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Faraday's Induction and capacitance circuits

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t.wade
#1
Oct15-09, 06:31 AM
P: 10
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 Images
File Type: bmp generator.bmp (150.0 KB, 11 views)
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Andrew Mason
#2
Oct15-09, 09:17 AM
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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
t.wade
#3
Oct22-09, 06:19 AM
P: 10
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

t.wade
#4
Nov5-09, 09:48 AM
P: 10
Faraday's Induction and capacitance circuits

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/M...m_high_GM00010 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
Andrew Mason
#5
Nov5-09, 11:40 AM
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Quote Quote by t.wade View Post
Hi again,

So I managed to source a huge capacitor, 10F, but couldn't get anything to even show signs of life,
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
t.wade
#6
Nov6-09, 08:50 AM
P: 10
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
Attached Thumbnails
coil.jpg   front.jpg  
Andrew Mason
#7
Nov6-09, 10:53 AM
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Quote Quote by t.wade View Post
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
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
t.wade
#8
Nov6-09, 11:01 AM
P: 10
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
Andrew Mason
#9
Nov6-09, 11:06 AM
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Quote Quote by t.wade View Post
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
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
t.wade
#10
Nov10-09, 06:46 AM
P: 10
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
Andrew Mason
#11
Nov10-09, 09:27 AM
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Quote Quote by t.wade View Post
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
It depends on how fast you can turn it. Why not measure the output voltage with a voltmeter as you turn it.

AM
t.wade
#12
Nov10-09, 10:37 AM
P: 10
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?
Andrew Mason
#13
Nov10-09, 11:48 AM
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Quote Quote by t.wade View Post
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?
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
t.wade
#14
Jan8-10, 06:07 AM
P: 10
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
t.wade
#15
Jan8-10, 06:08 AM
P: 10
silly me, here's the circuit diagram...
Attached Thumbnails
circuit 1.jpg  
Andrew Mason
#16
Jan8-10, 11:55 AM
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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
t.wade
#17
Jan11-10, 05:00 AM
P: 10
Quote Quote by Andrew Mason View Post
Each half turn the capacitor charges and in the next half turn it discharges so you can never charge the capacitor.

AM
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?

Thanks for your continued help
Andrew Mason
#18
Jan11-10, 10:29 AM
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Quote Quote by t.wade View Post
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?

Thanks for your continued help
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


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