# more current needed in an electrical circuit

Currently, I'm doing my wind turbine electrical project. Let me get straight to the topic, wind turbine, with a small generator in it. Lets say it can produce 2V and less than 1A . It is connected to an inverter to increase its voltage. Now my idea is to connect the output to a load, where I will need more than 30A. Do you guys have any idea on how to make this happen? Or not necessarily to connect it to an inverter, anything to increase the current. I have many ideas to increase the voltage.

• I guess you need more wind for more power. What amount of power does your load need? Can you give any other information?
– c-a
Dec 16 '15 at 12:27
• So you want to convert 2V/1A to 220V/30A ? Maybe just develop some free energy snakeoil to lubricate that thing... Dec 16 '15 at 12:34
• So the one output from the wind turbine is to be connected to a fan to blow on the turbine to make it turn ... ? You do know that perpetual motion doesn't work in real physics don't you ... ? Dec 16 '15 at 12:59
• Well, you can cheat by using time... Use your wind turbine to charge a suitable battery for - I don't know, say 23 hours and 59 minutes a day, and then power your load through the battery (and a suitable converter) for a minute. This might do the trick. Dec 16 '15 at 18:40

You can't cheat physics. You can convert between different voltage and current tradeoffs, but you can't make more power out than in. Power is Volts x Amps, so whatever Volts x Amps you convert to can't exceed the Volts x Amps you have coming into the converter. Actually the output will be a bit less due to inevitable loss. No converter is 100% efficient.

You have 2 V and 1 A available. (2 V)(1 A) = 2 Watts. That is your input power, which means it's also the upper limit on your output power. Let's say you make a converter that is 80% efficient. 80% of 2 W is 1.6 W, which is how much your converter can put out. That could be, for example, in the form of 10 V and 160 mA, 5 V and 320 mA, etc.

To get 30 A out, the voltage would only be (1.6 W)/(30 A) = 53 mV. In practice, such a large conversion ratio (30:1 in this case) and very low output voltage will make the converter very inefficient. You're not going to convert 2 V and 1 A to 30 A at whatever voltage with 80% efficiency.

By using storage, like a battery, you can have temporary peak output power exceed the input power. However, in the long run, you're still limited by the same laws of physics. You could, for example, make a circuit that charges a 12 V car battery from the 2 V at 1 A input. Let's say you can do that at 80% efficiency, so the current into the car battery would be (1.6 W)/(12 V) = 130 mA. If you draw 30 A from the car battery for a short time, it would take (30 A)/(130 mA) = 230 times longer to recharge the battery, and that's not taking into account the efficiency of the battery itself. For example, 30 A for 1 minute would take 3 hours 51 minutes of charging time to get back to where you were. Taking the battery efficiency into account, it's probably closer to 4½ to 5 hours in reality.

• What have you done with Olin ? :-). +1 Dec 16 '15 at 14:18

Let say it can produce 2v and less than 1A

That's a power of 2 watts in

It is connected to an inverter to increase its voltage. Now my idea is to connect the output to a load, where I will need more than 30Amps

If voltage increased to 2.0000001 volts the power required by the load is 60 watts.

It won't work. Power cannot be created from nothing in other words power out cannot exceed power in unless you output the power in short burts so that: -

Average power out was a bit less than average power in.

I think you have to take a look at the power. The relation is P = V*I, where P is power, V is voltage and I is the current.

The power output from your example is: 2*1 = 2 Watts. The only way to increase the power in your system is by using a bigger wind turbine.

Because of the power being constant, you see that by increasing the voltage, the current has to decrease. And vice versa.

Things you can use to increase/decrease the voltage is by using coupled inductors (transformers) or step-down voltage regulators, but you have to study the datasheets to know the lower limit for the voltage. https://en.wikipedia.org/wiki/Transformer

If your 30 Amp load has a limited voltage to run, this also needs to be considered.

By calculating P/I = V, it gives a voltage at ~67 mV to get 30 Amps.

Hope this cleared some things up!