I'm trying to get my head round the wiring diagram for my 1965 B40F. he power generated by the alternator, I believe, goes to the rectifier, where it is converted to from AC to DC, and then to the battery. Could some patient person please explain to me, a clueless novice:
1. What route does it take?
2. When the engine is running and lights on, does all the generated power go to the battery, which then powers the lights, or does some of it go directly from the alternator to the lights and some to top up the battery? Again, which routes?
I have the wiring diagram in the Haynes manual which I can refer to to try to understand any explanations (and I assume where it says ignition coil it means ignition switch?)
Thanks very much. Phil (novice)
Here is the wiring diagram.
Hi Phil,
Imagine the battery and alternator as pumps.
When the engine is running at slow speed the alternator is not pumping enough to run the system so the battery pumps out, this is shown as negative on the ammeter. When the engine speed increases slightly the alternator pumps say the same as the battery, this is shown as zero on the ammeter and the system is in equilibrium. When the engine is running at an increased speed the alternator via the rectifier pumps more than the battery the alternator feeds the system and charges the battery, this is shown as positive on the ammeter.
The rectifier has to convert to DC to charge the battery
With this in mind follow the cable routes on the diagram.
I hope this helps somewhat.
Best wishes...
Two very basic questions:
1. When the alternator powers the lights, is it always on a route via the battery, or does it sometimes send some power direct to the lights without going via the battery?
2. When the alternator feeds the battery, via the rectifier, through which wire does that power enter the battery? Other than the earth (via frame?), the only wire in contact with the battery is the NU wire at the -ve (live) terminal. If the EMF in that wire is going away from the battery, to power lights etc, then how can the alternator send power to the battery through it?
Phil (novice)
Actually there is a second wire at the -ve terminal, but that goes to the horn and stop lamp, so is obviously irrelevant to this discussion.
Anyway, if somebody could answer my two questions, in an idiot-proof way, then I'd be grateful. Thanks in advance.
Phil (novice)
https://www.amazon.co.uk/Classic-Motorcycle-Electrics-Manual-James/dp/1847979955
This should explain everything you need to know.
Steve
The rectified (DC) output voltage from the alternator and the wiring harness for the headlights, horn, stop light, coil, etc are all effectively connected together at the battery -ve terminal. The way the system works depends on the instantaneous battery voltage.
If the engines revs are high enough, and the rectifier output voltage is greater than the battery voltage, current will flow into the battery from the rectifier. At the same time, current will flow out of the battery to supply the wiring harness load demanded at that time depending on what is switched on. The actual current flowing out of the battery will therefore be the difference between that demanded by the load and the current being provided by the alternator/rectifier.
At lower revs, more current may flow out of the battery than the rectifier is providing but, hopefully, over a period of many minutes riding, there will be a net average flow into the battery and it will maintain its charge.
Hi Phil,
I cannot see the colours too well on your drawing, however answering your last questions:
1. What route does it take?
From the rectifier middle wire (NU), this passes to the system, through the ammeter onto the battery via wire (NU) and gives a +ve reading on your ammeter.
2. When the engine is running and lights on, does all the generated power go to the battery, which then powers the lights, or does some of it go directly from the alternator to the lights and some to top up the battery? Again, which routes?
When the engine is running and the voltage from the alternator is greater than the battery it travels from the rectifier middle wire (NU), this passes to the system, through the ammeter onto the battery via wire (NU) and gives a +ve reading on your ammeter. If the engine is off or the revs are not enough to generate more than the battery stored voltage the voltage travels from the battery through NU to the system giving -ve on the ammeter.
The wire NU is a two way street.
Note:
(i) when electric current was first discovered people didn't know which way to choose and they assumed that it flows from positive to negative. Later it was proved that it is the other way, electrons seeking the positive terminal.
(ii) the ammeter is in series so power passes through it
Steve, I already have the book, but there are still some basics such as this that I'm not grasping. Thanks for your patience, Bess and AWJDT. I am understanding some, but not all, of what you say. For example, I now know that current passes both ways in wire NU. But for example I'm not sure what "passes to the system" means exactly. I think I'm not making very clear what it is that I don't understand. If I could bother you a bit more, please tell me if this is right:
When, at high revs, the alternator is producing plenty of voltage, more than the battery, it will send current directly to the headlight etc, via the left hand wire (GB) from the rectifier, and via switch contact number 7, without it going via the battery. The alternator will then send what is left over along the middle wire from the rectifier, NW, through the ammeter and on via wire NU to recharge the battery, if it needs it. If the battery doesn't need any charge, nothing will be sent to it.
Is that right so far?
Wiring diagram attached. Hard work this, isn't it?! Phil (novice)
Hi Phil,
From the wiring diagram it has a emergency start position, the ac side of the alternator is used to start the bike. This is where the battery is flat and using the emy ignition position you can start and run the bike, the battery is used as a voltage stabiliser. When the battery has some charge rotate the switch back to normal or the battery will be damaged.
Best wishes...
This wiring diagram from the Lucas service manual may assist by showing how the internal connections in the lighting and ignition switch function.
You will see that the GY from the alternator first goes to the 17 terminal on the ignition switch, then with ignition on ,via the 18 terminal to the lighting switch terminal 5 by a black cable. When the headlight is on the 5 connects to 7 and the alternating current goes to the GB connection on the rectifier giving full alternator output.
The second photo also from Lucas explains howthe alternator output is controlled using coil switching, fairly crude and sometimes leads to overcharging, but effective without any voltage regulator in the circuit.
Hi,
Sorry the frame is the 2 way street not NU cable. Got mixed up with +ve not pushing but actually receiving ::)...
Totally confused again!
Hi Phil,
I would just ride the bike and worry if things go wrong...........
To simplify the understanding of the way the wiring works, pretend that the ammeter has been removed and all 3 brown and white wires that connected to the ammeter now connect straight to the battery -ve terminal. This means that there are now 4 wires connected to the -ve terminal if you include the brown and blue wire to the horn and stop light.
Of these wires, current always flows from the rectifier to the battery (it can't flow the other way) but it then flows out of the battery along all the other 3 wires to power the lights, horn, coil, etc depending on what is switched on. In each case, the amount of current flowing in each wire depends on the battery voltage. The power for the lights goes to lighting switch T3 and the power to the coil goes to ignition switch T12A.
The amount of current that has to be provided by the battery is then the difference between the amount of current flowing out of the battery in 3 of the wires minus the amount of current flowing in from the rectifier. If the battery receives more current than it is being asked to provide then this excess current will 'charge' it. When the battery is being charged, you can then view the current being drawn to power the lights, coil, horn etc as coming primarily from the alternator/rectifier.
Unfortunately, there is no regulator built into the wiring harness and, even if the battery is fully charged, the excess current from the rectifier will continue to flow into the battery; in the worst case, cooking it. In order to match the alternator/rectifier output current with the amount of current that needs to be provided to the lights, horn, coil etc, the lighting switch also controls how the alternator coils are used. Thus, when the headlights are on, the full output (3 pairs of coils) of the alternator is used whereas, with no lights on, only 1 pair of coils is used.
All the ammeter does is to show the combined current flow in the 3 brown and white wires so that you can tell when the battery is being charged and by how much.
Okay, thanks AWJDT, that makes sense, even to me. So the wire NU can have current flowing both ways in it at the same time (the two-way street Bess described.) And the current for the lights etc always flows from the battery, never straight from the alternator/rectifier. Please tell me that's right!
Also Julian I get the bit about the alternator output control via the switches (kind of!) - thanks for that.
By the way, just before this I thought I had posted a message asking three questions, but I don't see it posted - I hope it didn't come across wrong in any way and so was removed - can't think how it could have, and certainly didn't mean it to.
Thanks everyone for your efforts and patience.
Phil (novice)
Hi Phil,
Hope all works out for you, it's great to see another person keen to keep a BSA on the road.
ps: please remove the novice tag, you're in the fold
Best wishes..........
Quote from: Phil C on 28 Mar, 2018, 18:33
Okay, thanks AWJDT, that makes sense, even to me. So the wire NU can have current flowing both ways in it at the same time (the two-way street Bess described.) And the current for the lights etc always flows from the battery, never straight from the alternator/rectifier. Please tell me that's right!
Unfortunately, the answer to both questions is NO!
Please see diagram which explains things graphically. From this you should be able to see that the current in wire NU can flow in either direction but in only one direction at a given instance in time. You should also be able to see that the current to the lights is either straight from the alternator/rectifier, or from a combination of the current from the alternator/rectifier and the battery, or from just the battery if the engine isn't running.
Thanks very much indeed, AWJD Thumper - I get it! Thanks very much - that's made my weekend!
Phil.
PS Bess suggested I drop the (novice) bit after my name. I thought about changing it to Phil (clueless), but thought I would try to build my confidence by leaving it out. Thanks again!
Good morning everyone,
When I looked at AWJDThumper's diagram (over a year ago now) it made perfect sense, and I printed it off for future reference if needed. I just happened to come across it again this morning, and it is now not quite so clear to me.
As I see it, AWJDT's equation Ic = Ir - Ii - Il (sorry, lower case rather than smaller capitals, as I don't know how to do the smaller capitals)means that the alternator/rectifier send a certain number of amps to the lights etc, and any not needed for that go to charge the battery. Correct?
AWJDT's final equation, It = Is - Ic, made sense to me a year ago. Why not now, I wonder. I thought the amps returning to the battery should equal the amps leaving it. In which case It would = Is + Ic? Just out of interest, could someone explain please?
Phil.
As AWJDThumper is now around again, I thought I would test people's patience and reintroduce this. Any explanation, please? Phil
Blimey - did I draw that diagram a year ago??
I think the answer to the last question needs to take into account the fact that Ic can flow in two possible directions: when the alternator is not supplying enough current (eg headlight on and bike ticking over), Ic will flow out of the battery; however, when the alternator is providing sufficient current to supply all the bike's electrics (eg headlight off and bike revving at 5,000 rpm), Ic will flow the opposite way into the battery. The diagram shows arrows for Ic in both directions.
Therefore, strictly, we have It = Is + Ic for the first scenario but It = Is - Ic for the second which is normally the way the system should be operating.
Okay, yes, thanks AWJDT. I guess the equation It = Is + Ic holds true either way, because if c is negative then in the equation the plus will effectively change to a minus. Thanks again for be all your help with this. Phil