If I want to get the wavelength of a radio wave, I must divide the speed of light by the frequency? So having a 125 kHz RFID means an estimated 2km wavelength?

If its wavelength is 2km long, why does this low frequency RFID have a short read range?

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    \$\begingroup\$ Coil diameter must be bigger than the gap to the RFID in order to capture most possible of the magnetic flux, after which inverse squared losses apply to spacing loss. \$\endgroup\$ – Tony Stewart EE75 Nov 7 '16 at 16:13
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    \$\begingroup\$ Wavelength is not the same as range \$\endgroup\$ – Chu Nov 7 '16 at 16:34
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    \$\begingroup\$ Because the reader doesn't have an antenna the size of a baseball stadium. \$\endgroup\$ – David Schwartz Nov 7 '16 at 19:09

Because RFID doesn't work based on wave propagation. It's thus not actually a radio system (despite working at "RF"=Radio Frequency).

Think of an RFID tag more as the secondary side of an air-core transformer, where information is transmitted by the tag changing the amount of power it draws from the primary side of the transformer, or by charging a energy storage (capacitor), and then exchanging the role of secondary and primary side of the transformer.

Because we're not talking about a wave propagating away from antenna, but about a coil coupling into a magnetic field, the decay in power is even worse than the distance² for free space loss, and after a couple of cm, practically no effect of the tag on the reader can be made.

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    \$\begingroup\$ Nice and simple explanation. \$\endgroup\$ – Rev Nov 7 '16 at 15:17
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    \$\begingroup\$ In short, it is a near field, not a far field device. \$\endgroup\$ – user110971 Nov 7 '16 at 15:21
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    \$\begingroup\$ With the obvious proviso that this is only true for low frequency RFID. The 900MHz systems are true far field radio systems which is why they can achieve far greater ranges. \$\endgroup\$ – Andrew Nov 7 '16 at 15:38
  • \$\begingroup\$ @Andrew I agree. I think this is only valid for NFC, not all RFID systems.. \$\endgroup\$ – m.Alin Nov 7 '16 at 15:58
  • \$\begingroup\$ Follow up q: if RFID works on modulating its waveform to transmit data, does it safe to say that 1 cycle is equal to 1 bit? (Just wondering, Im kinda new with this stuff) Thanks in advance. \$\endgroup\$ – Black Nov 9 '16 at 1:16

You're confusing radio communications through the air with coupled inductors.

To communicate through the air using a 125 kHz carrier you would need an antenna of around 1/4 lambda so about 600 meter long in order to transmit that 125 kHz effectively.

Obviously RFID does not work this way.

RFID uses coupled inductors which means there are two coils (optionally with a magnetic material in its core) which couple to each other magnetically.

For this coupling to be effective, the distance can only be a few centimeters.


Simply put, the antenna does not radiate - it operates more as a coupled inductor. The coupling drops with \$r^3\$ rather than \$r^2\$ (if I remember right).

You have calculated the wavelength, so you can see how small the antenna is as a fraction of the wavelength.

With larger antennas, maybe a few meters, and higher transmit powers, its possible to get ranges of hundreds of meters using near-field communications. This is used for example in cave-radio - since the near-field does penetrate rock reasonably well.


I'm kind of surprised the correct answer has still not been given. Yes, it is important to understand that RFID actually operates via changing magnetic fields and that the magnetic fields drop off with \$r^3\$.

However, the reason low frequency (e.g. 125kHz) works at a shorter distance than high frequency (>1Mhz) is due to Faraday's law:

\$V_{emf} = -N\frac{d\Phi}{dt} = -NA\frac{dB}{dt}\$

Where N is the number of loops in the inductor coil, \$\Phi\$ is magnetic flux, \$B\$ is the magnetic flux density (i.e. magnetic field strength), and \$A\$ is the area of the coil loop. Therefore, for fixed \$N\$ and \$A\$, the generated voltage for signal recovery/circuit powering is directly proportional to the speed at which the magnetic field is changing -- i.e. the frequency.

So for example, a 1MHz signal would generate 8x the received voltage as a 125kHz signal at the same distance. Combining this with the \$r^3\$ decay of \$B\$ means that the 1MHz signal could generate the same voltage as a 125kHz signal at twice the distance.

Note: this ignores the parasitic capacitance of inductors, which also comes into play at higher frequencies.


Low frequency RFID as others have said relies on the magnetic vector not the electric vector of the EM radiation emitted by the coil. The magnetic energy falls off with the cube of the distance. The energy that can be transmitted is limited by regulatory authority. The tags themselves have two types passive and active. Passive tags as the name suggests harvest energy from the received field, and power themselves from that field then modulate their receiver coil. The signal emitted by the tag then also falls off according to the cube law and has to be recovered from the transmitting antenna in a separate circuit to that transmitting the energy to power the tag. In both tag and reader there is generally one coil transmit and receieve. Active tags have longer read range because they only have to detect energy above a threshold to switch on thier receive and transmit cicuitry and their range is determined by the power available from their battery.


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