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I am fairly new to electronics and I recently made an inductive charger that transfers 18 volts over a couple mm from 120 volts. The problem I am having is that while the voltage transfer is good, the current transfer is minimal. I tried putting the voltage through a step down transformer to increase the current, but this got rid of the voltage as while as the current. Is there any way to increase the current? enter image description here

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    \$\begingroup\$ Please share your circuit diagram. With which people will be able to understand the situation better for right solution. \$\endgroup\$ – Umar Jan 25 '17 at 1:12
  • \$\begingroup\$ Show the physical arrangement. Your leakage inductance is probably through the roof. What frequency are you running? \$\endgroup\$ – winny Jan 25 '17 at 8:33
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Short answer: It doesn't work this way. U1*I1 == U2*I2 is only meaningful for the ideal transformer. No losses and a coupling of 100%.

You have neither of these. Because your coils don't share a common iron core, the coupling is in the range of 5% at most and your losses depend on the amount of current you want to transfer through this loose coupling.

You have to reduce the air gap for a tighter coupling and will see the amount of current you can draw from the secondary without a loss in voltage increases greatly.

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  • \$\begingroup\$ How would inductive chargers made by Samsung and such work even if they don't share common iron core \$\endgroup\$ – user136941 Jan 25 '17 at 1:41
  • \$\begingroup\$ They work because the air gap is tight between the two cores. Plus, they use a higher frequency which allows a better coupling with less iron. Your setup is more like that of an old-fashioned electric toothbrush loader. That one transfers about 10mA@2.4V over about 3mm. \$\endgroup\$ – Janka Jan 25 '17 at 1:51
  • \$\begingroup\$ Ok thank you. My last question is how am I able to get a lot of voltage but no current and why can't I put the voltage I am getting through a step down transformer and get more current and less voltage? \$\endgroup\$ – user136941 Jan 25 '17 at 1:57
  • \$\begingroup\$ You can measure the voltage you expect from the winding ratio because your voltmeter draws no current . Your transformer is ultra-soft because of the big air gap. As soon you draw more than a few µA, the voltage drops to near-zero. Your step-down transformer doesn't help here because it has losses, too, and draws more than a few µA. The mechanic equivalent is putting another gearbox behind an already overloaded gearbox and expecting the axles to start moving for some magic reason. \$\endgroup\$ – Janka Jan 25 '17 at 2:02
  • \$\begingroup\$ So in theory, increasing the frequency and making the air gap smaller will fix this? \$\endgroup\$ – user136941 Jan 25 '17 at 2:09
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To get enough output from a loosely coupled transformer the resonance is a good help and it's practically a must to keep the system economical and lightweight. Your schematic has no oscillator circuit. So you operate at 60 Hz mains frequency. At 60 Hz resonance circuits would be impractically massive. The oscillator makes AC that has a thousand times higher frequency - where the resonant circuits can have a practical size.

I wonder what are your coils? They must have a heavy iron core or quite a lot of wire - much more than 60 Hz transformers have. Otherwise at 60 Hz they would take so much current that they either get hot or the fuse blows.

Anyway, the induction charger that haves only loosely coupled coils is in practice impossible at 60 Hz. You may have seen some voltage output without load. That's because a good voltmeter takes only 1 uA or less at 18 V.

The following link shows a plausible homebrew inductive charger: http://www.nutsvolts.com/magazine/article/august2013_Bates I have not proven the circuit, but this one at least is promising. In my former edit I presented a circuit that really can't be recommended seriously.

The next is a link to the basic theory behind the inductive charger https://en.wikipedia.org/wiki/Resonant_inductive_coupling

You should keep away from mains voltage projects until you have some good local help that can prevent the damages that are waiting for the beginner. Mains woltage can

  • kill you
  • kill somebody else
  • cause a fire.

In all three cases you are in trouble for the rest of your life.

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  • \$\begingroup\$ I put the basic schematic I am using in the question body note: I rectify after the transformer \$\endgroup\$ – user136941 Jan 25 '17 at 3:41
  • \$\begingroup\$ Induction can work without resonance else Faraday wouldn't be so important. Think again -1 \$\endgroup\$ – Andy aka Jan 25 '17 at 9:10
  • \$\begingroup\$ @Andy aka yes, it works. See my recent writing about induction. electronics.stackexchange.com/questions/282053/… But loosely coupled coils can generate much higher output voltage via resonance. That voltage is needed to get something into the battery. BTW thanks for the downvote! \$\endgroup\$ – user287001 Jan 25 '17 at 9:23
  • \$\begingroup\$ Yet you say "cannot work without resonance circuits"! \$\endgroup\$ – Andy aka Jan 25 '17 at 9:31
  • \$\begingroup\$ @Andy aka It's the principle that starts from having resonance. I edit its full name into my answer. OK Thanks. \$\endgroup\$ – user287001 Jan 25 '17 at 9:35
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Transferring power magnetically implies a device called a transformer. Regular transformers have closely coupled primary and secondary windings. Close coupling means that all the flux generated by the magnetization current is coupled to the secondary and, as per the implications of Faraday's law of induction, the output voltage is proportional to the input voltage via the turns ratio.

Faraday's law of induction; voltage = \$N\dfrac{d\Phi}{dt}\$

Magnetic Flux is common to both windings hence, the output voltage equals the input voltage (modified by the turns ratio).

When only 50% of the flux is coupled, the output voltage drops to 50% (under no load conditions). If there is only 10% coupling then the output voltage is reduced to a value of 10% compared with full-coupling.

As soon as the coupling reduces from being 100%, the equivalent circuit of the transformer incorporates leakage inductors - these leakage inductors do produce flux but, that flux does not couple with the other coil hence, all they do is impede the flow of current should there be any. So, you try and take current from a 50% coupled secondary and the output voltage drops from 50% straight away.

More current means lower output voltage. Badly coupled coils with lots of turns on the secondary may look able to provide a decent output voltage yet, under load, that voltage drops significantly and pitiful amounts of power are transferred.

The next option is to series resonate the secondary winding and here is an example circuit fed from a sine wave of 40V p-p at 1 MHz: -

enter image description here

You can see that the secondary voltage (blue) is pretty close to 40 Vp-p when loaded with 1 ohm : -

enter image description here

Power in is green and power out is red. This shows a pretty efficient power transfer of about 96%. However, if the coupling changed to (say) 30%, the waveforms would look like this: -

enter image description here

Just reducing the coupling from 50% to 30% has ruined the output voltage level. Instead of 40 Vp-p it has dropped to about 6 Vp-p. This is because the resonant point has moved and the 8.4 nF capacitor is no longer perfect for voltage transfer.

Operating at 60 Hz is more problematic; the ability to series tune the output is going to require large value capacitors. The output coil inductance might be in the realm of 10 mH and this will need 1000 uF to tune it to 50 Hz (as an example).

So, reduce your expectations and look at what is really happening inside a transformer with low coupling. Also, be very careful with AC power voltages - they can be lethal.

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