# Wireless power transmission though magnetic coupling

#### athulascc

Joined Aug 15, 2014
94
Hi
I am transmitting power to receiver coil at resonance frequency.
I am facing a difficulty of transmitting power to receiver. I can only transmit mW power. I think problem is with the driver circuit.
I am using this driver circuit.

I use pulse input at frequency of system resonance frequency. I noticed that when I bring receiver closer to the transmitter, transmit power increases. but still in mW range.

I calculated power from power supply current and voltage. system draw max 100mA from the power supply. I can't believe it because at resonance frequency impedance of capacitor inductor unit should be 0. thus I expected large current , but mA range?

Why I can't send more power?
How to design parallel LC circuit driver to send about 100W power?

here is my driver design

Thanks

#### ronv

Joined Nov 12, 2008
3,770
Hi
I am transmitting power to receiver coil at resonance frequency.
I am facing a difficulty of transmitting power to receiver. I can only transmit mW power. I think problem is with the driver circuit.
I am using this driver circuit.

I use pulse input at frequency of system resonance frequency. I noticed that when I bring receiver closer to the transmitter, transmit power increases. but still in mW range.

I calculated power from power supply current and voltage. system draw max 100mA from the power supply. I can't believe it because at resonance frequency impedance of capacitor inductor unit should be 0. thus I expected large current , but mA range?

Why I can't send more power?
How to design parallel LC circuit driver to send about 100W power?

here is my driver design
View attachment 107356

Thanks
Have a look at this:
http://www.discovercircuits.com/dc-mag/Issue-1/issue-1-exciter-circuit.htm

#### shortbus

Joined Sep 30, 2009
9,050
I calculated power from power supply current and voltage. system draw max 100mA from the power supply. I can't believe it because at resonance frequency impedance of capacitor inductor unit should be 0. thus I expected large current , but mA range?

Why I can't send more power?
How to design parallel LC circuit driver to send about 100W power?
Don't understand how you expect to put 100mA into something and get 100W out.

#### crutschow

Joined Mar 14, 2008
28,507
A parallel resonant circuit will not draw a large current from the supply since, at resonance, its impedance is high.
The only current drawn is that required to make up the losses in the inductor resistance and for any power coupled from the inductor to the receiving coil.

#### DickCappels

Joined Aug 21, 2008
8,156
You will need to drive a loop antenna with a push-pull circuit to get any substantial power. The resistive load gives very poor efficiency.

I suggest that you get some experience at 1 watt or less. That's enough to do some damage. Once you understand how everything works you can start to increase the power. 100 watts can do a lot of damage and might even cause injury. Most of the wireless power products I have seen are under about 15 watts. There must be a reason for that.

While at it, you might want to check into regulations that affect the use of radio in your area. Having a radio interference investigator show up unannounced at your front door can be very hard on the nerves. Take if from somebody who experienced it ;-)

There used to be a large radar station on Mt. Umunhum overlooking Silicon Valley from 1950's through the 1980's. It was part of the Early Warning radar system to spot Soviet missiles coming from the North. It was said that many birds that flew close to it met their fate on the top of that mountain. It might be of benefit to keep that image in mind.

Just to set thing straight, a parallel tuned circuit when viewed from the outside impedance approaches infinity. From the point of view of the capacitor and inductor, they see what approaches the total ESR of the components -close of zero with very good parts. That means you can get very high circulating currents and a strong magentic field until that field is loaded down. Another reason to avoid resistance in the output circuit.