Electronics Forum / Repair / August 2008

"Without them the core gets very hot" and the coil is cool? If the
core is hotter than the coil, it would suggest high hysteresis loss or
the diameter of the core wires is too large.  I'd use wire diameter of
~ 1-2 mm to limit core loss.  Welding rod?  

The core wires should also be insulated from one another.  If the wire
is galvanized - an acid treatment or just leaving them out in the
weather for a few months (like electric fence wire - cheap small
diameter and galvanized).  Throwing iron wire in a BBQ will help form
an oxide coating, and will anneal it -soft wire has lower losses than
hard drawn wire.

Hard stiff wire has higher hysteresis loss.  The wire doesn't switch
poles easily so it acts like it isn't there until the magnetic force
is high enough to turn it over.  A test is to see how much magnetic
remanence it has:  Stroke a piece of wire with a magnet and see if it
becomes a magnetized.  If it does, the hysteresis loss will be higher
- annealing the wire helps there.

The iron has higher permeability than air so most of the flux should
go though the iron but some will still be dissipated as heat in the
brass tubes - larger diameter tubes, higher loss.

Last but not least, the core should protrude from the ends of the coil
- the magnetic lines of force expand as they get to the center of the
coil (on the outside of the coil).  Stretching the length of the core
raises inductance for the same number of turns.  I like at least one
diameter of core to protrude from each end of the coil and 2D is
better - for induction coils.

"Without them the core gets very hot" and the coil is cool? If the
core is hotter than the coil, it would suggest high hysteresis loss or
the diameter of the core wires is too large.  I'd use wire diameter of
~ 1-2 mm to limit core loss.  Welding rod?  

The core wires should also be insulated from one another.  If the wire
is galvanized - an acid treatment or just leaving them out in the
weather for a few months (like electric fence wire - cheap small
diameter and galvanized).  Throwing iron wire in a BBQ will help form
an oxide coating, and will anneal it -soft wire has lower losses than
hard drawn wire.

Hard stiff wire has higher hysteresis loss.  The wire doesn't switch
poles easily so it acts like it isn't there until the magnetic force
is high enough to turn it over.  A test is to see how much magnetic
remanence it has:  Stroke a piece of wire with a magnet and see if it
becomes a magnetized.  If it does, the hysteresis loss will be higher
- annealing the wire helps there.

The iron has higher permeability than air so most of the flux should
go though the iron but some will still be dissipated as heat in the
brass tubes - larger diameter tubes, higher loss.

Last but not least, the core should protrude from the ends of the coil
- the magnetic lines of force expand as they get to the center of the
coil (on the outside of the coil).  Stretching the length of the core
raises inductance for the same number of turns.  I like at least one
diameter of core to protrude from each end of the coil and 2D is
better - for induction coils.

I doubt that "losing capacitance" is the only symptom of the capacitor failing
in this circuit - it's just that we don't know what else is happening to the
capacitor in the application. The only data available for the GE97F9002 cap
doesn't give details of rated or peak current or dV/dT rating. All we know is
that it is specced as a motor run capacitor. The normal use for this capacitor
type is in series connection with a motor winding having considerable inductive
reactance across an ac supply where the peak inrush current period is limited
and infrequent in nature, and normal operating current is not likely to be
anywhere near 6A. The OP is endeavouring to use it in a circuit where it
receives frequent high peak inrush current and a large inductive kick when the
circuit is broken, at each swing of the pendulum. I doubt that these conditions
are ideal for maximising the life of this type of capacitor. While the original
GE capacitors might have done the job for 10 years in the OP's first coil
design, these later incarnations made by a different process by Regal Benoit may
not be as good as the originals - ie. they are built down to a price.

While using a better capacitor might not be the most elegant solution in
engineering terms, if it works without having to change the current coil design
then it is still an effective solution. My guess is that the GE97F9002 cap
doesn't like the constant hammering of pulsed ac operation where the inrush
current is surely quite large and the current at resonance is still rather high
for a capacitor which is meant to be permanently connected across an ac supply
in series with  a higher impedance winding than the 0.75H coil, and having a
relatively small current swing. The alternative cap I suggested does come with
the necessary data and is designed to handle high currents and with a large
dV/dT rating. Surely this is a good starting point.

Even if we were to find exactly what the failure mechanism for the GE cap is, it
would most likely turn out that a better specced capacitor is the only solution
anyway, so why not anticipate this necessity. It might give the electronics
perfectionist a warm glow to know exactly why a component fails but in practical
terms the end user only cares whether the item works and that it keeps on
working.

If the circuit performance doesn't deteriorate over time with the use of a
better specced cap it will have proved by empirical method to be an effective
solution. If the OP later decides to design a better coil then this cap will be
even better suited since it will possibly have to endure even lower stress.

Using that logic, then while the circuit drifts towards resonance, it
tends to degrade faster towards resonance, until it crosses over the
resonance point, then degradation slows. (or it fails at resonance!)

I'm having difficulty visualising the arrangement, too.

As for why the circuit is detuned, it may be to avoid the huge
voltages and currents at resonance. Even so, I calculate that a 9.7uF
capacitor would be subjected to a voltage of 3200V and a current of
12A. Surely that can't be right.

- Franc Zabkar

As the OP said, the circuit is tuned "off resonance" when the pendulum
is not close. Then when it approaches, its proximity alters the
resonance in such a way that there is more attraction while it is
approaching, and slightly less while it is receding. The net result is a
transfer of momentum to the pendulum. Sort of like the way a "slingshot"
orbit works.

Isaac

Oops, I see my mistake.

Z = jWL + 1/jWC = jWL - j/WC = j (WL - 1/WC)

So the magnitude of the impedance of L & C, assuming R=0, is given by
...

|Z| = WL - 1/WC  =  XL - XC

There should be no squared terms. Sorry.

PI = 3.14159265#
C = 9.7 * .000001
L = .75
F = 60
W = 2 * PI * F
XL = W * L
XC = 1 / W / C
Z = XL - XC
I = 110 / Z
VC = I * XC
VL = I * XL
E = .5 * L * I * I
PRINT C, L, XL, XC, Z, I, VC, VL, E

The results are now ...

.0000097      .75           282.7433      273.4621      9.28125
11.85185      3241.032      3351.032      52.67489

So the voltage is 3200V, current is 12A, and the energy in the coil is
53J (I think). The current for a C of 10uF is 6A as Ross calculated.
Sorry again for my mistake.

- Franc Zabkar