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Measuring air transformer at resonant frequencyby voidnoise
Tags: air transformer, frequency, magnetic resonance, measuring, mutual inductance, resonant, transformer, wireless power 
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#1
Jul313, 10:22 AM

P: 4

Hi,
I am in the process of trying to develop an air transformer but encountering unusual behaviour which I hope can be explained with some fundamental theory I don't yet know. My set up is as follows;  Two identical wound 13.3uH inductors designed to resonate at 4.364MHz. They are 10 turns, 2mm pitch, 100mm diameter, 1mm core. Each coil is designed to be part of a series LCR circuit where the capacitance is 100pF, and the resistance is 100ohm.  The primary coil is powered by a 20Vpp signal generator, 50 ohm output, using a coax cable with measured capacitance between the lines of 90pF.  I am not sure of the full effect of the transmission line capacitance which would be parallel to the LCR circuit, but to compensate I have used a variable capacitor (instead of the designed 100pF) to tune the primary back in to 4.364MHz.  The secondary side is a closed LCR, 13.3uH, 100pF and 100ohm.  I am using two separate oscilloscopes so that no grounds etc. are shared to confused measurements. The confusion: I am trying to measure current through each of the 100ohm resistors to infer the mutual inductance, coefficient of coupling and efficiency. This is done by measuring the voltage across the resistor and then using ohms law. Measurements show that at close range less voltage and so less current passes through the resistor in the primary circuit than the resistor in the secondary circuit. This would hint at more power out than being put in would it not? 


#2
Jul313, 02:34 PM

PF Gold
P: 766

It might be easier to talk in terms of watts. P=IV.
Compare the two wattages. Does the secondary use more wattage thru the 100 ohm resistor? And if you are comparing wattages, don't you need to compare the wattage of the entire loop....L, C, and R? Actually, you would have VA with the L and C.....but since at resonance, perhaps it would be just Watts across the resistor....???? What does "close range less voltage" mean in your last paragraph? 


#3
Jul313, 04:46 PM

P: 2,535

I think you may have a few misconceptions about transmission lines and networks in general. If you have a 50 ohms resistive load fed with a 50 ohm transmission line, then it makes NO DIFFERENCE how long the line is (neglecting loss). The length only makes a difference when there is a mismatch.
 You have set yourself right off the bat for a mismatch when trying to drive 100 ohms with a 50 ohm source. 


#4
Jul1613, 04:53 AM

P: 4

Measuring air transformer at resonant frequency
Thanks psparky. I have been playing around with the circuit a bit more and getting a better understanding about the equipment. Power is definitely the correct way to go. I think my original confusion is due to my ignorance of transmissions circuits, but I am a bit better read up on now. Averagesupernova I am sure you are correct about need match the load to the source impedance, but how would you deal with circuits that have varying impedance with a change in frequency?
One thing I have noticed is that there is an optimal load on the secondary side, would any body know a way to calculate what that would be? I would assume that it can be inferred from the calculated/estimated emf? Which would also mean that it would vary with distance between the two coils. 


#5
Jul1613, 12:55 PM

P: 1,822

voidnoise, are the components on the each side connected in series or parallel? What is the purpose of the 100 ohm resistor? When you speak of load on the secondary, are you referring to the 100 Ω resistor or an additional load? In what sense do you mean optimum load, maximum power dissipation or some other criteria?



#6
Jul2913, 10:51 AM

P: 560

The easiest way to measure the mutual inductance is to excite one coil with a known current / frequency and measure the voltage induced in the second.
Using the following equipment, build a circuit with five nodes: 1  Ground and input of an oscilloscope. 2  Ground and output of a 200kHz sine wave generator. 3  a low inductance 10 ohm resistor, R12 (i.e. 1/4w is good) The Nodes are associated as follow: node 0  (ground) scope ground (chan 1), signal generator ground, R1 (pin 1) node 1  (signal source) signal generator source, primary (pin 1) node 2  (current measurement) primary (pin 2), R1 (pin 2), scope input (chan 1) node 3  scope ground (chan 2), secondary (pin 1) node 4  scope input (chan 2), secondary (pin 2) Adjust the signal generator as high as practical at 200 kHz and measure the current using R1 (V chan 1 / R1) In this condition, measure the voltage of the coupled sine wave on the secondary (chan 2). L = V(Chan2) * R1 / [ V(chan1) * 2 * pi * f ] 


#7
Jul3113, 05:58 AM

Sci Advisor
Thanks
PF Gold
P: 12,251




#8
Feb1514, 07:19 PM

P: 560

Sophie,
You're correct, there is an optimum load which is inclusive of shunt capacitor on the secondary coil to tune out its leakage inductance. When working with implants, the EEs built a huge variety of coils and tested them over various orientations. I found you can approximate the same work (within 10%) using MathCad and treating the leads as filaments. Once you have the self inductance and mutual inductance of the coils, just pop it into LTSpice to get a feel of how the system behaves.  Mike 


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