Vintage motorcycle 6V dynamo regulator

[Shark]

Hi all,

I'm building a dynamo regulator for my father's vintage motorcycle. The system is 6V and I need the regulator to adjust field coil voltage so that
output from the dynamo will be between 6.5V and 7V (not above to avoid damaging the Lead-Acid battery).

Found a circuit online which does the job for a 12V system and tried to adapt it, but it didn't quite cut it.
Circuit provides power to field coil through Q2 when voltage in the system is below threshold set by zenner D1 and trimmer R1.
When voltage in the system exceeds the set threshold, power to field coil is cut and dynamo's output is reduced.

My main problem is that field coil does not get enough voltage because transistor Q2 has a voltage drop of over 2 volts.
When Lead-Acid battery is not full, field coil needs to get around 6V for dynamo to produce enough power to charge battery, supply headlight etc.

To make my explanation clearer, I'm attaching the circuit, hoping it will be clear what I'm trying to do.
Any suggestions what I should modify to diminish the voltage drop and still keep the same functionality?

Thanks!

[kezat]

I suspect you can't  due to the way the coil is wired but if you could move Q2 on the ground side bet it would work the way you expect. Because you have a NPN transistor on the high side it will be hard to drive the transistor in to saturation. Perhaps redesign with a PNP?

Or so it looks to me, anyone feel free to correct me:)

[KJDS]

Hfe min on a 2n3055 is 20. To get an amp out, you need 50mA in. As it warms up that will get worse.

[woodchips]

Unless you are really fussed about originality then fit a 12V dynamo regulator off a car. You should find that the 6V dynamo will happily produce 12V. There was an article about this decades ago in, I think, Motorcycle Sport. Also change to negative earth, makes life easier.

[kezat]

But that would regulate it to ~12v right? Unless it can be adjusted?

Unless you are really fussed about originality then fit a 12V dynamo regulator off a car. You should find that the 6V dynamo will happily produce 12V. There was an article about this decades ago in, I think, Motorcycle Sport. Also change to negative earth, makes life easier.

[langwadt]

Hi all,

I'm building a dynamo regulator for my father's vintage motorcycle. The system is 6V and I need the regulator to adjust field coil voltage so that
output from the dynamo will be between 6.5V and 7V (not above to avoid damaging the Lead-Acid battery).

Found a circuit online which does the job for a 12V system and tried to adapt it, but it didn't quite cut it.
Circuit provides power to field coil through Q2 when voltage in the system is below threshold set by zenner D1 and trimmer R1.
When voltage in the system exceeds the set threshold, power to field coil is cut and dynamo's output is reduced.

My main problem is that field coil does not get enough voltage because transistor Q2 has a voltage drop of over 2 volts.
When Lead-Acid battery is not full, field coil needs to get around 6V for dynamo to produce enough power to charge battery, supply headlight etc.

To make my explanation clearer, I'm attaching the circuit, hoping it will be clear what I'm trying to do.
Any suggestions what I should modify to diminish the voltage drop and still keep the same functionality?

Thanks!

taking power for the circuit on the other side of D3 would give you another ~0.7V and not drain the battery

[Zero999]

Hfe min on a 2n3055 is 20. To get an amp out, you need 50mA in. As it warms up that will get worse.

Why would it get worse at higher temperatures?

Hfe increases with temperature which is why thermal runaway can be such a problem when BJTs are used in the output stage of a class AB amplifier. If anything, it's likely to improve as it warms up.
http://pi-amp.com/Pi-Amp/amp.html
http://vaedrah.angelfire.com/rf_pa.htm

[Circlotron]

One problem I see with the circuit is that when the engine is not running, the battery voltage will be lower and so the generator field will be turned on and this will flatten the battery. Q2 collector and R3 need to be connected to the generator side of D3. For that matter, D3 could be a schottky diode for lower forward voltage drop. Wouldn't hurt to put a big electrolytic cap across the generator output to reduce noise and ripple so the reg may behave itself better.

[Circlotron]

If you can get to the bottom end of the field winding, try this:

Use a mosfet to switch the bottom end of the field to ground.
Run the top end of the field to dynamo + , not to battery.
Run a 10K pullup resistor from mosfet gate to battery +.
Use Q1 collector to pull mosfet gate to ground when over-voltage.

Put a 470R resistor in series with, and close to, the mosfet gate lead to keep it all calm and stable.

[KJDS]

Hfe min on a 2n3055 is 20. To get an amp out, you need 50mA in. As it warms up that will get worse.

Why would it get worse at higher temperatures?

Hfe increases with temperature which is why thermal runaway can be such a problem when BJTs are used in the output stage of a class AB amplifier. If anything, it's likely to improve as it warms up.
http://pi-amp.com/Pi-Amp/amp.html
http://vaedrah.angelfire.com/rf_pa.htm

Hfe has a negative temp co at high currents, though it doesn't alter the fact that there just isn't enough drive regardless of temperature.

[Zero999]

I suspect you can't  due to the way the coil is wired but if you could move Q2 on the ground side bet it would work the way you expect. Because you have a NPN transistor on the high side it will be hard to drive the transistor in to saturation. Perhaps redesign with a PNP?

Yes, I suspect the 2V loss is an inherent problem with using an emitter follower.

Switching to a PNP or moving the NPN to the low side would help to reduce the voltage drop but then you have a common emitter configuration which has a lot of gain and a high risk of oscillation.

[Moriniman]

This is one of those times when you have to ask yourself the question...

Am I doing this for the experience, or to get the job done?

There are off the shelf solid state dynamo regulators in both 6v and 12v versions.

If the aim is a more usable bike, then a 12v conversion is often helpful. You can even buy an alternator that looks just like a dynamo.

There's lots of information out there; http://webspace.webring.com/people/qp/pravg/solidsta.htm

[Shark]

Hi again, thanks for all of your input! Gave me a lot of good ideas.

I will stick to 6V because all other components on the motorcycle are 6V and would be a pain to go changing everything.

Indeed the field coil is tied directly to dynamo's minus, but I will separate it because putting transistor Q2 on low side seems like a good solution 

I will also try to stop powering field coil from battery, if dynamo will have enough power to 'start itself'; avoiding draining the battery when engine is stopped.

Thanks again, will report back about progress!

[Moriniman]

Dynamos have enough residual magnetism to self start. Sometimes you have to 'flash' a dynamo to restore that magnetism, or to convert from positive earth to negative, or vise versa.

[Zero999]

I don't see any need for modifying the dynamo.

You could use a PNP and do high side drive.

The attached circuit has above has 50mV of hysteresis which will avoid Q1 operating in its linear region.

[Fank1]

If you are getting 2 volts drop across the 2N3055 you are not driving the base hard enough.
Max current into the field should be 6.5 battery minus .7 transistor drop divided by the 5 ohm field resistance or 1.16 amps.
Using an hfe of 20 this means you have to supply 5.8 ma base current to turn it full on.
Again using the 6.5 battery minus the base emitter drop of .7 divided by 5.8 ma the max resistance from the collector to the base is 100 ohms, not the 150 shown on your diagram.
I would use an 82 ohm resistor here.
Good luck.
Jerry

[KJDS]

If you are getting 2 volts drop across the 2N3055 you are not driving the base hard enough.
Max current into the field should be 6.5 battery minus .7 transistor drop divided by the 5 ohm field resistance or 1.16 amps.
Using an hfe of 20 this means you have to supply 5.8 ma base current to turn it full on.
Again using the 6.5 battery minus the base emitter drop of .7 divided by 5.8 ma the max resistance from the collector to the base is 100 ohms, not the 150 shown on your diagram.
I would use an 82 ohm resistor here.
Good luck.
Jerry

20 * 5 = 100 and we need 1000

[Zero999]

If you are getting 2 volts drop across the 2N3055 you are not driving the base hard enough.
Max current into the field should be 6.5 battery minus .7 transistor drop divided by the 5 ohm field resistance or 1.16 amps.
Using an hfe of 20 this means you have to supply 5.8 ma base current to turn it full on.
Again using the 6.5 battery minus the base emitter drop of .7 divided by 5.8 ma the max resistance from the collector to the base is 100 ohms, not the 150 shown on your diagram.
I would use an 82 ohm resistor here.
Good luck.
Jerry

20 * 5 = 100 and we need 1000

No, it needs to be less than that and ou'll never saturate a transistor when it's configured as an emitter follower.

In the original schematic, the transistor is configured as an emitter follower, so the voltage drop will be equal to V_BE + V_R3.


The output impedance seen at the transistor's emitter is roughly equal to R3/Hfe so in order to have an output impedance of something low like 1Ohm then it needs to be 20 times that which is 20R.

[KJDS]

If you are getting 2 volts drop across the 2N3055 you are not driving the base hard enough.
Max current into the field should be 6.5 battery minus .7 transistor drop divided by the 5 ohm field resistance or 1.16 amps.
Using an hfe of 20 this means you have to supply 5.8 ma base current to turn it full on.
Again using the 6.5 battery minus the base emitter drop of .7 divided by 5.8 ma the max resistance from the collector to the base is 100 ohms, not the 150 shown on your diagram.
I would use an 82 ohm resistor here.
Good luck.
Jerry

20 * 5 = 100 and we need 1000

No, it needs to be less than that and ou'll never saturate a transistor when it's configured as an emitter follower.

In the original schematic, the transistor is configured as an emitter follower, so the voltage drop will be equal to V_BE + V_R3.


The output impedance seen at the transistor's emitter is roughly equal to R3/Hfe so in order to have an output impedance of something low like 1Ohm then it needs to be 20 times that which is 20R.

Perhaps I should have put

20 (Hfe) * 5mA = 100mA and we need 1000mA out.

I should have been clearer with my post, but was still on my first sip of coffee.

[Shark]

Hi again,

I really appreciate all the replies and suggestions! During the weekend I had time to do some experiments.

I wanted to keep the original circuit (because it was already assembled), if possible. I had the possibility of
separating field coil negative from dynamo negative.

Moved the Q2 transistor to the low side and, as suggested, it improved the voltage drop significantly.
The voltage drop is now around 1 Volt (@6V, 1A). I think this will be good enough for the time being.

Changing resistor R3 to a lower value did not make any significant change in the voltage drop.
Will probably keep it at 150 Ohm. After tweaking trimmer R1 the circuit gives a very nice cutoff at
around 6.9V, to protect the battery.

Need to wire everything up on the motorcycle and test it.

Thanks for all your help 

[Circlotron]

Don't forget to make sure the field is not still energised when the engine stops and the battery voltage drops a little, otherwise the battery will be totally flattened.

[jonshaw]

Does anyone out there have a positive earth version? I am involved with a restoration of an old Ferguson TE20 tractor that was out in the open for many years and the original Lucas RF 97 regulator was full of water and rusted beyond repair. I would like to put an electronic circuit into the original case, can anyone help with a schematic please?

[soldar]

Simple reverse-engineered schematic of a very old alternator with regulator.

[floobydust]

Does anyone out there have a positive earth version? I am involved with a restoration of an old Ferguson TE20 tractor that was out in the open for many years and the original Lucas RF 97 regulator was full of water and rusted beyond repair. I would like to put an electronic circuit into the original case, can anyone help with a schematic please?

Lucas RF97 is the same as RF95 (LRT9 mechanical regulator guts inside) but with sealed metal cover and no fuses inside. RF97 mostly for tractors and marine use. One schematic is for positive ground.

The solid-state dynamo regulator circuits out there, seem not so great. They don't beat the Lucas lol.

[learner323]

Hi Zero999,

I have an old BSA that's in need of a regulator and found your circuit diagram on the web. I'd love to have a go at making one but i'm an electronics beginner and have a couple of questions.

1. Could you please suggest part numbers for D1 and D2?
2. I don't understand RL/5R, could you please explain. Is this a resistor? What value/power is it?
3. What power does the 10 ohm resister need to be?

This is my first post in the forum - hope I'm doing it the right way - I'm sure someone will tell me if I'm not.

Many thanks.