Author Topic: Test coil output  (Read 3032 times)

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Offline artv

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Test coil output
« on: February 20, 2012, 07:53:45 PM »
Hi All ,...
Been busy reading,testing and of course.. working..
A test coil ,of a given # of turns, a certian guage of wire, in a specific amount of magnetic flux,.
Will only give a max amount of voltage out ,or power out???
Or does the power depend on the amount of resistance you put in the circut to get to the load??

If you interupt the phase feeds to the bridge ,  you can catch more power....charge caps alot faster ..
And since they pretty much dis-charge instantly..........the faster they charge,... the more amps you can use??

Also I think steel in the design needs to be limited to the strength of the magnets.....too much steel is bad.....
The bare minimum seems to work best.........so far.

I've tested where , a coil or a phase will only put so much into a cap bank..it reaches it limit,...by tapping the phase lead you can triple that #.......

Just wondering what the rules are for coil output............thanks ......artv

Offline rossw

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Re: Test coil outout
« Reply #1 on: February 20, 2012, 09:06:25 PM »
A test coil ,of a given # of turns, a certian guage of wire, in a specific amount of magnetic flux,.
Will only give a max amount of voltage out ,or power out???
Or does the power depend on the amount of resistance you put in the circut to get to the load??

The voltage induced will be fixed. The amount of resistance you put on the output will determine how much load it has, which will reduce the output voltage (depending largely on the resistance of the coil).

The POWER you get out will be a function of the voltage, and the current delivered, at each instant. So a small resistor will only draw a small current and thus a small power.


Quote
If you interupt the phase feeds to the bridge ,  you can catch more power....charge caps alot faster ..

I can't see how this would be. Disconnecting the wire to the bridge will result in ZERO current flowing. Charge in a capacitor is a function of amps/volts/time. If the current drops, and the voltage drops (because you disconnected the lead!), the cap simply CAN'T get even as much power, much less MORE.


Quote
And since they pretty much dis-charge instantly..........the faster they charge,... the more amps you can use??

Also I think steel in the design needs to be limited to the strength of the magnets.....too much steel is bad.....
The bare minimum seems to work best.........so far.

I've tested where , a coil or a phase will only put so much into a cap bank..it reaches it limit,...by tapping the phase lead you can triple that #.......

The only thing  I can think that might be happening is that interrupting the circuit, you're somehow creating a collapsing magnetic field... but for the life of me I can't see how a "properly connected machine" (coils, rectifiers, caps) couldn't outperform another similar machine with the wires to the bridge rapidly connected and disconnected.

Got a diagram?

Offline artv

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Re: Test coil output
« Reply #2 on: February 23, 2012, 04:28:28 AM »
Thanks Ross..."the voltage induced will be fixed"..
This confirms what I've seen when testing a normally connected machine...

A single phase ,a diode bridge, 2 headlights, and a blocking diode feeding the caps..
At full rpm of my drill,..  the lights shine and the caps charge to 13volts..

I don't disconnect the phase leads, I run parallel connections to the secondary coils and interrupt or open and close these connections.
The caps reach ~35 volts instantly then continue to climb ,the caps are 56wv,  I've  pushed them to 75 volts and am sure they would continue to rise ,but I'm scared of them exploding...

The voltage across the lights  fluxuates in relation to the interrupting of the phase....
I can discharge the caps  into an electro-magnet as fast as my hands will let me and get a continued magnetic pulse out..

Currently building a new rotor with less steel ,to see if I can use the pulses to drive the rotor....artv



Offline WooferHound

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Re: Test coil output
« Reply #3 on: February 23, 2012, 02:36:28 PM »
My question is ,  Why ?
Can't you just make a regular charger like everyone else.
Interrupting the current to charge a capacitor so you can discharge it into a battery seems like a long way to go to get there.
There are better ways, I left you 4 links  to pulse chargers complete with schematics to make them.
Basically you hook the current straight to the capacitor and never disconnect it. The capacitor charges and when it reaches a set voltage it is connected to the battery for a very short time through semiconductors, It is discharged to battery voltage and disconnected again, repeat as necessary. Usually this happens at 60-400 hertz but can be lower or higher depending on design and battery.
Did you ever look at those links  ? ? ?
The answers are there . . . Really . . .
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Offline rossw

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Re: Test coil output
« Reply #4 on: February 23, 2012, 03:45:48 PM »
I don't disconnect the phase leads, I run parallel connections to the secondary coils and interrupt or open and close these connections.

Now I'm more confused.

Are the parallel coils "in" the machine?
If so, you are probably in effect, making a transformer.
You're opening and closing the second coil, increasing the flux cutting the primary coil. *I SUSPECT*.
Without a clear diagram of what's going on, I'm only guessing - are I suspect are others here.

Put simply, you can't get more power out of a machine by opening and closing the circuit.

You MAY be able to get more power if you're doing something that is changing the flux density around your main coil,
but in that case - I'd think that simply connecting the second coil, in the right phase relationship, would achieve even better results - unless it's in the wrong physical position and can't contribute - in which case, there are other (electrical) means to achieve the result.

Offline artv

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Re: Test coil output
« Reply #5 on: April 15, 2012, 07:07:52 PM »
It's been a while,.....if you short a single coil, the voltage spikes off the scale ,.....but only for an instance
you have to break the connection instantly,....you can charge caps higher and faster...

My question is what voltage and how many micro-farads should I discharge into a 12 volt car battery...
to make it charge safely??
crazy busy with work.....thanks artv

Offline A of J

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Re: Test coil output
« Reply #6 on: April 15, 2012, 07:17:41 PM »
   1234

Offline WooferHound

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Offline ghurd

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Re: Test coil output
« Reply #8 on: April 15, 2012, 07:49:28 PM »
if you short a single coil, the voltage spikes off the scale

No.

If you short a coil, the voltage drops to 0.

If you open a coil, the voltage goes off the scale, "but only for an instance".
There is NO power in that microscopic pulse.

Amps x Volts = Watts, right?

When you 'open' the circuit, by the definition of 'open circuit', there are NO amps. 0A. Zero. Zip. Nadda.

0A x <infinite> Volts = 0W.

There is no power there to be harvested.  Period.

Lets say there is 50A at 12V for 5ms.
50A x 12V = 600W
600W for 0.5ms = no watt-hours of power worth typing that many characters.
Maybe 0.33 milliwatt hours (mental math), which is nothing.

Not sure how many ways we can explain it to you...

You are trying to catch smoke in a fishing net.
G-

Offline artv

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Re: Test coil output
« Reply #9 on: April 16, 2012, 05:56:08 PM »
Ross , I'm not so sure about the secondary coils now, but they are wound around the tops and bottoms of the coils so as to be cutting the flux the same as the sides.
Woofer , I do have a regular charger and I do read the links you provide, that stuff is beyond me.
Aof J ,Not sure what 1234 means.
G- , Thanks that makes more sense about the removing the short to get the spike.
But the caps will only charge to a certian level under normal conditions.
By switching a shorting connection 100's of times per revolution, the caps will charge beyond that certian level and faster.
What I want to do is dis-charge the caps into the battery more often , in a given time frame.
But at what levels??
Since the caps charge faster and can be dumped more often , this should make for a better generator?
I don't know maybe it doesn't work this way...?
Thanks to All for trying.......artv

Offline WooferHound

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Re: Test coil output
« Reply #10 on: April 16, 2012, 07:59:22 PM »
Why ?
Would your method be better than what everybody else is charging with ?
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Offline MadScientist267

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Re: Test coil output
« Reply #11 on: April 16, 2012, 09:29:40 PM »
Quote from: artv
By switching a shorting connection 100's of times per revolution, the caps will charge beyond that certian level and faster.

I've looked this one over fairly thoroughly, and have come to one conclusion.

Effectively, what you have is called a "magneto", just like the one in your handy-dandy lawnmower that provides the juice for the spark plug to fire.

The voltages are a little different, but the principle you're seeing is almost certainly (from what I can gather) exactly the same.

In a magneto, a magnet comes by, polarizes an armature with two coils on it in one direction, while holding the "primary" coil shorted. As the magnet passes, the coils are arranged in such a way that the magnetic polarity in the coils suddenly flip *hard* the other way (Disclaimer: Depending on the exact design, it may or may not actually flip just prior to the next part). At that moment, a switch (called the "points") holding the primary shorted opens up, and the magnetic field collapses violently, cutting across the secondary coil and thus firing the spark plug.

So, what you've "discovered" is an age old trick with magnetic fields used for creating very high voltages with a very simple mechanism.

But as a generator for charging batteries, the technology is just about useless, unless you have something like a 15kV 2mAh battery laying around to charge with it.

Probably not the case. But I do however admire your perseverance... :)

Steve
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Offline artv

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Re: Test coil output
« Reply #12 on: April 19, 2012, 03:30:13 AM »
Woofer, with a regular pmg at a given rpm, connected via a bridge to the battery, your putting x-amount of charge into the battery. By putting caps in parallel with blocking diodes and charging to 14 volts ,then dumping into the battery through at firing switch, you get the regular x-amount of charge plus the charge from the caps.
Or am I completely wrong in thinking this way?

Steve, the high votage spikes charge the caps faster. I have 12000 uf at 16 volt caps. If I charge that cap to 14 volts and dump it into a battery , that puts charge into the battery, although very small....right??
Would it be better to increase the voltage by series caps or parallel to increase the capacitance, or maybe both since I have 100 of these caps to play with??
Thanks for your time.....artv

Offline WooferHound

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Re: Test coil output
« Reply #13 on: April 19, 2012, 07:19:31 AM »
Think of it this way . . .
A battery is a huge capacitor
you are trying to charge a capacitor with capacitors
Watts are watts, If you are charging with 150 watts, then charging through capacitors will not increase those watts, will probably reduce the watts due to the inefficiencies of the capacitors and the switching circuits.
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Offline off the wall

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Re: Test coil output
« Reply #14 on: April 11, 2015, 04:35:45 PM »
I found this thread looking for information about exploding batteries . . .

But I think I know what's going on here - charging from pulses from discharging capacitors into a battery is said to be one way of desulphating batteries.

I wonder, however, if there may be more gas produced in doing this.

Best wishes

OTW