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OverUnity Research > Electrical / Electronic Theory and Learning Center > Induction > Current produced to rate of change of magnetic field within an inductor
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Author Topic: Current produced to rate of change of magnetic field within an inductor  (Read 31967 times)
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img  Itsu
Re: Current produced to rate of change of magnetic field within an inductor « Reply #25 on: 2017-07-01, 18:12:01 »

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Hi Brad,

i have no idea, to answer your questions.

I just made some screenshots of something, i think, verpies asked me to do.
No idea why, but as mentioned, he probably will further explain it when he is able to connect to the net.

It could be that the setup i used measuring across the cap/resistor is wrong, but verpies will tell that soon enough.


Itsu
   
img  Allcanadian
Re: Current produced to rate of change of magnetic field within an inductor « Reply #26 on: 2017-07-02, 04:28:00 »
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@tinman
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My forward peak potential voltage available from the coil at the low speed pass,was 6 volt's,but where that voltage is clamped by the voltage across the capacitor + resistor+V drop of the diode.
So that's 2.16v + 1.36V +.2V=3.72V,which is still well under the potential voltage of the coils 6 volts.

Therein lies the problem, as the cap voltage rises it resists the current flow by creating a Cemf opposing the induced or applied Emf. Then the resistor dissipates part of the energy and the diode also creates a 0.2V threshold blocking the current. Now if you switched the inductor like a boost converter then any resistance/impedance would create a proportional voltage rise solving all your problems.

Quote
Maybe you would answer the question everyone else seems to be avoiding ?
If a 220uF cap is charged to say 2.5 volts in one second,what would be needed to charge a 470uF cap to the same voltage in half a second?.

I don't think anyone is avoiding it ... we just don't care.

I think you demonstrated in your scope shots that the fast moving magnet produced more power because the voltage thus current to the cap was higher for fairly obvious reasons. Higher voltages produce less losses and an inductive discharge has almost no losses. So while the slow/fast magnet inductive action may be similar the slower magnet having a lower induced voltage will always lose. The thing to remember is that every circuit component wants to dissipate the energy flowing in that circuit. The trick is to find ways to prevent it from doing this.


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Comprehend and Copy Nature... Viktor Schauberger

“The first principle is that you must not fool yourself and you are the easiest person to fool.”― Richard P. Feynman
   
img  TinMan
Re: Current produced to rate of change of magnetic field within an inductor « Reply #27 on: 2017-07-02, 09:00:42 »

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Quote from: Allcanadian on 2017-07-02, 04:28:00
@tinman
Therein lies the problem, as the cap voltage rises it resists the current flow by creating a Cemf opposing the induced or applied Emf. Then the resistor dissipates part of the energy and the diode also creates a 0.2V threshold blocking the current. Now if you switched the inductor like a boost converter then any resistance/impedance would create a proportional voltage rise solving all your problems.



Higher voltages produce less losses and an inductive discharge has almost no losses. So while the slow/fast magnet inductive action may be similar the slower magnet having a lower induced voltage will always lose. The thing to remember is that every circuit component wants to dissipate the energy flowing in that circuit. The trick is to find ways to prevent it from doing this.

Quote
I don't think anyone is avoiding it ... we just don't care.

Thats nice.

Quote
I think you demonstrated in your scope shots that the fast moving magnet produced more power because the voltage thus current to the cap was higher for fairly obvious reasons.


Actually,in the fast moving magnet test,the voltage across the coil was lower,as the cap was larger,and clamped the coil voltage + the voltage across the CVR and diode. If i remove the CVR,then the end cap voltage would be slightly higher,and the current flowing into that cap would also be higher  O0


Brad

So you agree that the faster moving magnet pass the coil,would produce more power?.


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img  Allcanadian
Re: Current produced to rate of change of magnetic field within an inductor « Reply #28 on: 2017-07-02, 15:14:04 »
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@Tinman
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Actually,in the fast moving magnet test,the voltage across the coil was lower,as the cap was larger,and clamped the coil voltage + the voltage across the CVR and diode. If i remove the CVR,then the end cap voltage would be slightly higher,and the current flowing into that cap would also be higher


I was simply using the data you provided in post #21. You had shown a circuit diagram with a 220uF cap then two scope shots where the faster magnet had a higher voltage/current and the slower magnet had a lower voltage/current. It is what it is in my opinion.

Quote
So you agree that the faster moving magnet pass the coil,would produce more power?.

Based on the data you provided yes I would agree. Logically if the slow moving magnet moved at 1 meter/minute almost no voltage/current would be induced. If in the same setup a fast moving magnet moved at 1 meter/second a fair amount of voltage/current would be induced. As well with a MPPT (Maximum Power Point Tracking system) you could determine the optimal speed which produces the most Power (Volts x Amps).


---------------------------
Comprehend and Copy Nature... Viktor Schauberger

“The first principle is that you must not fool yourself and you are the easiest person to fool.”― Richard P. Feynman
   
img  TinMan
Re: Current produced to rate of change of magnetic field within an inductor « Reply #29 on: 2017-07-02, 15:20:41 »

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Quote from: Allcanadian on 2017-07-02, 15:14:04
@Tinman





Quote
I was simply using the data you provided in post #21. You had shown a circuit diagram with a 220uF cap then two scope shots where the faster magnet had a higher voltage/current and the slower magnet had a lower voltage/current. It is what it is in my opinion.

Ah,ok.
I thought you were referring to the video.


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img  Kator01
Re: Current produced to rate of change of magnetic field within an inductor « Reply #30 on: 2017-07-03, 00:25:48 »
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Itsu,

can you please use the inverted mode for one of the load-processes so that one can see better the difference ?

thank you

Mike
   
img  Itsu
Re: Current produced to rate of change of magnetic field within an inductor « Reply #31 on: 2017-07-03, 09:59:22 »

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Hi Kator01

do you mean inverting the ch1 (yellow) trace?   I can do that, but it will flip over the yellow signal from top to bottom, not from left to right,  i think.

I will check on that later today,  thanks.

Itsu
   
img  Itsu
Re: Current produced to rate of change of magnetic field within an inductor « Reply #32 on: 2017-07-03, 16:06:34 »

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Ok, i now used the same components as TM did, meaning a 220uF cap, a 1n5819 schottky diode and as load the same 100 Ohm resistor.

When i invert ch1 (yellow), it flips from top to bottom, so not right to left, so therefor i have put up both signals at the same time.

Yellow (across the cap/resistor) and blue (across the input) are one pair (sawtooth with peak right), and
white R1 (across the cap/resistor) and white R2 (across the input) are another pair (sawtooth with peak left).

FG was set at 5Hz, 4Vpp (ac)

We see a clear difference in output signal (yellow / R1), but there also is a difference in input signal (blue / R2), so a 1:1 compare is not easy to make.


Itsu   
   
img  Allcanadian
Re: Current produced to rate of change of magnetic field within an inductor « Reply #33 on: 2017-07-03, 17:51:52 »
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@itsu
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When i invert ch1 (yellow), it flips from top to bottom, so not right to left, so therefor i have put up both signals at the same time.

Yellow (across the cap/resistor) and blue (across the input) are one pair (sawtooth with peak right), and
white R1 (across the cap/resistor) and white R2 (across the input) are another pair (sawtooth with peak left).

I use an Allegro hall effect current sensor to isolate the current sensing channel on my scope. Without isolation the common ground between the probe inputs becomes a real pain in the rear.


---------------------------
Comprehend and Copy Nature... Viktor Schauberger

“The first principle is that you must not fool yourself and you are the easiest person to fool.”― Richard P. Feynman
   
img  TinMan
Re: Current produced to rate of change of magnetic field within an inductor « Reply #34 on: 2017-07-04, 08:54:09 »

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Quote from: Itsu on 2017-07-03, 16:06:34
Ok, i now used the same components as TM did, meaning a 220uF cap, a 1n5819 schottky diode and as load the same 100 Ohm resistor.

When i invert ch1 (yellow), it flips from top to bottom, so not right to left, so therefor i have put up both signals at the same time.

Yellow (across the cap/resistor) and blue (across the input) are one pair (sawtooth with peak right), and
white R1 (across the cap/resistor) and white R2 (across the input) are another pair (sawtooth with peak left).

FG was set at 5Hz, 4Vpp (ac)

We see a clear difference in output signal (yellow / R1), but there also is a difference in input signal (blue / R2), so a 1:1 compare is not easy to make.


Itsu

Itsu

The 100 ohm resistor is not a load,it is the CVR.

One channel of your scope should be across the capacitor,and the other channel across the 100 ohm resistor.
You could also put a third channel across the resistor/diode series,so as to see the V drop across the diode.



Brad


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Never let your schooling get in the way of your education.
   
img  Itsu
Re: Current produced to rate of change of magnetic field within an inductor « Reply #35 on: 2017-07-04, 09:21:53 »

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Brad,

i understand,  in your case.

But it is not what i understand from verpies what he wanted me to measure/show.

In due time verpies will tell.

Itsu
   
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