How many amps can I generate with a dc motor?

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SUMMARY

The discussion centers on using a 12V DC Brushless permanent magnet motor as a wind-powered generator. The motor is rated at 12V and 1.20 amps, but due to inefficiencies, the actual output is significantly lower. The participants conclude that the motor operates at approximately 50% efficiency, yielding an output of about 6V and 0.6 amps under ideal conditions. Additionally, they emphasize the importance of managing load resistance and preventing battery discharge through the use of diodes and switching circuits to optimize energy efficiency.

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  • Understanding of DC Brushless permanent magnet motors
  • Knowledge of electrical efficiency and power calculations
  • Familiarity with diode functions in electrical circuits
  • Basic principles of load resistance and its impact on voltage and current
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This discussion is beneficial for electrical engineers, renewable energy enthusiasts, and hobbyists interested in optimizing wind-powered generation systems using DC motors.

Trimacon
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I want to use a 12V DC Brushless permanent magnet motor as a wind powered electrical generator for a project. The motor is rated DC 12V and 1.20 amps. Once I reach an output of 12V does that means that I also have 1.2 Amps?

Thank you.
 
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Tried to find answer to this years ago and was never able to, so made following assumptions.
Small Motors are very inefficient, say only 50%.
So effective power to motor is only 6V and 1.2 amp.
Motor operating as generator is only 50% efficient, so output from motor is only 0.6 amp.
So your output is only 6V and 0.6 amp.(assuming motor/generator is turnin at rated motor speed.
Test on one motor roughly agreed with preceding.
Would really like for someone to give a scientific explanation of the actual output.
 
The faster you spin the motor, the more volts you'll get - somewhat!... it depends on the circuit you attach it to! A low resistance will increase the current and decrease the volts.

Think more about the power you are putting into rotate the rotor of this motor. What happens to that power when you attach a load?

You'll end up with [pure theory, discounting losses, &c.] Power-in=Volts x Amps, and Load resistance = V/I. Two variables (V and I), two knowns (Power and load). [This is very idealised.]

So you can see that volts and amps are dependent on power and load.

One thing to remember about using a DC generator to, e.g., charge a battery is that you have to include some means to prevent the battery from discharging back into the generator when it is no longer 'generating'. The output volts have to stay above the volts of the battery (and not too far above either, else the battery will overcharge).

Preventing the discharge can be done with just a simple diode, but if you are, for example, charging a 6V battery with a diode with a 0.6V forward voltage drop, you are losing 10% of your input power to the diode. Similarly, if you end up generating 10V because your generator is running fast, then if you are running a simple linear circuit you'd need to dump 40% of the power.

Clever switching circuits are needed to efficiently manage the power coming out of, particularly, wind generators for reasons of their (obvious) speed variability...and the emphasis is, indeed, usually on 'efficiency', because 'energy optimisation' is the name-of-the-game if you're already bothered to build a wind generator!

I would imagine that what you'd want to do for an efficient DC wind generator is to find one that generates high volts, thus low current and the diode wouldn't suck up such a big percentage of losses, then feed capacitors from which you use a switching scheme to draw the power off them. As the generator drops below the level where it is feeding volts onto the capacitors, it'd 'unload' a little and the output volts would go up again. This should therefore reach some balance where it runs pretty much at a rate that generates the volts enough to cross the diode to the caps, and current to supply the power demand on the other side of the switching circuit. This is just a guess, I am sure there are heaps and heaps of pages on DIY wind power management on the www.
 
Last edited:
Trimacon said:
I want to use a 12V DC Brushless permanent magnet motor as a wind powered electrical generator for a project. The motor is rated DC 12V and 1.20 amps. Once I reach an output of 12V does that means that I also have 1.2 Amps?

Thank you.

A 'brushless' DC motor still needs a commutator function - it's just done electronically rather than with a commutator and brushes. I would suggest that, (unless someone can tell us they have used one successfully! haha) this could be a problem, as the switching circuit circuit would need to be measuring voltages and switching appropriately.

If you have this sort of motor, though, it wouldn't be too difficult to remove the switching circuit and use it as a multiphase Alternator. Follow this with suitable rectification and you're back with DC and no discharge problem. Look at google images of a brushless DC motor and an alternator. You will see what I mean.
 
I did that and removed the circuits. I will add a full bridge rectifier to have dc output, but my question is still how many amps would come out if the output voltage is 12V.
 
It will depend on the rotation rate and the field - and the resistance of the coils. I suspect the output power would be similar to the output power when used as a motor?
 

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