Refitting a 6 volt shunt regulated system to a buck regulated system

Question

I have a 1981 Yamaha XT250H. This might seem like an automotive or motorcycling question, but what I'm trying to answer is a question about refitting an older motorcycle's electrical system.

Here is an explanation of the electrical system in stock form. It is massively antiquated compared to the modern 3 phase system used by almost all motorcycles.

There is a rotor coupled directly to the engines crankshaft with a bolt. There are 3 coils

1. The "pickup" coil, this coils location is critical as it controls when the spark plug is fired and thus ignition timing.
2. A "magneto" coil with approximately 500 ohms of resistance.
3. A "lighting" coil with approx. 4.1 ohms of resistance.

The "pickup" coil is essentially just a signal source, the amount of current extracted from it is neglible. The "magneto" coil is just called so due to historical reasons. Both of these coils connect to a CDI box. This still works after 30+ years so I have no reason to change it. I intend to leave it as is.

The "lighting" coil is a coil with two taps. One tap is used to feed AC voltage directly to a regulator and the 6 volt 35 watt headlamp. The lower tap was used to feed a single rectifier which fed numerous lights and a battery. Here is the complete listing of all the equipment

• 6 volt 35 watt headlamp
• 6 volt 5.3/25 watt tail light
• 4x 6 volt 17 watt signal lamps
• 4x 6 volt 3 watt indicator lamps

There was also a 6 volt battery in the system that was there literally just to prevent the lights from pulsing. The motorcycle does not have an electric start.

Since the ignition circuit of the motorcycle is completely separate, I can disconnect the lighting circuit and run the engine to measure the voltage. What I found was over 14 volts AC at idle from the lighting circuit. This obviously will only increase as the engine RPMs go up.

The signal lights on this motorcycle were the "always on" variety that were pulsed off to act as turn signals. Summing the original wattage we find:

35 + 25 + (4*17) + (4*3) = 140 watts

Or over 100 watts of current. The reality is the voltage regulation is incredibly poor. The headlamp is useless at idle and usable at moderate engine RPM. I'm almost certain the voltage regulator only worked on "half" of the AC waveform it was connected to. The headlamp was fed from AC, but with the other lights being fed from current through a single diode they really do not pull their rated current. If anything current was sourced from the battery for 50% of the time, which I've removed.

It does not take much math to determine the total current available at idle from the lighting circuit is

14.0 volts / 4.1 ohms = 3.42 amps 3.42 amps * 14.0 volts = 47.88 watts

So the 140 watt figure is just an upper threshold. If the bulbs were connected to a nominally 6 volt AC system but only getting half the waveform we know the equivalent RMS voltage is computed as

6 * sqrt(2)/ 2 = 4.25

This is quite a bit below the nominally 6 volts the bulbs were designed for. The filament of each bulb has an obviously non-linear relationship between current and impedance. So I'm not going to try and compute the real power each bulb was consuming.

I've removed the indicator and signal lamps entirely. I only need the tail lamp and the head lamp. What I want to do is to run 12 volt electrics on this motorcycle. My thought is that since this motorcycle was originally feeding a shunt regulator and I have removed much of the load this should free up quite a bit of capacity. 12 volt electrics are both common and cheap. 6 volt electrics are expensive and often take weeks of waiting on shipments to arrive.

My thoughts are that I can feed the single coil into a full bridge rectifier. I'll use Schottky barrier diodes. They are common, cheap, and available with a high current rating. I'm expecting a voltage drop across them of 0.45 * 2 = 0.9 volts. I can smooth this with a capacitor and then feed it into a buck-boost regulator. There are plenty of cheaply available units on Amazon and eBay. There were not really meant for this, but I am going to wire a mechanical switch in before the recitifer. I only need the lights on during state inspection and when I'm on public roads. With a buck-boost regulator I should be able to set the voltage output to what I need. My intent is to take advantage of the fact that 12 volt electrics are actually engineered around a 13.8 DC voltage. So by setting it a bit lower than 13.8 volts I can make the bulbs use a bit less current.

From what I understand the available power from the stator should increase as engine RPM increase. So if I can make it work at idle things should be no problem at all at higher engine RPM. My only concern is that the voltage may be far too high at high engine RPM. At top speed in the highest gear the engine rpm is around 7500 RPM. This is at least 10x the idle RPM. Is there a cheap shunt regulator circuit I could buy or build that I could put on the input side of the buck-boost converter? It would just need to shunt the circuit down to something like 50 volts DC.

Are there any obvious flaws in this idea? Are there are any ways to make the system more effecient? Should I consider switching to an LED tail lamp which draws less current?

Answer 1

I do not understand clearly your goals, however, I can answer your questions concerning electronics.

First, if you care about efficiency, do not use shunt regulators. They just eat extra power and produce extra heat (by themselves and inside your generator).

Second, I suggest you to buy 100 V Schottky rectifiers (they are not that easy, the voltage is a bit too high) and 100 V aluminum capacitor about 1000 uF. Then you can connect the circuit you described and measure the voltage on aluminum capacitor when the engine runs at idle RPM and when you add some. You need to measure twice: with no load and loaded to 25 Ohm resistor that can dissipate 100 W.

Based on these measurements, you make the decision on the type of regulator you need.