I'm not sure we're on the same page. If I run a 6 volt 500ma halogen bulb direct off of AC, the peak value of voltage is 6 times the square root of 2, or 8.5 volts. If I want to limit the voltage the bulb is exposed to 110%, I would want a slicer circuit (say, using back to back zeners) that "slices" the peaks off the sine wave at positive 9.3 volts and negative 9.3 volts. My real question is, what is the allowable overvoltage percentage, in this context?

Where does your 3.5 volts come from?

Sorry if I'm not showing a sense of humor, but I am REALLY looking for some input from people who might understand how to protect a halogen bulb.

Or I just don't get it ...

But let me be a little clearer if I can: I'm not making an LED light right now. I am converting a bike lamp that uses a conventional hard-vacuum tungsten bulb to halogen, and I am concerned about the effect of overvoltage on the bulb. Based on my experience in automotive lighting, a little overvoltage shortens bulb life by a lot. An AC slicer would be the simplest circuit, and a rectified and regulated DC voltage driver would be one of the most complex. Less complex but more expensive is to run the light off a battery and use the dyno just to trickle charge the battery.