Air Coupled EV Charging

That it isn't practical to go any significant distance unless you're going to stay there several hours. That's not always the case. But it has a gas engine too I suppose.

Reply to
Tom Del Rosso
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??? Why can't you drive back? I'm going to drive 120 miles today. I'll s till have some 150 miles of range left. What are you talking about???

Win uses 120 volts at 12 amps to charge his car because he has a tiny batte ry in his plug in hybrid. Battery EVs have much larger batteries and there are many other charging options available. With ranges up to 300 miles mo st trips require nothing other than home charging from a 30 amp, 240 volt c ircuit. In about 10 hours this would fully top off my EV.

Are you totally unaware of EVs? They've been discussed here quite a bit.

--

  Rick C. 

  -- Get a 1,000 miles of free Supercharging 
  -- Tesla referral code - https://ts.la/richard11209
Reply to
gnuarm.deletethisbit

They're not comparing against the plug itself, though; they're comparing against the whole existing charging system. When you effectively have a transformer spanning the gap between car and charger, that can do part of the work of converting the voltage from 240VAC to 400VDC (or whatever the car battery runs at). That transformer is competing with the transformer inside a DC-to-DC converter or inside the power supply of the plug-in charging system. It can't be quite as good, but if they get the rest of the power conversion electronics to be more efficient they can beat the plug-in system -- though not at the same price, of course.

The bad coupling that the air gap imposes mostly manifests itself as leakage inductance, which being an inductance doesn't waste energy in and of itself. They'd have designed the system to recover that energy, which is what the "resonant" buzzword is about. Of course if there is something conductive that's close enough to couple with the leaked field they'll lose energy via eddy currents. Probably the main thing that'd be close enough is the skin of the car. One would want to look closely at the context of the claim to see whether they specify that the efficiency number is for a unit installed in a car or whether it might be a number measured on a lab bench.

Reply to
norman.yarvin

On Saturday, 23 March 2019 19:21:17 UTC-7, snipped-for-privacy@gmail.com wrote: ...

really. They have a video that compares their 10 kW system to the "standar d" 3.3 kW system. I've never seen a 3.3 kW L2 charger. That would be 240 volts, ~15 amps. The lowest I've seen is about 7 kW. ...

The Volt, SparkEV and early Leafs, were all 3.3kW. The Prius plug in hybrid s are ~2kW.

kw

Reply to
keith

On Sunday, 24 March 2019 03:46:53 UTC-7, Tim Williams wrote: ...

...

Standard SAE J2954 specifies ~81-87kHz although strictly that only goes up to 11kW.

kw

Reply to
keith

:

t really. They have a video that compares their 10 kW system to the "stand ard" 3.3 kW system. I've never seen a 3.3 kW L2 charger. That would be 24

0 volts, ~15 amps. The lowest I've seen is about 7 kW. ...

ids are ~2kW.

You are confusing what charging units provide with what the cars will use.

--

  Rick C. 

  -- Get a 1,000 miles of free Supercharging 
  -- Tesla referral code - https://ts.la/richard11209
Reply to
gnuarm.deletethisbit

:

te:

ort really. They have a video that compares their 10 kW system to the "sta ndard" 3.3 kW system. I've never seen a 3.3 kW L2 charger. That would be

240 volts, ~15 amps. The lowest I've seen is about 7 kW. ...

brids are ~2kW.

.

...

.

I'm not sure what you are referring to.

The AC to DC conversion internal to the vehicles that I mentioned can only work up to those levels (3.3kW or less) even if the EVSE advertises higher capability.

I have measured a few of the cars while charging and usually the input powe r is constant at the maximum rating until close to the end of the charge.

The J1772 EVSE advertises its capability by means of the PWM duty cycle on the Pilot line. The charger internal to the vehicle only consumes power up to that amount or to its internal limit.

kw

Reply to
keith

Ricky often talks about confusion. "You are confused" and "I'm confused" are both his ways of saying "you are an idiot."

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics
Reply to
John Larkin

On Sunday, March 24, 2019 at 10:12:19 PM UTC-7, snipped-for-privacy@gmail.com wro te:

:

te:

ort really. They have a video that compares their 10 kW system to the "sta ndard" 3.3 kW system. I've never seen a 3.3 kW L2 charger. That would be

240 volts, ~15 amps. The lowest I've seen is about 7 kW. ...

brids are ~2kW.

.

Obviously, you haven't been to many places.

  1. Bakersfield Amtrak: L1 on solar
  2. Exeter City Hall: L1 on PCG.Q
  3. Sanger City Hall: L1 on solar ...

-- Get 1,000,000 miles of free L1 (3.3KwHr) charging

-- Referral code - Anything you want

~
Reply to
edward.ming.lee

rote:

te:

rote:

sort really. They have a video that compares their 10 kW system to the "s tandard" 3.3 kW system. I've never seen a 3.3 kW L2 charger. That would b e 240 volts, ~15 amps. The lowest I've seen is about 7 kW. ...

hybrids are ~2kW.

se.

Now I'm confused :-)

L1 is 120V (as opposed to 230V in US)

L1 is limited to 1.4kW (120V @12A) in the US.

And "3.3KwHr" is an incorrect way of specifying a power.

The 'k' should be lower case to indicate '1000'. (An upper case 'K' is used for Kelvin). The 'w' should be upper case as it is short for a person's name - Watt. If you mean a power then the 'Hr' should not be present.

The correct presentation should be "3.3kW". However if it is really Level 1 then the maximum is 1.4kW.

kw

Reply to
keith

rote:

ny sort really. They have a video that compares their 10 kW system to the "standard" 3.3 kW system. I've never seen a 3.3 kW L2 charger. That would be 240 volts, ~15 amps. The lowest I've seen is about 7 kW. ...

n hybrids are ~2kW.

use.

ed for Kelvin).

1 then the maximum is 1.4kW.

Yes, you are correct. These are L1(1.4kW). I have seen L2(3.3kW) as well. Just don't remember which one for now.

Reply to
edward.ming.lee

Seems clear enough upon re-reading it.

Of course I'm unaware of what kind of cars you and he have. You don't know what my car is either.

Reply to
Tom Del Rosso

Why? The mutual inductance between two coils doesn't change if you connect a cap across either one.

--

John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  

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

Current goes up, though. The optimal Q probably isn't very high.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC / Hobbs ElectroOptics 
Optics, Electro-optics, Photonics, Analog Electronics 
Briarcliff Manor NY 10510 

http://electrooptical.net 
http://hobbs-eo.com
Reply to
Phil Hobbs

On the transmit end, if a black box drives the coil, there's no way to know if there is a resonating cap inside the box, as long as the coil voltage is what you want.

I think a similar argument can be applied at the receiver. Resonating is just impedance matching, not magic. But spraying h-field all around the garage seems inefficient to me.

I have done resonant power transfer (to read out an electric meter if mains power is off, as it is often in India) but I didn't care about efficiency.

--

John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  

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

The power source delivers energy into a resonant circuit, not into the load. The load removes power from that resonant circuit slowly because of low flux coupling, but that doesn't matter; the resonance on the primary means that larger magnetic flux is available, because several power-source cycles (Q of them) have contributed. Q of six and 50% flux coupling gives you 90% power efficiency.

I'd think an 80% + flux coupling would be achievable (with moving parts)

An analogy would be a bicycle generator: the inertia and flywheel effect of a moving bike holds all the power you put into the pedals, even an inefficient coupling in the generator will run the bike lights just fine, it doesn't waste mechanical energy to leave some of the rotor flux un-gathered.

Reply to
whit3rd

If the terminal voltage across the transmit coil is, say 100 volts RMS at 100 KHz, what difference does it make if the driver amp has a capacitor inside? How could you even tell?

--

John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  

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

On Monday, 25 March 2019 13:14:28 UTC-7, John Larkin wrote: ...

...

Not directly , however if the current was also measured it would be apparen t that a large proportion of the power was reactive. The mismatch between t he input power to the to the driver and the reactive power at its output wo uld also indicate that output impedance is reactive implying a capacitor.

The voltage across the transmit coil can reach thousands of volts and the h igh circulating currents also require use of Litz wire with hundreds or tho usands of strands to keep the losses acceptable.

kw

Reply to
keith

te:

rote:

sort really. They have a video that compares their 10 kW system to the "s tandard" 3.3 kW system. I've never seen a 3.3 kW L2 charger. That would b e 240 volts, ~15 amps. The lowest I've seen is about 7 kW. ...

hybrids are ~2kW.

se.

se.

y work up to those levels (3.3kW or less) even if the EVSE advertises highe r capability.

Yes, exactly.

wer is constant at the maximum rating until close to the end of the charge.

Ok, so what about it?

n the Pilot line. The charger internal to the vehicle only consumes power u p to that amount or to its internal limit.

Yes, I'm familiar with the J1772 mechanisms. Why is the power drawn by a f ew vehicles significant?

--

  Rick C. 

  -- Get a 1,000 miles of free Supercharging 
  -- Tesla referral code - https://ts.la/richard11209
Reply to
gnuarm.deletethisbit

Right, but if it's near resonance, there's a circulating current inside the tank that makes the magnetic coupling stronger. IIRC it goes like sqrt(Q) or something like that. Loading it down to get power transfer destroys the Q, so I expect that the optimum will be somewhere like Q ~

1 or 2.

Doesn't matter unless something absorbs it significantly, or you have a pacemaker. ;)

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC / Hobbs ElectroOptics 
Optics, Electro-optics, Photonics, Analog Electronics 
Briarcliff Manor NY 10510 

http://electrooptical.net 
http://hobbs-eo.com
Reply to
Phil Hobbs

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