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Forum Index : Windmills : Newbie - Voltage too high??
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| terrorr Newbie  Joined: 23/07/2010 Location: CanadaPosts: 2 |
Hi, everyone, first time posting, looks like a great forum here. I built my first wind turbine using a Shimano Dynohub for the generator. Written on the hub it says 6V - 3 watt. I am assuming the amps are fixed at .5 amp My turbine is connected to a bridge rectifier to convert the AC output of the dynohub to DC. Initial tests with no load, just a voltmeter attached, show avg. 30 volts and with a good strong wind 65 volts. A light wind produces 15 volts. The purpose of the wind generator is to charge a 12 volt deep cycle battery. So, should I be worried about the high voltage?? Also, it seems like the amps are low, does this mean it won't do much except act as a trickle charger even with decent wind? I've been doing lots of research but I am still confused about the voltage vs the battery voltage. Cheers! |
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| VK4AYQ Guru  Joined: 02/12/2009 Location: AustraliaPosts: 2539 |
Hi Terror The voltage you are getting is high bur remember there isnt a load on it, its a bit like leaving your foot on the accelerator in your car without it in gear, the engine will scream to high revs, but when in gear the engine will take the load and pull at lower revolutions of the motor for the same throttle setting. When you connect to your battery the generator will match its self to the battery voltage, and load up and produce power [amps] into the battery, your device is very low rated and will only produce an amp or so, only test will tell, that is over its rated power out put because the wind turbine will turn it faster than it would go in a bike wheel. Have fun playing with it then make a bigger one if you want serious power. All the best Bob Foolin Around |
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Don B Senior Member  Joined: 27/09/2008 Location: AustraliaPosts: 190 |
Hi Terrorr, You are probably going to be disappointed with the output from your Shimano hub when you attempt to put a charge into a 12V battery. There are two important things to note with your hub. The first is that it is designed to output 6V when connected to a 6 Volt 3 Watt globe, and you are trying to charge a 12V battery (presumably through a bridge rectifier that changes the AC output into DC). As Bob says, you might get something, because you will hopefully be able to spin it at at least twice the speed that it would rotate on a bicycle. The voltage output will be directly proportional to the speed that it rotates at. You should not be misled by the voltage that you measure when there is no load connected. The unloaded voltage will always be higher, sometimes substantially so. If you connect it via a rectifier to a 12V battery, then the voltage will only ever reach 12V, or maybe just a little more into a charged battery. Current can only flow in a closed loop (ie from your generator, through the rectifier and battery, then back to the generator). This is where the Amps part comes in. The Amps that can flow in your circuit will be driven by the available voltage, and will be restricted by the resistance inside your generator, the voltage drop across your rectifier (usually about 1.5V), and the internal resistance of the battery, less 12 Volts or so. Watts measure the amount of energy flowing in a circuit. You calculate the Watts by multiplying the driving voltage by the resulting current that flows. The fact that your hub is rated at 3 Watts does not mean that it will output 0.5 Amps no matter what. This rating is the maximum that it can deliver long term into a 6 Volt load with a resistance of 6/0.5 Ohms (ie 3 Ohms) when it is spinning at its designed operating speed as a bicycle hub. If you can spin it at twice its designed (ie bicycle wheel) speed, then it should be able to deliver 0.5 Amps into a 12 Volt load. At this time it would be delivering 6 Watts, but it should not overheat because the current is mainly responsible for the heating that determines its rating. One of the problems with windmills is that the more power you want from them, the bigger the blade diameter that you need, and the slower they will turn for any given wind speed. If you are only looking for 6 Watts or so then a small blade diameter is no problem, but, if you later look to upgrade, then you will need large diameter blades (and a better generator). Regards Don B |
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| davef Guru Joined: 14/05/2006 Location: New ZealandPosts: 499 |
> So, should I be worried about the high voltage?? Yes, when the battery is fully charged if you have enough current being delivered by the generator you will eventually "cook" the battery. Even 0.5Amp continuously into a 13.5Volt car battery might be too much, over a long period of time. I keep a normal car battery at "float" with about 0.05A. >Also, it seems like the amps are low, does this mean it won't do much except act as a >trickle charger even with decent wind? Could be more than a decent trickle charger, in high wind conditions. *** Do a bit of research on: Sepic converters Then on to MAXIM or TEXAS INSTRUMENTS web sites for parts. |
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| terrorr Newbie Joined: 23/07/2010 Location: CanadaPosts: 2 |
Thanks for the responses. Still murky regarding the power produced and what is going to work for a battery charger. I know the formula watts = volt x amps. What is more important volts, amps or watts? So assuming my windmill with a load connected to the battery produces only 20 volts, that would be about 10 watts. However because it is connected to a 12 volt battery the volts will equal the battery it is charging so then the generator will be pumping 12 volts. What happens with the rest of the power that is generated? Is it still only 6 watts? In my research for a suitable motor/generator what I consistently found was to try and look for a motor/generator that produces high volts/ low rpm (or 1 volt per 25 rpm or less). So the dynohub generator that I have seems to fit the bill. Yet when attached to a battery the amps are too low therefore the watts are too low therefore it really won't do much as far as recharging the battery, correct?? |
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Downwind Guru Joined: 09/09/2009 Location: AustraliaPosts: 2333 |
Correct, The generator will make volts as it spins and the faster it spins the more volts. The generator volts must get to equal or above the battery voltage before any current (amps) will flow into the battery, this is called the "cut in". When the battery is low in charge it will soak up all the amps your mill can make, while the volts will remain at the same level as the battery voltage, as the battery charges the voltage will rise with it. When your battery reaches full charge or close to it, it stops soaking up the amps and the voltage rises much higher. This is when you need to disconnect the mill from the battery, but you must not let a mill operate unloaded, so when it is disconnected from the battery it needs to be connected to a dummy load or dump load as it is called. A dump load is normally just a big resistor bank that turns the power generated into heat or perhaps light, to keep the mill loaded and under control. Pete. Sometimes it just works |
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| davef Guru Joined: 14/05/2006 Location: New ZealandPosts: 499 |
> When your battery reaches full charge or close to it, it stops soaking up the amps and the voltage rises much higher. It will rise up to about 15 to 15.5V (equalisation voltage) then start boiling off water. When enough water boils to expose plate material big damage starts to occur. >However because it is connected to a 12 volt battery the volts will equal the battery it is charging so then the generator will be pumping 12 volts. What happens with the rest of the power that is generated? Is it still only 6 watts? *** The rest of the power is either not generated due improper loading of the generator (most likely) or dissipated in the internal impedance of the generator (depends on on the resistive loss in the windings and how you have loaded it). If the internal impedance of the generator is equal to R = E/I, 6Volts/0.5A = 12Ohms then you could get I = E/R, 12V/12Ohms = 1 amp. N.B. R is not a pure resistance , it is really impedance, which varies with frequency or RPM. The above assumes it's internal impedance does not change with the generated voltage or power demand or RPM, ie I am assuming it is behaving in a linear fashion. Let's assume the wire can handle the added current but will the generator saturate due to inadequate flux. This is why I suggested a test at a loaded voltage of say 12Volts and see how much current you can suck out of it. You may or may not get 1Amp. A lathe setup that has lots of gear ratios and a big selection of power resistors will allow you to draw up curves of current versus RPM versus loaded voltage and work out the optimum operating point at different RPM. After reading about SEPTIC converters then read about MPPT to get better loading versus RPM. I hope you reach the point where you find that saturation is the limiting factor as you didn't tell me how money you have invested in this hub! |
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| Don B Senior Member Joined: 27/09/2008 Location: AustraliaPosts: 190 |
Terrorr, You ask which is more important, Volts, Amps, or Watts. To charge your battery, you need Watts, which is a measure of the rate at which energy is flowing. As you note, Watts = Volts X Amps. If you have Volts, but no Amps (as you do when you measure the output voltage of your generator with no load except the miniscule one posed by your voltmeter)then you have no Watts. If you short circuit your generator then you have Amps but no Volts and hence again no Watts. It helps to picture Volts as the level of water in a tank. The higher the level, the more Volts. Amps are equivalent to the rate at which water from the tank flows through a hose from the tank into a bucket. Watts are equivalent to how quickly the bucket fills. Obviously, the higher the water level in the tank (Volts), the more quickly water will flow through the hose to fill the bucket. The bigger the hose for a given water level, the more quickly water will flow (Amps) to fill the bucket. You need both level and flow to get water into the bucket (Watts). The level of water in the tank can be measured as pressure. If you measure the pressure at the stert of the hose when water is running, it will be higher than the pressure at the end of the hose. This is because some of the tank level or pressure is used just to push the water through the hose. Just as a hose resists the flow of water through it, so wire resists the flow of electricity through it. This is the property of resistance, and it is measured in Ohms. Your generator has wire inside it that has resistance. When current flows through this wire to get to the generator terminals, and hence through the load, some of the 20 V open circuit voltage that you have been measuring is lost. This is what happens to the "missing Volts" that you were wondering about in your last post. Regards Don B |
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| davef Guru Joined: 14/05/2006 Location: New ZealandPosts: 499 |
terror, Maybe this article will give some insight into "what the generator looks like". Some background material |
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