Series voltage regulator for motorbike

Hi

I have an old school Royal Enfield motorbike which uses a blackbox rectifie r regulator unit driven by a 4 wire alternator (permanent magnet rotor, two separate pairs of coils). By default one set of coils drives the headlight s via a shunt regulator and the other set charges the battery via the rec-r eg unit.

The consensus among enthusiasts is to switch to a DC headlight circuit, by connecting the alternator coils in parallel and using them only to charge t he battery. Many people successfully do this, this avoids running separate AC wires to the front of the bike and lets the headlight work brightly even at low RPMs in the city.

However I don't like the fact that they use a shunt regulator : The alterna tor produces about 180 watts and maximum of about 50 volts - at full RPM, t he shunt will be dumping all that wattage and it will dissipate as heat in the alternator coils, inside the primary case which is already burning hot from the clutch and primary drive friction.

I want to make a simple series regulator with minimal component count, and also have the ability to run without battery if need be (this is a points b ased bike). Perfect regulation is not critical, it's more of an overcharge protection.

This is the circuit I have in mind :

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Note that the alternator will produce upto 6200 Hertz, peak voltage is abou t 50V, maximum current output is about 15 AMPS - peak intermittent load fro m the bike (horn, lights, ignition) may be upto 20 amps.

I wonder if maybe the 10K is much too high or whether a 2n3055 can take tha t voltage and current. Do I need a darlington pair? The output should be 13.6 volts for a lead acid, as far as I know.

How would one build a similar circuit around an IRF540 instead? Does it make sense to put the capacitor as shown? My intention is that if t he battery goes kaput (it happens often!) , i can simply disconnect it and the capacitor should be able to handle the ignition load at least.

Reply to
Vivek N
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You'll just move the heat from the alternator coils to the pass transistor. There is still something to get burning hot.

The 2N3055 will need up to 1 A to the base for your currents, and IIRC, it is far too small for 20 A (and about 600 W).

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Tauno Voipio
Reply to
Tauno Voipio

A shunt regulator draws current from the supply all the time while a series regulator only draws current when the load does. There can be a huge difference under any condition other than full load.

--

Rick
Reply to
rickman

It may be better yet to build a switching regulator. That would minimize wasted power, meaning less heat and more power available at the wheel.

You need more circuits chops to make it work, though.

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Tim Wescott 
Control system and signal processing consulting 
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Reply to
Tim Wescott

You need to characterize your alternator first.

You need curves of voltage at various load currents versus RPM. I tend to agree with Tauno Voipio, a series regulator will simply move the heat to a different device.

In fact I suspect a simple series regulator may produce more heat... if the alternator can output 15A AT 50V, for dissipation, you would have (50V-12V)*15A = 570Watts

Probably won't be quite that bad, but that's why I suggest characterization first.

A possibility that just occurred to me is phase controlled rectifiers driving thru an inductance to your battery... sort of a buck switcher. I think I've seen that scheme being commercially available. ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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Reply to
Jim Thompson

Not trying to give you a hard time, but you are aware that a shunt regulator also has a series element that drops the voltage, no? The shunt regulator, instead of varying the series resistance, draws a parallel current to adjust the drop on the series element. So you not only get the same power dissipation in the series element, you get an additional dissipation in the shunt current element. If the alternator has significant series resistance, some of this power will end up in the alternator, but the total power dissipated will always be more with the shunt device, if for no other reason because the current drawn from the alternator will always be greater, no?

--

Rick
Reply to
rickman

...

nator produces about 180 watts and maximum of about 50 volts - at full RPM, the shunt will be dumping all that wattage and it will dissipate as heat i n the alternator coils, inside the primary case which is already burning ho t from the clutch and primary drive friction. ...

out 50V, maximum current output is about 15 AMPS - peak intermittent load f rom the bike (horn, lights, ignition) may be upto 20 amps.

Don't forget that the output impedance of the alternator is highly inductiv e and shunting its output does not necessarily cause as much dissipation in side the alternator windings as you suggest. The energy will be stored in the inductance during part of the rotation and then returned as mechanical energy for the remainder. Shunt regulation is common for permanent magnet alternators.

kevin

Reply to
kevin93

I was under the impression that field current pre-regulates the output voltage so it is not all dumped from max power. At least this is how it is done in cars with 14.2 nominal Voltage.

Reply to
Anthony Stewart

Lots of alternators, even ones used on aircraft, use shorting (scr/triac) regulators. It's like a phase control dimmer, except it shorts the source. If an alternator has a lot of inductance, it doesn't get too hot when it's shorted. A motorcycle alternator, designed to drive a headlight at about constant-current, is probably very inductive. And if it is, a series regulator might see a huge voltage input at light loads, like your 50 volts. You might fry the transistor.

This calls for a switcher, really, one that can handle a high input voltage and current-limits into the battery.

--
John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    
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Reply to
John Larkin

Hey

Thanks for your responses - I didn't realize that the coils would have enou gh impedance to restrict the dissipation - In the light of that it seems like a shunt regulator is more sensible, othe r than using a chopper or switching mode device.

It's not so much the design of the stock regulator unit that I worry about but the fact that these are poorly made, and I have had the rectifier fail on this black box rec/reg unit. Since shunt regulators are OK as per you guys, maybe I'll just get a better quality one, or build my own which is always more fun...

How about a simple MOSFET as a switch in the charging circuit, driven by a reference voltage? If the battery voltage goes above a certain level, it cuts the charging. I could add some hysteresis so charging will only restart if the voltage drop s below another threshold. Surely lead acid batteries should be fine with r apid intermittent charging and discharging close to their normal 12V, Right ?

Reply to
Vivek N

In a PM alternator there's no field current.

--

-TV
Reply to
Tauno Voipio

On a sunny day (Sat, 21 Sep 2013 11:46:39 -0700 (PDT)) it happened Vivek N wrote in :

Very long time ago, early sixties perhaps, there was a circuit in a Motorola handbook with a thyristor, very simple circuit. I did build it once with some mods (no longer had the book) in the eighties and it worked.

But I found this:

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page 26

Reply to
Jan Panteltje

Not a shunt regulator, but a shorting regulator. When the output voltage or current are too high, short the alternator winding *before* the bridge rectifier. That is ideally 100% efficient.

A shunt reg would work, but it has to dissipate power.

You could use a series switch, on/off, if you handle all the flyback cases.

--
John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    
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Reply to
John Larkin

Whether you have a permanent magnet alternator or electromagnetic field regulated alternator this link will describe some solutions.

formatting link

Reply to
Anthony Stewart

Tauno, Face it, we'll trying to educate the "low-information" crowd. ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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Reply to
Jim Thompson

yeah, but the alternator is a current source, not a voltage source.

if you put a series regulator in there you risk burning out the recitifiers or the regulator at high RPM.

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Reply to
Jasen Betts

This only true when used with a voltage source. it's not applicable here.

Most don't, it's ballpark 1 ohm. you waste more power in the SCRs in the regulator.

No!

model the altenator as a poorly coupled current transformer fed from a variable frequency fixed current AC source.

  • AC because the field rotates. * variable frequency because the RPM is variable * fixed current because the field strength is contant. * poorly coupled because there is a gap between the rotor and the stator. If you short the secondary nothing spectacular happens.

As you turn the frequency up the voltage also rises, but not the available current.

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Reply to
Jasen Betts

Executive summary...give up.

Check your requirements/assumptions before you go any farther. Don't know anything about the bike, but you don't mention the ignition running off it's own winding. Run without battery? How are you gonna get it started? Unlikely the kick-starter will let you generate enough energy to run the ignition.

The ignition is set up to run a delicate balance. You may not have enough regulator headroom at low rpm and enough voltage to zap your transistor at high rpm.

Put lots of protection around your pass element. Nasty spikes in that environment.

Need much more base current...bigger zener...probably higher voltage one too. Don't forget the huge heat sink.

This is not a simple project. You don't appear to have the skills to pull it off. If I were gonna do this, I'd probably start with a P-channel FET and a single-chip microcontroller to manage the switcher. Probably smaller and cheaper and more efficient than the alternatives. If your alternator isn't grounded anywhere, you could float the ground end and use N-channel FETs.

Did I mention...give up??? Life is too short...

Reply to
mike

I have an old school Royal Enfield motorbike which uses a blackbox rectifier regulator unit driven by a 4 wire alternator (permanent magnet rotor, two separate pairs of coils). By default one set of coils drives the headlights via a shunt regulator and the other set charges the battery via the rec-reg unit.

The consensus among enthusiasts is to switch to a DC headlight circuit, by connecting the alternator coils in parallel and using them only to charge the battery. Many people successfully do this, this avoids running separate AC wires to the front of the bike and lets the headlight work brightly even at low RPMs in the city.

However I don't like the fact that they use a shunt regulator : The alternator produces about 180 watts and maximum of about 50 volts - at full RPM, the shunt will be dumping all that wattage and it will dissipate as heat in the alternator coils, inside the primary case which is already burning hot from the clutch and primary drive friction.

I want to make a simple series regulator with minimal component count, and also have the ability to run without battery if need be (this is a points based bike). Perfect regulation is not critical, it's more of an overcharge protection.

This is the circuit I have in mind :

formatting link

Note that the alternator will produce upto 6200 Hertz, peak voltage is about

50V, maximum current output is about 15 AMPS - peak intermittent load from the bike (horn, lights, ignition) may be upto 20 amps.

I wonder if maybe the 10K is much too high or whether a 2n3055 can take that voltage and current. Do I need a darlington pair? The output should be 13.6 volts for a lead acid, as far as I know.

How would one build a similar circuit around an IRF540 instead? Does it make sense to put the capacitor as shown? My intention is that if the battery goes kaput (it happens often!) , i can simply disconnect it and the capacitor should be able to handle the ignition load at least.

++++++++++++++++++++++++++++

Hello Vivek N,

What year is your Enfield? I have several old Triumphs that use the PM alternator. In my 1966 Bonneville I have just a 6800 uF 25 volt electrolytic and the Lucas 15 volt 50 watt zener and it works just fine. At very low idle, the headlight does slightly dim but not too badly.

You might take a look at the Koller gas generator regulator as it uses an SCR to shunt the stator to regulate the output.

Be careful when you parallel the windings. They are not in phase. Maybe you could feed each winding into a bridge and then sum the DC outputs.

The most common problem with these is the magnetic rotor looses strength over the years because of the heat and vibration it is exposed to. I made an electromagnet device once that I used to recharge the rotor magnet. I dumped a large capacitor into to coil with the rotor poles in the field of the electromagnet.

Reply to
tm

ifier regulator unit driven by a 4 wire alternator (permanent magnet rotor, two separate pairs of coils). By default one set of coils drives the headl ights via a shunt regulator and the other set charges the battery via the r ec-reg unit.

Snip..

Life is not too short... I have plenty of years left in me...

This is a very old design for a bike and it starts without battery if there is a 10000 microfarad capacitor instead - there are no electronic gizmo, o ther than a tachometer

This is not the first time I'm doing electronics stuff - I have built simpl e stuff like solar charging controllers and etched my own PCBs and all that , It's just been a long while and I need to refresh myself.

Even if I screw up, theres hardly anything I can damage on the bike.

Reply to
Vivek N

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