Where the OVERUNITY using INDUCTION COILS comes from

[pfrattali]

http://overunity.com/17297/where-the-overunity-using-induction-coils-comes-from-eg-joule-thief/#.WUbdOMaQxPY

[Zephir]

The answer may depend on whether this coil has ferromagnetic core or not.

[antijon]

The answer may depend on whether this coil has ferromagnetic core or not.

What do you mean, Zephir? To avoid hysteresis?

[lancaIV]

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

[lancaIV]

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