# Faraday's Induction and capacitance circuits

1. Oct 15, 2009

1. The problem statement, all variables and given/known data
Have been given the task of creating a circuit powered by a rotating magnet in a copper coil that will charge a capacitor long enough to light an LED for 600 seconds after the rotating magnet has stopped.

In all the magnet charge up torches I've opened, there are two 3V Lithium Cells (CR 2032) between the generator and the circuit. These cells aren't rechargable. The current from the generator is AC.

I know this shouldn't be too difficult, but can't figure out how to approach it with so many unknown variables. Just looking for any help possible for value of capacitor or resistor.

Voltage for LED = 1.5V
Power for LED = 4mW
Current for LED = 2.7mA
t for capacitance discharge = 600s

2. Relevant equations
EMF = N(dФB)/dt
P = IV
C = Q/V
I = Q/t
F = (A.s)/V

3. The attempt at a solution

I know that in Faraday's Law of Induction, the magnetic flux in Webers is defined by the equation. But what does the change in time represent? What time is it referring too in the process of passing the magnet through a single loop?

I've entered the values into the capacitance equation and come out with 1.08F. However all capacitors I've found are measured in pF.

Any help at all will be much appreciated.

#### Attached Files:

• ###### generator.bmp
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2. Oct 15, 2009

### Andrew Mason

First of all, you are going to need something to rectify the current from the generator. Then you have to calculate the energy needed by the LED.

I think it is just a matter of determining the size of the coil, magnet and capacitor needed to generate and store the required energy.

You will have to measure the magnetic field of the magnet and then determine the frequency that it will turn. That gives you enough to work out dB/dt (the time rate of change of the magnetic field enclosed by the coils).

I am not sure how you would regulate the voltage on the capacitor's discharge in order to maintain the correct voltage for the LED. Is there a voltage range for the LED?

AM

3. Oct 22, 2009

Hi Andrew,

Thanks for your response. Just getting all the bits together to try and figure out some more of these variables, will come back when I've got a little more to contribute. Thanks again

Tom

4. Nov 5, 2009

Hi again,

So I managed to source a huge capacitor, 10F, but couldn't get anything to even show signs of life, so took to testing the generator straight onto a 2.5V LED but nothing! I tried it on three different LEDs, I doubled up the magnets, tried a different wire (in case the copper isn't insulated) and even built a container to stabilise the voltage - before, the magnets were on a rod being turned back and forth, the motion used to make fire with a stick, now it is spinning constantly.

I'm using these powerful magnets: http://www.modelshop.co.uk/product/Magnets_neodymium_15mm_diameter_8mm_high_GM00010 [Broken] with the flat surfaces out, not down. The wire is wrapped perpendicular to rotating motion.

Any ideas on what I'm doing so wrong?

If the description isn't clear, I can attach photos.

Thanks for any help,

Tom

Last edited by a moderator: May 4, 2017
5. Nov 5, 2009

### Andrew Mason

You have to test everything to make sure it works.

Test the capacitor. Put a 3 volt DC source (eg. 2 1.5 volt batteries) across the capacitor (be careful about polarities or you may wreck the capacitor). Let it sit there for a few minutes. Then remove the battery and put a voltmeter across the capacitor leads. Do you get 2.5 v?

If the capacitor stores charge, then check your magneto generator. Put a voltmeter on it while you are cranking it.

etc.

AM

6. Nov 6, 2009

The problem is with the generator, but for the life of me can't figure out what's wrong. I've tested all the variables as far as I can tell, read everything I could find online, watched all the videos of people building their own generators. What is so obviously wrong?

Thanks for any help, please see photos attached for generator set up.

Tom

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7. Nov 6, 2009

### Andrew Mason

See my first comment. You need to generate DC. Unless you have a diode or a commutator in that generator, you will generate AC. What you have is an alternator. This will charge the capacitor in the first half turn and then discharge it (ie apply opposite voltage) on the other half turn. The simplest solution would be to put a diode in series with the output of the generator.

AM

8. Nov 6, 2009

Thanks Andrew,

In an isolated set up, ignoring the capacitor for a sec, with just the alternator and an LED, would the LED still not flicker with the AC current?

Tom

9. Nov 6, 2009

### Andrew Mason

Not with a 10 F capacitor! A half turn on the alternator will not give the capacitor enough charge to reach the minimum voltage to run the LED.

If you take the capacitor out, and crank hard it may or may not light. You would have to determine how much voltage you are generating in your coil. You may have to increase the number of coils to get the output voltage high enough.

AM

10. Nov 10, 2009

I'm just using it without the capacitor, and it's definitely falling on the 'may not' side.

Here's just a quick speculation - if the LED won't light when attached directly to the generator, then wouldn't cranking up the 10F capacitor to the necessary charge to keep the LED alight for 600 seconds be very difficult by hand?

If the copped wire wasn't insulated, would this effectively reduce the no. of coils to 1?

Sorry for the torrent of questions, thanks again,

Tom

11. Nov 10, 2009

### Andrew Mason

It depends on how fast you can turn it. Why not measure the output voltage with a voltmeter as you turn it.

AM

12. Nov 10, 2009

I've had a go and it doesn't pick up more than a couple of mV. But attaching it to professionally made magnetic charge up torches doesn't give much more either. Could this be because it's only a cheap multimeter and doesn't instantly adjust to a constantly changing AC current?

13. Nov 10, 2009

### Andrew Mason

I doubt that it is a problem with the meter. You will need to add more coils in order to step up the voltage. The voltage is directly proportional to the number of turns in the surrounding coils.

AM

14. Jan 8, 2010

Hi all,

Managed to get an AC generator working to light the LED, but can't get it to work within the context of the circuit. Is it the size of the capacitor that's causing the problem? The generator is just hand driven. Or am I making some blindingly obvious mistake?

Thanks for any help in advance,

Tom

15. Jan 8, 2010

silly me, here's the circuit diagram...

#### Attached Files:

• ###### circuit 1.jpg
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16. Jan 8, 2010

### Andrew Mason

What is in the top left corner of this circuit? It makes no sense to me. It looks like an AC source that is always shorted out!

Where is your diode in this? You are generating AC in your generator (when it is connected properly, which it is not here). Each half turn the capacitor charges and in the next half turn it discharges so you can never charge the capacitor. You need a diode to rectify the AC so that current flows only in one direction.

AM

17. Jan 11, 2010

I thought capacitors could convert AC to DC? Or is that only specific capacitor types? If the generator is connected like a battery would be, then the discharging current has to go through the entire copper spool; will this not hugely reduce the current?

18. Jan 11, 2010

### Andrew Mason

Capacitors do not rectify current. You need a diode. If you do not have a diode your capacitor will never charge.

Also, your circuit is wrong. In your schematic, the AC generator is shorted out.

It would be better to disconnect the AC generator when the capacitor is fully charged so the capacitor discharges only through the lamp. I don't think it will make a huge difference though.

AM

19. Jan 18, 2010

### Relay

I can't tell from your pictures but I believe your biggest problem is with the magnets. It looks like you have two round magnets side by side turning within your coil. This is fine if the magnetic poles are perpendicular to the coil. Looking at your pictures, it looks like the magnetic poles are parallel to your coil. If I'm right you will not get enough voltage to make it through the bridge rectifier to charge the capacitor. Because you are using magnets, the magnetic field doesn't collapse and expand as it does in transformers. This makes pole orientation a critical design consideration. As stated before test this circuit in steps. Connect your generator to a voltmeter and measure your voltage. Then change your magnet orientation so the flats of the magnets are perpendicular to your coils. Measure the voltage out again. If you see a huge difference, you found your problem.
Once you solve this try to center tap your coil. Make the tap your negative terminal and the ends will have diodes with their anodes connected together to give you the positive terminal. This way you get full wave rectification with only .7 voltage drop. Good luck.