I'm restoring a old motorcycle and the headlight is very dim. Upon inspection, there appears to be a shunt regulator. I infer that the part they call the 'regulator' in the diagrams is just a zener diode in parallel with the light bulb. I am basing this on my observation that the regulator has 2 leads which just go between the headlight + terminal and ground. My measurements seem to indicate the series resistance in the shunt regulator is about 1 Ohm.Anyway Vin comes from the magneto and is AC so we would expect the voltage to the headlight to be between Vz and about -.7V.
The light bulb is 6v and 25watts. If I just pull out the 'regulator' (what I think is simply a zener) the light gets bright and the voltage at idle is ~5v and it goes up much higher at higher rpms (>8). With the zener in there, voltage is limited to about 2V (should be 6).
I bought a new zener because it seemed obvious that the current one was permanently conducting too much current. The one I bought was 50W and 6.3V. I tried it and it is barely better! It limits the voltage to about 3V. How is that possible? How can the output voltage actually be lower than the zener voltage? I thought the zener wouldn't even start conducting until we reached the zener voltage, but it appears to be conducting even below the Vz. Is there anything I can do to fix this?
Best Answer
Just a guess, the original Zener diode was a bipolar device and the replacement that you purchased is unipolar. That means that it looks like a forward-biased diode on one direction and a Zener diode in the other.
You can fix it (sort of) by interposing a bridge rectifier between your replacement Zener and the magneto connections. You will want to build the bridge yourself using Schottky diodes to minimize the added voltage drop (which will show up as higher-than-desired headlamp voltage.
Or: purchase another of those Zener diodes and connect the two Zeners in series back-to-back.