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# How can i make 6 volt and 10 amps with just single coil ?

How many turns of winding and the what number of guage of wire do I need to make 6 volts and 10 amps from a single coil with 12 magnets at 600 to 800 rpm?

85Views5Replies

How many turns of winding and the what number of guage of wire do I need to make 6 volts and 10 amps from a single coil with 12 magnets at 600 to 800 rpm?

## Discussions

I no longer go nuts for a zero poster with a weird request - not worth the hassle ;)

Faraday's Law for a single coil is, V = -N*(dPhi/dt), where N is the number of turns in that coil, and dPhi/dt is the instantaneous rate of change of the flux through that coil.

I'm using "Phi" in place of the capital Greek letter with the same name. This is the usual symbol used for magnetic flux. Also Phi = Bavg*A, where A is the area of the coil, and Bavg is the average value of magnetic field in the direction normal to the coil.

So I don't have an exact answer for you, but I can tell you that I expect the peak-to-peak voltage of the coil to be proportional to N, number of turns of the coil.

Also I expect V to be proportional to the angular speed, since this determines how fast the flux through the coil is changing. For example, I would expect the voltage at 800 rpm to be approximately twice the voltage at 400 rpm.

Regarding current, this depends on the resistance in whatever load you attach to the coil, plus the resistance in the wire of the coil itself.

Actually, when you put a load on the coil, the voltage from Faraday's Law, calculated above, divides itself across the resistance of the coil, and the resistance of the load.

For the output you've described, 6 volts at a current of 10 amperes, well that kind of feels like a 0.6 ohm load, since 6 volts = (10 A)*(0.6 ohm). This says to me, that the resistance of your coil should be less than 0.6 ohm. The reason I say this, is because the efficiency of your alternator is approximately Rload/(Rcoil+Rload). In other words if the resistance in your coil was greater than Rload, then more than half of your power would be wasted, just in heating the coil resistance Rcoil.

The size of Rcoil is proportional to the length of wire in the coil, and inversely proportional to the cross-section area, which depends on its "gauge" or thickness.

Going back to Faraday's law above, that equation says you can get more voltage just from making a coil with more turns, N. The problem is, more turns, especially if its more turns using smaller gauge wire, is going to quickly lead to a coil with more resistance, and as I've already mentioned, making Rcoil too large, means losing power to ohmic heating of Rcoil.

If you start with the assumption that your coil has a finite volume, and the wire you wrap onto it fills that volume, then you can derive an equation for the length of the coil, depending on the width of the wire; e.g. Volume = L*a*a, or something like this, and you're expecting Rcoil to be proportional to L/a*a, right?

Anyway, this doesn't tell you how the story ends, but I am hopeful it gives you some ideas how to get there.

You might have to wind a few test coils, with known N, and try those out with your moving magnets, to see how much flux (measured as open circuit voltage) you're getting. I mean, that's probably the place to start.

You want to generate 6 volts at 10 amps?

That's 600 watts.

I think we need more information. HOW do you intend to generate this power?

no i think its 60 watt 6*10=60 watts ( volts * amps = watt)

woops finger stutter sorry

It's 60 W (as the OP said). He also wrote, "from a single coil with 12 magnets at 600 to 800 rpm?" So he wants to build a very simple generator using permanent magnets. I'm not sure his goal is reachable, but that's a quantitative question, which depends, among other things, on how strong his magnets are.