Best way to get high current from dirtbike magneto? 3

[tabletop]

Hi, I have struggled with this problem for many years now and Im almost ready to conceed defeat...
I want to generate aprox 12 amps @ 12VDC average from my dirtbikes for a powerful headlight (3 50W halogens). At least 5 Amps at idle. I can easily generate say 10 amps at 12V even at idle by rewiring the stators to suit but the problem is regulating the voltage when the RPM goes up. At around max RPM I end up with over 150Volts. Other than it being lethal voltage/current, no bulb will work over such a broad voltage range. I have got around it in the past by using bypass power mosfets around a (special hi input voltage) 7812 voltage reg. Works, but the 7812 is not durable enough.
btw, yes I know there are aftermarket solutions but they are very expensive and still quite low output. Also I enjoy the whole DIY challenge. I dont want to run a battery so a (self regulating)alternator is out of the question.
There must be a better way. I have been thinking of ways to regulate by way of limiting the induction with rpm increase.
I have designed a very simple mechanism that increases the gap between the stator coils and the magnets in proportion to RPM rise (using centrifugal force). Can anyone tell me (before I build it) will increasing the gap with RPM rise lower the *VOLTAGE*? Im pretty sure it will lower the current, but I can increase wire size and add winds to compensate. Will this work, or is there a better way still?
Thanks.

 

[waross]

Tabletop
Usually current causes heat and excess current causes damage and failure.
Higher voltage causes higher currents.
Current does the damage.- Voltage causes the higher Current.
I am sure that your familiar with Ohm's Law.
Current times Resistance equals Voltage.
or Voltage divided by Resistance equals Current.
Simple example: Connect a resistor of ten ohms across your alternator and at 1000 RPM you get 10 volts.
10 volts divided by 10 Ohms equals 1 amps.
Now wind it out to 10000 RPM and at 10 times the speed you get 100 volts.
100 volts divide by 10 ohms equals 10 amps.
In the first example the watts lost in the resistor are current squared times resistance or 10 Watts.
In the second example the watts lost in the resistor are current squared times resistance or 1000 Watts.

Now inductance;
The inductance of a coil increases proportionally to the frequency. If we use the formula to convert the Inductance to Inductance Reactance, we can measure or express it in Ohms and we can use use those Ohms in Ohm's Law.
Here comes the good part; One of the factors in the equation that converts Inductance to Inductive Reactance is frequency. As the frequency rises, the Inductive reactance rises. So we may have the following induction coil characteristics;
10 Ohms at 1000 RPM. 10 Volts, 1 Amp.
At 10000 rpm, the impedance increases to 100 Ohms (Inductive Reactance.)
So we have 100 Ohms and 100 Volts and 1 Amp.
Now what about the Watts loss in the coil. Well the current through Inductive Reactance does not cause heating. The current through Resistance causes heating.
Our induction coil will have some Resistance.
For this example let's assume a resistance of 1 Ohm.
In the first example, 1 Amp squared times 1 Ohm Resistance equals 1 Watt loss in the induction coil.
In the second example the coil still has 1 Ohm Resistance and the Current is still 1 Amp so the Watts loss in the coil is still 1 Watt.

Now, in real life it isn't quite so simple. Because you are losing some of the voltage across the Induction Coil, You will need more voltage from your alternator before the lamp is bright enough to use.
The circuit is not linear; We have an Inductance in series with a Resistance (The Lamps). The voltage across the inductance will be changing with frequency while we will be trying to hold the voltage across the lamps at a constant value.
I'm starting to really like you inductor suggestion skogsgurra.
Trying different values of inductive reaction will probably yield a better value to use in practice.
If you let me add my Zener, I think we have the solution. Intuitively the rating of the Zener will probably be much less with the help of your induction coil.
tabletop, I googled Zener Diodes and found a rating of 300 watts quite quickly. When I tried to find it again, I found myself in the middle of a bunch of surge limiting Zeners with ratings of 1500 Watts. I am going to have another look and try to find the best steady state ratings. I don't know if the previous rating was peak surge or steady state.
respectfully

 

[itsmoked]

Surge of 1500W is nil for continuous...

 

[waross]

I know, that's why I'm rechecking, I'll let you know what i find.

 

[Skogsgurra]

Thanks for the detailed discussion. I think that tabletop has got it now.

I checked your stats and find that you are quite new to this site. That may explain your rather verbose posts. After some months, you will find that people do not read your posts in full - they are sometimes simply too long to be read. I suggest that you try to condense your text a bit. You will develop a sense for it as you continue writing your otherwise excellent postings. I hope that you can take this advice without being offended - I'm not so sure that I could... ;-)

Gunnar Englund

http://www.gke.org

 

[waross]

skogsgurra
No offence taken. I will take it as good advice from a friend.
Thanks for your patience.
Re the diodes.
Found 75-100 Watts (5000) Peak.

http://www.littelfuse.com/data/en/Data_Sheets/BZY91.pdf

yours

 

[tabletop]

waross, thanks for taking the time to detail that out! It makes complete sense. Thanks also for the zener find. Beefy looking things, built to last.Good.
Seems I have my 'concept' now: inductor in series and zener in parallel. Can anyone save me many hours by giving me a ballpark value for the inductor?
Thanks so much all.

 

[Skogsgurra]

Fine waross, that's a relief.

Remember that the zener doesn't have to be extremely powerful any more. I would gues something like 50 watts and good heat-sinking. The inductor keeps current down. Also remember that you need two zeners back-to-back. Zeners have a break-down voltage in one direction only. They behave as normal diodes in the other direction.

Gunnar Englund

http://www.gke.org

 

[fsmyth]

I would suggest that you quit screwing around and use a
battery-based system. The battery does not need to be very
large; you could use Ni-Cad cells distributed about the
frame in convenient locations. The electronics are more
stable and much more available.
An added bonus is lighting at idle, or with the motor stopped.
I have found that extremely helpful in some situations
(fixing a flat, finding dropped tools, etc).
I hate shunt-regulators - if you lose the regulator, you lose
anything attached. With pass-regulators, all you lose is the
regulator.
You did not mention what type of bike/engine you were dealing
with, but bear in mind that the magneto on some models is
barely enough for the ignition; there is not enough extra
to supply an additional lighting system.

<als>

 

[MikeHalloran]

I'm thinking eleven AA NiCd cells and that cute little alternator from a Mazda Miata could work nicely, and wouldn't look out of place on a dirt bike, nor slow it down as much as the shunt regulator.





Mike Halloran
Pembroke Pines, FL, USA

 

[tabletop]

fsmyth,
A battery system doesnt compare to 200W of light. Tried it. Great as a backup to limp home but riding a dirtbike is difficult enough with GOOD visability. Oh and btw, I ENJOY "screwing around". The challenge of trying to tame all that wasted energy stimulates me.
Bikes (at the moment) are 250 RMX, 290 Sherco, KX100.

Mike, I need to look at this "cute" alternator. Do you know how many amps? Have inbuilt reg?

 

[LionelHutz]

I was thinking along the lines of a simple transistor current regulator. Use a transistor, zener (or LED) and 2 resistors. If you regulate the current in the headlight then the voltage will take care of itself.

You can also try changing the circuit to a simple voltage source by switching the load from the collector to the base of the transistor.

http://en.wikipedia.org/wiki/Current_source#Simple_transistor_current_source

You can (likely should) put a seperate current regulator in series with each headlamp. This will lower the current/power requirements of each circuit and also provide redundancy in case of a lamp or circuit failure.

 

[aolalde]

The simplest solution is Skosgurra proposal; add an inductance in series with the bulbs. For 150 watts at 12 Volts, the nominal current is In = 12V/.96 = 12.5 amps.

To limit 20% over current at 150 Volts approximately 10 mH inductance must be added. So it will be connected in series with the 0.96 ohms of three 50 watts, 12 Volts bulbs in parallel.

The total impedance at 167Hz (neglecting the dynamo internal impedance)
Z = (0.96^2 + (2*3.1416*167*0.010)^2 )^0.5 = 10.54 Ohms
The current, I = E/Z = 150/10.54 = 14.23 Amperes
The bulbs power W = 0.96*14.23^2 = 194.4 Watts.

When the dynamo frequency is 20 HZ and delivers 12 Volts:

Z = (0.96^2 + (2*3.1416*20*0.010)^2 )^0.5 = 1.581 Ohms

And the current, I = 12V /1.581 = 7.6 amps

The bulbs power W = 0.96*7.6^2 = 55.5 watts (1/3 of nominal)

Not a real regulator, but avoids killing the bulbs.

 

[waross]

Hi aolalde
I like your numbers. Can I add a few more.
Looking in my old repair manual for a 1983-84 Honda Shadoo 500 I find these specs;
Nominal voltage of bulbs. 12 Volts
Output voltage of the alternator 14 to 15 volts. (The range of the voltage regulator.)
The old Honda went through a couple of bulbs a year, so 14 volts may be as high as we want to go.
Also when a lamp burns out, the current in the inductor will drop, and so also the voltage drop across the inductor. This will give us a higher voltage across the remaining lamps.
If you let me add my diode, oops, diodes, (thanks skogsgurra).
Then we can drop 5 or 6 volts from maximum voltage and protect the lights. Two 13 Volt diodes should give us 13.7 volts. 13.8 was a standard automotive voltage during my hot rod years.
If we use your figure of 14.23 amps, times 6 volts we get
86 Watts for the zener. That's if all three lights go out at once and you run full RPMs. Much less with all three lights.

Or you could wind your alternator Star and buy a Honda regulator.
No don't do that, this is more challenging.
yours

 

[Skogsgurra]

And still. A rectifier and cap and a switcher that can work with a high DC link voltage seems to be the right way to go. Small, efficient, standard solution.

The only problem I can see is low lamp voltage when idling. On the other hand, there are buck/boost regulators.

Gunnar Englund

http://www.gke.org

 

[Mobius44]

There is another method to regulate power output of motorcycle magneto, it uses a couple SCR's to "short" the AC output of the magneto.

The same circuitry is used on motorcycle battery regulators, you might be able to just use one of those rather than build your own.

More details for those interested:
When the SCR's are triggered the magneto will output 10-20A across the small voltage drop of an "on" state SCR, so very little power is generated/dissipated. A bridge rectifier isolates the SCR circuit from the lamps. The SCR's gates are driven from zener diode/resistor combination powered from the rectified DC output of the bridge. Cap's are required across the SCR gates and on the output of the bridge rectifier to eliminate lamp flickering.

 

[Skogsgurra]

That's it Moebius! I saw something like that on an Enduro bike once. Couldn't find out the working of it. Do you have any references/diagrams?

Gunnar Englund

http://www.gke.org

 

[waross]

Hi Mobius44
Thanks. My diagram for the old Honda regulator just shows the control circuit as a box labelled "IC" and the power diagram has a couple of obvious typos, but what I trust of the power diagram of the regulator is expained perfectly by your explanation. Thanks.

 

[mc5w]

Your problem is that a magneto uses a permanent magnet field. One solution would be to switch an inductance in shunt across the stator coild which would draw reactive power power from the rotor producting the sane effect as reducing the field current of an automotive alternator.

A buck converter such as a switching transistor-inductor-shunt diode acts as a variable autotransformer for direct current which would decrease the current drawn from the magneto as the speed and output voltage go up. You would also need a rectifier and a big capacitor before the the switching transistor and also another big capacitor at the output.

However, by the time that you build all of that you might be getter off to get a small automotive alternator and belt that to your dirt bike. You might need to change to output sprocket or pulley to one that will take 2 chains or belts. There is nothing wrong with replacing the alternator pulley with a sprocket. 150 watts or so is a lot of power to expect from a cheap permanent magnet magneto.

 

[ScottI2R]

Has anyone considered that (from my experience) automotive alternators are usually overdriven? eg, 1 driving pulley rev = 2 driven pulley revs. Consider a shrapnel guard before turning one of these outsourced and rebuilt by the lowest cost supplier before trying to spin one at 20,000 rpm's.

I apologize, but I dont know the actual pulley ratio. I just wouldnt want one coming apart between my legs while I'm going WOT up my favorite jump hill!

Just a thought.

Scott

In a hundred years, it isn't going to matter anyway.