# Induction Coil powering a load (inductive reactance problem)

1. Mar 27, 2009

### orphefs

Hello all!

I am a student of electrical & electronic engineering, in my final year; for my dissertation, I am exploring the possibility of scavenging energy from a magnetic field in order to power a wireless sensor..I am using a coil of rectangular cross section with a magnetic core occupying the interior volume of the coil. However, due to the core being of permeability around 30000 (which i need, to maximise power) at the values the B-field is oscillating in, the inductive reactance rises incredibly much, (19mH * 30000) making the power highly reactive, with only a VERY small amount of real power going through the load (not mention matching the coil impedance with the load impedance, which is another problem I am facing).

I am sure that out there, there are some power electronics that could solve my problem (what do transformers do when they have a high inductive reactance and output less real power?)...Would anyone know how to confront this problem?

I had the idea of placing a capacitor in series with the coil, so that the inductive reactance is cancelled out by the capacitive reactance, but that would create a resonant circuit, and then we are on to other complications, right??

Thank you very much for taking the time!!

2. Apr 1, 2009

no ideas? :(

3. Apr 1, 2009

### Bob S

What frequency do you plan to pick up with your wireless sensor? The 19 mH sounds reasonable for an air-core 400 turn coil 20 cm dia by 20 cm long. If you plan to pick up a single frequency, using a series capacitor to lower the output impedance is a good idea. But the permeability of 30,000 sounds very high except for some special hi-mu iron alloys which are usually available only in tape form, and then will require annealing after the the final core shape is made. Eddy currents need to be calculated for AC excitation.

See http://www.66pacific.com/calculators/coil_calc.aspx for air core inductance calculator

4. Apr 2, 2009

### orphefs

Hey!

Im trying to pick up power frequency, so ideally 50 Hz, assuming no harmonics. Thanks for the idea, I tried putting a capacitor in series before the load, however didn't see much difference, what value should the capacitor have?

I am not worrying much about the construction yet, this study is more theoretical for the time being, thanks for the concern though!

Kind Regards

5. Apr 2, 2009

### Phrak

Um. Yeah. 30,000 what?

6. Apr 2, 2009

### orphefs

30,000 is $$\mu_r$$, namely the magnetic permeability of a given magnetic material, in this case a core.

By definition, $$B={\mu_0}{\mu_r}H$$, where $$B$$ is the magnetic flux density quantified in Teslas, and $$H$$ is the magnetic field quantified in Amperes/metre. $$\mu_0$$ is the magnetic permeability of free space ($$4{\pi}{\times}10^{-7}$$)and is a dimensionless quantity like $$\mu_r$$.

7. Apr 2, 2009

### orphefs

@ Bob S,

I forgot to mention that the wireless sensor is not of any concern in my study, since it is a separate project carried out by another student in my department. I only need to find a way to power it through a magnetic field at power frequency.

By the way, the biggest problem I am encountering right now is that the load (wireless sensor) is very small (about 10 ohms) and there is minimal power transferred to it (nevermind the reactive, I am talking about apparent power to simplify things a bit). Is there a way I could possible increase the power transferred to it without altering the impedance of its load?

8. Apr 2, 2009

### Bob S

orfefs-
To maximize the power transfer at 50 Hz, you need to zero the reactive part of the circuit by using a series capacitor. So

jwL - j/wC = 0 or LC = 1/w2 = 1.0 x 10-5

So if L = 30,000 times 19 mH, then C = 0.018 uF.
If L = 19 mH, then C = 530 uF

The best thing you can do is put a large resistor (try 10k) in series with the L plus C in series and measure the voltage drop across the L plus the C in series as a function of frequency (say 10 Hz to 10,000 Hz), and then change the capacitor to get a minimum at 50 Hz. The relative voltage drop across the L and C in series compared to the 10k resistor is the output impedance.

You then should use an op amp with large gain to measure the signal you get. Try a voltage follower with gain (say 1000:1).

Last edited: Apr 2, 2009
9. Apr 2, 2009

### uart

orphefs, what is the nature of the stray field that you're trying to extract energy from. Is it leakage from a transformer or is it from a power line or something else?

Do you know the magnetic field strength?

Also, what are the power requirements (in watts) of the sensor.

10. Apr 2, 2009

### Staff: Mentor

You are not going to power a 10 Ohm load via Energy Harvesting. Tell your wireless partner that s/he needs to get down into the microwatt range. That's where wireless sensors operate most of the time. You will want to use capacitive storage to accommodate the milliamp-range transmit bursts.

11. Apr 3, 2009

### orphefs

It is a three-phase rigid power line in a substation.

Field magnitudes range from 30 uT to 100 uT (at least that's where the position of the scavenging mechanism can be).

However, the induced voltage in the coil will not be sinusoidal, since the three phases lag each other by 120 degrees, and the induced waveform will be a superposition of the three fields, with magnitudes of each waveform depending on the radial distance from each conductor. However, I am assuming for simplicity that the induced voltage is sinusoidal and with no harmonics, just to tackle the problem of powering the sensor first.

The sensor requires about 15Vdc and 100 mA = about 1.5 W (unfortunately it doesn't seem that I can adequately power it, as our friend berkeman said).

12. Apr 3, 2009

### orphefs

I tried what you said, however I need some more help with the simulation, because it seems like I am not doing something right, having some troubles with measuring the voltage drops you mentioned correctly; I am probably referencing the wrong nodes...

In the attached .jpg, the time-domain simulation operates with a voltage source of 230 Vac @ 50 Hz. Can you please tell me if the reference nodes at the voltmeter window are chosen correctly? (it currently operates at peak voltage)

In the freq. sweep, the curve is voltage at node 5 minus the voltage at node 3...isn't that the voltage drop across the inductor and capacitor together?

A million thanks!

#### Attached Files:

• ###### sim.jpg
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13. Apr 3, 2009

### Phrak

Some people might think of encircling some permeable material around the wire. Is this possible?

14. Apr 3, 2009

### orphefs

well, as I mentioned earlier, the coil will contain a magnetic core in its interior that will concentrate the flux lines...the problem is that this maximises the inductance and limits the power output of the mechanism when the load is small..

15. Apr 3, 2009

### Bob S

You need to do the measurement on the real coil and series capacitor with a signal generator and ac voltmeter, not just in similation. You have selected the correct nodes, now do it with an ac voltmeter on the bench. This is the way you can determine the real inductance and self-resonances in the coil. The coil will have a substantial real (lossy) component so the resonance will not be sharp as shown in your simulation. By the way, I was unable to read any component values or frequencies in your simulation (picture was too grainy).

Where does the 15 volts and 100 ma come from? You have not told us about any external circuit. by the way, if you have 6o Hz three phase power lines as your signal source, then the signal you pick you will always also be 60 Hz.

16. Apr 3, 2009

### orphefs

I know the real thing will differ, however at this stage I am not able to build the coil yet. :( (plus I dont have the equipment since I am not anywhere near the university)

The lossy component is quantified as a resistance (0.7 ohms) in series with the inductance. These results I got from doing a finite element method simulation. From past experience, the particular method is quite accurate, so in that specific matter I don't expect big differences from theory to practice. (resistance is a function of number of turns, wire thickness, material and mean diameter of the turns) (the coil impedance will also contain a stray capacitance created by the windings)

Yes I have :) it is the sensor I am looking to power, of impedance 10 ohms.

Yeah, since it is a periodic signal, but the waveform will not be perfectly sinusoidal.

Last edited: Apr 3, 2009
17. Apr 3, 2009

### uart

Due the nonlinearity of your iron circuit perhaps, but not for the reason you stated previously (that was, the combination of different phases at different amplitudes).

Any linear combination of sinusoids at a given frequency results in a perfect sinusoid at the same frequency. (a sinsusoid being a sinewave shape of arbitrary phase shift).

18. Apr 3, 2009

### Bob S

Make a 375 turn coil 20 cm diameter by 20 com long, and measure it with a series capacitor, in series with a 10 k resistor, like in the simulation. What is the Q? How does it compare to the simulation?

The effective series resistance also includes AC losses, which are probably not in your model.

I believe that your output signal source impedance is a lot more than 10 ohms, and your signal might be a few millivolts.

The sum and difference of 60 Hz sinusoildal signals, even if out of phase, will always be 60 Hz and sinusoidal.

19. Apr 3, 2009

### orphefs

yes, but they are not in phase! 120 degree phase difference...this means that their superposition (although all of them are at 50 Hz) will yield a different waveform!

http://en.wikipedia.org/wiki/3_phase" [Broken]

depending on the positioning of the scavenging mechanism in the plane of the conductors, one of the phases will be greater in amplitude, so the waveform will vary!

see what I am saying?

Last edited by a moderator: May 4, 2017
20. Apr 3, 2009

### orphefs

Precisely. The modulus of the output impedance is $$|R+j{\omega}L{\times}{\mu_r}|$$, where $${\omega}=2{\pi}f$$, and if we substitute the values, this yields $$|0.7+j{\times}314.16{\times}0.019{\times}5000|=29.8k{\Omega}$$ roughly...

By the way, I think we mean the same thing regarding the waveshape; what I am saying is that the waveshape will not be perfectly sinusoidal as defined by the function $$\sin$$, though it will contain three phase-displaced $$\sin$$ terms, resulting to a different waveshape than a pure $$\sin$$.