1975 cb125s single 12v instructions please

wrighty

New Member
Hello everyone I'm mark iv just restored a 1975 cb125s. The bike was in a very poor state when I collected it but I took pitty and gave her a new lease of life. I fitted a 190cc big bore kit and made a custom twin exhaust to make her a little different. My next task is to swap from 6v to 12v iv swapped battery/bulbs/flasher, my questions are would a 4 pin regulator from 12v honda c90 do the job???? And would the original cb125s stator cope with charging the battery ???????

Here is my bike

Before



After







Many thanks
 
Not sure about a regulator from a c90 (I thought they were also 6v?), but we have what you need:

http://www.sparckmoto.com/products

The 12V Single Phase R/R will suit your needs.
 
Don't you need to swap to a stator from a 12v bike (I know a CB350 stator bolts right up) for the upgrade?
 
VonYinzer said:
Don't you need to swap to a stator from a 12v bike (I know a CB350 stator bolts right up) for the upgrade?

TL;DR - Alternators produce the same amount of power regardless of whether they're hooked up to a 6V or 12V system. It's the components within the system that determine whether or not the alternator is up to the job.

The amount of power produced by an alternator is relational to the number of windings in the coils and the speed at which the magnets pass by the coils. Alternators are rated in Watts, a combination of Volts and Amps (W = V x A), but that rating is usually determined by calculating the Amps from a set Voltage rating. This is important because of the relationship between Wattage and Ohm's Law (I = V / R).

Basically, think of an alternator as a device that creates current (Amps). It can only supply so much current, and after that, no more. This is why we see people with charging problems when they put a 65W bulb on their CB350. And because of the relationship between current and Wattage, we can see a lack of power production as a drop in voltage across the entire system.

For example, lets assume an alternator supplies 100W at 12V and we hook up a 65W bulb (and only that bulb) into the system. Presumably, our 65W bulb is also rated at 12V, so this means the resistance of that bulb is about 2.2 Ohms. W = A x V, so 65 = A x 12, therefore current = 5.42 Amps. Since Amps = 5.42 and Volts = 12, Ohm's Law means resistance is 2.21Ω.

In this scenario, we're bound by the resistance of the bulb only. In basic components, resistance never changes. The alternator is still producing 100W, so we have a spare 35W that has to go somewhere. In this case, it will manifest as a rise in Voltage. Because of all of the relations between current, Voltage, and Wattage, this affects the rest of our values as well. If Voltage in a circuit increases and resistance stays the same, more current will pass through the circuit.

A rise to 12.5V (with resistance remaining steady at 2.21Ω) will mean our system's current draw has increased to 5.66A and 70.70W. This will continue until the system reaches about 14.87V. At this point, the total current draw is about 6.73 Amps and Wattage is right about 100W. Now, instead of being bound by the resistance within the circuit, we're bound by the amount of power the alternator is capable of producing. At this point, the power demands of the circuit and the power being produced are in balance.

Next example: Our alternator produces 50W and we still have a 65W bulb hooked up to it. Again, our bulb is rated to operate at 12V, so resistance of the bulb is still 2.21Ω. We can follow the math through in the opposite direction and we end up with our equilibrium point being at around 10.51V. Same bulb, just a different alternator. The alternator will put out as much juice as it can, but it's not enough in this scenario.

Final example: Same alternator from scenario one, but now we're using a 65W bulb rated at 6V. 65W at 6V means we're now drawing 10.83 Amps, which gives the bulb a resistance of .55Ω. In this example, we still have an excess power production of 35W, but following the math through, you'll notice that the final voltage sticks at a lot lower value: About 7.41 volts. Only an increase of 1.4 volts over our rated voltage instead of the 2.87 volts over the 12V system.

Now this is important. It's not the voltage that causes damage to your circuitry, it's the current. Generally speaking, components with low resistance are better at handling fluctuations in current because they're higher current devices to begin with. That's part of the reason older systems tend to be rated at 6V instead of 12V. A 6V battery will have a lower internal resistance than a 12V battery, and so it will pull more current from the system than a 12V battery would. It acts like a regulator for the entire circuit. As voltage regulation technology improved, the move to 12V systems made sense because this lets you use thinner copper and you get less total loss in the system due to differences in resistance. I would not be surprised to see vehicles move to 48V within our lifetimes. Military vehicles already use 24V.
 
Hi
keen to know more about the 190cc big bore on the CB125S

im building a 72 CB100 custom and phase two will be an upgraded engine.

I have a complete SL125S engine that I will build up and swap out.


cheers
Adam
 
Here you go Roots....

http://www.dotheton.com/forum/index.php?topic=18879.0

At the end of the day probably easier and less heartache to shoehorn an XR200 or Lifan 200 and call it a day
 
thanks man
ill check it out.

this is my build thread
http://www.dotheton.com/forum/index.php?topic=65546.0

as for going the easy way? that would be too easy. ::)
the best way get a swag of good stories to tell while drunk is to repeatedly do things the HARD way.

cheers
Adam
 
That's a fact. Just be prepared to do a lot of research and machining if you want to go over 145cc. I think based on that thread you can use a CB750 piston to get to 145, might still need a little work on the piston to use it though. Or you may be able to find some old Powroll performance parts (good luck!)
 
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