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what are some math problems to make a efficient circuit? Answered

i am trying to make my own wind mill using a stepper motor, capacitor,led, and a resister. My problem is that my dad wants me to make it more efficient meaning making the led go brighter and longer. I curently have 11 capacitors in parallel ,one resister. what are some math equations i can use to find out how to make this more efficient and how do i use the problems? 



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Jack A Lopez
Jack A Lopez

Best Answer 8 years ago

From a mathematical perspective, I would say that your goal is to deliver power to your LED, and that's the easy part. 

The challenging part is to avoid delivering power to other components which dissipate power: like the the resistance in the motor windings, component resistors, and also diodes that do not emit light, if your circuit uses any of non-LED diodes.

Model for the motor/alternator:
You can model your stepper motor/alternator as a sinusoidal AC voltage source, with voltage Vmotor=omega*Phi*sin(omega*t) in series with a resistor Rw. The way this works is the amplitude of the voltage produced by the motor generator increases as omega increases.  omega is the alternator's angular speed. Rw is the resistance of the winding itself.

I am guessing that there is a minimum speed that your alternator has to turn, or else you max(Vmotor)= omega*Phi is less than the voltage needed to forward bias the LED.

Model for diode power dissipation:
You can model your LED as a diode, and any rectifier diodes should be modeled as diodes too.

The power lost through a forward biased diode is approximately the current that flows through it multiplied by its characteristic forward voltage drop.

Pdiode = Idiode*Vdiode.

The power lost through a reverse biased diode is approximately zero, because the current is zero.

Model for resistor power dissipation:
The resistance in the motor/generator winding can be modeled as a resistor, and of course if you have any actual component resistors those are modeled as resistors too.

The power lost through a resistor is: Pres =(Ires^2)*R = (Vres^2)/R

Zero time averaged power loss for capacitors:
Any capacitors in the circuit do not contribute to power losses.  I mean instantaneously the capacitors can be absorbing, or releasing, power, but you can assume that the time-averaged power lost to the capacitors is zero. Probably the only reason you are using capacitors at all, is to sort of smooth out the pulsing AC from the motor.

Pcap = 0


I think the easiest best circuit would just be two anti-parallel LEDs, wired across the output of the motor/alternator.   That way, you can get a pulse of  light on both the positive and negative voltage swings of the alternator.

If you are actually getting enough power out of the alternator that the LEDs are in danger of being overpowered, you know, burned out from the current,  then maybe you have enough power available to justify some kind simple switching regulator (like a joule thief) to protect the LEDs.  Guessing that you would have some sort of rectifier stage (whose diode, or diodes, waste some power) in between the output from the alternator, and the input to the joule thief.


8 years ago

Bigger or more efficient propeller/fan.

Higher output motor - in general the higher voltage required to run the motor the more you will get out.

gearing up the drive system so the motor runs faster.


8 years ago

The first key thing to understand

Lunch <> Free

The energy you have is ENTIRELY limited by the wind, and by the motor you have chosen - more wind = more power.

Capacitors make no difference.

How have you wired your stepper to your LED ? I don't hear you mentioning rectifiers.