Regulator for Constant Voltage, Dynamic Current?

[SEnergy]

An MPPT is the combination of a more-or-less standard DC/DC switching regulator

like this one? http://www.gme.sk/img/cache/doc/332/008/l4970-datasheet-1.pdf
or should I look for something different?

[Hideki]

L4970 is better than nothing. There are so many DC/DC converters on the market that it can be hard to make a sensible choice.

I think the problem here is this: For a given wind speed you assume that the windmill outputs a certain voltage and current, which when multiplied together gives you a certain power that you need to transform.

It doesn't work that way!

There is no current without a load, and the load determines how fast the windmill spins and what voltage you get out. The load in this case is the DC/DC converter, which in turn feeds the battery.

- If the load is small, current is low, the voltage is high and it spins fast. This gives you a certain power value.
- If the load is large, current is high, but it will now spin slower and so the voltage will be lower. This gives you a certain power value too. It could even be the same as before. Load it way too much and might practically stop. You won't get much power out by doing that.

If you plot a curve for different loads you will find that the maximum power peaks somewhere between the two extremes. You load it just enough so that you don't slow it down too much, thereby maximizing the combination of current AND voltage. The windmill is not a "constant power" system.

What an MPPT controller does is to continually measure the voltage and current and multiply them together. It then uses some algorithm to adjust the output voltage of the DC/DC converter, which in turn affects how much current is going to the battery. If the calculated power drops compared to the previous measurement it will have to make some semi-intelligent decision about increasing or decreasing the load. It is a continuous trial and error to make the calculated power as high as possible.

[SEnergy]

The load in this case is the DC/DC converter, which in turn feeds the battery.

BUT the propeller is connected to the DC motor, which, when spinning is producing voltage AND current even without load connected to it

[Hideki]

BUT the propeller is connected to the DC motor, which, when spinning is producing voltage AND current even without load connected to it

Oh really? Where exactly does that current go when nothing is connected?

[sdg]

The load in this case is the DC/DC converter, which in turn feeds the battery.

BUT the propeller is connected to the DC motor, which, when spinning is producing voltage AND current even without load connected to it

There. Is. No. Current. Without. A. Load.
None.
Never.

[retrolefty]

BUT the propeller is connected to the DC motor, which, when spinning is producing voltage AND current even without load connected to it

Oh really? Where exactly does that current go when nothing is connected?

 OP is just refusing to get a handle on Ohms law before trying to design his system. Many have tried, many have failed. Perhaps your comment  will turn on a light bulb for him. 

[suicidaleggroll]

The load in this case is the DC/DC converter, which in turn feeds the battery.

BUT the propeller is connected to the DC motor, which, when spinning is producing voltage AND current even without load connected to it

Once again, Ohm's Law...

I=V/R

If there's no load, R=infinity and I=0.

[SEnergy]

The load in this case is the DC/DC converter, which in turn feeds the battery.

BUT the propeller is connected to the DC motor, which, when spinning is producing voltage AND current even without load connected to it

There. Is. No. Current. Without. A. Load.
None.
Never.

--
 -sdg

so you are saying my multimeter is broken?

[SEnergy]

BUT the propeller is connected to the DC motor, which, when spinning is producing voltage AND current even without load connected to it

Oh really? Where exactly does that current go when nothing is connected?

 OP is just refusing to get a handle on Ohms law before trying to design his system. Many have tried, many have failed. Perhaps your comment  will turn on a light bulb for him. 

I'm not refusing anything, Im just stating the fact

doesn't DC motor have copper wires in it? oh wait, it does...
don't copper wires have resistance? oh wait, they do...

so, who's the one ignoring the ohm's law and ignoring the resistance within the motor? oh wait, that's you

[Hideki]

so, who's the one ignoring the ohm's law and ignoring the resistance within the motor? oh wait, that's you

If you measure the current by connecting your multimeter in amps range... guess what the load is...      -->       it's the multimeter!

We are saying: WITHOUT a load

Did you measure the voltage at the same time, WITH the load connected? If not, the measurement is mostly meaningless if you want to calculate the power.

[SEnergy]

so, who's the one ignoring the ohm's law and ignoring the resistance within the motor? oh wait, that's you

If you measure the current by connecting your multimeter in amps range... guess what the load is...      -->       it's the multimeter!

We are saying: WITHOUT a load

Did you measure the voltage at the same time, WITH the load connected? If not, the measurement is mostly meaningless if you want to calculate the power.

I think the problem here is this: For a given wind speed you assume that the windmill outputs a certain voltage and current, which when multiplied together gives you a certain power that you need to transform.

It doesn't work that way!

but wait, if there's load it actually produces voltage AND current! and there's load ALL the time, otherwise the wind turbine would just spin around without any effect, which could be achieved just by placing the propeller on a stick faced towards the wind -> that is not a wind turbine, that is a propeller on a stick, isn't it?

also what about the copper that's DC motor full of? isn't THAT considered a LOAD? it has a resistance, motor is producing voltage, how come suddenly in this case an ohm's law doesn't take place?

[SEnergy]

Did you measure the voltage at the same time, WITH the load connected?

since you said the multimeter IS the load in this case, I did the same thing but measured the voltage with second multimeter, guess what? blowing slightly on the fan was producing ~100mV and ~5mA, at least those are the numbers I was able to catch at the same time

[Hideki]

If the two terminals of the moter is open - nothing connected to them - the circuit is broken. The current can not jump through the air unless there's several kilovolts  between them. This is the "no load" situation.

If you short the two terminals together, it is the resistance of the copper that limits the current through the motor. In addition you're using the magnetic field produced to actively brake it and slow it down.

You have also determined that slightly blowing can generate 0.5 mW in your multimeter shunt resistor.

[SEnergy]

If the two terminals of the moter is open - nothing connected to them - the circuit is broken. The current can not jump through the air unless there's several kilovolts  between them. This is the "no load" situation.

If you short the two terminals together, it is the resistance of the copper that limits the current through the motor. In addition you're using the magnetic field produced to actively brake it and slow it down.

You have also determined that slightly blowing can generate 0.5 mW in your multimeter shunt resistor.

so this statement

I think the problem here is this: For a given wind speed you assume that the windmill outputs a certain voltage and current, which when multiplied together gives you a certain power that you need to transform.

It doesn't work that way!

in fact is not true, because a wind turbine without a load is absolutely useless and is just a "tower" with propeller attached to it, NOT a wind turbine, correct?

[Hideki]

Do you REALLY want to use all your energy on attacking people who try to help your understanding? It looks like the answer is yes. Well, it's your choice.

It is no more useless than a lightswitch that's turned off because you don't want the light to shine at the moment. There will still be voltage, but no current.
Once you turn the switch on, hopefully some kind of light is connected (a load) and current starts flowing through it.

[SEnergy]

Do you REALLY want to use all your energy on attacking people who try to help your understanding? It looks like the answer is yes. Well, it's your choice.

It is no more useless than a lightswitch that's turned off because you don't want the light to shine at the moment. There will still be voltage, but no current.
Once you turn the switch on, hopefully some kind of light is connected (a load) and current starts flowing through it.

i'm not using all my energy on attacking people who try to help me here, as the posts that were actually helpful could be squished to one page, instead of 3, while the rest are just replies from ignorant people that live and enjoy the fact that they know something more (e.g. retrolefty), and instead of actually providing an information with explanation why I'm wrong and why they are correct so I can understand it they are just arrogant

the ppl who actually helped me here and taught me something are suicidaleggroll and you, while some of the rest were here just to boost their ego...

hopefully last question; I still don't quite understand how the regulator (at least the one I posted above) handles current -> linear regulator outputs the same current as it inputs, how about the switching one? how is the output current managed?

[sdg]

The load in this case is the DC/DC converter, which in turn feeds the battery.

BUT the propeller is connected to the DC motor, which, when spinning is producing voltage AND current even without load connected to it

There. Is. No. Current. Without. A. Load.
None.
Never.

--
 -sdg

so you are saying my multimeter is broken?

No.
Your multimeter is a load.
As soon as you connect it, the "even without load connected to it" part isn't true anymore.

[suicidaleggroll]

hopefully last question; I still don't quite understand how the regulator (at least the one I posted above) handles current -> linear regulator outputs the same current as it inputs, how about the switching one? how is the output current managed?

You're working in the wrong direction.  The output current is not "managed" by the regulator in either case, it's managed by the load.  The regulator controls its input current.

A linear regulator, wired normally, outputs a voltage (V).  When you connect a load (R), the load draws a current equal to I=V/R.  In order to supply this current to the load, the regulator draws the same amount of current from its input.  The efficiency is calculated as Pout/Pin, but since Iout=Iin, you can get the efficiency just by Vout/Vin.  Linear regulators can only drop the voltage down, they can't raise it up.

A switching regulator also outputs a voltage (V).  When you connect a load (R), the load draws a current equal to I=V/R.  In order to supply this current to the load, the regulator must draw AT LEAST as much current from its input to satisfy Pout <= Pin.  It does this by pulsing current from the source and storing it in a magnetic or electric field before delivering it to the load, which allows it to step the voltage up or down.  So the current drawn from the source isn't continuous, it's pulsed, but stick a big cap on there and you can smooth most of that out.  The best way to estimate the average current drawn from the source is to use the manufacturer's provided efficiency numbers.  You should find some chart or table in the datasheet that gives you the approximate efficiency as a function of input/output voltage and output current.  Once you know the efficiency, then the power input will be Pout/eff, and the current input will be Pin/Vin.  So for example with a voltage input of 30V, output voltage of 15V, current output of 2A, and efficiency of 90%, you get 30W out, 33.3W in, and 1.11A in.

[SEnergy]

hopefully last question; I still don't quite understand how the regulator (at least the one I posted above) handles current -> linear regulator outputs the same current as it inputs, how about the switching one? how is the output current managed?

You're working in the wrong direction.  The output current is not "managed" by the regulator in either case, it's managed by the load.  The regulator controls its input current.

A linear regulator, wired normally, outputs a voltage (V).  When you connect a load (R), the load draws a current equal to I=V/R.  In order to supply this current to the load, the regulator draws the same amount of current from its input.  The efficiency is calculated as Pout/Pin, but since Iout=Iin, you can get the efficiency just by Vout/Vin.  Linear regulators can only drop the voltage down, they can't raise it up.

A switching regulator also outputs a voltage (V).  When you connect a load (R), the load draws a current equal to I=V/R.  In order to supply this current to the load, the regulator must draw AT LEAST as much current from its input to satisfy Pout <= Pin.  It does this by pulsing current from the source and storing it in a magnetic or electric field before delivering it to the load, which allows it to step the voltage up or down.  So the current drawn from the source isn't continuous, it's pulsed, but stick a big cap on there and you can smooth most of that out.  The best way to estimate the average current drawn from the source is to use the manufacturer's provided efficiency numbers.  You should find some chart or table in the datasheet that gives you the approximate efficiency as a function of input/output voltage and output current.  Once you know the efficiency, then the power input will be Pout/eff, and the current input will be Pin/Vin.  So for example with a voltage input of 30V, output voltage of 15V, current output of 2A, and efficiency of 90%, you get 30W out, 33.3W in, and 1.11A in.

great explanation, thanks!