# How do I convert a small AC current into DC current?

Discussion in 'The Projects Forum' started by b2386, Oct 16, 2010.

1. ### b2386 Thread Starter New Member

Feb 9, 2010
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I am constructing a small wind generator for a college project. I want the 0.1 V AC output of the generator to charge a battery. To do this, I need to convert the current into DC. I am familiar with things like bridge rectifier circuits, but the smallest diodes I can find have a forward "turn on" voltage of about 0.25 V, which will not even turn on for my small 0.1V signal. What are my options for getting the AC current output into the battery? Thanks

2. ### blueroomelectronics AAC Fanatic!

Jul 22, 2007
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Bigger generator.

How much current does the generator supply?

3. ### marshallf3 Well-Known Member

Jul 26, 2010
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Generators produce DC, alternators produce AC. What are you trying to use?

4. ### eblc1388 Senior Member

Nov 28, 2008
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Step up transformer, then rectify.

5. ### R!f@@ AAC Fanatic!

Apr 2, 2009
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what would 0.1V produce......to step..is this even practical

6. ### b2386 Thread Starter New Member

Feb 9, 2010
13
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I guess the correct term is alternator. Sorry.

Well, the project is to build our own wind flutter generator. (http://www.popularmechanics.com/science/energy/solar-wind/4224763)

The voltage is created by small magnets moving near a metal coil so it is not going to output a large value of power (or voltage). The output of 0.1 V was just a single run with a rather sloppy coil of 500 turns using 36 guage wire. Our real coils are probably going to have at least 5x more turns. I was just trying to look at some of our options so we would have an idea in mind when it came to actually assemble the output circuitry.

7. ### R!f@@ AAC Fanatic!

Apr 2, 2009
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by practical I meant tht 0.1V is useless. U got to have a higher voltage to do anything

8. ### R!f@@ AAC Fanatic!

Apr 2, 2009
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See...he generate at least few volts to light the LED's.

That is pretty awesome.

9. ### b2386 Thread Starter New Member

Feb 9, 2010
13
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In some other article or interview (not this one), he says that he boosts the low voltage generator output with a device that can be found in any old radio. However, he never mentions what it was specifically. Does anyone have any idea what the device may be that he is referring to?

10. ### wayneh Expert

Sep 9, 2010
12,405
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Audio transformers were used to match high impedance circuits to low impedance of the speaker. Not sure what the winding ratio would be, but perhaps at least 10 to 1. Put the low voltage on the "speaker" side and you might get something useable out the high end.

But, aren't there "energy harvesting" modules these days, intended exactly for this type of application? They're not super cheap - I think \$10-15 each but probably worth it if it saves a load of hassle.

11. ### capnray Member

Jul 3, 2010
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Power in equals power out. Your AC generator must first be capable of providing the power to rectify and then charge the battery. The .1V could be amplified to provide RPM, however, charging a battery with that kind of output is out of the question. You also need to consider that power generation will impede the rotation of the fan/windmill.

12. ### b2386 Thread Starter New Member

Feb 9, 2010
13
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Hi all,

Just a quick update with some additional questions. We have improved our design so our output for a single belt is now a 2.5V AC output @ 8mA. This should be enough voltage to rectify with a full wave bridge rectifier using Schottky diodes, correct?

Also, the 2.5V and 8mA are only averages and are highly variable even at a constant wind speed. For instance, the output voltage can range from 0.5V to 3.5V over a period of 2-3 seconds. If I rectify this signal, then I will have a DC output with a similarly varying voltage, correct?

I am still wanting to store this rectified output in a battery (two AA's, possibly). Can a varying voltage DC current be delivered to a battery or does the voltage need to be constant? I guess I am wanting to know if the rectified generator output will need to be "cleaned up" somehow before sending it to a battery.

13. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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You would want to increase the voltage with a transformer, otherwise the rectifier losses would account for 25% of your voltage, maybe more due to the fluctuations.

The 8mA @ 2.5V would be 2mA@10V (5x voltage = 1/4 power in ideal case). Rectified, this would still be enough to power an LED, but not be useful as a battery charger, unless you had a few dozen in parallel.

14. ### b2386 Thread Starter New Member

Feb 9, 2010
13
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Instead of using a transformer, we can achieve a higher voltage by using a higher gauge wire and fitting more turns into our chosen coil dimensions. (We are using 32 right now but can go up to 36. Any higher than that and the coils become difficult to wind without breaking the wire accidentally.) The only reservation with going to a smaller wire is the increased resistance. I am assuming that the total wire resistance of the coil can be considered the source resistance and the smaller the source resistance, the more voltage will be dropped across our load.

If I did use a transformer, would I have to step it back down after rectification?

Also, this single windbelt will be paralleled with at least 10 others once we get the single design optimized.

15. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
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It appears he has FAR more than 36 turns on each coil, there are many layers to each one, with over 20 turns per layer.

The voltage would need to be stepped up via a transformer, then rectified and filtered to DC AT EACH TURBINE before putting them in parallel. Otherwise the phase differences in the AC would cancel out a great deal of the power generated by individual ribbons. This is unless you have found a way to have them all vibrate exactly the same.

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16. ### b2386 Thread Starter New Member

Feb 9, 2010
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Sorry for not being clear. 36 and 32 refer to the wire gauges, not the number of turns. We have about 2500 turns in our coils. (Our design is a little different, too. We have attached a small spring to the underside of the belt along with a small magnet hanging from the spring. The magnet hangs into a cylindrical coil so it is always inside the coil with a very small air gap while it is moving)

Yes, we have already decided that rectification must occur at each individual belt before paralleling them together.

17. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
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How much of an air gap do you have between the magnet and coil? It should be as small as possible, e.g. hundreths of an inch if you can keep alignment.

18. ### b2386 Thread Starter New Member

Feb 9, 2010
13
0
You guys are awesome! Keep the suggestions coming.

Right now the plan is to parallel at least 10 of these together for our final prototype so our total output current that goes to the battery will be more than what we have right now.

Just a question about our voltage and current measurements, though. When we are measuring the voltage and current of a single coil, we connect the multimeter to both of the open-circuited coil ends and measure both the AC voltage and AC current in that position. So it seems like we are measuring the open circuit voltage and the short circuit current produced by the coil. If we connect this coil to a 1.5V battery (after rectification), will our current going into the battery be much less than the short circuit current we measured with the multimeter?

Another question. Assume we take the AC output from the generator, rectify it, and then send it to the battery. Where does the linear wire resistance (ohms/meter) of the coil come into play during this process? Looking at my attached image of the simple circuit diagram, is R_s the resistance of the coil (all ~100ft of wire in the coil) or is it something else?

Regarding the design of the generator itself, I think there are some improvements we can make. We found out last week that our belt can withstand larger magnets so we will be testing those out later today to see how much our output improves.

Our magnetic circuit may be as good as we can make it right now. As you can see in the attached photos, we have tried to improve the magnetic circuit by attached a small spring to the underside of the belt along with a small magnet hanging from the spring. The magnet hangs into a cylindrical coil so it is always inside the coil with a very small air gap while it is moving. If we make the air gap any smaller, I am afraid that the magnet may start rubbing the inside of the tube while it is moving and impede its motion. Additionally, we have to wind the coil around something (a very thin plastic tube in our case) and finding something just barely smaller might be difficult. We thought about doing some sort of iron core but we realized that the main advantage of the windbelt is that it is dirt cheap. Adding iron to each coil would increase the cost and, unless the gains were quite significant, the power/\$ ratio would probably decrease. We are still trying to optimize the generator by testing magnet placement and other variables. For the windbelt Shawn Frayne was using in his video, he was only able to generate 40mW with 10 MPH winds. We are using 8 MPH for our testing.

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19. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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The coil resistance will be the internal resistance of the source, just as batteries have internal resistance (though usually under 2 ohms for a non-dead AA battery).

To make this useful, you'd want to step it up, rectify it, then filter it in a large cap, optionally store it in a super-capacitor instead of a battery (Several Farads). Out of that, you would have a voltage source with very low internal resistance.

A few LEDs drawing 20mA would run for quite a while off a fully charged 3V supercapacitor, at the same time, the same energy wouldn't make much of a dent in a discharged AA NiMH Battery's charge state.

--ETA: What is the wall thickness of your core? Reducing the air gap will have a good effect on performance. Go so far as winding it on a wax covered core, then potting the coil in epoxy, resulting in the very thin layer of epoxy being the "core".

Last edited: Nov 16, 2010
20. ### wayneh Expert

Sep 9, 2010
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You can prove on paper that you'll get maximum power into the load when its impedance matches the impedance of your coil. That will be its DC resistance plus an inductance effect proportional to frequency. If you're at only a few hundred Hz, the DC resistance will dominate, but will definitely underestimate your impedance.

If your load is too high (resistance too low), you'd get more power by using a larger gauge wire. Of course your voltage drops if you do that. It's a tightrope.

I'm just trying to warn you that thin wire will give a very satisfying higher voltage but will severely limit your output.