got it now.. u are generating more power so that u can even use things which run on higher wattage.. the current drawn by the existing components would be same, but if required they can be changed to more powerful ones.. when you will use the higher rated ones, u'll have to change the wires as they may burn...
am i right? i think so

Yeah, you got that right!

Yea, that's exactly what I was worried about. So all options, except a shunt regulator, will need the "car-type" alternator that allows for varying the field strength.

Even an autotransformer setup will let the primary voltage rise to high levels when the secondary load is small and the engine RPM is high, causing the same problems with insulation.

This is probably off the wall, but how about multiple taps on the stator itself? If the insulation of the stator can take the induced voltage, it should be possible to pick the tap that has the voltage required.

No, the insulation would not be such a problem. At around 10k rpms, voltage generated is around 90V or so. Multiple taps would be a solution. I had thought about that. But that would have meant using a micro-controller in something as simple as a regulator. You could say, lazy me! Maybe someone could use this idea for a college project etc.

Okay, had no idea about the unloaded voltages. 90V you say. In that case a even a linear voltage regulator might be able to do the job. Since the coil voltage is anyway sort of inversely related to the load at a given RPM, it would not have to dissipate too much of heat. (The shunt regulator will anyway dissipate plenty of heat in that case too.) The only challenge would be to build a voltage regulator that is capable of dropping so much voltage. (Worst case would be high RPM with a partial load.)

Or for the really technical a PWM buck converter capable of handling the system load (a couple tens of amperes at most I think) would even eliminate most of the power dissipation that a linear regulator faces.

It would be really interesting to see if it could be done cheaply. But I don't think the gains are worth the effort. But I find it an interesting thought experiment.

A linear voltage regulator works on the same shunting principle. Both are the same thing! A linear voltage regulator has a inbuilt voltage sensing circuit. I had to build one myself. That's it.

The problem is that, in case of PWM, the source is disconnected from the load (open circuited) whenever it exceeds certain limits. Hence, no wastage as heat. But the stator coil can not be open circuited without resultant damage. So, strike out that idea. I had thought of this too!

What exactly are you using? A shunt regulator that dumps excess power by "shorting" the stator? Or a series regulator that drops the excess voltage? Because they are not the same thing.

And no, a PWM regulator never open circuits the stator. You would use a rectifier and capacitor setup to get the power from the stator. So the stator will always be feeding into the capacitor.
The voltage from the capacitor is then fed into a chopper circuit that feeds power into a properly sized inductor and diode. This again feeds into an output capacitor that smooths out the output voltage. Atleast that's how it appear on wikipedia: Buck converter - Wikipedia, the free encyclopedia
On top of that a protection could be added to always present at least some load to the alternator to keep the voltage down. This could be a resistor switched in the circuit under low load conditions and switched off when the bike electricals present enough load.

But as I said, I don't think the gains are worth the effort.

Oops, was not reading your post carefully. Yeah, you are correct. I am using a shunt regulator, not a series regulator like LM317 etc.

That wiki diagram had no capacitor on the generator side and does open circuit the source from the load. Where exactly did you think about placing the cap? Parallel to the source and before the switch?

You're right the wiki doesn't have a cap there. However if you look at say the schematics of a PC power supply. You'll find there's a bridge rectifier followed immediately by some large 320V or higher capacitors to provide smoothing of the rectified voltage. (I'm glancing over some details like PFC circuits and other nasties.) So the capacitor is parallel to the source after the rectifier. You also need the capacitor to be there, since the switch does not synchronize with the peaks of the AC you're rectifying.

So the switch is generally a MOSFET with enough voltage and current rating to hold the full high side voltage and the current required by the secondary side. The secondary side also has a capacitor to help smooth the choppy voltage that comes from the inductor. (Looks a bit like a triangle wavefrom.)

These things are mostly controlled by a dedicated PWM chip that has all the bits onboard to adjust the pulse width to the load. The only things needed on the outside are some caps and resistors etc to provide for power and voltage/current sensing. The datasheets would have the details.

Then if the load is too small, there can be a high wattage resistor or even a high wattage transistor to dissipate some base load, which can be switched off when there's enough load coming from the bike electricals.

Ok, got it. The main thing is that, the current flowing through the coil (inductor) should not decrease abruptly. That can give us a nice spark! To prevent that we should place a capacitor just across the switch. (parallel to the switch, series with the source.) That would allow a gradual switching OFF of the switch, just like how a condenser works in a point ignition system in cars. Or how relays have a capacitor in parallel to prevent arcing. But then, the rate of switching of the switch would be somewhat limited by the cap. That would mean some ripples at the PWM output. But it would still work, I think with a nice big cap at the output.

(The capacitor in parallel to the source would not really prevent an abrupt decrease in current flow once the switch opens. It will only prevent abrupt change in source voltage with varying loads while the switch is closed. And filter out the diode ripples. )

Good discussions here!

Great discussions going on.. i'll have to finish my engineering first to understand all these.. since iam from electronics department, these things look similar to my subjects..

Actually the capacitor parallel to the source will make sure it doesn't stop abruptly, since it will absorb the current coming from the stator, causing just a slight voltage rise. Putting it across the switch will burn up the switch the moment you turn it on, since the capacitor will dump it's charge into the switch. (Which will be a mosfet or something similar.) Also for these PWM circuits to work well, you need it to switch as sharply as possible.

I could give you the electronics theory behind the whole thing with inductor and capacitors, but that'll take some writing and I'll need to find some time for it. Anyway, just remember that an inductor generally wants to keep it's current going, that's the reason you can't open circuit the stator. A capacitor when it charges will have a rising voltage across its terminals. So if you connect them together through a rectifier, the rising voltage will cause the current from the stator be counteracted and thus kept within limits. However the voltage can rise very high, that can be stopped by making sure there's a base load provided, it will keep the voltage down by absorbing some of the energy. This base load can be switched on/off based on the voltage across the capacitor or something.

Also you want the capacitor to be no bigger than needed, because otherwise you'll end up with large overshoots and possibly oscillations. (Inductor/capacitor circuits are used in frequency generators and other resonating circuits.)

All in all designing a proper PWM buck converter is quite an undertaking, more so if you're not a skilled electronics designer. (It would definitely be mostly out of my skill set. I'm just a software engineer with an interest in electronics.)

IMO, you got this part wrong buddy. Let me attempt to explain my reasoning. Consider the system at a steady state. (Ignore the diode ripples for now.) There is a constant voltage generated by the stator (constant engine rpm) and there is a constant load. So, current flow is constant. The capacitor is also at a constant level now. (Same voltage as the source.) Now I open the circuit. The inductor, being the stubborn fella he is, will try to keep the flow going. Can the capacitor in parallel help here? No, it is already at a constant voltage level and hence can not cause any current flow through it. So, arcing can potentially occur across the switch contacts.

Now, if I have a capacitor across the switch, at steady state, the capacitor will have no charge. The 2 legs of the capacitor are at the 2 contacts of the same switch. So as there is no potential difference, the capacitor is discharged. As soon as the switch opens, the contact at the load side goes to ground, while the other contact at the source side is at source voltage level. Now, the inductor gets to charge up the capacitor for some time. So, an abrupt stop in current flow is avoided. The larger the capacitor is, more the time current can flow through the source coil (inductor), hence slower and more gradual the stop in current flow is.

Phew! I hope you are clear now! Anyways feel free to look up in the net why a cap is used across the relay contacts.

This is pretty much how a capacitor acts as a ripple filter. Got nothing to do with the switch. This capacitor is a standard part of any rectifier.

Check out the diode in the circuit. It's connected in a way that it cannot conduct any electricity when the switch is on. But what happens when the switch is off? It completes the current loop through the capacitor on the secondary side and carries the current from the inductor.

The primary side capacitor will absorb any spike from the stator, you're wrong considering the steady state. When in a steady state, you're right the cap has a constant voltage. But consider the load dropping, that means the stator has current flowing, and because it's an inductor it's voltage will rise, thus the voltage on the capacitor will be less than the stator voltage and it will absorb the excess energy.

Anyway, try reading up on PWM voltage converters from the net and you can find the same knowledge. I'm not just saying things for the heck of it. I've build quite some electronic circuits and also work with microcontrollers. So I'm quite sure of what I've been telling you. Read up on the subject, and then tell me again. For now I'm getting out of this discussion.

Oh, and the wiki image shows a switch, but as I said it's not a switch is generally a MOSFET. That means a silicon switch similar to a transistor.