Air Coupled EV Charging

Sure, if you keep pumping AC current into a parallel tank, it will store energy. DC voltage into an inductor will store energy too. But we don't want to store energy in the transmitter coil, we want to charge a car battery.

At K=0.1, most of the transmitted mag field is going into places that are not the receive coil. Your sim has over a 10:1 ratio of coil currents, which will be over 100:1 ratio of copper losses.

In your sim, the voltage across the transmit coil settles to about

2.5KV p-p. The coil dosen't know where that voltage came from.

Your load gets about 500 watts, 6.7 amps RMS. The capacitor current is

96 amps RMS and the drive into the transmit coil is 90 KVA.

Resonating is just a way of impedance matching sine waves. Given some driver technology and some coil geometry, it might make sense to invest in some giant film caps, series or parallel, and maybe cooling for them. But resonating is not magic, except that it seems to be for raising money. If there is practical amounts of power transfer, Q must be low. That's one definition of Q.

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John Larkin
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That's utter BS

Resonance keeps the Class D driver from having switching losses, AND provides power over long distance by more or less removing the leakage inductance

Cheers

Klaus

Reply to
klaus.kragelund

Are you denying the 90 KVA into the transmit coil?

Are you denying that most of the "long distance" mag field is going somewhere else than the receive coil?

Crank your sim up to, say, 5KW load power and we'll re-do the numbers. Sounds like around half a megawatt into the transmit coil. A lot of volts and amps into those poor film caps too.

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John Larkin

In classic RF transmitter design class-C was preferred (when possible. in constant amplitude modes such as FM). The active output stage kicks the resonant circuits by a pulse in fully saturated mode only for a few tens degrees of the RF cycle.

For variable amplitude modes (such as AM/SSB) a linear class A/AB/B output stage is required with a significant dissipation.

With PWM/Class D operating at the RF frequency, there is no need for a tank circuit. For transmitters, only some low pas filtering is needed and not that for inductive power transfer.

Reply to
upsidedown

On Tuesday, 26 March 2019 10:20:51 UTC-7, snipped-for-privacy@downunder.com wrote: ...

The designs I've seen use a full bridge feeding a square wave into the coil with a series capacitor to resonate. There is no PWM just an even duty cycle square wave - the resonant circuit does all the filtering needed.

kw

Reply to
keith

t

vides power over long distance by more or less removing the leakage inducta nce

I didn't run the sim, is busy have limited time. But later tonight I may ha ve an opening, so I will dig up some information (lots of APEC papers the l ast 10 years on wireless power transfer, at least one of them must have an analytic approach, that will demonstrate that the cap is needed :-)

As a reference, the LLC converters rely on the leakage inductance (about 20 % of the magnetising inductance). If you do not have it, it will run hard s witched, and will have poor efficiency. That is the main reason that resona nt converters have higher efficiency than half bridge converters, even thou gh the resontant converters has higher tank current than the half bridge ma gnetising current

Cheers

Klaus

Reply to
klaus.kragelund

Well, or high impedance for a current-fed circuit.

Or somewhere inbetween.

The small-signal impedance really doesn't matter any; you can make an amplifier with any characteristic you like, and it's independent of the losses.

Nonlinear systems are annoying to design, they violate the impedance matching theorem constantly. :-\

Why? Induction heaters routinely deliver hundreds of kW at high Q. What does it matter that it's kW? The reactive power could be 100s kVA, or MVA, just as well (and often is, in these situations!).

Incidentally, while a load like, say, cold steel is pretty low Q, in and of itself -- it /still has/ self inductance.

That is, consider the fields around, and within, a workpiece. It is essentially a shorted turn on a transformer -- but a turn nonetheless, always and necessarily posessing self inductance.

Induction heating works equally well with cold copper as well as hot graphite. The Q factor of the work itself depends on its resistivity, and that Q is reflected in the work coil Q.

Well, "well" to the degree that it works at all, of course -- a cold copper workpiece doesn't heat as much as the work coil itself, with most of the magnetic field being reflected, in phase, without much power loss -- the efficiency might be 20 or 30%. But that doesn't magically change the physics, it's still getting hot. Of course as it gets hotter and hotter, efficiency rises.

Mind, such numbers are for tubular coils and solid workpieces -- Litz for both will have far lower coil losses! (Assuming a purposeful load, not merely a shorted circuit.)

High Q does not magically increase the range of a single coil. The heating available, at a distance, to a non-resonant load, is minuscule.

High Q only somewhat magically increases the range of a coil pair, but it's a magic severely hampered by those pesky laws of physics.

Tim

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Reply to
Tim Williams

A low impedance in parallel with an LC tank is a damping resistor. The beauty of class C drive is that you can let the impedance be high almost all the time, and low only when you boost the oscillation.

Sure have. Grid dip meter, probably first in the early 1970s. Why do you ask?

Reply to
whit3rd

Using the energy definition of Q,

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increasing loss (useful power transfer to a car battery or to your metal things) reduces Q. In the car case, we want to transfer a lot of power, and that kills Q. High Q then requires huge stored energy in the transmit coil, with corresponding losses.

Don't you water cool your induction heater coils? I wonder how efficient induction heating is. I'd guess not very. We'd want a car charger to be very efficient, which is the point of electric cars.

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John Larkin

If you are transferring 75 kW, 98.7% efficiency would dissipate 1 kW which would need to be cooled. What's your point???

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Reply to
gnuarm.deletethisbit

To get enough power across a weakly coupled transformer, you need to up the flux until there is enough of it in the receive coil. The point of resonating the transmit coil is that you don't need an excessively powerful driver to get that. The driver merely needs to make up for the power absorbed. Of course there will be losses in the resonant circuit, but I'm sure the losses would be greater still if the driver were saddled with providing enough flux /without/ employing resonance.

The point of resonating the receive coil is that it will optimize the power drawn from the field set up by the transmit coil. The whole setup is basically a coupled resonator bandpass filter. That's a well-understood subject. Every radio used to be full of them.

The principle works, but I don't know if it can be turned into a practical car battery charger getting a few tens of kW across with acceptable loss. My guess is yes, but it might require some serious engineering. I'd think that people designing long wave radio broadcast transmitters should feel right at home doing this.

Jeroen Belleman

Reply to
Jeroen Belleman

"Very" is relative. Like I said, it ranges from maybe 20% for awful geometry on high-conductivity loads, to 80%+ on low-conductivity and magnetic loads.

It goes up even further if you make the coil out of Litz, but not enough to do away with water cooling (except for lower power cases, i.e. induction hobs), just because you're doing so many amps in the industrial case.

I guess what he's trying to get at is, you need Q >> 1/k to get the efficiency, since merely equal would be, presumably, somewhere around 50% efficient. So to transfer 75kW at 1kW dissipated, with a 1/k of 10, you need a Q of more like 750.

Again, not insurmountable. Inductors can get that good. But it's somewhat heroic to pull off, and it makes even the mere use of ferrites (as shields/flux concentrators) somewhat questionable.

A required Q over 1000 would be a bit of a show stopper, in terms of trying to achieve that kind of efficiency over a huge distance. So if the coil efficiency is actually that good, we can determine the k isn't too low (i.e., k > 0.07 or so).

Tim

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Reply to
Tim Williams

I would break this into two separate questions

1 resonance of the Tx coil 2 resonance of the Ex coil

  1. Resonance of the Tx coil may have the practical advantage of making it e asier to design a driver, i.e. it may be easier to get X Amps to flow throu gh the Tx coil. But I don't think it increases the coupling to the Rx coil in any way. Resonance of the Tx coil just makes it easier to drive the T x coil.

  2. Resonance of the Rx coil may provide a more fundamental advantage. Intu itively it seems a resonate AM loop antenna picks up more power from a giv en power density field compared to a non resonate loop. But is it true?

If it is true, that means resonating the Rx loop makes it have effectively a larger capture area and therefore less of the Tx field (power) escapes wh ere it may be absorbed by other objects or just radiate away.

Both of these are similar to actual antenna issues. Mark

Reply to
makolber

Sure. But the bigger picture is whether a small coupling, like K=0.1, namely a large distance between the charger and the car, will allow usably efficient charging. I don't think it will, and a lot of people have raised a lot of money claiming that it can.

Sure, at K=0.1 you need a lot of current in the transmit coil, and resonating can help do that. Numbers in the 1 MVA range maybe. Power losses will be horrific, but it would keep your garage warm.

The way to get high Q is by *not* extracting power from the transmit coil! Hoard all that energy.

RF transmitters are (or at least used to be) happy running at low efficiency, well below 90% maybe. Electricity used to be cheap. We need high charging efficiency to Save The Earth.

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John Larkin

Utility power transformers can be over 99% efficient at 60 Hz and are not resonated.

If the *load* kills the coil Q, that's not loss, it's doing the job you wanted to do. There's no sense in having a *loaded* Q be 1000, or even 2. High unloaded Q is just spinning unused energy around.

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John Larkin         Highland Technology, Inc 

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John Larkin

Yes. Resonating may be one way to optimize some driver to some coil.

I don't think so. It's again just impedance matching.

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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

On Wednesday, 27 March 2019 08:08:06 UTC-7, John Larkin wrote: ....

...

Products at the 7kW level have been available for a number of years as after market add-ons for existing EVs.

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As far as I know they use Litz wire in the coils but only have air-cooling. They claim an efficiency "12% less than conductive charging".

from other papers the unloaded Q of the transmit coil is in the 200-700 region with a working coupling coefficient of 0.1-0.2.

kw

Reply to
keith

If you model the low-K coupling as a 3-inductor pi, with the coupling inductor much bigger L than the transmit/receive coils, it occurrs to me that for maximum power transfer you don't want to resonate the coils, you want to series resonate the coupling inductance. Gotta think about that some more.

Of course that equivalent leakage inductance isn't a real inductor, it's space.

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John Larkin

If this technology isn't suitable for wireless charging of EVs, how do you think Momentum Dynamics did it?

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Reply to
gnuarm.deletethisbit

hich

%

So? Geese fly south in the winter. What's your point?

hat

ing

There is some BS here. John keeps talking as if *using* any of the power d rops the Q to a level where it is not useful. But I know transmitting loop antennas are designed exactly this way, tuned to a high Q resonance so the emitted field is maximized and the most power is transmitted, i.e. all of the power entering the antenna is transmitted to the extent possible.

So what is John trying to say here? Obviously high Q can be used to maximi ze power in the "load", ask any ham. Is John just trying to say this is no t "easy"???

Rick C.

Reply to
gnuarm.deletethisbit

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