to leakage inductance in motors.

I've only put a few hundred dollars into this generator. Perhaps I
should let go of it. I have a DC generator that has produced loads of
power with little drag.

Doug

wrote:


But "where" is the leakage inductance in a motor? Restated, in what way
is a motor like a transformer?
In many ways.

You take a primary winding (stator) excite it with voltage to create
magnetic fields. THe magnetic fields then (across the air gap) induce
currents in the rotor (secondary). You have an air gapped
transformer, with a shorted secondary.

Now, a 3 ph motor running as a rotary phase converter, is acting even
more like a transformer. THe rotor currents (once the rotor is
turning) create a field, that interacts (again across the airgap) with
the other two stator windings, phase displacing the voltage you
already have applied across them.

jk

On 12 Jun 2005 11:17:27 -0700, wrote:

I understand that leakage inductance is a defect in an otherwise
perfect transformer, and it is clear why toroidal transformers would
have low leakage inductance.

But "where" is the leakage inductance in a motor? Restated, in what way
is a motor like a transformer?

I've looked all over the web and haven't found this information. I did
find many documents referring to leakage inductance in motors.

I've only put a few hundred dollars into this generator. Perhaps I
should let go of it. I have a DC generator that has produced loads of
power with little drag.

Doug

An induction motor IS a transformer. The primary is the stator
winding(s) and the secondary is the rotor bars with the short circuit
acting as a zero resistance load.

The induced current in these rotor bars produces the magnetic
field pattern which generates the rotary torque. Exactly similar
forces occur in a heavily loaded or short circuited convential
transformer but these are not normally noticed because the primary and
secondary windings are firmly anchored in their correct positions.

Leakage inductance is best thought of in terms of flux which
is generated by one winding and which closes in on itself without
intersecting the field pattern of the associated secondary.

The leakage inductance of a motor winding is higher than that
of a similarly rated convential transformer because of the much larger
physical separation between primary and secondary. An example of
leakage flux is the flux that passes from stator tooth tip to tooth
tip without intersecting the rotor iron circuit.

The above ignores the effect of rotor rotation which changes
both the frequency and the amplitude of the rotor currents.
Nevertheless it's still a pretty good working guide to what's
happening. A fuller explanation is of course possible but not really
appropriate for a metalworking group.

Jim

To properly use an LCR meter, I believe the following would be
appropriate.

1) Realize the meter measures total impedance and computes inductance
using the impedance figure in the place of reactance. (I read this on a
web page)
2) Measure the inductance and resistance, or capacitance and
resistance.
3) Find the impedance associated with the inductance, using the meter's
test frequency.
4) Recombine the impedance and resistance to find the reactance using
Pythagoras.
5) Once again using the meter's test frequency, find the new inductance
or capacitance associated with the reactance.

Right? Assuming the real component is strictly L-R or R-C.

Doug

On 12 Jun 2005 16:59:08 -0700, wrote:

To properly use an LCR meter, I believe the following would be
appropriate.

1) Realize the meter measures total impedance and computes inductance
using the impedance figure in the place of reactance. (I read this on a
web page)
2) Measure the inductance and resistance, or capacitance and
resistance.
3) Find the impedance associated with the inductance, using the meter's
test frequency.
4) Recombine the impedance and resistance to find the reactance using
Pythagoras.
5) Once again using the meter's test frequency, find the new inductance
or capacitance associated with the reactance.

Right? Assuming the real component is strictly L-R or R-C.

Doug

Most LCR meters are pretty crude devices and simply indicate
the scalar impedance of the test device. However the dial calibration
is based on the assumption that the test device is a pure lossless L
or C.

IF the loss component of an inductance is pure series R
loss AND you know the value at the test frequency (which may be
considerably higher than the DC value), as you have assumed, the true
value of L can be arrived at by quadrature addition. Based on AC and
DC LCR measurements on an air cored coil this method can give
reasonably accurate results.

However, if the test piece is an iron cored component the
measurement does not take into account the shunt losses (iron eddy
currents, hysteresis etc.) and permeability variation both with
frequency and flux density. These are all second order effects but
mean that the "true" inductance of an iron cored device is a pretty
variable quantity unless the measurement conditions are closely
defined.

Fortunately, with power frequency electric
motors, the low working frequency and the significantly air gapped
iron circuit reduces the effect of these second order components.
However measurement accuracies are likely to be limited to one, or at
best, two significant figures,

Jim

wrote:

IF the loss component of an inductance is pure series R
loss AND you know the value at the test frequency (which may be
considerably higher than the DC value), as you have assumed, the true
value of L can be arrived at by quadrature addition.

Wouldn't that be quadrate subtraction, Jim?

Subtracting the resistance at 0 degrees from the impedance associated
with the mismeasured inductance at an unknown phase angle...

That is,

Xt = 2 * pi * f * L (measured)
Xl = sqrt( Xt^2 - R^2)
L ( compensated ) = Xl / ( 2 * pi * f ) ?

Doug

I have found a candidate motor for use as a generator:

http://surpluscenter.com/item.asp?UI...tname=electric

I believe this motor has 8 poles, a synchronous speed of 900 rpm and
full load slip of 12%. So it should be around 70-75% efficient,
shouldn't it?

Would this be a suitable motor for further experiments with
self-excited induction generators? It's used, but it's cheap, small,
and seems efficient enough.

Doug

In article ,
wrote:

On 12 Jun 2005 16:59:08 -0700, wrote:

To properly use an LCR meter, I believe the following would be
appropriate.

1) Realize the meter measures total impedance and computes inductance
using the impedance figure in the place of reactance. (I read this on a
web page)
2) Measure the inductance and resistance, or capacitance and
resistance.
3) Find the impedance associated with the inductance, using the meter's
test frequency.
4) Recombine the impedance and resistance to find the reactance using
Pythagoras.
5) Once again using the meter's test frequency, find the new inductance
or capacitance associated with the reactance.

Right? Assuming the real component is strictly L-R or R-C.

Doug

Most LCR meters are pretty crude devices and simply indicate
the scalar impedance of the test device. However the dial calibration
is based on the assumption that the test device is a pure lossless L
or C.

The only ~$200 handheld LCR meter that doesn't make the assumption of
lossless (pure) inductance or capacitance that I know of is the Extech
model 380193 LCR Meter. I have one, and it works well. It measures L,
C, or R at 120 Hz or 1000 Hz (but not R at DC), and for L and C also
reports parasitic R.

The B+K model 875B LCR meter did not work for me because they only
worked with very pure inductances. I have heard that the Wavetek meters
also have the problem. The test is to take a relatively pure
inductance, like one winding of a power transformer, and put a
potentiometer in series. Does the reported inductance vary as the
series resistance is increased? With the B+K, it just explodes, with
the indicated inductance becoming a large factor bigger than the true
value, so a 2-henry inductor was reported as 45 henries. Complete
nonsense, rendering the meter useless.

Note that one-frequency and two-frequency LCR meters cannot detect
self-capacitance in an inductor, or self-inductance in a capacitor.
Only parasitic resistance can be detected.

Joe Gwinn

My meter uses a test frequency of 100 Hz.

I measured the two windings and got R1 = 38.5 ohms, L1 = 0.149 H, R2 =
123.5 ohms, L2= 0.457 H.

So I got X1 = 94 ohm, X2 = 287 ohm.

And X1v = 85 ohm, X2v = 259 ohm, with sqrt(X1^2 - R1^2) etc...
And L1v = 0.136 H, L2v = 0.413 H.

Doug

On 13 Jun 2005 04:58:15 -0700, wrote:

wrote:

IF the loss component of an inductance is pure series R
loss AND you know the value at the test frequency (which may be
considerably higher than the DC value), as you have assumed, the true
value of L can be arrived at by quadrature addition.

Wouldn't that be quadrate subtraction, Jim?

Subtracting the resistance at 0 degrees from the impedance associated
with the mismeasured inductance at an unknown phase angle...

That is,

Xt = 2 * pi * f * L (measured)
Xl = sqrt( Xt^2 - R^2)
L ( compensated ) = Xl / ( 2 * pi * f ) ?

*********************

sorry " quadrature addition" was just a careless reference to the fact
that the impedances were in quadrature.


*********************

Doug


My meter uses a test frequency of 100 Hz.

I measured the two windings and got R1 = 38.5 ohms, L1 = 0.149 H, R2 =
123.5 ohms, L2= 0.457 H.

So I got X1 = 94 ohm, X2 = 287 ohm.

And X1v = 85 ohm, X2v = 259 ohm, with sqrt(X1^2 - R1^2) etc...
And L1v = 0.136 H, L2v = 0.413 H.

Doug

**************

This is an example of how easy it is to attribute spurious accuracy to
inductance measurements made with an LCR meter. There is no way that
the measurement is accurate to three significant figures and even the
second figure is pretty dubious.

***************

I have found a candidate motor for use as a generator:

http://surpluscenter.com/item.asp?UI...tname=electric

I believe this motor has 8 poles, a synchronous speed of 900 rpm and
full load slip of 12%. So it should be around 70-75% efficient,
shouldn't it?

Would this be a suitable motor for further experiments with
self-excited induction generators? It's used, but it's cheap, small,
and seems efficient enough.

Doug

*****************

Certainly a more suitable than your existing 36 pole device but I'm
still pretty doubtful. Small multipole induction motors suffer from
low efficiency and comparatively high leakage inductance. Given a free
choice I would go for a 2 or 4 pole motor with full load load slip of
no more than 5%. Even with a motor as good as this there would still
be a small chance that the rotor residual magnetism be too low to
initiate self excitation.

However the advertised machine is a nice little motor at an
attractive price Even if it doesn't work it's all useful experience
and a handy motor to add to your stock.


Jim

I wrote:

I measured the two windings and got R1 = 38.5 ohms, L1 = 0.149 H, R2 =
123.5 ohms, L2= 0.457 H.


So I got X1 = 94 ohm, X2 = 287 ohm.


And X1v = 85 ohm, X2v = 259 ohm, with sqrt(X1^2 - R1^2) etc...
And L1v = 0.136 H, L2v = 0.413 H.

I do understand that L1v is accurate to about two digits. I can show
the tolerance buildup if anyone cares to see it. For example 2 is an
exact number, pi can be brought in to any precision, but L is, judging
from what Jim has written, accurate to maybe 10%. So with a tolerance
on each number, the tolerance or precision on the result can be had.

And L is of course, suspect as well.

However, I have computed the resonant capacitor for the 1 winding as
51. blah blah blah microfarads, and understand that I can pick out 50
ufd of capacitor and even measure the cap and still be off by 10%.

But I did include an extra digit. So sorry. Mea culpa.

Correct me if I am wrong, but it seems, looking at a Bode plot, that
generator operation should be arranged at the nose of the curve, the
inflection point between positive and negative curvature, where
bandwidth is measured. Yes? Behavior is linear there.


Doug

wrote:

On 12 Jun 2005 16:59:08 -0700, wrote:


To properly use an LCR meter, I believe the following would be
appropriate.

1) Realize the meter measures total impedance and computes inductance
using the impedance figure in the place of reactance. (I read this on a
web page)
2) Measure the inductance and resistance, or capacitance and
resistance.
3) Find the impedance associated with the inductance, using the meter's
test frequency.
4) Recombine the impedance and resistance to find the reactance using
Pythagoras.
5) Once again using the meter's test frequency, find the new inductance
or capacitance associated with the reactance.

Right? Assuming the real component is strictly L-R or R-C.

Doug


Most LCR meters are pretty crude devices and simply indicate
the scalar impedance of the test device. However the dial calibration
is based on the assumption that the test device is a pure lossless L
or C.

IF the loss component of an inductance is pure series R
loss AND you know the value at the test frequency (which may be
considerably higher than the DC value), as you have assumed, the true
value of L can be arrived at by quadrature addition. Based on AC and
DC LCR measurements on an air cored coil this method can give
reasonably accurate results.

However, if the test piece is an iron cored component the
measurement does not take into account the shunt losses (iron eddy
currents, hysteresis etc.) and permeability variation both with
frequency and flux density. These are all second order effects but
mean that the "true" inductance of an iron cored device is a pretty
variable quantity unless the measurement conditions are closely
defined.

Fortunately, with power frequency electric
motors, the low working frequency and the significantly air gapped
iron circuit reduces the effect of these second order components.
However measurement accuracies are likely to be limited to one, or at
best, two significant figures,

Jim
If you want to call RF/AF bridges crude. Just not uP I guess. Some are.
Owner and long time user of low tech to high tech versions.

Martin

--
Martin Eastburn
@ home at Lion's Lair with our computer lionslair at consolidated dot net
NRA LOH, NRA Life
NRA Second Amendment Task Force Charter Founder

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On Tue, 14 Jun 2005 00:00:42 -0500, "lionslair at consolidated dot
net" "lionslair at consolidated dot net" wrote:

snip

If you want to call RF/AF bridges crude. Just not uP I guess. Some are.
Owner and long time user of low tech to high tech versions.

Martin

My comments referred only to LCR "METERS". A good RF/AF Bridge
could well handle the highlighted problems.

Jim

OK, I went to my hoarding/clutter group tonght and put a deadline on
the motor/generator project.

Basically, what I am hearing is:

Slow motors don't generate well.
Small motors don't generate well.
Both, because the poles are small.

Well, this is a small, slow motor, so I am going to try to see if the Q
at the nose of the curve is greater than one with math and measurement.
And I will try to make it work, but at some point I am going to have to
let this one go. Pity.

Doug

Well, I received two motors and eleven capacitors from Surplus Center.

I used a short 1/8 pipe nipple to join the motors and added a handle
made from another nipple and a bit of 1/4-20 threade rod between the
mounting brackets. I ran another rod through the hollow shafts and
added finishing washers on the ends to keep the rod concentric. I
topped the rod with cap nuts. The motors turn together with a
pronounced squeal.

So now I am sitting here with a $143 cap sub box I don't dare use
because another is coming UPS from ebay today. If that box tests out,
I'll return the expensive one ordered in haste. I plan to run the
motors together and use various capacitances on the driven motor with
the oscilloscope to look for generator action, one value at a time.

Doug

I got the capacitance sub box from ebay! Lenny, our UPS driver, brought
it by at 5:37 PM.

The sub box checked out with a few slightly unreliable switches.

With the sub box in place the motor/generator pair produced an odd
looking wave on the scope. With more capacitance, the waveform adopted
a more nearly sine wave form. There was a very small change in
amplitude.

I need to add more of the capacitors. I think I have to wire up a 10,
20, 20 uf sub box.

Doug

The waveform produced by the motor/generator spans 15 milliseconds.

That's 67 Hz.

I was expecting a slower wave, a lower frequency. But of course! The
motor isn't running under a full load, so the speed is closer to the
synchronous speed. Still, the output should be a lower frequency than
the input.

Doug

OK, the waveform might span 16, 17, or 18 microseconds. That would be
about right.

I have found something quite heartening, an easy DOS program to analyze
RCL/LCR parallel tanks:

http://www.smeter.net/software/lcr.exe

"Introduction
C is in parallel with L and R in series. Z is the impedance measured
across C. This deceptively simple circuit is found in narrow-band tuned
amplifiers, wideband video amplifiers, impedance-matching L-networks
and in filters, etc.
When R is less than 1.554*Sqrt(L/C) there is a hump, Zmax, in the Z vs
frequency response. As R decreases, Zmax increases and moves to higher
frequencies.
When R Sqrt(L/C) the angle of Z is always -ve and unity power,
commonly defined as the resonant condition, does not occur at any
frequency.
For smaller values of R the angle of Z is +ve at low frequencies and
passes through zero (unity power factor) on the LF side of the
frequency of Zmax which is itself lower than the LC series resonant
frequency. At much smaller values of R, giving high Q, all three
frequencies converge on a common value."

When I try my values for R, L, and C, the program says, in effect, "No
resonance".

Still with 30 ufd across the main winding, 1 VAC is present, and a
mighty clean sine wave at that.

I have ordered a 1 pole, 12 position switch and will add 10 ufd per
position. It's a shorting switch. I may build a second switch up with
50 ufd per position. We will see....

Doug

On 17 Jun 2005 15:29:32 -0700, wrote:

OK, the waveform might span 16, 17, or 18 microseconds. That would be
about right.

I have found something quite heartening, an easy DOS program to analyze
RCL/LCR parallel tanks:

http://www.smeter.net/software/lcr.exe

"Introduction
C is in parallel with L and R in series. Z is the impedance measured
across C. This deceptively simple circuit is found in narrow-band tuned
amplifiers, wideband video amplifiers, impedance-matching L-networks
and in filters, etc.
When R is less than 1.554*Sqrt(L/C) there is a hump, Zmax, in the Z vs
frequency response. As R decreases, Zmax increases and moves to higher
frequencies.
When R Sqrt(L/C) the angle of Z is always -ve and unity power,
commonly defined as the resonant condition, does not occur at any
frequency.
For smaller values of R the angle of Z is +ve at low frequencies and
passes through zero (unity power factor) on the LF side of the
frequency of Zmax which is itself lower than the LC series resonant
frequency. At much smaller values of R, giving high Q, all three
frequencies converge on a common value."

When I try my values for R, L, and C, the program says, in effect, "No
resonance".

Still with 30 ufd across the main winding, 1 VAC is present, and a
mighty clean sine wave at that.

I have ordered a 1 pole, 12 position switch and will add 10 ufd per
position. It's a shorting switch. I may build a second switch up with
50 ufd per position. We will see....

Doug


Long range diagnosis is pretty dangerous but the following
comments may help.

The fact that there is any output means the rotor is
retaining significant residual magnetism. The amount depends on its
previous history. It could be maximised by temporarily DC energising
the stator with the rotor stationary. Use about twice the rated full
load current.

With output generated by residual magnetism the output
should rise roughly directly proportional to speed and be at
synchronous frequency - it's a "permanent" magnet alternator.

Regenerative generation is signalled by a much more rapid
rise in output with speed as stator resonance is approached.

Significant regenerative generation can only occur if the
stator resonance is of high enough Q. The necesary minimum Q depends
mainly on motor efficiency. As a pure guess I would expect Q2 to be
essential and it may need to be as high as 5.

Jim

Thanks, Jim, that's enormously helpful. I am in search of Q, be it with
this motor or another.

I have computed with

http://www.smeter.net/software=AD/lcr.exe

using input values 136000 microhenries, 51000000 picofarads, 23 ohms,
and 0.00006 MHz,

that there is "no impedance hump".

Can you or any reader verify this by running the program?

Can anyone explain why, since R 1.554 * sqrt (L/C),
there is no impedance hump? I must be missing the fine print....

Doug

In article ,
wrote:
On 17 Jun 2005 15:29:32 -0700, wrote:

[ ... ]

When I try my values for R, L, and C, the program says, in effect, "No
resonance".

Still with 30 ufd across the main winding, 1 VAC is present, and a
mighty clean sine wave at that.

I have ordered a 1 pole, 12 position switch and will add 10 ufd per
position. It's a shorting switch. I may build a second switch up with
50 ufd per position. We will see....

Hmm ... consider the following:

1) One four-pole 12 position switch

2) One 10 uF capacitor.

3) Two 20 uF capacitors.

4) one 40 uF capacitor

Wire switch so in the following positions you have the following caps
connected:

0 None

1 Single 10 uF

2 Single 20 uf

3 One 20 uF and one 10 uF in parallel (30 uf total).

4) Two 20 uF in parallel (40uF total)

5 Two 20 uF and one 10uF in parallel (50 uF total)
or
One 40 uf and one 10uf capacitor in parallel (50 uF total)

6 One 40 uF and one 20 uF in parallel (60 uF total)

7 One 40 uF and one 20 uF and one 10 uf in parallel (70 uF total)

8 One 40 uf and two 20 uf in parallel (80 uF total)

9 One 40 uF, two 20 uF and one 10 uF in parallel (90 uf total

So -- you have 0-90 uF in 10 uF steps from four capacitors. This is
pretty much how capacitor decade boxes are made.

If you want to extend the range a bit, add another 20 uf (and another
deck) and you can get up to 110 uF from 12 steps.

Essentially, each capacitor has its own deck, and which are
connected in is determined by the wiring to the terminals of the switch.

I would suggest avoiding the switching while it is powered, as
it will burn the contacts each time you switch.

An alternative way would be to install one toggle switch for
each capacitor, and simply sum the values of the switches which are on.

Enjoy,
DoN.
--
Email: | Voice (all times): (703) 938-4564
(too) near Washington D.C. | http://www.d-and-d.com/dnichols/DoN.html
--- Black Holes are where God is dividing by zero ---

I do understand that halving the resistance requires halving the number
of turns and this could change

R 1.554 * sqrt(L/C)

Doug

On 18 Jun 2005 11:00:06 -0700, wrote:

Thanks, Jim, that's enormously helpful. I am in search of Q, be it with
this motor or another.

I have computed with

http://www.smeter.net/software?lcr.exe

using input values 136000 microhenries, 51000000 picofarads, 23 ohms,
and 0.00006 MHz,

that there is "no impedance hump".

Can you or any reader verify this by running the program?

Can anyone explain why, since R 1.554 * sqrt (L/C),
there is no impedance hump? I must be missing the fine print....

Doug

You're being seduced by illusory accuracy again! With iron cored
inductors you are working with only reasonable approximations to the
behaviour of a complex component and this is compounded by pretty
dubious measurement accuracy.

Your DC value of R assumes that this is the only loss
mechanism and totally ignores the shunt losses arising from eddy
currents and the iron circuit losses.

In the same way as Q=(rootL/C)/R for series losses
Q=r/(rootL/C) is for shunt losses where r is the effective value of
the shunt loss component. The combined effect of both types of losses
must be taken into account.

With the setup you have there is no method of a
accurately measuring the separate or combined losses and, since you
are only taking part of the series loss into account for your Q
calculation, your result is necessarily pretty optimistic.

Jim

Even if I run the resistance down near zero, the LCR.EXE program still
says "No impedance hump."

Is the program broken? I doubt it.

Am I doing something wrong? I don't know.

L = 136000 uH, C = 51000000 pf, R = 23 ohms, f = 0.00006 MHz.

Doug

That's a good design, Don.

Doug