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Author Topic: Where the OVERUNITY using INDUCTION COILS comes from  (Read 1145 times)


Free Energy | searching for free energy and discussing free energy


Offline Zephir

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Re: Where the OVERUNITY using INDUCTION COILS comes from
« Reply #1 on: June 23, 2017, 05:31:36 PM »
The answer may depend on whether this coil has ferromagnetic core or not.


Offline antijon

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Re: Where the OVERUNITY using INDUCTION COILS comes from
« Reply #2 on: June 25, 2017, 08:35:24 PM »
The answer may depend on whether this coil has ferromagnetic core or not.

What do you mean, Zephir? To avoid hysteresis?

Offline lancaIV

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Re: Where the OVERUNITY using INDUCTION COILS comes from
« Reply #3 on: June 26, 2017, 01:56:43 PM »
Watt(average) and Wattpeak and Wattless power ( no-load and low/..../ half/..../full load conditioning):

about peak/average savings by pulse power circuit arrangement :
https://worldwide.espacenet.com/publicationDetails/description?CC=US&NR=5130608A&KC=A&FT=D&ND=3&date=19920714&DB=EPODOC&locale=en_EP

https://worldwide.espacenet.com/publicationDetails/originalDocument?FT=D&date=20130606&DB=EPODOC&locale=en_EP&CC=WO&NR=2012065719A3&KC=A3&ND=4#

about Volt/Ampere/Frequenz-relationship :
https://worldwide.espacenet.com/publicationDetails/biblio?DB=EPODOC&II=1&ND=3&adjacent=true&locale=en_EP&FT=D&date=20121011&CC=US&NR=2012256422A1&KC=A1#

    [009   At 1000 rpm2] , Vmax=100 volts, so PWM duty-cycle (Ton)/(Ton+Toff) is essentially zero. Therefore, losses=Imax<2>(RL+Rs)+2 VfImax=(10 amp)<2>(0.25 ohm)+(2 volt)(10 amp), amounting to 45 watts loss. Output power=(Imax)*(Vmax)=(Imax)*(VDC)=(10 amp)(100 volts)=1000 watts. So, generator efficiency at maximum speed and maximum power is about 95% for this example of generator and integrated electronics parameters.

 [0093] At 500 rpm, Imax=(10 amp)/(4)=2.5 amps; and Vmax=(100 volts)*(0.5)=50 volts. So PWM duty-cycle=1⁄2. Average pulse power generated=(Imax)*(Vmax)=(Imax)*(VDC)/2=(2.5 amp)(50 volt)=125 watts. Losses to maintain inductor current=Imax<2>(RL+Rs+Ron)=(2.5 amp)<2>(0.26 ohm)=1.6 watts. Fly-back diode losses=2 Vf*Imax/2=(0.6 v)(2.5 amp)=1.5 watts. So total losses=3.1 watts. Therefore, mid-speed generator efficiency is about 97%.

 [0094] At 100 rpm, Imax=(10 amp)/(100)=0.1 amp; and Vmax=(100 volts)/(10)=10-volts. So PWM duty-cycle= 9/10. Average pulse power generated=(Imax)*(Vmax)=(Imax)*(VDC)/10=(0.1 amp)(10 v)=1 watt. Losses to maintain stator and inductor current=Imax<2>(RL+Rs+2*Rdson)=(0.1 amp)<2>(0.27 ohm)=0.0027-watt. Fly-back diode losses=(2*Vf)*(Imax)/10=(0.6 v)(0.1 a)/5=0.012 watts. So total losses=0.015-watt. Thus, generator efficiency at low speed is about 98%.

                                                              PWM duty-cycle

and the Volt-bank and Ampere-bank connection :
https://worldwide.espacenet.com/publicationDetails/biblio?DB=EPODOC&II=5&ND=3&adjacent=true&locale=en_EP&FT=D&date=20160812&CC=DE&NR=202016004514U1&KC=U1#

translated :
http://translationportal.epo.org/emtp/translate/?ACTION=description-retrieval&COUNTRY=DE&ENGINE=google&FORMAT=docdb&KIND=U1&LOCALE=en_EP&NUMBER=202016004514&OPS=ops.epo.org/3.2&SRCLANG=de&TRGLANG=en


Offline lancaIV

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Re: Where the OVERUNITY using INDUCTION COILS comes from
« Reply #4 on: June 28, 2017, 10:26:58 AM »
Watt(average) and Wattpeak and Wattless power ( no-load and low/..../ half/..../full load conditioning):

about peak/average savings by pulse power circuit arrangement :
https://worldwide.espacenet.com/publicationDetails/description?CC=US&NR=5130608A&KC=A&FT=D&ND=3&date=19920714&DB=EPODOC&locale=en_EP

https://worldwide.espacenet.com/publicationDetails/originalDocument?FT=D&date=20130606&DB=EPODOC&locale=en_EP&CC=WO&NR=2012065719A3&KC=A3&ND=4#

about Volt/Ampere/Frequenz-relationship :
https://worldwide.espacenet.com/publicationDetails/biblio?DB=EPODOC&II=1&ND=3&adjacent=true&locale=en_EP&FT=D&date=20121011&CC=US&NR=2012256422A1&KC=A1#

    [009   At 1000 rpm2] , Vmax=100 volts, so PWM duty-cycle (Ton)/(Ton+Toff) is essentially zero. Therefore, losses=Imax<2>(RL+Rs)+2 VfImax=(10 amp)<2>(0.25 ohm)+(2 volt)(10 amp), amounting to 45 watts loss. Output power=(Imax)*(Vmax)=(Imax)*(VDC)=(10 amp)(100 volts)=1000 watts. So, generator efficiency at maximum speed and maximum power is about 95% for this example of generator and integrated electronics parameters.

 [0093] At 500 rpm, Imax=(10 amp)/(4)=2.5 amps; and Vmax=(100 volts)*(0.5)=50 volts. So PWM duty-cycle=1⁄2. Average pulse power generated=(Imax)*(Vmax)=(Imax)*(VDC)/2=(2.5 amp)(50 volt)=125 watts. Losses to maintain inductor current=Imax<2>(RL+Rs+Ron)=(2.5 amp)<2>(0.26 ohm)=1.6 watts. Fly-back diode losses=2 Vf*Imax/2=(0.6 v)(2.5 amp)=1.5 watts. So total losses=3.1 watts. Therefore, mid-speed generator efficiency is about 97%.

 [0094] At 100 rpm, Imax=(10 amp)/(100)=0.1 amp; and Vmax=(100 volts)/(10)=10-volts. So PWM duty-cycle= 9/10. Average pulse power generated=(Imax)*(Vmax)=(Imax)*(VDC)/10=(0.1 amp)(10 v)=1 watt. Losses to maintain stator and inductor current=Imax<2>(RL+Rs+2*Rdson)=(0.1 amp)<2>(0.27 ohm)=0.0027-watt. Fly-back diode losses=(2*Vf)*(Imax)/10=(0.6 v)(0.1 a)/5=0.012 watts. So total losses=0.015-watt. Thus, generator efficiency at low speed is about 98%.

                                                              PWM duty-cycle

and the Volt-bank and Ampere-bank connection :
https://worldwide.espacenet.com/publicationDetails/biblio?DB=EPODOC&II=5&ND=3&adjacent=true&locale=en_EP&FT=D&date=20160812&CC=DE&NR=202016004514U1&KC=U1#

translated :
http://translationportal.epo.org/emtp/translate/?ACTION=description-retrieval&COUNTRY=DE&ENGINE=google&FORMAT=docdb&KIND=U1&LOCALE=en_EP&NUMBER=202016004514&OPS=ops.epo.org/3.2&SRCLANG=de&TRGLANG=en



https://worldwide.espacenet.com/publicationDetails/biblio?DB=EPODOC&II=0&ND=3&adjacent=true&locale=en_EP&FT=D&date=20070716&CC=TW&NR=200727310A&KC=A#
Over 90% no-load consume and more than 25% under load power consume reducement,
                                                                           but no "overunity"
                                                                 but  Power Factor improvement

Free Energy | searching for free energy and discussing free energy

Re: Where the OVERUNITY using INDUCTION COILS comes from
« Reply #4 on: June 28, 2017, 10:26:58 AM »
Sponsored links:




Offline lancaIV

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  • Hero Member
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  • Posts: 1974
Re: Where the OVERUNITY using INDUCTION COILS comes from
« Reply #5 on: June 29, 2017, 01:52:55 PM »
a.core saturation (no low high free charge capacity )
b. b-/emf = generatoric/motoric orientation
c. PWM= inrush current phase

wattless/-free "Blind",displacement,dielectric,radiant,cold current

                           same-similar-different :

Tibor Kemeny   www.rexresearch.com/kemeny/kemeny.html


Dr.Pavel Imris  https://worldwide.espacenet.com/publicationDetails/biblio?DB=EPODOC&II=5&ND=3&adjacent=true&locale=en_EP&FT=D&date=20160812&CC=DE&NR=202016004514U1&KC=U1#

A device for an electric drive for electric vehicles, characterized in that in an accumulator (1), in which galvanic cells are connected in series electrically and in a second accumulator (12), in which galvanic cells are electrically connected in parallel and in the first accumulator (4, 5, 10, 11) are connected alternately between the two accumulators (1, 12) and are connected with a band - condenser (6), which is connected in series with the second accumulator (12) Is wound on the magnetic core of a transformer (7) as a primary coil, in which electric energy is converted into a magnetic field, and electric energy induced in the secondary coil (8) is connected to electrical load through electrical terminals (22) as a first half-period.

 

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