Discussion in 'The Projects Forum' started by pizzaboy150, Aug 20, 2015.

1. ### pizzaboy150 Thread Starter New Member

Aug 20, 2015
5
0
Hi I all,

I am currently making a small homemade wind turbine to power some outside lights but I have I think a basic question.

Now if my turbine turns at 1000rpm and is creating lets say 20 volts and i pushed the output through a resistor of say 1K would the current be calculated using the usual ohms law?

20 / 1000 = 0.02Amps?

The reason I ask is that I have been looking at videos of people making there own generators and the current output they have been getting is milli amps compared to the size of the voltage?

2. ### Dodgydave AAC Fanatic!

Jun 22, 2012
5,155
772
The voltage / resistance = current, but the max value of current able to draw from your generator depends on the size of wire guage used to make it, thicker wire means bigger currents can be maintained.

3. ### pizzaboy150 Thread Starter New Member

Aug 20, 2015
5
0
hmm interesting, is there any special mathematical equation that will take me a year to understand that will predict draw and wire gauge?

4. ### Dodgydave AAC Fanatic!

Jun 22, 2012
5,155
772
google is your best mate for that one, its all to do with the magnetism , number of turns, rpm , area of the coils etc.....

5. ### Kermit2 AAC Fanatic!

Feb 5, 2010
3,852
967
Watts, or power output of a generator has a maximum value that is pretty much fixed for a generator by size alone. There is a peak power output that cannot be exceeded for all generators based on there size if they can receive sufficient input power, or driving torque to reach it.
If you use larger Guage wire in your generator you will get a lower voltage because the generator has a fixed size for the wire winding spaces, and larger Guage wire takes up more space per turn. It CAN carry more amps, but the reduced number of turns means a lower voltage.
Conversely, if you use a smaller Guage wire you will get a much higher voltage but the max current will be much smaller.
Example: you wind with, let's say 16 Guage wire and get 10 volts and 10 amps at max RPM. If you change the wire to about 26 Guage you might get 100 volts, but only have 1 amp of current. Both of these would give you 100 watts of power. To get more power requires a LARGER generator, because you require more room to put more wire in the windings. So, all this simply means power is really directly related to SIZE of your generator.

6. ### pizzaboy150 Thread Starter New Member

Aug 20, 2015
5
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Thanks for the help guys a bit more reading required on magnetism though if I used small gauge wire and lots of coils to get a really high voltage wouldn't it be better to use a transformer to reduce that voltage down to get the increased current?

7. ### #12 Expert

Nov 30, 2010
16,704
7,351
Let me try to dumb this down. Wire has resistance. You should try to use a wire fat enough to be pretty much irrelevant to the current you expect to make. There are lots of wire size tables available on the internet but I have a wall chart which says:

10 AWG = 1001 Ft/ohm
11 AWG = 793 Ft/ohm
12 AWG = 629.8 Ft.ohm
13 AWG = 499.7 Ft/ohm
14 AWG = 396.2 Ft/ohm
15 AWG = 314.4 Ft/ohm

20 AWG = 98.7Ft/ohm
30 AWG = 9.64 Ft/ohm

The thing to notice is that for every 3 gauge numbers, the volume of the copper wire doubles, for 5 gauge numbers, the copper about triples, and for every 10 gauge numbers, the volume of the copper changes by a factor of 10. Memorize about 3 of these and you can do mental calculations about wire gauges.

So, let's say you calculate 100 feet of wire to get enough turns for 120 volts and you expect 10 amps?
One ohm per hundred feet will cost you about 10 volts for this hundred feet of wire. Flip it upside down and you have 100 feet per ohm. Look at the chart...20 AWG. For residential wiring, 10 AWG, 60 degree Centigrade wire, is good for a 30 amp breaker, so 20 AWG must be good for 3 amps...but this isn't residential wiring. You have forced air cooling and you can buy insulation rated to run a lot hotter. Suddenly, 20 gauge wire looks like a reference point to start your design compromises.

This post is a lot of guessing. There are real wire charts for real alternator/motor windings. The point is that a bit of math will get you in the ball park. If you Google a motor winding chart and it says 17 AWG rated at 75 C, you will have some confidence that it's about right. That would come out to 0.5054 ohms at 20 C for a starting loss of 5.054 volts off your 120 volt goal. Heat that up to 75 C @ 0.393%/C for a hot resistance of 1.0924 ohms @ 10 amps. If you don't like losing 10.9 volts as the generator heats up, you can already do in your head, "Three AWG difference will cut the resistance in half, and that makes a hot 14 AWG wire look like a cold 17 AWG wire.

I bet I'm overestimating the size wire you will need, but the goal is to get you familiar with juggling AWG numbers and knowing what heat to expect and what heat will do to the wires.

Last edited: Aug 21, 2015
8. ### wayneh Expert

Sep 9, 2010
12,389
3,245
Generally, no. Transformers have losses, so ideally to minimize losses you wind a generator to mate up the generator to the load. If you need 12V, you wind the largest possible wire that will still give you enough turns to hit the 12V voltage target. If instead you wound for 6V or 24V, you could use a transformer to bring either back to 12V, but these would be a bit less efficient than having the ideal windings in the first place.

The problem with wind is that it is very variable, and the power you can get out goes up with the square of the wind speed. Or was it cube? Anyway, the majority of your power is made at the highest rpm the rig can tolerate.

Fortunately, modern technology gives us DC-DC converters that can be very efficient over a wide range of input voltages. Forget transformers and start looking at a modern controller to give you a steadier output.