Hi guys. Just a question about my wiring on my B40. 1963. It has a boyer bransden box, and a replacement SS rectifier. I replaced the ignition switch and the whole electrical system looks very simple now.
My question is really about the alternator. I have 2 leads going to the rectifier, and one going to the light switch. It's my hazy understanding that this is to provide extra power for the lights, once switched on. But is this then a AC current? Should this not go through a rectifier and if so, can you fit 2 alternator leads together at the rectifier ( via light switch)? I've seen wiring diagram of a BSA B40 WD which has 2 alternator wires to one port of the alternator but I'm not sure if this would work?
I've never fully understood how the Lucas light switch works, so I only guess that the third alternator wire (GY) goes to #17 on the switch. By magic it's connected to #7 when headlamp switched on and then is wired (GB) to the rectifier.
If I'm barking under the right tree, the additional third wire is also rectified to DC. So no AC to the lamp.
The original design is made to give the right supply to ignition and charging (without overcharging) the battery. With headlamp on, additional power was switched in.
The B40 WD which has a 12V two wire alternator (always full power) uses a zener diode to dump excessive power, not to overcharge the battery or blow lamps.
A common way of improving the charging system is to fit a regulator/rectifier unit (both 6V and 12V available) replacing rectifier, zener and the detour via the light switch. Then two of the alternators three wires are joined, making it work as a two wire alternator.
I've not made that mod on a 3-wire stator, so i suggest you get advice from a more knowledgeable person.
Both Paul Goff, http://www.norbsa02.freeuk.com/ and Alan Osborn, http://www.aoservices.co.uk/ are known for giving good advice.
Hi Mike, Thank you for your help with this. Haven't had much time to look at this, but your advice is very much appreciated. :)