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Ignition Coil Troubles

D

DillonCo

Jan 1, 1970
0
I'm trying to bulid a high voltage 'supply' using an ignition coil.
Unfortunatly, whenever I try to spark it (by disconnecting the power to the
primary) I get nothing from the secondary and a fat spark where I
disconnected the power. How can I remedy this?
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that DillonCo <[email protected]>
I'm trying to bulid a high voltage 'supply' using an ignition coil.
Unfortunatly, whenever I try to spark it (by disconnecting the power to the
primary) I get nothing from the secondary and a fat spark where I
disconnected the power. How can I remedy this?
Put a 10 nF 500 V capacitor across the points where you interrupt the
primary current. There is such a capacitor in every car distributor that
uses coil and contact-breaker ignition, and if the capacitor fails, the
car doesn't go.
 
D

DillonCo

Jan 1, 1970
0
I'm trying to bulid a high voltage 'supply' using an ignition coil.
Put a 10 nF 500 V capacitor across the points where you interrupt the
primary current. There is such a capacitor in every car distributor that
uses coil and contact-breaker ignition, and if the capacitor fails, the
car doesn't go.

Thanks, that worked perfectly.
 
D

Don Klipstein

Jan 1, 1970
0
I read in sci.electronics.design that DillonCo <[email protected]>

Put a 10 nF 500 V capacitor across the points where you interrupt the
primary current. There is such a capacitor in every car distributor that
uses coil and contact-breaker ignition, and if the capacitor fails, the
car doesn't go.

Don't be afraid to try different capacitor values. I played around with
this sort of thing 25 years ago, and according to the fuzzy faded traces
in my memory, values larger than .01 uF (maybe more like .047 or .1 uF)
worked better for me.

Also I would use a capacitor with an AC voltage rating or rated at least
600 volts DC, or else it may fail from heavy use.

- Don Klipstein ([email protected])
 
T

Tim Shoppa

Jan 1, 1970
0
John Woodgate said:
I read in sci.electronics.design that DillonCo <[email protected]>

Put a 10 nF 500 V capacitor across the points where you interrupt the
primary current. There is such a capacitor in every car distributor that
uses coil and contact-breaker ignition, and if the capacitor fails, the
car doesn't go.

I agree with you, that's the way to get spark out of the secondary of the
ignition coil. (I was playing with that stuff in first grade! My parents
figured that with just a lantern battery and some old car and radio parts I
couldn't do anything dangerous... It wasn't until junior high that
I discovered car radios with working vibrators, and by that point I
had figured out TV flybacks.) IIRC having a bad
capacitor also ate the points. (I think when I was a kid an old rasp
and a dragged wire often served as the points after I destroyed the
originals.)

But electrically, what does that capacitor do? Does it
make a tank circuit on the primary such that it doesn't spark there,
so it has to spark on the secondary? The capacitor will limit dI/dt
which will potentially limit the spark if it's too big, though I guess
it's more important for the spark to come out of the right place :)

In the perfect world, where the two windings were 100% coupled,
resonating one would resonate the other, but I would guess that the
coupling in an ignition transformer is way less than 100%. Am I anywhere
near the proper theory?

Tim.
 
P

Paul Burridge

Jan 1, 1970
0
Don't be afraid to try different capacitor values. I played around with
this sort of thing 25 years ago, and according to the fuzzy faded traces
in my memory, values larger than .01 uF (maybe more like .047 or .1 uF)
worked better for me.

Also I would use a capacitor with an AC voltage rating or rated at least
600 volts DC, or else it may fail from heavy use.

Isn't the cap there solely to protect the points from sparking and
thereby getting pitted, coked-up and going hi-res owing to poor
contact?
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Tim Shoppa <shoppa@trailing-
But electrically, what does that capacitor do? Does it
make a tank circuit on the primary
Yes.
such that it doesn't spark there,
so it has to spark on the secondary?

Not really.
The capacitor will limit dI/dt

No. When the points are closed, the cap is discharged. As the points
open, the current switches to charge the capacitor, briefly. As the
current collapses, the back-EMF of the coil drags the cap plate
connected to the coil hundreds of volts negative. It's this voltage that
gets transformed up by the coil. Without the cap, the back-emf just
makes the current arc across the open points.
which will potentially limit the spark if it's too big, though I guess
it's more important for the spark to come out of the right place :)

In the perfect world, where the two windings were 100% coupled,
resonating one would resonate the other,

No. For both to resonate, they would both have to be tuned to the same
frequency.
but I would guess that the
coupling in an ignition transformer is way less than 100%.

No, it's quite high.
Am I anywhere
near the proper theory?

About half way?
 
B

Boris Gjenero

Jan 1, 1970
0
I agree with you, that's the way to get spark out of the secondary of the
ignition coil. (I was playing with that stuff in first grade! My parents
figured that with just a lantern battery and some old car and radio parts I
couldn't do anything dangerous... It wasn't until junior high that
I discovered car radios with working vibrators, and by that point I
had figured out TV flybacks.) IIRC having a bad
capacitor also ate the points. (I think when I was a kid an old rasp
and a dragged wire often served as the points after I destroyed the
originals.)

But electrically, what does that capacitor do? Does it
make a tank circuit on the primary such that it doesn't spark there,
so it has to spark on the secondary? The capacitor will limit dI/dt
which will potentially limit the spark if it's too big, though I guess
it's more important for the spark to come out of the right place :)

In the perfect world, where the two windings were 100% coupled,
resonating one would resonate the other, but I would guess that the
coupling in an ignition transformer is way less than 100%. Am I anywhere
near the proper theory?

Tim.

When the primary current is interrupted and there's no capacitor you
almost instantly get a pretty high voltage accross the points/switch
as the current tries to keep flowing because of the coil's inductance.
Because it happens so fast the contacts are still very close to each
other (they just opened). A spark forms there and it uses up a lot of
energy. This seriously weakens the spark on the secondary and damages
the switch/points.

If there is a capacitor the current that keeps flowing through the
coil first has to charge that capacitor. Now by the time a high
voltage is developed across the switch/points the contacts are further
apart and there's no arc there. Instead a much higher voltage arc
happens from the secondary and the energy goes there (where we want
it).

Of course the capacitor takes some of the energy but that's a lot less
than is lost when there's a primary-side arc and most of it is
returned to the secondary-side arc as the circuit resonates.
 
T

Tim Shoppa

Jan 1, 1970
0
John Woodgate said:
I read in sci.electronics.design that Tim Shoppa <shoppa@trailing-


Not really.


No. When the points are closed, the cap is discharged. As the points
open, the current switches to charge the capacitor, briefly. As the
current collapses, the back-EMF of the coil drags the cap plate
connected to the coil hundreds of volts negative. It's this voltage that
gets transformed up by the coil. Without the cap, the back-emf just
makes the current arc across the open points.

How many "rings" of the tank occur before the energy
is all out in the spark? I didn't have a lab full of fancy scopes when
I was a kid playing with ignition coils, and now I have all the scopes
but no ignition coil :-(. I would hazard a guess that the primary has
inductance of something like 1-10mH's, and this combined with the 10nF
capacitor means a resonant frequency in the low 10's of kHz range. That's
not too different than a usual flyback transformer, but I may be
comparing apples and... pickles.

In my mind this is different than a flyback transformer, where the
whole thing (including the driver) "sings" at the resonant frequency,
while in an ignition coil it's more like a hammer hitting a bell.

Tim.
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Tim Shoppa <shoppa@trailing-
I would hazard a guess that the primary has
inductance of something like 1-10mH's, and this combined with the 10nF
capacitor means a resonant frequency in the low 10's of kHz range. That's
not too different than a usual flyback transformer, but I may be
comparing apples and... pickles.

It's a long time since I looked at breaker ignition with a scope but low
tens of kHz is right.
In my mind this is different than a flyback transformer, where the
whole thing (including the driver) "sings" at the resonant frequency,
while in an ignition coil it's more like

Sorry, snipped a bit short. Yes, very like a hammer and bell.
 
M

MikeM

Jan 1, 1970
0
Paul said:
Isn't the cap there solely to protect the points from sparking and
thereby getting pitted, coked-up and going hi-res owing to poor
contact?
No,

Its there to resonate with the primary inductance (The capacitor/primary
inductance form a parallel resonant tank circuit)

MikeM
 
M

MikeM

Jan 1, 1970
0
Tim said:
How many "rings" of the tank occur before the energy
is all out in the spark?

Only one! The air in plug ionizes during the first peak; becomes
a low impedance, and 95% of the energy is used up.

MikeM
 
D

Don Klipstein

Jan 1, 1970
0
I read in sci.electronics.design that Tim Shoppa <shoppa@trailing-


Not really.


No. When the points are closed, the cap is discharged. As the points
open, the current switches to charge the capacitor, briefly. As the
current collapses, the back-EMF of the coil drags the cap plate
connected to the coil hundreds of volts negative. It's this voltage that
gets transformed up by the coil. Without the cap, the back-emf just
makes the current arc across the open points.

The way I understand it, the capacitor briefly conducts the primary
current so that when the points are opened, there is not much back EMF
across the points. The voltage across the points rises as the capacitor
charges.
There would be even more voltage without a capacitor, except you don't
want high voltage across the points until they are far enough apart to not
spark from the primary voltage. The capacitor "delays" the high EMF until
the points are far apart enough to not spark from the high EMF.

- Don Klipstein ([email protected])
 
T

Tony Williams

Jan 1, 1970
0
Don Klipstein said:
The way I understand it, the capacitor briefly conducts the
primary current so that when the points are opened, there is not
much back EMF across the points. The voltage across the points
rises as the capacitor charges. There would be even more voltage
without a capacitor, except you don't want high voltage across
the points until they are far enough apart to not spark from the
primary voltage. The capacitor "delays" the high EMF until the
points are far apart enough to not spark from the high EMF.

Yes, with a resonant primary the voltage across the
contacts is shaped into a rising quarter-sine.

The exchange of energy from the L into the C means that
1/2.L.I-squared could become 1/2.C.V-squared at the peak
of the quarter-sine. That defines the maximum Vpk that
could occur across the contacts (under an open-circuit
fault condition on the secondary side).

Vague memories of a quarter-sine of about 100uS and a peak
voltage of about 200V.
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that MikeM <[email protected]
tah.edu> wrote (in said:
Only one! The air in plug ionizes during the first peak; becomes
a low impedance, and 95% of the energy is used up.
You have a poke around with a scope probe, without actually touching
anything. You'll see a LOT of ringing.
 
P

Paul Burridge

Jan 1, 1970
0
Yes, with a resonant primary the voltage across the
contacts is shaped into a rising quarter-sine.

The exchange of energy from the L into the C means that
1/2.L.I-squared could become 1/2.C.V-squared at the peak
of the quarter-sine. That defines the maximum Vpk that
could occur across the contacts (under an open-circuit
fault condition on the secondary side).

Vague memories of a quarter-sine of about 100uS and a peak
voltage of about 200V.

Crumbs! And there was me thinking it was just to prevent arcing across
the points! Resonance, eh? <groan>
 
B

Bob Stephens

Jan 1, 1970
0
I read in sci.electronics.design that MikeM <[email protected]

You have a poke around with a scope probe, without actually touching
anything. You'll see a LOT of ringing.

And try not to get your metal watchband across the terminals of the coil -
like I did once - gets your attention right away ;)
 
T

Tony Williams

Jan 1, 1970
0
Crumbs! And there was me thinking it was just to prevent arcing
across the points! Resonance, eh? <groan>

In this case though resonance is your friend.

Somebody-or-other's Law says that (in air, at NTP)
an arc cannot strike unless the voltage between the
contacts exceeds about 350V, whatever the distance
between the contacts. Plus there is small delay
between the application of a voltage and an arc
actually striking.

The C slows the initial rate-of-rise and the LC
resonance limits the Vpk that can occur (to <350V).
 
B

Bill Sloman

Jan 1, 1970
0
Tony Williams said:
In this case though resonance is your friend.

Somebody-or-other's Law says that (in air, at NTP)
an arc cannot strike unless the voltage between the
contacts exceeds about 350V, whatever the distance
between the contacts. Plus there is small delay
between the application of a voltage and an arc
actually striking.

http://www.fortunecity.com/greenfield/bp/16/paschen2.htm

It is Paschen's Law, which can be explained - as the URL says - on the basis
that electrical discharge through gases depends on avalanche breakdown.
Specifically, random electrons, produced by cosmic rays or potassium-40
decay or whatever, have to pick up enough energy from the electric field in
the gap to ionise a gas molecule when they hit it, knocking off another
electron.

This is an inelastic collison. Elastic collisions just divert the electron.
Because the mean free path of the electron depends on gas pressure, the
necessary electric field goes up as the mean free path decreases with
increasing pressure.

If the mean free path of the electron is longer than the gap between the
electrodes, you don't get any avalanche multiplication either, so your
Paschen curve has a minimum which depends on the gas, of about 327V for air.

The delay before you get a discharge depends on numbers of random electrons
around. The discharge evolves from a glow discharge to an arc discharge, if
your source can supply enough current, over a period of the order of a
microsecond as the surfaces involved heat up enough to produce
field-assisted thermionic emission.

It is a fairly complicated process, but the physics was worked out about a
hundred years ago.
 
T

Tony Williams

Jan 1, 1970
0
Bill Sloman said:
It is Paschen's Law, which can be explained - as the URL says -
on the basis that electrical discharge through gases depends on
avalanche breakdown.
[snip]

Yes, that's him, thanks Bill.
If the mean free path of the electron is longer than the gap
between the electrodes, you don't get any avalanche
multiplication either, so your Paschen curve has a minimum which
depends on the gas, of about 327V for air.

Yes, I used the wrong terminology, "whatever the distance
between the contacts" was incorrect.

As you say, there is a specific distance (for that temp/
/pressure) at which Vbreakdown is a minimum. ISTR calculating
it many years ago as about 0.01mm for a pair of relay contacts.
The delay before you get a discharge depends on numbers of random
electrons around. The discharge evolves from a glow discharge to
an arc discharge, if your source can supply enough current, over
a period of the order of a microsecond as the surfaces involved
heat up enough to produce field-assisted thermionic emission.

I've always assumed that Soddes Law says that there will
always be some amount of unsnubbed inductance in there,
with enough energy to drive Cstray up to well above
Vbreakdown. An avalanche will happen (with Cstray supplying
the current) but there will only be a few 10's of uJ of
available energy, so it should only last a uS or so.
 
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