Power Generation, Voltage fluxuation and regulation.

[MadMechanic]

Hello All,

I am designing a 12vdc power generation system using an R/c DC brush-less 236Kv electric motor. My goal is to input enough RPM to produce 12v which I will then regulate to 5vdc for USB charging.

From what I have researched I will need to add a bridge rectifier to the motor and then a voltage regulator. I am trying to keep this as simple of a circuit as possible and I have found a few modular things on ebay including this for automotive use:

http://www.ebay.com/itm/261075648343

The input on this module is DC8-20V. My question is: For good clean DC power for charging things like an ipod, would a rectifier plus this regulation module be substantial given the change in voltage due to RPM fluctuation of the motor/generator? Or is there something in the plumbing that I am missing or need to add?

Any input greatly appreciated. And as always...have nice day.

 

[Harald Kapp]

A rectifier alone will be insufficient. You will need:
- motor/generator
- rectifier (bridge type to get the most from the motor eh.. generator)
- smoothing capacitor (electrolytic) after the rectifier. As a rule of thumb 1000µF per Ampere load current, at least rated for 25 V nominal voltage (even if you're using it with 12 V only, electrolytic capacitors should never be operated near their limits).
- 12 V to 5 V converter.

The specs on the module you refer to are not brilliant
(- Input voltage: DC8-20V, (12V changes to 5V)
- Output parameter: DC5V3A, 15W)
but one can assume that the module delivers 5 V output for any input voltage between 8 V and 20 V. Insofar this module is a good choice because it will be tolerant to fluctuations in the input voltage (due to changing speed of the generator or changing load).

A known problem with this kind of cheap converters is that the output voltage is often not filtered very well. This can lead to funky behaviour of the USB powered device. An additional line filter may be required between the adapter and your device. Only a practical test can show what's needed.

 

[MadMechanic]

Thanks Harald, this is quite useful information. I am trying to figure out the perfect range of circuit to build where I can get the nice clean 5v for usb charging and also a raw 12V for other applications. The motor will max out at 37V, 60A, 1850W. I doubt my design will produce that much, but I will have to test the output once I have the motor and gearing setup. At that point I should be able to figure out how to design the circuit to have a broad range of output for most battery devices.

 

[MadMechanic]

Edit from original post,

I found this website That has helped me understand:

http://www.kpsec.freeuk.com/powersup.htm

So.....Looking at this I understand that the smoothing capacitor is connected in parallel to the rectified DC-out leads on both the + and - and then I should have a smoother wave. The next circuit shows the addition of the voltage regulator which then shows a smooth solid DC-out. Will any basic regulator further filter the signal?

Also you mentioned (An additional line filter may be required between the adapter and your device). I really want to overbuild this thing. What simple components could I add to further filter the signal? Thanks again -MadMechanic

 

[Harald Kapp]

No. A second regulator after the first will not improve the overall behaviour. It will only cause losses since no circuit can be 100% effective.
Don't put too much into your circuit from the start unless you know you well need ut - and you don't know. Start with the rectifier, capacitor and regulator. Add components only if the circuit doesn't fulfill your expectations. Any unnedded component is a potential source for trouble.

 

[MadMechanic]

No. A second regulator after the first will not improve the overall behaviour. It will only cause losses since no circuit can be 100% effective.
Don't put too much into your circuit from the start unless you know you well need ut - and you don't know. Start with the rectifier, capacitor and regulator. Add components only if the circuit doesn't fulfill your expectations. Any unnedded component is a potential source for trouble.

True, I understand this, But is there any other component besides a voltage regulator that will convert the signal to a pure DC signal? You had mentioned "An additional line filter " What does a line filter consist of? -Thanks

 

[Harald Kapp]

This is what a line filter is about. In your case it would be inserted between generator and rectifier, but it is not clear whether you need one. If so, use a ready made one. E.g. mouser has them, other retailers too.

 

[MadMechanic]

Cool, thanks. This is good to know just in case I do need one. I also wanted to ask about the capacitor. If I were to use a huge capacitor....such as a (1 farad), would there be any disadvantage other than charge-up time? And would using a larger capacitor reduce fluctuation in more so than a smaller capacitor?

 

[Harald Kapp]

Yes, the larger the capacitor, the less fluctuation (we call it ripple). But you have the regulator to take care of that. The smoothing capacitor is required to keep the input voltage of the regulator above the output regulator so the regulator can operate. The ripple will be removed by the regulator.
As a rule of thumb, use 1000µF (=1mF) per Ampere of current. Use multiple capacitors in parallel, if required.

You'll have a hard time finding a 1F/25V capacitor. The capacitors in the Farad range you can find (at affordable prices) are usually rated 3V to 6V. Also this type of capacitor (Goldcap, Supercap) is not suited to large ripple currents (that is the AC that remains after recifying your generator's voltage).

 

[MadMechanic]

Great, thanks again for the info. I should be able to build the circuit now. I will try and post some video of the entire project once it is completed. It should be pretty cool if it works. Danke! -MadMechanic

 

[BobK]

Yoiu can think of the smoothing capacitor as follows. If you simply rectify AC with a bridge, the voltage will go to zero twice in each cycle. To get DC you have to be able to provide power all the time. The capacitor charges when the AC waveform is above the design voltage, and it discharges when the AC waveform is below the design voltage.

If you have a regulator, it will need a certain voltage above the regulated voltage, this is the dropout voltage. Your smoothing capacitor must be large enough to hold the output at above the design voltage + the dropout voltage while drawing the design current. You can actaully calculate it from there.

Bob