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Basic Honda charging circuit...

22601 Views 10 Replies 3 Participants Last post by  66Sprint
Here is a simplified diagram of the charging circuit..... The black lead pointing left carries power only when the ignition key is on, and can power the ignition, lighting, and horn sub-circuits.... Normally,(in the stock harness) a brown and/or a brown and white wire provide lighting power from the ignition switch, but they are (usually)simply parallel or alternate connections to the red B+ wire coming from the battery.....

This diagram shows the white to yellow connection usually made only when the lights are on , as a permanent connection (constant high-charge mode)

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But, for what bike?

Are all the twins the same basic design?

I STILL didn't think the smaller/older twins had a voltage regulator. Even WITH the education I got on this the other day... :D

Kirk.... The diagram is a compilation of the basic Honda twin system.... It will work on ANY Honda with a magnetized rotor/wire-wound stator assembly regardless if they originally had a regulator or not.....If you remove the regulator (and its three wires) from the drawing, you have the basic older single cyclinder bike circuit (like my 1963 90cc for instance), that was also used on some early twins (not many...., My 1968 CL 175 had a regulator)).... The regulator only works (or is needed) if the alternator output is sufficient to "trigger" it.....but it can't hurt to be in the system, unless it is defective.... Steve
The CM 185T from 1979 has 6 Volt and the same diagram.
It has only the bridge rectifier and no regulator.
Some older Bikes like the cb 250 k and cb 250 g have 12V systems with rectifier and regulator like this diagram.
Maybe the stator from the cb 250 g will fit on my CM 185T to get 12 Volt.
Excuse my bad english, I'm a german ^^. :oops:

Bernd, The alternators of most bikes puts out between 30 and 90 VAC....... The majority of voltage loss occurs during the rectification process when the AC is converted to DC...... I'm currently working on an "amplification" modification to boost the AC voltages higher, so the DC output (to recharge the battery and run the bike) will be in the 13 to 14 VDC range (from an original "6V" system on my 90cc bike) so I can use a 12V system without changing and/or modifying the alternators...If it works, I'll gladly share the specs, parts, and proceedures involved.....
Unaltered, the 90's "6V" system puts out 34VAC, yielding a maximum of 8.8VDC ... I'm hoping for about 70VAC and 14+ VDC with the mods....
I BELIEVE (not sure) the CM200T used a 12V system that may fit the 185...By the time it evolved into the "rebel" 250, it most likely is a 12V set-up....... Anyone know for sure?

Ich ein bissen Deutsche sprechen konnen....... Spelling it, nicht so gut.....
Forgive MY poor German...... Steve
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Interesting... I thought that rectifiers were diode bridges that allowed AC to conduct one way but not the other. I didn't realize they had a resistive value such that they could drop voltage.

I always thought the difference between the, say, 30 VAC output and the DC value was a trick of measurement - kind of like water pressure on one side of a valve. If closed, the pressure (voltage) can eventually build up pretty high. However, once opened and water is allowed the flow (amperage), the pressure can drop way down...

Elevated AC voltages are obtained when measuring across disconnected wires, i.e. lots of pressure built up. As soon as the wires are connected and lots of current can flow (into the battery, into the various loads, etc), the voltage (the pressure pushing the current) drops way down... A function of the strength of the permanent magnets on the rotor, the number of winding coils cut by that magnetic field and the resistance of those winding coils.

Anyway, for a while, I owned a '78 CM185T Twinstar and it was a 6-volt system. However, the pvs. owner had replaced the battery with a new 12-volt battery (unknowingly...) and blew all the bulbs in the system. I got the bike for nothing because of this. At the time, I didn't realize the bike was meant to be a 6-volt bike, so I just scrounged all 12-volt bulbs, etc. from junkyards and put it all together. I commuted 60 miles a day for over a year on that bike, and it was MONTHS before I realized the 6-volt deal. Meanwhile, the system kept the battery charged just perfectly, including constant lights-on, with my longish, high-rpm commute. I kinda wondered at the zippy electric starter... :) But it DID work.

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Steve, have you measured the AC Voltages with Load on the Alternator ?
I think, if you put Load on the Alternator the AC Voltage will drop down.
The Alternator has a inner Resistance.
The more Load is connected to the Alternator, the higher is the Voltage wich
lies on the inner Resistance of the windings from the Alternator.
This has to do with the higher Current wich flows in the System.

The Voltage drop over the bridge rectifier is typicaly 1.4 Volts.

Kirk, do you have changed the Coil from the ignition to a 12V Type ?
Nope. Never changed anything at all. Ran perfectly for 10K+ miles with just swapping out the battery (and the bulb replacement that I did).

That's what made it such a surprise.


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Bernd, I'm sure SOME of the difference of opinion comes from intracacies "lost in translation", but here's MY take on the matter....
The length of the wiring that is wound into coils inside the alternator does NOT cause "internal resistance" that "traps some of the energy" when charging..... Actually, the greater the ohm reading of the coils (resistance caused by MORE windings), the greater the output VOLTAGE. ( when induced by a specific magnetic field per unit of time).... More windings would yield higher Voltages under otherwise alike circumstances (same magnet, RPM's, etc)... The reason you will read less voltage with a load connected is because the load CONSUMES some of the power, leaving less available to be read by the meter....
For those who don't believe the rectification process CONSUMES much voltage, please connect the yellow and pink alternator wires ONLY to the rectifier, and do NOT connect the other rectifier wires to anything else, not even ground through the mount bolt...(except the meter during the DC test)........ Read/test AC volts across Yellow and Pink at a specific rpm value (disconnected from everything), then compare to both the AC volts (yellow and pink), and the DC volts across Red and Green at the same RPMs with only the rectifier connected to the alternator.. (if you use the SAME meter, the micro amperages it consumes in either the DC test or AC tests will be equal (as a load), and negligable (as far as these tests go)......
The simple fact is that it requires some power (consumes) to "push" the remaining power through the diodes

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Thx Kirk, if my Battery will go down, i try this cheap way to convert from 6V to 12V.

Steve, you are right.
The DC-Resistance wich you messure over the coils is very low.
The more Windings the coil has, the greater is the ouput voltage.

But the coils have also a inner AC-Resistance wich will eat up power
on Load.

The rectifier will take more power on 6V as on 12V.
This has to do with the Current.
The current in a 12V System is half as in a 6V System.

If you have a Load with 60 Watts

The Current on a 6 Volt system is 60Watt / 6 V = 10A
The Power wich will lost on the Bridge Rectifier is 1,4V * 10A = 14W

The Current on a 12 Volt system is 60Watt / 12 V = 5A
The Power wich will lost on the Bridge Rectifier is 1,4V * 5A = 7W

I have found a way to step up from 6V to 12V at there:

But the used Toroidal core transformer has also power losses.

At the moment i can't run my bike.
I think it has a damged Oil seal on the Crankshaft or something else.
The Case where the alternator and the points are in (left crankcase cover) fills up with oil.
Then the oil runs down the side stand.
I must wait for 'bugs' to buy oil-seals and gaskets to repair this.

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I thought we might be argueing the same point, only using differing terminology, and viewpoints.......LOL... Kind-of like Standard vs Edison theory......
I think of the alternator as the "source" of the energy, and ignore that it also has resistance, taking only the "net" output into consideration..... :D Steve
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