Secondary capacitance 2

[enfield]

Hi there

I am usually found lurking over in the mechanical section, but was asked a question the other day which is beyond me ]

In a Spark igintion , coil points ignition system. What is secondary capacitance

Someone was using this calculator and didnt know what the last term was

http://www.bgsoflex.com/igncoil.html

The only thing I can think of is the spark plug. But then the question arises, what is a good value and how do you arrive at it.

If you use the posted value of 100 pF then it gives reasonable current and secondary voltage...

Any Ideas??

Thanks in advance

Stephen

 

[alehman]

That's probably the intrinsic capacitance between the HV winding of the ignition coil and ground. I don't know if that info is available from manufacturers or not. You may have to measure directly with a capacitance meter.

 

[jbartos]

Suggestion: Visit

http://www.hammondmuseumofradio.org/spark.html

http://www.celox.co.uk/advanced_automotive_ignition_sys.htm

etc. for more info

 

[enfield]

Thanks All
Yes on further reading its interwinding capacitance that I need to eastimate,

I have trawled the net ..but to no avail,
I did find a rough guide of 10pf at 100 turns and about 35 pf at 1000 turns

I just need a rough guide in order to estimate a current

Thanks in advance

Stephen

 

[jbartos]

Suggestion: Visit

http://www.explosionresearch.com/research.htm

etc. for more info
The capacitance will vary according to a circuit design

 

[enfield]

Hi there all well getting closer thanks for all the input

, I am getting some where!!, Ok I will start from the top, as I (Think) I know whats going on.
Firsts let me describe the system, it’s a Kettering points operated, coil Ignition system with 12v source. Very simple affair operating at 4000 rpm.

An average coil spec look something like this

Turns Ratio 100 to 1
Secondary turns 25 000
Primary turns 250
Primary Inductance 10 mH
Primary resistance 1.5 ohms
Secondary resistance 10 K Ohms
Secondary Inductance 40 Henries

And its an ordinary core with lots of losses!! Not an E core

Now If let assume I need a Spark current of 0.06 A and a spark time in the order of 1000 x 10^-6 sec and a voltage of 15000V
I can work out the Energy in Joules = 15000 x 0.06 x 1000e-6 = 0.9 Joules???
Now here I am ( or am!!!) off the plot now

Some formula for capacitance I found were C = 1 / ( 2 x pi x freq)^2 x L
And Q ( watts ) = ( C x V^2) /2

Now if I use the first formula I get 0.56 x 10 ^-6 farads 0.56 Pico farads when L = 40 and Frequency = 33

The other result I get is C= .08 Pico farads when V = 15000 and Q= 0.9

Am I on the right track here folks

Remember I am after Secondary capacitance in a Kettering coil ignition circuit. ( I assume that is the capacitance effect between the secondary windings, insulation ( dielectric ) and the iron core

You all have been most helpful to me so far in an area that I didn’t know existed!!

Thank you

Stephen