12 vdc Relay

Basic concept sounds OK to me. You would need to look at the relay spec. to find the actual pull-in voltage.

Your other problem is being careful not to fry the relay coil. You'd have to make sure your relay could stand the generator output voltage at maximum speed. You may need to put a resistor in series with the relay coil, & a zener diode across it.

Also, not sure what your series diode is for - after the rectifier, you have DC anyway. A series diode and a capacitor across the coil may be needed if the voltage gets high enough to pull the relay in while the frequency is still low enough for the relay to chatter.

This concept doesn't deal with the rpm issue. How will the machine know when the rpm's are high enough? A simple system found on a small 1/2-2 hp motor could help. The system employs a set of weights and springs to disengage a capacitor after start up, this could work in reverse. When your equipment has reached an rpm high enough to throw out the weights, it could be used to create a contact point for your field coils or relays.

The system could be used for the over speed situation. When the system has increased to higher rpm's the weights would over come the next set of springs and engage a braking system. An old fashion system of weights and springs rather than an electronic feed back circuit that requires constant power.

Worth a thought.

I dont think this would require constant power. The wind turbine will be turning an alternator that is making the ele.When it starts to turn and you have a load on it ,it will start making a small amount of ele. The faster the rps the higher the voltage goes.But with any kind of load on the alternator ,it will cause more [wind] power to start to turn the blades. and the main objective it to produce about 14.8 volts.So now well say that at 100 rpm's it will produce 14.8 volts.What I want to do is not to have a load connected to the alternator UNTILL it reached the 100 rpm s. But seeing that at 75 rpm it will be making 12 volts ,that would be good enought to Start to charge battaries and also it would have more momentum.I was looking at a few cir. on the net useing " LM741 opamp voltage indicator" this cir just turns on a small light to let you know when you are at 12 volts,instead of the light,I would use that to feed the coil on a 12 volt dc [n.o] relay and then connect the load to the realy contacts.I just want a varying Input voltage say to about 25 volts and regulated to 12 volt output.ALSO after it reaches the 100 rpm I wouldn't need the relay to keep the cir closed becaues then the wind T. would be putting out enought voltage.

I agree you're going in the right direction - the cost & time involved in setting up a mechanical governor (even with switching outputs for control of the circuits) is entirely disproportionate.

For your voltage level detection, try looking at a voltage comparator, e.g. the LM311. This can sink 50mA at the output - enough to drive a small relay which can be used to switch a decent sized contactor for your main load. The 741-type op-amps can only drive about 25mA, and you can't find many relays that will switch with that.

Again, unless you incorporate some sort of regulator for supplying the relays, you'll need to be careful of the max. voltage applied to the coils.

Tks for the info on the LM311 and I was looking at this cir.battary charge cir.

http://www.aaroncake.net/circuits/charger1.asp

You are right about watching the input voltage.I found one cir that will take up to 26 volts in and with a regulated output of 12 volts dc.Or the outher way that I was thinking of was with rpm.Useing a cir something like a digtal bicycle speedometer.When it reaches a certain rpm it will close the circuit. They just use a magnet and sensor and go to a digitial meter.Then another way I just thought of is to make another But VERY SMALL alternator that will only put out just a small amount of ele. As I always say"there is always more ways then one to get the job done."

But the first would be the best.all I need is to worry about the radical input voltage.But their must be some ic,transistors of something to take care of that.

I think I found what might work and it has a high input voltage.

I just want to make sure that I won't get any backfeed from the supply realy as not to make Input realy chatter.

http://www.elecfree.com/circuit/power-supply/lm2577-5v-to-12v-dc-converter-step-up-voltage-regulator/

Johnbg, I found a cir, that will let the input voltage go as high as 60 volts.

http://www.siongboon.com/projects/datasheet/voltage%20converter%20and%20regulator/dc-dc%20step%20down%20regulator%201A,LM1575%20LM2575%20$3.pdf

You have to scroll down the page .It put out a regulated voltage of 5 volts.I can then get a small relay to switch on a high amp. relay.

what do you think? I dont think that the wind turbine will go above 60 volts.It might be close.Dont know that yet.

Looks good at first glance - you could protect the PSU chip with a series resistor & 56V zener diode.

Your on the right track, use an OPTO 22 driver (odc) to interface the small current output of the LM unit you described later. These OPTO, or similar there are numurous manufactures, can easily drive up to 5 A which you can use to turn on any mechanical rely.

Good idea.

Some have already answered your question and added other infos, but depending upon the characteristics of the 12 volt relay, it could pull in as low as 6 volts and not drop out till less than 3 volts. It needs electronics to make it more accurate.

Furthermore, 12 volts will only partially charge a battery, you need to get to at least 14.4 volts to charge quickly, but to cut the charge off when the battery is full. Also, depending upon the type of LA battery (I assume lead acid at this time), you need to keep the batteries voltage above 12.6 to reduce ageing to a minimum. Car batteries are a good example here. Leisure LA batteries are more forgiving, but more expensive.

A suggestion was to incorporate a LM317, that would work.... you do need to build your whole control system electronically, or buy something specially made.

Furthermore, in a windy period, you need a way to "NOT" overcharge your battery. That will also be quite easy with some fairly simple electronics.

Then you need also to be able to stop the Turbine in overly windy conditions (or when repairing or maintaining) in a safe way. Brakes are only good enough to stop, but should not be trusted to hold. Make sure you have some other mechanical system to ensue that the blades cannot rotate if you do not wish them to.....even if its something simple that you need to do like putting a rope over something.....and tying a knot of course! (Might be an over simplification....! But I am sure you get my gist.)

Best wishes and good luck.....

Andy G. As you stated about the rpm. getting too high. Im building or buying a voltage controler just made for that.When the battaries get full or are full then the Voltage Controler switches all the electricty to a DUMP LOAD. That could be a ele.hot water heater or any thing that draws a lot of amp's.

One other person said drawing too much amps will hurt the first Small relay that will be switching on and off the main 2nd relay that will carry the amprage. This is why I want to make sure that I isolate the small relay from any backfeed.

I guess now the Hardest part not knowing alot about changing circuit's and useing different components is when all these people that know and just say "oh just change this or do that," lol Then I have to figure that out .

Yes about the mechanical break, yes I agree with you.That is easy to do.

Sure is nice to have so many helping out with this small project.

I have several 12 volt relays that were salvaged from an industrial AC controller panel. I thought about doing just what you propose. However the wire that makes up the electro-magnet in the relay isn't very heavy. Unless you are going to prevent higher voltages from passing through this magnet, you risk burning it up. It's like putting 110 volts through a 12 volt light bulb.

I saved my relays for another project.

TMF

Hi,

You need to have a regulator (constant voltage). I would use a series regulator with an output of 13.8 volts if you want to charge batteries. Your light bulb would just be on the output of the regulator. If the light comes on, you can apply a load. If the input voltage is too low, the regulator won't draw any current, so the generator will turn easily. No muss - no fuss.

-S

As you have already noticed the output voltage of the alternator (permanent magnet or recycled automotive unit) varies with the speed. Thus, until the unit reaches a speed at which it is generating in excess of the battery volts it will have no load.

Given that, is it worth playing this game? I think not.

Incidentally if the rotor has a lot of inertia when the unit connects, and the batteries are low, you stand a chance of damaging your shaft or the connection from rotor to shaft with the sudden shock loading.

If you really want to do this anyway then the easiest way to control the relay is to use the aux output, usually marked as D+, 61 or L, dependant on mfr. This is usually only available on automotive units but if it is not present then it can be added simply with three diodes with each anode connected to a coil and the cathodes commoned to form the D+. You need not worry about overvoltage damaging the coils of your relays, the battery will hold the voltage down until it is fully charged. Once the battery is charged then you have the problem of stopping that from overcharging. If that happens then it is a much more expensive loss.

kind regards

Chas

Looks like everything is getting OFF track.Im tring to make everything simple.Also all i want to do is enegize a Small relay coil ,I know that their are MANY types of relaies out their with all kinds of contacts, and everything else.And the cost of each could be small or large. This has Nothing to do with charging the battaries,nor a break system.Another thing this is on a Vertical with turbine.Not a horizontal wind turbine.

What Im tring to do is with the "CUT in speed".The 2nd part would be the "Voltage control regulator with dump load" to charge the battaries. I just want to close the cir. of the "voltage control regulator with dump load" when the alternator is making ABOUT 8 to 12 volts.I """ know that it has to get atleast 12 to 14.8 volts to charge a 12 battary. But with this cir. their will be No resistance on the turbine,and that means it will spin up to the correct rpms Easier. Thats all.I just dont want any "back feed" from the 2nd relay. I know that the input voltage will fluctuate .

If back feed of current is your only issue, simply install a directional diode on both sides of your relay. This way the flow of energy can only pass in the direction that you select.

There are numerous automotive relays on the market that can handle DC current. If you are unsure of the type to get then I suggest that you start heavy with a relay suitable for activating head lights.

Good Night,

TMF

tomuchfun.,Thanks.I kind of was thinking of that in the start.

An automotive alternator in modern versions, since about 1967, are self regulating as to output voltage and will stop charging the battery when the batter is charged. Internal regulators do the job. Only problem is the load on the wind turbine will decrease and the unit might overrev.

mickl/ you can or not many use an auto alternator of generator,or their rectifier in wind turbines because they have to reach such a high rpm.Over [1000]

Mike,

we dont know whether he is using an automotive unit or some other unregulated unit, he doesnt make it clear although he does refer to a dump regulator, intimating an unregulatable field.

regards

Chas

Your comment:-

Incidentally if the rotor has a lot of inertia when the unit connects, and the batteries are low, you stand a chance of damaging your shaft or the connection from rotor to shaft with the sudden shock loading.

Is not true, at least not for any car alternator that I have seen or used (it could possibly be true for something selfmade perhaps!).......the voltage and the take up of load current is quite a smooth transition, otherwise we might have damage to the connecting mechanism between the alternator and the car engine......or at least we would hear a "shock" sound.....we do not on any car that I have had.....up to now at least!!

Andy,

A car alternator does start up smoothly, you are correct but this is not a car alternator, and he looks as if he is trying to connect it to the load once it has spun up to a good output speed, so the comparison isn't safe, since this would never happen in an automotive situation. It can happen on the test bench if someone is careless and I have seen the results.

Manufacturers of high output regulators always make them start on a ramp to the field so that they do not get problems of mounting damage or belt slip/ breakage.

regards

Chas

You mentioned both in your original post that I answered, I covered both car alternators and home made stuff.....but not commercial units.....

Using a relay is still wrong on both counts, totally unnecessary......the car alternator does not need anything extra, its already built in, the home made stuff needs proper electronics....commercial equipment that does that which you mentioned is badly designed.....

cap. the start up rpm's that Im looking for is only 10 to 12 vdc which is only about 20 rpm's. This is a vawt not a hawt,big difference. I have a 1to 6 ratio from the vawt to the alt.Also at 12vdc the battary would have to completly dead to draw a huge amount of amps.My coils are 1 inch thick because I have 200 windings for each coil, so the air gap is large,and that alone will cause it not to stop suddenly.

The battery will need to almost fully discharged to accept any charge at all from 12 volts....there must be a voltage difference, for a current to flow at all. All silicon Diodes will introduce a loss of around 0.6 volts if used to stop the battery discharging....Germanium diodes have less voltage drop, usually around 0.4 volts, Schottky can be as low as 0.2 volts.....

Please look at this link for the full text with regard to 12 volt (nominal) Lead Acid Batteries, a small part is also shown below:-

http://en.wikipedia.org/wiki/Lead_acid_battery#Measuring_the_charge_level

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Voltages for common usages

These are general voltage ranges for six-cell lead-acid batteries:

• Open-circuit (quiescent) at full charge: 12.6 V to 12.8 V (2.10-2.13V per cell)
• Open-circuit at full discharge: 11.8 V to 12.0 V
• Loaded at full discharge: 10.5 V.
• Continuous-preservation (float) charging: 13.4 V for gelled electrolyte; 13.5 V for AGM (absorbed glass mat) and 13.8 V for flooded cells

1. All voltages are at 20 °C, and must be adjusted -0.022V/°C for temperature changes.
2. Float voltage recommendations vary, according to the manufacturer's recommendation.
3. Precise (±0.05 V) float voltage is critical to longevity; too low (sulfation) is almost as bad as too high (corrosion and electrolyte loss)

• Typical (daily) charging: 14.2 V to 14.5 V (depending on manufacturer's recommendation)
• Equalization charging (for flooded lead acids): 15 V for no more than 2 hours. Battery temperature must be monitored.
• Gassing threshold: 14.4 V
• After full charge the terminal voltage will drop quickly to 13.2 V and then slowly to 12.6 V.

-----------------------------------------------------------

These are the most important voltages to keep in mind when designing your system.

Another important aspect that we have not mentioned is the safety aspects, for example the acid is very dangerous and will attack people and their clothes if you are not careful. Also the gas produced is extremely explosive and has blinded and burnt many people over the years....

Best wishes.

cap.here is what the alt. is .you can get a picture of what they look like.

http://www.instructables.com/id/DIY-1000-watt-wind-turbine/
http://www.otherpower.com/bdwm53.html
http://www.otherpower.com/wardalt.html

Apparently I didn't make myself clear. You do NOT need a relay at all. You DO need a regulator. If the alternator you have has one built in, you're done. Since you have batteries connected, a sudden load will be supplied in part by the batteries. If the mechanics can't take a sudden load, then they are poorly designed as Andy has said. What good is a wind powered power source if you can't suddenly switch in a load? NO GOOD at all!

All set,Thanks all

Would you know if this is a 9 v reg that they are useing ,and I cant find it in allied electronic bood or web site.It is the 7808 on the top left of the board.Here is the link

http://www.fieldlines.com/story/2004/9/20/0406/27488

tks

I assume the 7808 is an 8V regulator, as 7805 is +5V and 7812 is +12V. It would be limited to 1A, I believe.

7808 is an 8V regulator, just as 7805 is 5V, 7812 is 12V etc. Here is the generic datasheet.

thanks.Just want to make sure.Don't want to see toooo much Smoke.

I have just noted that I was a little late, but there seems to be no point in erasing it now!

------------------------------------------------------------------------------

An LM 7808 is an 8volt positive regulator.

See the following link:-

http://www.fairchildsemi.com/ds/LM/LM7808.pdf

I hope this helps......

Thanks alot. Now I just have to build the 2 different cir. that I need.The first is the voltage control regular and 2nd is the Dump Load cir. And then put them together.

Tks

John

Andy,maybe you can help me on this. In this link when you scroll down you will see the parts for what Im building.I cant find a place to buy the "IFR510" it takes two of them.In both pictures I cant read the print to good.Above the ifr510 is that two 560k or 580k resistors.I know that their so close that I don't think it would matter.

tks

John

Its a fairly standard part, but you need to tell us where you are in the world first!!!

Assuming USA, Radio Shack has them.....I googled them and that was the first entry...

There are plent of them (MOSFET) around, there must be an equivalent easy to find....

Tks ,I found it after I posted it.

Don't suppose you'll find an "IFR510" - but try looking for IRF510, and you may be more lucky