Over voltage

[G-Man]

Guys

I like these threads. :-)

Whilst I agree that the output of an alternator depends upon the number of turns in the coils this does not mean that a particular alternator will produce 'only' 12 volts. The voltage in these Hondas is kept to around 12v by the battery and the other loads - bulbs, ignition coils etc. Most alternators will generate a much higher voltage than the nominal output. If you let the voltage rise then you can get 'something for nothing' (never strictly true).

I managed to 'convert' my old BSA Bantam to 12v just by fitting 12v bulbs and 12v battery with a solid state rectfier and Zener diode. Because of the output curve of the alternator, 12v is generated at a higher rpm than 6v but it did, indeed work with this simple old alternator.

To do the same trick with a Gilera 150, I had to reconnect the coils in series to generate enough voltage but it did still work. In Wilf's words, I was doubling the number of turns on that one. The nice thing with 12v compared with 6v is that the current is much lower and the electrical resistance of connectors less significant. How about a 24volt Super Hawk............

As for the resistance issue. If the battery terminals are corroded then if you put a meter on the cable terminals (when the bike is running), the meter will see the volts across the battery AND the high resistance, fooling the meter and you into thinking that the battery is seeing that voltage. In reality the battery is seeing rather less voltage and a reduced current flow in and out.. Keep the terminals nice and clean and the battery gets what you measure.

G

[Wilf]

Quite so!

Wilf

[Snakeoil]

Like I said, I'm no expert here. My comment regarding the strength of the magnetic field determining the voltage is based on large power generation equipment where the field is an electro-magnet with a DC current applied to create (excite the field) the magnetism. When a generator (alternator, really) is about to be synchronized to the grid, the frequency must match the grid as must the phase rotation. The generator output voltage must also match the line voltage. It's been quite a while since I watched an operator sync a generation unit to the grid, but I thought he adjusted the generator output voltage by adjusting the excitation which in turn adjusts the strength of the field. There is no real load on the generator at that point, other than auxilliary loads so that might make a difference. Once the unit is connected to the grid, the excitation control becomes VAR control.

Now, I'm a flangehead and not an electrical engineer so I spent my career with the turbines and tried to stay clear of the generator end and all those nasty high voltage gizmos that have a tendency to fry people. So, although I'm a little off topic, the theory still applies.

Are you saying that if the magnetic field were weak, that the stator would put out the same voltage for a given RPM? If so, that would say that only current would be reduced. I'm not sure I understand that because I thought current was a function of load and the resistance in the coils would not have changed. Please lead me from the darkness here.

One last point. Way back when I started, we used a unit of measure called "ampere-turns" and it referred to how an ancient control system would be calibrated. It's been too long to remember the details but I think the specs had the number of turns in the various coils we had in the system that were driving solenoids and we would measure the current on meter in the panel as we tweaked big Ohmite pots to calibrate the system.

regards,
Rob

regards,
Rob

[Wilf]

Hi Rob,

If the magnets in a generator got weak the voltage in the coils will be reduced because it's the coils cutting the magnetic fields that generates the voltage. With the same load resistance, the current flowing through the coils would also be reduced.

Now I'm going to have to read about ampere-turns!

Wilf

[Snakeoil]

Okay Wilf, I'm not thru with you yet. I think I remember you telling me that you teach nuclear physics to 1st graders or somthing like that. So you should be well equipped to deal with me.

Now, you said that the coil wire passing thru magnet field creates the voltage and the number of turns in the coil determines how many volts are created. I think I remember that now that you mention it. There was a thread in another forum about a guy that builds more powerful stators for 305 Honda and the result is he uses bigger and a few more turns so the resultant stator is bigger and requires a minor mod. But I digress. So, if the field strength is fixed and the number of turns are fixed and the speed of the rotor is fixed (for discussion purposes), then I would think that the stator voltage is fixed. I realize that the current would be near zero with no load on the stator so power out of the stator would also be near zero. I assume that some current flows due to the resistance in the coils.

But you also said that the stator voltage varies with the load. I don't think that is right. I think that the current changes, but the voltage remains the same. If you have a battery in place of a stator, the voltage is fixed and the current varies with the load to vary the power out of the battery. Why wouldn't teh stator act the same. The resistance/load has no effect on factors that determine the voltage in the stator. They just determine the necessary current. I would think that if you were to put a volt meter across a stator with no load on it, it will read a fixed voltage and if you put a huge load on it, as long as you did not exceed the capabilities of the stator/rotor, a meter across those same stator leads would read the same fixed voltage.

I think that the only thing that varies the voltage in a stator is rotor speed variation.

Okay, your turn.

On the subject of ampere-turns, I may have an old controls course book buried in my archives that talks about ampere-turns. I'll take a look tomorrow.

regards,
Rob

[Wilf]

Hi Rob,

So, field strength, coil turns and rpm are all fixed, therefore the unloaded voltage (what you measure across the brown and yellow wires connected only to your meter) is fixed. No current (aside from a very small draw by the meter) is flowing in the coils.

Now add a load (a tail light bulb for example) across the brown and yellow wires and watch that voltage drop and the bulb light up. Current is flowing through the coils and the filament. The reason for the drop in stator voltage is that the coils have resistance, and as current flows through a resistor, there is a voltage drop across the resistor.

If you added another bulb in parallel with the first one, you would be doubling the load on the stator and the voltage would drop again because now there is twice the current flowing through the stator coils so the voltage drop in the coils would be doubled. You could keep on adding more of a load until your load resistance was effectively zero (i.e. a short circuit between the brown and yellow wires) and your voltage reading would also be effectively zero. All the while, the field strength, coil turns and rpm have remained the same as when we started.

As an aside, with no load on the stator, no power is being consumed (except for frictional losses and ineffiencies in the relationship between magnetic field and coil). It's quite easy to spin the rotor by hand to produce its no load voltage. But it gets really hard to spin when you add a load to it!

I took a quick look at ampere-turns--it appears to be the inverse of the stuff we're talking about here.

Wilf

[G-Man]

Wilf

I agree. The load (bulbs etc..) on the alternator changes its voltage as they draw more current. In the extreme - if you connected the two output wires together the voltage would drop to near-zero and if you disconnected them the voltage would rise to a large value.


That's why bikes like these have a complicated ignition / lighting switch which cuts in extra coils when there is extra load to drive. If the headlamp bulb blows the voltage on the other bulbs will rise. Most simple systems without proper regulators are like this.


G