I am designing a high frequency passive RFID tag-like device that receiving 27MHz wireless power, use it to power a onboard MCU and a low power transmitter. The estimated current drain is about 50mA. You may say this is a huge amount power for a passive RFID tag, moreover, the distance between the reader and tag will be 46cm maximum. However, in our application, we will have a powerful(25W) reader for this application. My specific question here is about the power receiver. I have two available options. First one use diode and capacitor for a voltage multiplier, which is very common seen on RFID article, then follow by overvoltage protection and linear regulator, like this in following picture: option 1 Comments:

  1. Because of the high frequency of incoming power, I can use rather small(100nF) capacitor in voltage multiplier.

  2. I add an overvoltage protection on linear regulator input(LT3065's Vin max up to 45V) for situation where the tag is occasionally brought too close to reader. M1 will start to draw excessive current when LT3065's input larger than about 20V.

  3. I will use schottky diode whose trr<<40nS for this 27MHz input power.

  4. I don't know how to deal with the transmitter and receiver coil, so I just left the receiver coil to a sine signal with a roughly(guess) amplitude. How can I model its parameter, like Rser? I noted the output of voltage multiplier depends on it.

So, will this circuit work?

********updated on 2.13:

I found a signal generator with maximum 1W output power to simulate the transmitter. For convenience and test purpose, I used a NFC antenna transmitter coil(2.1uH at 13.56MHz) at hand with the signal frequency changed to 13.56MHz. For maximum transmitting current, I add a capacitor(actually three, according to this app note by Atmel, for better trimming) serial with the transmitter coil for serial resonance at 13.56MHz. So the transmitter circuit just looks like below. Because the signal generator's output impedance is 50Ohm, and ought to be used with 50Ohm load. 100mW output power means about 6.3V amplitude on sine wave voltage source(Am I right?). So I will get about 120mA(I actual measured about 100mA with a serial 5Ohm resistor with oscilloscope) transmit coil current for the transmitter. Is this approach for optimal(longest) operating distance right? Should I use a power amplifier whose output impedance is much smaller instead of 50Ohm for a larger transmitter coil current? transmitter model

  • \$\begingroup\$ If by "work" you mean "provide 50mA" ... probably not - measure the current you need from V1 to get 50mA out in your simulation, I think the performance of a multi-stage voltage multiplier will surprise you. \$\endgroup\$
    – user16324
    Feb 8, 2017 at 11:29
  • \$\begingroup\$ The linear regulator is shown with a 100 ohm load and configured to output 3.3V so that is 33mA. So power out is 0.109 watts. In a simulation of this circuit the voltage source is producing a peak to peak of 4.7VAC. The peak to peak sourced current is 1.06A. Converting these to RMS values gives 1.66VACrms @ 0.375Arms. So input power is 0.623Wrms. Net efficiency is about 17%. It will be a pretty impressive amount of transmit energy at 27mHz to induce this amount of circuit input power into an air coupled coil. \$\endgroup\$ Feb 8, 2017 at 13:51

1 Answer 1


I don't know how to deal with the transmitter and receiver coil

At 27 MHz, the wavelength is 11 metres so, at a read-distance of 46 cm, you are clearly in the near-field of any radiating antenna. This means that you cannot rely on the far-field advantages of H and E fields reducing linearly with distance.

Apart from that, to make a simple EM radio wave antenna needs some form of dipole and you appear to be opting for a transmit coil. At this frequency, a relatively compact transmit coil won't be able to produce a fully-fledged EM wave so, that looks like a done-deal on defining the field intensity you generate might pick up at some distance.

Here's a picture and some math taken from hyperphysics that describe this: -

enter image description here

The upshot of this is that the magnetic flux density at a point along the centre-line, is inversely proportional distance cubed (when z becomes more significant than R). When R is more significant than z, the flux density falls with distance. This tends to mean that it is preferable to make the coil radius of the same order as the maximum gap distance in order not to suffer from the inverse cube law.

So how much current can you drive around your transmit coil? The lower you can make the coil inductance, the more current you can pass but, if you are dealing with a 25 watt output that is intended for a 50 ohm antenna then you will likely need to use a step down transformer.

How much can a receive coil expect to generate is dependent on the surface area of the coil i.e. how much flux it "nets". The bigger the surface area the more flux that it "captures" and the bigger the voltage liberated.

So, on the receiver side you should aim for the same type of coil and then parallel or series tune it to liberate the required voltage you need. I don't think you will need to use caps and diodes acting as doublers.

If you go for significantly smaller coils then you will be facing difficulties that are high.

  • \$\begingroup\$ Hi Andy, I updated my question with more information. I know your are always right... however, I don't fully understand one of your statement. On receiver side, should I also make receiver coil resonance with capacitor for optimal operating distance like what I did on transmitter side? Should I also make more turns on receiver side like transformer for larger voltage? \$\endgroup\$
    – iouzzr
    Feb 13, 2017 at 11:53
  • \$\begingroup\$ And finally, I was told that NFC coil is just like transformer. What I didn't understand is that there's hardly no current on primary side when no load on secondary side. However, there's seems always the same current on transmitter coil regardless the receiver coil... \$\endgroup\$
    – iouzzr
    Feb 13, 2017 at 11:54
  • \$\begingroup\$ Make both coils use the same turns. Yes, resonate the receive coil to counter the leakage inductance due to bad coupling. Series resonance on secondary may work better under load but parallel resonance may work better at light loads. \$\endgroup\$
    – Andy aka
    Feb 13, 2017 at 12:00
  • \$\begingroup\$ When parallel resonating transmit coil it may seem that there is hardly any current feeding the parallel resonant coil but, it will be circulating a larger current due to resonating with the parallel capacitor. With series resonating the current fed to the series circuit is the same as the current in the coil so I'm not sure what you are saying here. \$\endgroup\$
    – Andy aka
    Feb 13, 2017 at 12:02
  • \$\begingroup\$ Maybe you haven't tuned correctly? \$\endgroup\$
    – Andy aka
    Feb 13, 2017 at 12:04

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