# Why resonance improves wireless power transmission?

#### Mariemation

Joined Sep 26, 2016
18
Hello,

I am trying to understand why resonance improves wireless power transmission, and I've found this in wikipedia:

"Because the Q factor can be very high, (experimentally around a thousand has been demonstrated[38] with air cored coils) only a small percentage of the field has to be coupled from one coil to the other to achieve high efficiency, even though the field dies quickly with distance from a coil, the primary and secondary can be several diameters apart."

Why only a small percentage of the field has to be coupled for high efficiency?

there is also this sentence: "provided the secondary coil cuts enough of the field that it absorbs more energy than is lost in each cycle of the primary, then most of the energy can still be transferred."

What does the part in Bold means?

Thank you!

Joined Mar 10, 2018
4,057
A high Q coil means it has very little energy loss. So takes less energy
transferred to it = more efficient in its use of energy.

The field is a 3 dimensional field, and the more the coil is inside this field the
more energy gets transferred to it. Think of dipping your hand into a bowl of
water, just the fingers not so much heat (energy) transfer, the full hand lots of
transfer.

Regards, Dana.

#### Mariemation

Joined Sep 26, 2016
18

But what i don't understand is this part: "only a small percentage of the field has to be coupled from one coil to the other to achieve high efficiency", I mean if only a small percentage of the magnetic field of the primary coil reaches the secondary one does this mean that when Q is high the secondary coil amplifies the power it receives, not just preserve it? Because if it only preserves it, then we can't say that the system can achieve high efficiency since only a fraction of the power reached the secondary coil in the first place.

#### panic mode

Joined Oct 10, 2011
2,669
your nightlight may be high efficiency even though it only takes tiny fraction of the energy that power plant produces. if it was not efficient, it would need much more energy from the grid (power plant).

#### nsaspook

Joined Aug 27, 2009
12,769

If you visualize the spinning wheels as the LC circuits in synchronization with the RF energy as the circular motion of its rotation you can see how a small coupling of the B field (like a small spring shaft) can still transfer energy as distance increases beyond just inductive pure near-field coupling because the systems are linked and operate as one system. High efficiency is a misnomer here because even with resonance much of the energy is lost in transmission.

https://spectrum.ieee.org/transportation/mass-transit/a-critical-look-at-wireless-power
A good way to understand why resonance helps is to imagine the mechanical analogue. Suppose you wanted to transfer mechanical energy across a room, but all you had coupling the power source with the load was a long and very weak spring. You’d have to pump the end of the spring you were holding vigorously, moving it back and forth as fast and as far as you could until sweat poured down your brow. It wouldn’t be very efficient, but only with such effort would the far end of the spring wiggle a bit.

To make life easier, you could attach your end of the spring to a pendulum swinging in a wide arc, for instance. Now your arm wouldn’t hurt so much, and the far end of the spring would still wiggle. But another difficulty would appear when you tried to attach that far end of the spring to a mechanical load. If you weren’t careful, you’d find the waves of energy being sent down the spring weren’t being absorbed—most of what little energy that got to the far end would just bounce back. To solve this new problem, you could attach the far end of the spring to a second pendulum, one that was built exactly like the first. Now, all you would need to do is give the first pendulum some gentle rhythmic shoves until the amplitude of its swinging became large enough to get the far end of the spring wiggling in time with it. And those little wiggles would in turn have the right timing to get the second pendulum swinging. Despite having only a weak spring as the conduit, you would have transferred power across the room. Then you could do something useful with it—maybe smash a window.

Last edited:

#### shteii01

Joined Feb 19, 2010
4,644
Physics

#### Mariemation

Joined Sep 26, 2016
18
Hi nsaspook thanks for the explanation

Do you know an equation that shows the necessity or importance of working at the RF in this case? or do you have a document where the theoretical aspect is explained?

Thanks again.

#### BobTPH

Joined Jun 5, 2013
8,659
The frequency used dictates the sizes of the coils you will need. Lower frequencies would require larger coils.

Bob

#### Mariemation

Joined Sep 26, 2016
18
Thanks, what i meant though is why the driver frequency in the primary coil needs to be equal to the resonant frequency of the two circuits (transmitter and receiver), because in experiments we say that we need to operate at the resonant frequency in order to see this effect, not any frequency.

#### nsaspook

Joined Aug 27, 2009
12,769
Hi nsaspook thanks for the explanation

Do you know an equation that shows the necessity or importance of working at the RF in this case? or do you have a document where the theoretical aspect is explained?

Thanks again.
The basic equations are the tuned transformer coupling equations normally used for RF.
https://www.minicircuits.com/app/AN20-001.pdf

For basic wireless power safety reasons we want the generated air-coupling fields to be mainly magnetic as to reduce possible human health hazards from RF electric field energy. This sets constrains on the physical to Electrical_length sizes of the resonator components to reduce the RF radiation (possible radiative far-field energy) component of the power transfer fields.