Project: Dynamo drive LED bike lights - question re: circuit design

[seanspotatobusiness]

Q1: I was hoping someone could explain what the purpose of the highlighted section of this circuit is for?
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The part in question is taken from circuit 6 on this website where it's described as "boosting power at intermediate speeds". What does that mean?? Is it increasing voltage or current or both?

Q2: If the dynamo is meant to produce 6 V RMS, and I suppose 1 V is lost via the rectifier (0.5 V on each diode), does that mean I will have about 5 V RMS over the XP-G LEDs? A voltage of 6 to 6.6 V would be ideal (these power LEDs can function up to 1.5 A).

Q3: Would this design work (to provide current when the bike is stationary (e.g. traffic lights/junctions)) if the cells are NiCds (four or five in series)?
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[SneaKSz]

Q1 = I think its a snubber network , to reduce the spikes of the dynamo.
Q2 = 6 V RMS (' root mean square ) the maximum voltage is 6V * Root of 2 = 8.5V - 1.2 ( 2 diodes ) = 7.3V over C3 .

So you want to charge your CR-AA; why place 2 LED's and a R in serie , the battery wont charge till its maximum voltage. why not place a diode to charge the battery and parallel the 2 LED's + R ?

Thats just my opinion

Q3 : You"ve forgot a resistor in serie with the 2 LED's . I think it will work that way .

 

[colin55]

VOLTAGE DOUBLER
This is a voltage doubler circuit from a bicycle dynamo. The dynamo produces 6v AC and charges a 3.3FARAD super cap via 2 diodes and an electrolytic.
This is how the circuit works.
The voltage at the mid point of diodes D1 and D2 can fall to -0.6v and rise to rail voltage plus 0.6v without any current being supplied from the dynamo.
When the voltage rises more than 0.6v above rail voltage, the dynamo needs to deliver current and this will allow the rail voltage to increase. We start with the dynamo producing negative from the left side and positive on the right side.
The left side will fall to -0.6v below the 0v rail and the right side will charge C1 and C2 will simply rise in exactly the same manner as we described the left side of the dynamo being able to rise.
Suppose C1 charges to about 7v (which it will be able to do after a few cycles). The voltage from the dynamo now reverses and the left side is positive and the right side is negative. The right side is already sitting at a potential of 7v (via C1) and as the left side increases, it raises the rail voltage higher by an amount that could be as high as 7v minus 0.6v.
The actual rail voltage will not be as high as this as the 3.3 Farad capacitor will be charging, but if energy is not taken from the circuit it will rise to nearly 14v or even higher according to the peak voltage delivered by the dynamo.
When the dynamo is delivering energy to the positive rail, it is "pushing down" on the C1 and some of its stored energy is also delivered. This means it will have a lower voltage across it when the next cycle comes around. C2, D3 and D4 are not needed and can be removed. In fact, C1 will always have rail voltage on it due to the 47 resistor, so the voltage doubling will start as soon as the dynamo operates.
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[SneaKSz]

Thanks for the info, but i can't see how C2 is getting charged . If the left side of the dynamo is -0.7V C1 can charge via D2. But if the right side is 7V and the left will increase . So the super cap will charge. I can't see how C2 is getting charged , there isn't a way.

I also cant see how the dynamo works. If you bicycle then how can the left side being negative and the right side positieve , and reversed??


EDIT : The sides of the dynamo will float to a voltage so the diode can conduct?
Thanks

 

[colin55]

I have already said C2, D3 and D4 are not needed and can be removed. Read my description.
A dynamo is a magnet with a north and south pole revolving next to a coil of wire. As the magnet turns, the ends of the coil firstly become positive/negative then when the magnet turns a half turn the same ends become negative/positive.