Schmidt SON powering Maha C401S charger

[wmaurer]

Firstly, apologies if this is the wrong forum, but I thought the bicycle forum would be best for questions regarding power from my Schmidt SON.

Secondly, please excuse my newbie question; I'm a software developer and reading all this electronic stuff hurts my head.

I will be taking an extended bicyle tour next year, and want to use the power from my Schmidt SON to charge AA/AAA NiMH batteries.

I'll be taking a smart charger such as the Maha C401S to plug into mains when available, and since the charger takes 12V DC input, I thought it might be possible to connect it to my Schmidt SON dynohub. From what I understand (which is very limited), I think I need:

• a bridge rectifier to convert AC to DC, e.g.
  http://www.reuk.co.uk/buy-W005.htm
  (or possibly build my own from four 1N4001 diodes)
• a capacitor to smooth the output from the bridge rectifier, e.g.:
  http://www.reuk.co.uk/buy-1000UF-CAPACITOR-16V.htm
• a DC-DC step-up converter / step-down regulator to set voltage to 12V DC, e.g.:
  http://catalog.digikey.com/scripts/partsearch.dll?Detail?name=PT5041N-ND
  http://catalog.digikey.com/scripts/partsearch.dll?Detail?name=PT5102N-ND
  (note: I don't really understand the difference between these two).

Just wondering if anyone could advise on whether I'm on the right track please? Any help would be greatly appreciated.

Thanks
Wayne

 

[Steve K]

I think you are close, but skip those voltage regulators. The best (most practical) solution is a 12v shunt regulator. Just put a 5 watt 12v zener at the output of the bridge rectifier. You might want to epoxy the zener to a heatsink, since it could be getting fairly warm.

There are other ways to rig up a shunt regulator. A better version is to use a power npn transistor with a low power zener. The zener's cathode and the transistors collector get tied to the output of the rectifier, the zener's anode is connected to the transistor's base, and the transistor's emitter is connected to ground (the negative side of the bridge rectifier). The transistor should be capable of handling 6 watts or so, which means you probably want a TO-220 package, and the tab should be attached to a heatsink. The zener should be 11.3v, or whatever is close.


Steve K.

 

[alexlockhart]

This is very interesting, as I've been thinking and planning to use power from a hub dynamo to do battery charging, although I want to charge a custom-built Li-Ion pack.

Steve, I'm sure you're right that the voltage regulator isn't really necessary, but I think a shunt regulator just dumps the extra voltage as heat and never provides more than the input amperage, whereas a proper buck regulator can, right? I.e. the input amperage (to the regulator) is always limited to around 500mA, but if the speed is high enough for the voltage to be, say, 24V, a buck regulator (ignoring inefficiency) can change 24V/500mA input to 12V/1000mA output, right? That would provide for faster charge rates when riding faster, which might be a big help on a long downhill. This is the effect I'm planning to make good use of for charging a big Li-Ion pack.

Alex

 

[Steve K]

This is very interesting, as I've been thinking and planning to use power from a hub dynamo to do battery charging, although I want to charge a custom-built Li-Ion pack.

Steve, I'm sure you're right that the voltage regulator isn't really necessary, but I think a shunt regulator just dumps the extra voltage as heat and never provides more than the input amperage, whereas a proper buck regulator can, right? I.e. the input amperage (to the regulator) is always limited to around 500mA, but if the speed is high enough for the voltage to be, say, 24V, a buck regulator (ignoring inefficiency) can change 24V/500mA input to 12V/1000mA output, right? That would provide for faster charge rates when riding faster, which might be a big help on a long downhill. This is the effect I'm planning to make good use of for charging a big Li-Ion pack.

Alex

well..... I sorta agree and disagree....

the shunt regulator dumps the extra current that the load isn't using, and thereby keeps the voltage at a safe level. If you don't have any regulator, and the battery charger only uses 100mA, then the dynamo voltage can be larger than the charger can handle, which could damage the charger.

Any series regulator has this same basic problem. The only solution is to use high voltage parts in the system. I designed a battery charger myself, which was set up as a series regulator. It was very efficient, but I did have to use 100v parts.

As far as using a buck regulator... it will draw xxx amount of power, depending on the load at its output. There are two spots on the dynamo's load line where it could end up (the load line is the graph you get when you plot the voltage versus current at a given speed). One spot is at a high current and low voltage position. The other spot is a high voltage and low current. Depending on how the buck regulator operates, it might end up at the high voltage spot. If this happens, there is a chance that the voltage is sufficient to destroy the regulator.

This is all the results of my experiences designing buck regulator systems to draw power from solar panels (which share a lot of similarities with dynamos), and a number of years of designing lighting systems for my bike dynamo. I don't expect you to just believe me, though. However, the alternative is to go ahead and wire up a buck regulator and see how it works out. Experience is definitely the best teacher, but it's not always the cheapest.

good luck,

Steve K.

 

[alexlockhart]

Wayne, I don't want to hijack this thread, but I hope the discussion can be useful to you. Have you tried the shunt regulator Steve suggested? If you need any advice beyond that, speak up! There's a lot of helpful and knowledgeable folks here.

Yes, a normal series regulator will just dump the extra power as heat, which is fine if you don't need the extra power, and in Wayne's case, just powering a battery charger, he probably doesn't need more than the 500mA the dyno provides.

What I'm wondering about is how to make use of that extra available power, to provide higher current at 12V. If you use a linear buck regulator rated for high voltage, will it be able to provide, say, 12V and 1000mA output from a 24V/500mA input from the dyno?

For my application, I'm planning to use a switching regulator rated for 60V max input (AP1512-K5, datasheet). It's adjustable, and I'm planning to set the output to 12V, then use that regulated power for a Li-Ion battery charger designed to accept 12V input. I'm hoping/assuming that I'll be able to make maximum use of the dyno's power, i.e. that the switching regulator will load the dyno in such a way that it provides as much power as it can for a given road speed, and that the output is a constant voltage with the available current going up with the speed.

What are the results of your high-voltage buck regulator - can you get higher amperage output than input? Do you think a switching regulator will be able to load the dyno properly for maximum power? Do you think it'll be able to output 1A or 2A at 12V with enough speed? I know I can make use of as much power as it can give me to charge my Li-Ions, so I don't want to throw away any more than I have to. I think Wayne could make use of more than 500mA also, but it may not be worth the complexity for him.

Alex