Charging a coil

[Mike6158]

Can someone point me in the right direction? I'm trying to figure out what I need to know to be able and how to calculate the charge time of an automotive (motorcycle) coil. The coil has two secondary windings (2 spark plugs per cylinder) but I'm not sure if a there is a single primary winding or two.

[Engineer1138]

OK, I'll have a go at this. Been a looong time since I took Signals & Systems <p>V= L * di/dt, so to find I, we integrate (V/L)dt, which gives: VT/L. Rearranging, we get T= I*L/V<p>Now, the steady state current (I) after everything is charged is simply the DC resistance of the coil divided into the supply voltage (V). This is easy to measure/calculate.
So, if my math is right, to find charge time T, it's (DC current * inductance)/Voltage.<p>Hope I got that right

[Mike6158]

Digging around in my trusty ARRL handbook I found one small mention. The time in seconds for the current to build to 63.2% of the max value is = L/R... After 5 time constants the current is considered to have reached it's max value.<p>t = 5L/R<p>and W = I²L/2

[Engineer1138]

I should have realized that the simple way to do it was using the time constant of the circuit. But even so, the result should have been the same. My answer shows it as one time constant, which as you point out is only 63% fullscale.
I should take a closer look and see where I went wrong.

[Mike6158]

Hopefully you or someone else will revisit this post. I have another formula but I'm a bit baffled by one of the terms. It's for the discharge time of a coil:<p>Discharge time (It=time for current to decay after removal of the source voltage.)<p>Damn... the formula isn't easy to put here. The first 1/2 is easy. It's E/R where E is voltage and R is ohms resistance. The second half, which the first is multiplied by, is the natural log (e) raised to the (-tR)/L. I'm at a bit of a loss to figure out what -t is supposed to be. Time obviously but what time... Charge time?<p>It = (E/R)(e^((-tR)/L)). Make sense? <p>Any help will be appreciated...<p>[ January 07, 2005: Message edited by: NE5U ]</p>

[rshayes]

The discharge formula that you have applies to a coil of inductance L with an initial current of E/R discharging into a resistance R. This isn't what happens in an ignition system.<p>When the coil current is interrupted by opening the circuit, the voltage across the coil rises. The rate of rise is limited by the stray capacitance of the high voltage winding. At first energy stored in the coil charges the stray capacitance. The waveform is a segment of a sine wave. At some point, the voltage rises enough to fire the spark plug. This stops the voltage increase and transfers energy from the coil to the spark, which will hold drop the voltage to a lower level and clamp it. Eventually the coil current reduces to a level where the spark extinguishes. After this point, the coil and its stray capacitance will oscillate for several cycles until the remaining energy is dissipated.<p>Under thses circumstances, estimating the discharge time isn't easy.

[dacflyer]

you might be easier to give us info on what bike it is etc..and perhaps find a lead chart to identify which wire is which..<p>also timing doesn't really matter i think..i mean the engine varies greatly in RPM's
so as long as the points are breaking you will get a spark,,even at low RPMs<p>are you trying to find a specific info for a project or????

[dacflyer]

oh, i did find out..some bikes with the dual secondary..one lead is "+" and the other is GND.
the coil fires both plugs at once..
another words the plugs are in series..
one cylinder gets fired for the combuston.and the other cylinder gets fired again for the exhaust. something about the emissions control ...
even a briggs and stratton fires on each RPM,,not every other one..<p>and some dual coils are center tapped..both sides fire at once also...the center tap is grounded to the coil frame..<p>hope that helped ya

[bodgy]

-t is the instantaneous time.<p>T would be the complete time period.<p>Colin

[Mike6158]

Stephen. Thanks. Now at least I have an excuse for not being able to pin it down <p>DAC- It's a Honda VTX1800R. 1800cc V-Twin. EFI. Shaft drive. Water cooled. Two plugs per cylinder but not for emmissions. The cylinders are huge and it takes two plugs to properly ignite the fuel/air charge. They both fire at the same time. Two coils with dual secondary's. One coil for the front cylinder. One coil for the rear cylinder. We seem to be stuck at 104RWHP no matter what we do to the engine (up from out of the box 85RWHP). The likely culprit is the induction system (air flow. Ie valves, heads, throttle body, and to some extent even the cam) but there is a belief by some that the ignition system is weak and that we can realize more RWHP by improving that. There is some incidental evidence that this is the case. Peope are seeing a 1-2 RWHP jump by changing to a specialty spark plug. Without getting into the details, there is a PHd, EE that has made a name for himself that says that a specialty plug will increase HP if the ignition system is weak (coil and wires). So 1+1=2. Personally I do not believe that the RWHP increase that they are seeing on a dyno will translate to the real world. I do not believe that there is enough HP gain to be found from tweaking the ignition. Not yet anyway. We're still looking for 10+ RWHP increments. The ignition offers an order of magnitude less than that.

[Mike6158]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by bodgy:
-t is the instantaneous time.<p>T would be the complete time period.<p>Colin<hr></blockquote><p>Ah... so t=1 for time cycle 1, 2 for 2, etc...<p>Note to any young person that is thinking that college is a bad idea or that you want to drop out of school or that math has no use in the real world. YOU ARE WRONG!!!! Get over being wrong, get your butt to school and pay attention in math. End of old guy speech

[Mike6158]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by dacflyer:
you might be easier to give us info on what bike it is etc..and perhaps find a lead chart to identify which wire is which..<p>also timing doesn't really matter i think..i mean the engine varies greatly in RPM's
so as long as the points are breaking you will get a spark,,even at low RPMs<p>are you trying to find a specific info for a project or????<hr></blockquote><p>No points. It's all electronic. Probably uses a Triac to fire the coils and injectors.<p>Timing matters from the standpoint that the time available to charge decreases as RPM increases. Granted we are talking ms here but at peak RPM the coils are asked to fire a bunch of times. Peak in this case is around 5500. Lot's of reciprocating mass so these things aren't like a 'Busa or CBR (turning 10k - 12k). That was the crux of the question. Do we achieve the maximum available power from the coil at max RPM. Is the coil designed for peak efficiency at max RPM or does it just get by.

[Mike6158]

You know something? We (VTX owners) have been chasing a dead end. The coils that the VTX uses were used on the Honda CBR1100 Blackbird. The Blackbird is an 1100cc, 4 cylinder bike with a 10,000 RPM redline. The VTX redlines at 5,500. If the coils were good enough for the Blackbird they are not going to be an issue for the VTX.<p>Thanks for all of the input...

[rocket scientist]

There have been a few good math questions asked that have not been answered yet, so I will take a try. <p>The differential equation, V = L * di/dt, that Engineer1138 proposed is correct, but he solved it assuming that V is constant. V is not constant, but depends linearly on i through the other impedances in the circuit. If you solve the equation with these included, you will get an exponentially decaying solution that leads to the answer NE5U found in the ARRL handbook.<p>The formula that NE5U asks about : It = (E/R)(e^((-tR)/L)) is almost the same as the exponential solution that I just mentioned, but for the case of discharging instead of charging. The variable 'It' is not a time; it clearly has units of current, same as E/R. In fact, it is the formula for the current in the circuit as a function of time t after discharging starts. You pick the time and you get the value of current. After t = L/R (one time constant), you get I = E/R (e^(-1)) = .37 E/R. Note that t has units of time, it is not 1,2,3 etc. <p>The exact solution to the original charging problem posed by Engineer1138 is <p>It = (E/R)(1 - e^((-tR)/L)). For t = L/R, It = .63 E/R or 63% of full charge, as NE5U found in the handbook.<p>This really supports NEU5 claim that math is important. Without a good background in complex variables, differential equations, and Fourier analysis, you can never really understand quantitatively what is going on, and you won't get this in high school.<p>[ January 08, 2005: Message edited by: rocket scientist ]</p>

[Mike6158]

Thanks Rocket Scientist (been on the net since '87 and I still can't get used to not using someone's name)<p>Believe it or not... I understood your response and I appreciate it. And I appreciate that you reiterated the importance of math. As far as I'm concerned it isn't stressed enough in ANY educational facility. But let's not get me started But I try...<p>Good stuff... Thanks a ton.<p>Mike