lower resistance coils ok????

[Anonymous]

Yep Scotty, thats really all an ignition system does is control the point at which the coils discharge and fire the plugs. It doesnt really matter whether its a breaker point system or an electronic/optical system, its nothing more than a timer.

Earl

thanks for replying Earl
do you know where the circuit is completed to make the plugs fire--is the ignitor box then completing the circuit on the secondary winding of the coil

I understsnd that Earl but the question i have is IS THE IGNITOR MAKING AND BREAKING THE CIRCUIT ON THE SECONDARY SIDE((THE SPARK PLUG SIDE)) OF THE COIL???

 

[kz]

Hi Slowpoke,

You are right according to Ohms law U=I*R where should not be any voltage drop on the + connector of the coil.

However....

1. Voltage drop

We assume:

bike is not running, ignition is on, ignition system is equipped with mecaniacal points,

Beacuse the current is supplied from a battery the battery-voltage will drop in relation to the total load of the battery, the voltage of a 12V battery with totally no load connected , is called EMK and is on a lead battery 2.1 Volts each cell times 6 equals 12.6V.

As soon as you load the battery with any load the Voltage will drop according to how the battery is charged.

If the battery is badly charged the voltage drop will be higher than if the battery is fully charged, the same goes for the size (Amphours) of the battery.

You can use this effect and measure voltagedrop on the primary side of the coil and the reading can tell you if the coil fits the bikes system

 

[kz]

JoJo wrote:

power = voltage times current
current = voltage devided by resistance

This means: if your coil resistance = 3 Ohms and your voltage is 12 volts, the current through your coils is 4 Amps. (12/3=4)

Now, if the current is 4 Amps, the power through your coils is 12 x 4 = 48 watts.

If it still is allowed to dsikusse electrical theory, here comes some more......

In a coil you have resistance R (Ohm) and induktans XL (Ohm),
the vectoriel sum of this to values equals Z Impedance also Ohm.

If you have DC-current the R and Z value are the same and the XL is 0 (engine not running).

The XL value is changeing according to the frecvency (enginerevs).

In this case XL = 2*PI*F*60*L, where F*60 is enginerevs and L is related to the turns the coil is winded with.

JoJo:s calculation only applys if the Ohms are related to the Z Ohms, beacuse the coil is feeded with a kind of AC-current during running, meaning that the power consumtion of the coil is different, if the engine is not running and different according to the revs.

So, sounds confusing, on a bike you normally have DC-current except for the coils, they are feeded whith a kind of AC-current (pulsating DC)generated by the points.

If you feed a coil with DC-current, you would only get one single spark then you turn on your ignition and an other single spark then you turn off the ignition.

Beacuse the current CHANGE on the primary side is generating the spark on the secondary (plug) side.

Then you ones have turned on your ignition the current woudnt change any more until you turn of the ignition, then you would get an other single spark.

Slowpoke, this is the reason for that the coils not are fireing all the time as soon as you turn on the ignition.

 

[earlfor]

I guess I was only clear in my own mind. The positive and negative leads on your coil both go to the ignitor box. The ignitor box is controlling the primary side of the coil. There, THERE...
I think I have made myself clear this time. On my coil, when I turn the ignition to on, the primary positive lead is hot when you put a meter to it and the neg lead of the meter to ground. The negative primary side of the coil runs to the ignitor box and the ignitor box completes the circuit with this wire.

Earl

thanks for replying Earl
do you know where the circuit is completed to make the plugs fire--is the ignitor box then completing the circuit on the secondary winding of the coil

[/quote]

I understsnd that Earl but the question i have is IS THE IGNITOR MAKING AND BREAKING THE CIRCUIT ON THE SECONDARY SIDE((THE SPARK PLUG SIDE)) OF THE COIL???[/quote]

 

[Anonymous]

yes it is very clear this time Earl--i thought that you originally said that when you tirned the ignition on and placed a meter across the plus and minus leads of the coil((the primary side)) that you would get 12 volts--now i understand that there is only 12 volts applied to the primary winding
when a signal is sednt to the ignitor box and thus com[letes the minus side to ground.
also the crangshaft must be turning in order to do this or in order to send the signal to the ignitor box and thats why i couldnt understand how putting a
meter across the primary windings could show a voltage without hitting the starter first--
Do I have it straight now???
Thanks for your ptience

 

[earlfor]

NO big deal Scotty, my fault for not offering a more concise explanation to start with. It would have been so much easier if
I could have just held up a coil and said power goes in here, comes out here, goes to this little box and the little box says yes or no.

Earl

yes it is very clear this time Earl--i thought that you originally said

 

[Hap Call]

JoJo wrote:

power = voltage times current
current = voltage devided by resistance

This means: if your coil resistance = 3 Ohms and your voltage is 12 volts, the current through your coils is 4 Amps. (12/3=4)

Now, if the current is 4 Amps, the power through your coils is 12 x 4 = 48 watts.

If it still is allowed to dsikusse electrical theory, here comes some more......

In a coil you have resistance R (Ohm) and induktans XL (Ohm),
the vectoriel sum of this to values equals Z Impedance also Ohm.

If you have DC-current the R and Z value are the same and the XL is 0 (engine not running).

The XL value is changeing according to the frecvency (enginerevs).

In this case XL = 2*PI*F*60*L, where F*60 is enginerevs and L is related to the turns the coil is winded with.

JoJo:s calculation only applys if the Ohms are related to the Z Ohms, beacuse the coil is feeded with a kind of AC-current during running, meaning that the power consumtion of the coil is different, if the engine is not running and different according to the revs.

So, sounds confusing, on a bike you normally have DC-current except for the coils, they are feeded whith a kind of AC-current (pulsating DC)generated by the points.

If you feed a coil with DC-current, you would only get one single spark then you turn on your ignition and an other single spark then you turn off the ignition.

Beacuse the current CHANGE on the primary side is generating the spark on the secondary (plug) side.

Then you ones have turned on your ignition the current woudnt change any more until you turn of the ignition, then you would get an other single spark.

Slowpoke, this is the reason for that the coils not are fireing all the time as soon as you turn on the ignition.

To clarify what Karl is saying here, there is a difference between
resistance and impedance. Resistance (R) is what most people think of when they see a resistor on a circuit board. Impedance (Z) is a combination of resistance and the effects of a changing voltage on capacitance (C) (measured in Farads) and changing current on inductance (L) (measured in Henrys).

Capacitors are made by having two plates with a dielectric between them. This dielectric can be air, insulation, oil, or many other materials. When voltage is applied across these plates the plates become charged with a voltage. Your battery is a close relative of the capacitor. Capacitors will store voltage for a short period of time and will resist any change in voltage. They are often used to smooth out choppy DC power. The greater the change in voltage over a period of time, the greater the impedance that is developed.

Inductors are made by winding wire into loops like in a transformer, the fields of a motor, or the stator on your motorcycle. Inductors will store current and will resist any change current flow. The greater the change in current flow over a period of time the greater the impedance

Most people would say that there is no changing voltage in a DC system but that is not quite true. Before any voltage is applied to a coil (which is a very small transformer) the primary side is at zero potential. When you apply voltage you then have a change in voltage from zero volts to 12 volts. Like wise changing current can be found anytime there is a switch thrown.

A transformer will not work if the voltage is not changing, thus a coil will not work if you do not have a changing voltage. This changing voltage comes from your electronic ignition, which has power transistors that are turned off and on to give a changing voltage to the primary side of the coils.

Basically, using constant DC theory alone cannot be used to determine the output current and voltage on the secondary side of a coil.

Okay, my head is starting to hurt.

Hap

 

[kz]

Thanks for your clarification Hap,

Your english is much easier to understand than mine, on the other hand my swedish is probably better................

 

[earlfor]

Thanks Karl, Hap for taking the time to explain. Much appreciated.

Earl

 

[Hap Call]

Thanks for your clarification Hap,

Your english is much easier to understand than mine, on the other hand my swedish is probably better................

Karl, I am sure your Swedish, German, French, Spanish, Russian, Finnish, Italian, Mandarin Chinese, Arabic, Bengali, Japanese, and Polish are all better than mine!

Hap
What do you call someone who speaks two languages – Bilingual.
What do you call someone who speaks one language – American.

 

[Anonymous]

Hmmm,

Does anyone know what "beer" is in mongolian?

I need one now

Jojo

 

[Hap Call]

Krash, another member of the GS Resources forum gang gently reminded me that you asked if the coils you have are appropriate to use on you bike. We did you a disservice by going off the deep end with a very technical discussion without answering your question.

I will attempt to give you an answer now... I would be very careful putting those Honda coils on. The best people to talk with are the folks at Dyna who can point you in the right direction.

All we did was discuss how to measure resistance in the coils - we do not know how Dyna does their measuring. Call them and get a firm answer. I would suggest going ahead and buying the coils from Dyna if you can (I know what it is like to be money limited) or getting a used set off of eBay.

Sorry about the run around.

Hap