Coil Circuit

Hi,

I am trying to make a wireless battery Charger. I made a coil ( primary coi l) with following Specs

  1. Number of turns :25
  2. Wire Guage: 24AWG
  3. Inductance : 426uH

I added a series Capacitor with the coil. I am driving this LC series Circu it with a Full H- Bridge driver ( HIP 4081A ) with four NMOSFETS. The power supply is +12 Volts. I am using Pulse Width Modultion to drive the driver from a microcontroller . The driving frequency is 100KHz.

I am in making of secondary coil system?

Questions:

  1. The multimeter shows that the coil is drawing about 500mA. Is this RMS v alue that meter is showing? Most likely it is, then will it be right to say that I have build a 6 watts system? ( .5 x 12 = 6W)? If not then how can I calculate how much power my system is actually dissipiating?

  1. I also want to measure the fied generate by this coil. I found the follo wing formula on the internet

Voc = 4.44xNxFxO

I am thinking of making a circular loop of radius 1 inch with N=1 and me asure the open loop voltage across the loop using Oscilloscope. This way I can find the magnetic flux generated by the prmary col. Am I gong in the ri ght direction?

melissa

Reply to
walravenmelissa
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Are you familiar with the idea of phase? Power is the product of voltage and current, but it needs to be a vector product to account for the relative phase. Current and voltage in an inductive circuit are seldom in phase. Is your circuit driven at resonance?

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Rick
Reply to
rickman

Yes. Its series LC resonant circuit. Melissa

Reply to
walravenmelissa

And the value of the C is?

Reply to
Tom Miller

A series LC is a very low impedance at the resonant frequency. The result is that your source (H-Bridge) will see only the resistance of the coil plus capacitor ESR at the resonant frequency. Is that what you want?

I don't know, are you?

It depends on several factors. Is the meter connected in series with the

12V source? Do you have a smoothing capacitor at the H-Bridge after the meter? If so, you might be correct.

Good luck.

Reply to
John S

short answer, you are going about this in the WRONG direction!

long answer. Back up, think about what you're doing. This is a SYSTEM. Tx coil makes a field, Rx coil receives a field. Convert to useable energy. Design EACH piece separately.

Ok, I'll start at the supply, you are first considering MAKING the field. [but that is NOT where to start!] make a coil that makes a field. The field is proportional to the current so it seems wise to increase the current, resonate the coil with a HIGH Q [translate that to extremely low Resr Cap]. Now you have a field proportional to complex distance and proportional to current. Assume Q of 10, the current went up by ten!

The REAL key is the Receiver coil. Putting the raw coil into the field you can read the voltage which will give you coupling coefficient of the loop, but that's going to be ALL over the place with the slightest position change. But you can find a minimum.

Voltage induced follows the old law e = B * 2pif * N * Area So many people see that N term and get excited, "let's increase THAT!" but later you will find that adding N increases the Req of the coil FASTER than you can pull more out, not the wya to go, rather increase that other term AREA! [Note skin effect in a coil is WORSE with more turnns than it is in a coil with only one turn]

Now, resonate the Rx coil very accurately like a Q of 100 to really pull the energy out of the field. put on a high eff SMPS to charge, and you're done, except for one more thing:

Magnetics is LOW impedance. Voltage is HIGH impedance. WE have 'spigots' of voltage around our home that we plug into, when not in use, we OPEN the circuit to that plug. Going from OPEN to ON makes a 'nice' transition. HOWEVER, in magnetics the OPPOSITE is true. to shut off the energy from a field SHORT the coil,killing the resonance, killing the energy going in. Yes, there is some residual leakage, but thee same is true for our open voltage spigots, just not nearly as much leakage. Now, to use an existing field, he Rx coil goes from SHORT to ON andsmothly transitions between the two. If, on the other hand, you went from ON to OPEN, well you've pulled a wire off a battery andwatched the arc, same principle.

One more note: EMC Compliance issues. *IF* you stay below 10kHz,no one cares until OSHA notices you're heating people up. At 100kHz, there are entities that mind what you're doiing, submarine communications, etc. And many of these bands are verboten.

Reply to
RobertMacy

So far so good.

Huh? Your grammar is a bit fractured here, and you only have one coil so far. Your question makes no sense.

Without knowing what your meter is and how its set, it's hard to say whether it's reading RMS or not.

The coil itself, or the H-bridge on the 12V side? Did you add a capacitor that is specifically resonant at 100kHz, or something else?

If you've got 500mA of AC current in the coil, then your actual power dissipation could be just about anything -- the terminal voltage of the coil will be more than 12V p-p (most likely) due to the cap, and the voltage on the coil will be out of phase with the current.

Perhaps you'll he happier if you slap another identical coil up against the one you have, and see how much power you can extract from it.

--
www.wescottdesign.com
Reply to
Tim Wescott

The REAL important statistics of a coil are N, Awg, and diameter, the inductance and parasitic resistance follows that.

Get a copy of free femm 4.2 finite element analysis program.

With it you can answer EVERY question youu have, and gain some real understanding of what's going on

It is VERY accurate when used in axisymmetric mode, which is what you have. It has built-in library of materials, like 24Awg wire.

After you start workind with femm you can place both coils in there and see what's going on, too.

Reply to
RobertMacy

The average multimeter will indicate way low at 100 KHz. Some start dropping off in the 3 KHz sort of range. Cross-check with your scope.

Measure "system dissipation" as the DC power into your oscillator.

An ideal coil dissipates no power. It will consume power when a lossy object enters its magnetic field, like the receive coil and load. Or any objects that have eddy current losses.

That should work.

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John Larkin         Highland Technology, Inc 

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com
Reply to
John Larkin

For a starting point, you might consider buying one of those toothbrushes which charge wirelessly. Then scale things. I have toothbrushes and nightlights which operate this way. They work very well.

Reply to
John S

The following link explains how to calculate the power for LC series circuit

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melissa

Reply to
walravenmelissa

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