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Series voltage regulator for motorbike

V

Vivek N

Jan 1, 1970
0
Hi

I have an old school Royal Enfield motorbike which uses a blackbox rectifier regulator unit driven by a 4 wire alternator (permanent magnet rotor, twoseparate 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 evenat 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 :

http://i.imgur.com/JCcHMfp.png

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.
 
T

Tauno Voipio

Jan 1, 1970
0
Hi

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 :

http://i.imgur.com/JCcHMfp.png

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.

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).
 
R

rickman

Jan 1, 1970
0
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).

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.
 
R

rickman

Jan 1, 1970
0
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

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?
 
K

kevin93

Jan 1, 1970
0
Hi ....
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. ....
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.

Don't forget that the output impedance of the alternator is highly inductive and shunting its output does not necessarily cause as much dissipation inside 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
 
A

Anthony Stewart

Jan 1, 1970
0
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.
 
V

Vivek N

Jan 1, 1970
0
Hey

Thanks for your responses - I didn't realize that the coils would have enough impedance to restrict the dissipation -
In the light of that it seems like a shunt regulator is more sensible, other 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 betterquality 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 drops below another threshold. Surely lead acid batteries should be fine with rapid intermittent charging and discharging close to their normal 12V, Right?
 
T

Tauno Voipio

Jan 1, 1970
0
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.

In a PM alternator there's no field current.
 
J

Jasen Betts

Jan 1, 1970
0
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.

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.
 
J

Jasen Betts

Jan 1, 1970
0
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?

This only true when used with a voltage source. it's not applicable
here.
additional dissipation in the shunt current element. If the alternator
has significant series resistance, some of this power will end up in the
alternator,

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

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.
 
M

mike

Jan 1, 1970
0
Hi

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 :

http://i.imgur.com/JCcHMfp.png

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.
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...
 
T

tm

Jan 1, 1970
0
Hi

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 :

http://i.imgur.com/JCcHMfp.png

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.
 
V

Vivek N

Jan 1, 1970
0
Snip..

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...

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 thereis a 10000 microfarad capacitor instead - there are no electronic gizmo, other than a tachometer

This is not the first time I'm doing electronics stuff - I have built simple 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.
 
T

Tauno Voipio

Jan 1, 1970
0
Tauno, Face it, we'll trying to educate the "low-information" crowd.

...Jim Thompson

Sorry - cannot always hold it. :)
 
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.

If there are only two windings with leads brought out separately, they can be wired in phase parallel quite safely.
 
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