CataM
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2) Typically a resonant circuit will result in higher power transfer efficiency (PTE), but like Brad pointed out its not required.
In your diagram, however, I don't see any load to deliver power to. Typically this is some rectifier circuitry, which could be a complex load. This could result in detuning of the resonant circuit.
In 2007 a team led by Marin Soljačić at MIT used
two coupled tuned circuits each made of a 25 cm self-resonant
coil of wire at 10 MHz to achieve the transmission of 60 W of
power over a distance of 2 meters (6.6 ft) (8 times the coil
diameter) at around 40% efficiency.
How do you know that the second case is not in resonance? the second circuit has the same frequency as the first one..First case is in resonance, second isn't.
Not at all. Setting K=1 makes the two inductors behave like one. Only separated by the 0.1 ohm resistor, both capacitors are effectively connected in parallel. First resonance circuit is 27.5 µH || 0.96 mF, second double the capacitance. Resonance frequency is altered respectively.the second circuit has the same frequency as the first one..
This is the current through the secondary coil with K=1. Seems to still be a sine wave with 1ms of period.But wouldn't this detune both inudctors equally?
Yes I know, but just saw that and was surprised.Of course K=1 isn't a reasonable model of your "wireless power transfer" problem. Choose K=0.1 or even lower.
Nothing for now, that is why I am using the simulator so much, OrCAD became my brother nowadays :grin:As far as measurements go, what sort of equipment do you have access to? Oscilloscope, arbitrary waveform generator, LCR meter?
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