I need help with my simple wind turbine, circuit provided - Page 1
Forum section: electronics / beginners

author: kafor1
author_rank: Contributor
timestamp: February 19, 2020, 12:23:04 pm
content: im a very beginner at electronics so i went here.i am given by my professor a circuit to work with my project that is converting a wind to electrical energy, so we are basically making a wind turbinemy materials was given below1- germanium diode1- 220 ohm resistor1-3300 electrolytic capacitor1 -6v dc motor1- LEDthe circuit was given at the picture belowim gonna explain what i found and what is my problemso first i tried the circuit without the motor with my 9v battery (first picture) the movement of the circuit is it lights up and then dims then turns off. Then i added the 6v dc motor. since this is wind to electrical energy, i rotated the motor the led wont light up (i even put in in front of my fan) still wont light UP. SO here is my 2 problems FIRST IS that my LED wont light up my assumption is that 6vdc is too low?, The NEXT problem is whenever i tried connecting my 9v battery in the 2nd circuit (with motor provided, I connected it in the black circles) it wont light up my led anymore compared to first circuit. my guess is that maybe the motor absorbs the voltage of the battery? i really need help, what should i do to fix it?

author: tunk
author_rank: Super Contributor
timestamp: February 19, 2020, 01:16:20 pm
content: If you have a multimeter, maybe you could measure some voltages and currents.

author: mikerj
author_rank: Super Contributor
timestamp: February 19, 2020, 01:53:27 pm
content: You need to make sure you rotate/connect the motor so the the output polarity is correct.Spin the motor with your fan etc. and measure the output with a multimeter set to DC volts, the motor terminal connected to the diode must be +ve. If not then reverse the connections.Does the motor spin when you connect up the battery to it (second digram)? If not then the battery may be dead, i.e. it can supply enough current to light an LED but not spin the motor.

author: kafor1
author_rank: Contributor
timestamp: February 19, 2020, 02:00:41 pm
content: Quote from: mikerj on February 19, 2020, 01:53:27 pmi made measurements during the rotation. it displays 0.3v , and yes the motor spins but the led doesnt light up, so does that mean that most of the voltage powered the dc motor and not the led?

author: kafor1
author_rank: Contributor
timestamp: February 19, 2020, 02:09:55 pm
content: i think i found the answer, i think i just needed to buy 9v or 12v dc motor to light up the circuit. and the problem of why my led doesn't light up is the motor gets all the voltage since it has lower resistance also my battery voltage is 8.2 maybe thats the case? correct me if im wrong

author: mikerj
author_rank: Super Contributor
timestamp: February 19, 2020, 02:16:56 pm
content: Quote from: kafor1 on February 19, 2020, 02:00:41 pmThe motor isn't generating sufficient voltage to light your LED. You need at least the forward voltage of the LED (depends on colour and chemistry, usualy 2-3v) plus the forward voltage of the germanium diode (~0.3v).Spin the motor faster!

author: atmfjstc
author_rank: Regular Contributor
timestamp: February 19, 2020, 04:06:00 pm
content: First you need to get your basic electronics terms clarified a bit. There's no such thing as a motor "absorbing" the voltage of a battery. When you connect things in parallel, all branches have the exact same voltage at their ends. What there is more or less of is thecurrentthrough each branch. Lower resistance, more current.While it is true that the motor presents a much lower resistance to the battery than the branch with the LED, this only affects the *ratio* of the currents going through the motor vs the LED. The absolute value of the current through the LED should stay the same as without the motor attached, as long as the battery is still actually providing 9V as before.I think what's really happening here is that the motor is too great a load for your 9V battery. If we assume about 3 ohms of resistance for the motor, the current it draws is around 9V / 3Ohm = 3 Amps. That's quite a respectable amount of current, and likely beyond what those typical consumer-grade blocky 9V batteries are intended to provide. Check the voltage at the ends of the battery while it is powering the motor. It will likely be much less than 9V, probably even down to 4-5V (especially if the battery is nearly depleted) which might not be enough to light the LED.The 0.3 Volts figure seems highly suspect. Is this with the battery connected, or is it when you are spinning the motor manually? If it's the latter, buying a 9V or 12V motor will not help much, since they'll only give you at best twice the efficiency, whereas it is clear you need 20 times more. As mikerj said, try spinning the motor faster, much faster.

author: rstofer
author_rank: Super Contributor
timestamp: February 19, 2020, 04:43:12 pm
content: If possible, turn the motor with an electric drill on the 'fast' range. In one direction the diode should conduct lighting up the LED and in the other direction it will generate reverse polarity and the diode won't conduct and the LED won't light up.There is no condition under which both the battery and the motor should be connected to the circuit simultaneously.The battery test is just to prove that the circuit works for a given polarity (+ to the diode). Note that if you connect the battery backwards, the LED won't light. Try it! That's the purpose of this test! It will work in one polarity and not in the other.The motor needs to spin fast to generate enough voltage to light up the LED. Check spinning in both directions, Again, the LED will only light up when spinning in one direction,At 12V output from the motor, you will need a larger resistor than the 220 Ohm variant. If the voltage drop across the diode is 0.7V and the voltage drop across the LED is 2.0V then you have (12 - 0.7 - 2.0) / 220 => 42 mA of current and this is well beyond the 20 mA usually selected for the common red LED. OTOH, when the LED starts to turn on, the generator voltage may sag substantially.ETA: The diode voltage drop for a Germanium transistor will be around 0.2V versus 0.7V for a silicon diode.

author: sarahMCML
author_rank: Regular Contributor
timestamp: February 19, 2020, 07:02:08 pm
content: Quote from: kafor1 on February 19, 2020, 12:23:04 pmTry this first part once again. Does it still do the same thing? If not, check that:1: The battery is the right way round.2: That it still measures 9 Volts when connected up.3: That both the germanium diode and the LED are the not now open circuit, since I suspect the initial >30+mA of current could have destroyed one or the other, given your description of "it lights up and then dims then turns off."Regards,Sarah

author: Jwillis
author_rank: Super Contributor
timestamp: February 19, 2020, 11:44:36 pm
content: Generators don't make an AC voltage. What you will get is a DC voltage pulse similar to an rectified AC voltage. Get rid of the diode. All it's doing is dropping the voltage. A DC motor used as a Generator will out put a very noisy voltage . Put a inductor in series and the voltage will be clean. The motor/ generator will need a fairly high rotation to get any significant voltage. A typical drill will have around 500 RPM. That's not enough to produce enough voltage and current to light an LED very well. The higher the rotation of your fan the more voltage and current will be produced. Here's a very basic schematic that will give you a fairly decent dc output.