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For antenna design, usually it is half wavelength or quarter wavelength antenna. I am trying to understand small loop printed antenna design for 434MHz frequency where even the quarter wavelength is too large to fit into the PCB. Some companies provide information on such antenna design. In such designs, the length of the loop is way to smaller than the quarter wavelength antenna.

Furthermore, the design of the antennas differ.

Small Loop Printed Antenna Design - MaximIntegrated - 1 enter image description here

Small Loop Printed Antenna Design - Infineon - 2 enter image description here

Can you make it clear what's the difference between the two designs and how to simulate an antenna with such components like capacitor, resistor and inductors in the antenna circuit?

Would I be right to assume as shown in the following image? enter image description here

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    \$\begingroup\$ There's many differences, but both are antennas. The answer to the question "what's the difference" is "if you simAulate the two antennas, you get different results, and their design is different". To simulate PCB antennas, you use EM solvers optimized for such use cases. CST microwave studio comes to mind. \$\endgroup\$ Jun 29, 2020 at 8:29
  • \$\begingroup\$ What puzzles me is that both are small loop printed antennas designed for 434MHz, and yet the circuits are so different. Can you tell the fundamental difference on design principle between them so that I can understand it better? \$\endgroup\$
    – spockshr
    Jun 29, 2020 at 9:13
  • \$\begingroup\$ no, I can't. That takes a 20 hour lecture series on wave theory, on equivalent amount of basics of transmission line theory, and another series on antenna theory. I presume you've studied electrical engineering, but different programs go differently deep into such topics, so I really wouldn't know where to start. General rule of thumb: If you can read Pozar's Microwave Engineering, antennas become easier to understand. \$\endgroup\$ Jun 29, 2020 at 9:17
  • \$\begingroup\$ You may have more information over here: electronics.stackexchange.com/questions/30822/… \$\endgroup\$
    – vu2nan
    Jun 29, 2020 at 14:19

1 Answer 1

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The Maxim document is better at explaining things so I'll mainly reference that document. It talks about radiation resistance and dissipative resistance: -

enter image description here

Both these resistances are components of the loop but only radiation resistance is useful to you because that resistance multiplied loop-current squared is the useful power transmitted into free-space. The dissipative resistance is just heat loss and wastes power transferred from the transmitter node of the PA amplifier.

Both the above resistances will be small (circa 1 ohm) compared to the (typical) 50 ohm output impedance of the PA driver so, in order to maximize power transfer efficiency, a matching network is used called an L-pad: -

enter image description here

The example one above converts a 50 ohm drive to a higher impedance but also works the other way round: -

enter image description here

The theory has the same math: -

enter image description here

\$R_{IN}\$ becomes the resistance of the loop as derived from the Maxim equations higher up. \$R_L\$ is the 50 ohm drive impedance of the PA.

So, start with the lower equation for C and calculate it based on knowing \$\omega\$ and the two resistors: -

  • \$R_L\$ is 50 ohms
  • \$R_{IN}\$ is circa 1 ohm (radiation and dissipative resistance) for example

$$C = \dfrac{1}{2\pi\cdot\text{ 433.92 MHz}\cdot\text{ 50}}\cdot\sqrt{\dfrac{50}{1}-1} = 51 \text{ pF}$$

Inductance is the above capacitance multiplied by 50 = 2.57 nH and should also factor in the loop inductance.

$$\color{red}{\boxed{\text{These are just example numbers for the L-pad I used}}}$$

Maxim uses a slightly different L-pad consisting of C1 and C2 and I would recommend that you use a simulation tool (freely available) to get C1 and C2 values but, at the end of the day you are trying to match 50 ohms to something around 1 ohm in order to maximize power into that 1 ohm resistor that represents radiation and dissipative losses of the loop.

As for the Infinion circuit, it looks like they are trying to drive the loop in a balanced way hence they have matching impedance circuits at either end of the loop: -

  • R8 and L1 form one impedance match from low to high and, I suspect R8 will be around 50 ohms
  • R2, C9 and L2 match the PA output impedance to the other end of the antenna
  • I believe C7 and C8 add a little tuning to the loop

Again, this can be evaluated using a standard simulation tool like micro-cap.

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  • \$\begingroup\$ One thing that puzzles me is: usually antenna design is concerned with finding the length of the antenna for resonance in desired frequency. But, this approach seems to have a loop antenna based on the size available on the PCB and then proceeds to matching the antenna and PA based on the dimensions of the loop antenna. \$\endgroup\$
    – spockshr
    Jun 29, 2020 at 11:34
  • \$\begingroup\$ @s This is also known as a "short" antenna and if has a different impedance characteristic making it difficult to match to. So, you either make the monopole length a quarter wave and get 37 ohms radiation resistance or you go smaller and have to match to something much much smaller and can be sub 0.3 ohm as the radiation resistance. See this answer for example. \$\endgroup\$
    – Andy aka
    Jun 29, 2020 at 11:40

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