# Mutual inductance/wireless power

1. Feb 16, 2008

### mkkrnfoo85

Hello,

I am trying to make two single-layer air cored coils in order to transfer power from one coil to other through mutual inductance. I've had a really tough time trying to come up with some theoretical guidelines for my experimentation. For example, I've found empirical equations describing how to measure the inductance of single layer air cored coils:

(1)$$\frac{0.001*N^{2}*r^{2}}{228*r+254*l)}$$
N = # of turns
l = coil length (m)
(l must be > 0.8r for this equation to be reasonably accurate)

I've found other conceptual information about needing to find the "self-resonant" frequency of the coil, in order to maximize the power transfer. This has something to do with the balance of the reactance of the ideal inductor (the coil) (w*L), which is in parallel with a "stray" capacitance developed between each turn of the coil (-1/(w*C)). From having those two things in parallel, there is a maximum frequency of the ac current that can be found.

I think using Eq. (1) to find the inductance of the coil is pretty straightforward. What is hard is finding the "stray" capacitance. I found another paper describing how this stray capacitance is found for a solenoid type coil, as mine is. The equations are:

$$C_{tt} = \frac{\pi*2r*\epsilon_{0}}{\ln(\frac{p}{2w_r}+\sqrt{{(\frac{p}{2w_r}})^{2}-1})}$$
p = pitch (distance between the centerlines of two adjacent coils) (m)
w = wire radius (it's gauge, basically) (m)
e_0 = permittivity constant ~ 8.85*10^-12 F/m

That gives you the turn-to-turn capacitance.

You then input that value into another equation that gives you the total stray capacitance:

$$C_{total} = \frac{C_{tt}}{N-1}$$
N = # of turns.

It is kind of hard to find the pitch value of the wire, because that takes into account the insulation thickness of the wire, which is a very small value for magnet wire.

Skipping all this theory, I tried just making two arbitrary coils, kind of keeping the # of turns in mind, and powered the primary coil with a function generator. I then put a secondary coil as part of a simple "diode bridge and filtering capacitor" set up to convert ac to dc, and put an LED as the load. I used the function generator on the primary coild and played around with the frequency until it maximized the voltage reading on the secondary coil. The maximum frequency came out to ~1.5 MHz. I think I can call this the "self-resonant" frequency of at one (or both?) of the coils? For the primary coil, I used a coil radius = .0305 m, coil length = .061 m, and 65 turns. For the secondary coil, I used coil radius = .0135 m, coil length = .027 m, and 70 turns. Different gauge wires were used (awg 18 and 26, respectively. The coils were placed upright ~1-2 cm from each other, and it barely lit the LED. I'm using a 20 V function generator, and I am able to get a induced voltage difference on the secondary coil close to about 14 V rms. However, I am confused on why the current (or at least, I think the current is the problem here) is so low going into the LED.

Here is a short list of questions and concerns I had about my experimentation:

(1) How do I increase the current going to the LED? I feel like I am getting maybe fractions of milliAmps going to the LED. I've tried decreasing the turns on the secondary circuit, because I thought it would increase the current (due to the rule, I1*V1 = I2*V2) going to the LED, but it doesn't seem to do much to the LED.

(2) I am currently finding this "maximizing frequency" between the coils just manually experimenting with the function generator. Is there some sort of - relatively simple to understand - equation or conceptual knowledge that can help me determine this maximizing frequency between coils?

(3) Would I be able to increase the current on the secondary circuit by hooking up two coils on the secondary side, and trying to bring current to the LED from two different diode bridge set ups?

(4) I would preferably want a voltage between 5-15 V, and a current of 50-150 mA going through my secondary circuit load, in order to power a small motor. Is this feasible with mutual inductance?

(5) I would be so grateful for any additional guidance on how to get more power into my secondary load through mutual inductance.

Sincerely,

Mark

p.s.

Sorry about such a long post. I don't know whether that information on equations will be all that useful for the reader.

Last edited: Feb 16, 2008
2. Feb 18, 2008

### zeitghost

A conventional power transformer has an iron core that closely couples the primary and the secondary... an air cored transformer lacks this... so the mutual inductance is much much less.

Electric tooth brushes use a form of contactless power transfer that uses effectively a split ferrite cored transformer...

You are attempting to the same thing with no core whatever...

You can increase the coupling to some extent by tuning the primary and secondary to the same frequency....

You realise that LEDs run off DC not AC?

You may need a bridge rectifier on your secondary, plus some sort of smoothing capacitor.

3. Feb 18, 2008

### mkkrnfoo85

Hey, thanks for the reply. Sorry about my long post, I am using a 4-diode bridge with a smoothing capacitor (100 uF). So, I guess my problem would be the need of a ferrite core to increase mutual inductance, which is of the form:

$$M = N_1*N_2*P$$
N1 = coils of L1
N2 = coils of L2
P = permeance of the space
(according to Wiki)

Would I need a ferrite core on both the primary and secondary coils, or is one at the primary coil potentially enough?

edit: Also, an additional question: If I am able to increase the mutual inductance, M, between two coils, this just seems like I am increasing the inducted voltage (through the equation, V2 = M*di1/dt). My problem seems to be that, although I am getting enough voltage (~10-15 V) on the secondary side, I don't seem to have enough current for the LED to light. Will increasing mutual inductance M also increase my current?

Thanks again.

-Mark

Last edited: Feb 18, 2008
4. Feb 19, 2008

### zeitghost

Sorry. I'd missed the part of your post mentioning the diode bridge & 100uF cap on the secondary.

I have no mathematical idea of how mutual inductance drops off with separation of the coils, but I'd expect it to be at least inverse square law, if not more rapid.

The primary and secondary would require ferrite... and as the air gap gets bigger, so the power transfer gets less & less efficient.

You may remember last year that someone transferred power over 6 feet to light a bulb using two coils.

From the photo of the experiment, the coils were 3 or 4 feet in diameter...

5. Feb 19, 2008

### mkkrnfoo85

Hey, thx for the reply. Yea, I heard about that mit lab being able to light a bulb. I had no idea the coils were so huge tho.. puts things in perspective.

Okay, I just saw the picture. The coils are huge. It doesn't seem like they require a lot of turns tho (~5). Doesn't look like they're using cores either. Interesting.

6. Dec 1, 2009

### dontommazo

Hi,

I would suggest that you calculate the inductances and stray capacitances in a FEM tool like Ansys or Comsol Multiphysics. That should give you the most accurate result.

How long is the distance between the coils, and how big are the coils? The coupling coefficient k will be proportional to the ratio air gap distance/coil diameter, so if you want to maximize the efficiency you should minimize the air gap.

How to you plan to feed the inductive power transfer with high frequency power? If using a MOSFET bridge, one advice is to try achieve soft switching. What is the level of power you plan to transfer?

/ Tomas

7. Dec 1, 2009

### famousken

You may want to try Googling "Tesla coil calculator" there are many websites that have calculators where you can input your coils number of turns, wire gauge, and size and get back resonant frequencies, self capacitance, inductance, resistance, etc. When you think about it, that is basically what you are building, a high frequency, resonant, air-cored transformer, right?