sign216,
Secondly, when the brake light burns out, the end of the second coil that should be grounded, is not grounded, and that is what stops the engine. The brake light switch grounds that end of the coil when the brake is released, allowing the engine to run.
Diver:
Thanks for jumping in. It's been lonely in here.
The shunt regulator around the brake light is an option, because when the bulb blows, the regulator would take the place of the 6V drop. You don't have to design it for 10 Amps, just a nominal value of 2.5A. It will only draw the full 2.5 A when the bulb blows. Otherwise it will be doing very little UNLESS you try to make a 6 VAC replacement LED bulb.
It can be. According to the 98ss diagram the main lighting switch receives the LT coil voltage either directly via the red (rosso) wire or indirectly via the choke IM and the green (verde) wire. The switch is shown as a 3-position one (centre off) and applies this voltage to the main/dip beam filaments (azzurro and marronne wires), the instrument light LPA (bianco wire) and the tail light LPP (giallo wire). All of these bulbs will be protected/dimmed by the choke presence if the indirect route is chosen. Perhaps the dimmed lighting is intended for daylight use?
You mean the brake light? Perhaps because the LT coil can't provide enough current for the extra 2.5A that the brake light draws, or for the reasons KISS gives, or even to drop the HT spark volts to reduce engine speed during braking.
I've been having a play with simulations of (a) a conventional car-type Kettering ignition system with the points in series with the ignition coil primary and (b) the system as per the 98ss diagram. I was surprised to find that (b) gives a higher HT voltage (I think because there's a greater current change when the points short the magneto coil). So the MC manufacturers are right and I was wrong on that score.
I've also come up with a possible way of regulating the brake-light voltage, using a couple of FETs. The 'regulator' (shown in the sim circuit below) clamps the voltage when it tries to go above ~6V (adjustable by resistor selection) and simply connects across the brake bulb/switch.
View attachment 88701
What's the purpose of putting the brake light as a part of that ground? Seems like inviting failure.
As I understand it: One end of the high tension (ignition) coil needs a ground, which is normally wired in. When the brake light switch is applied, that ground is broken, but the brake lamp is substituted as a path to ground. All is well, unless the lamp is burnt out and then there's no ground. Do I have it right?
Your understanding is correct, except that the ignition generator coil isn't a high tension coil. That coil generates spikes of a few hundred volts only when the contacts open, and that few hundred volts is transformed into 10 - 30 kV by the ignition transformer (usually called a coil). That is the high tensions coil, which is not part of the generator, and is close to the spark plug so that the fat spark plug wire, which takes the 10 - 30 kV to the spark plug, and needs to be fat to have lots of insulation, can be reasonably short.
What's the purpose of putting the brake light as a part of that ground? Seems like inviting failure.
Kiss, my first goal is optimize the current 6v AC system. To get it to run well. After it's running right, then the modifications/improvements come.
I'm holding off on LEDs because it's 6v AC, which are not LED friendly. In theory in LEDs are optimum, but real world motorcycle tests show it's hard for LED to beat an incandescent, and that's on a 12v DC system. See here: http://www.webbikeworld.com/lights/led-brake-lights-for-motorcycles/index.htm
Switching to an LED headlight and taillight package would be great, but that must be a future project.
Not quite. The left part (1) is a Kettering system with the points in series with the primary of the HT ignition coil (transformer). The middle part (2) is the system you presently have in the 88ss diagram, with the points shorting out (i.e. in parallel with) the magneto coil. The right-most part (3) is indeed the regulator, and although it's shown separately it is an active part of the simulation by virtue of the connection marked 'reg'. The waveform shows the regulation effect at high revs.Alec, your three diagrams are 1) the system as it is now, 2) system w a regulator, 3) regulator design. Am I right?
Diodes: any general purpose silicon, e.g. 1N4148, 1N400x, ...What's you design equations for the regulator?
In the sim, f is the magneto frequency. For the particular instance shown I chose f=125Hz (corresponding to 7500rpm, modelling one cycle of magneto waveform per flywheel rev). Open circuit output voltage vm of Mag2 is modelled as proportional to f. Wave period tp is 1/f. Points-open phase delay td and open time to are fixed fractions of tp.Theory of operation?
Aargh, I ran the bike last night just to check for blown bulbs, and sure enough, the brake is out again. So...I need to make Alec's regulator or else the bike isn't workable.
What amp rating should the diodes have?
Any other issues or design parameters I should be aware? Remember I'm a novice.
.... but would require electrical insulation kits for the two FETs in the proposed regulator.Some can hold two or more parts.
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