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I have built a small prototype transmitter, based on a schematic available online. It transmits on a 5Mhz carrier and uses BPSK modulation. It works well enough for my purposes.

The schematic for the TX output section looks like the below:

PCB0025 antenna section

My question is: if I reduce Vcc from 5v to 3.3v and also reduce R4/R5 to get the same current, will I get the same output from the TX antenna?


Further details

I would like to replace the current 9v battery & LDO with a single AA cell and boost converter. I have other existing circuits with a single AA cell but the IC I am using only goes to 3.3v from a single cell.

MOD1 and MOD2 in the schematic are the same for "quiet time" (when not transmitting) and always inverse when transmitting. When they change a carrier inversion happens.

The PCB trace inductor loops around the outside of the board, it can just about be seen in the image below. There are 10 turns on the top and bottom layers, winding in the same direction. I measured this to be ~25uH using a Peak LCR meter (I am aware that the test frequency is not the same as my carrier, therefore this is not perfect).

Rendered PCB

When doing some research I found a schematic in an FCC filing for a product similar to what I'm prototyping. Whilst the modulation is different the antenna and capacitors are quite similar. This suggests that the schematic I found online is at least mostly sensible. This design is fed from one end (the other is GND) and only has a 22R resistor.

FCC filing

I happily admit that I do not have much idea of how the antenna works, R4 and R5 in my design are there to ensure the 74AC86 current limits are not exceeded. Looking at it whilst writing this question I suspect R5 could be completely removed.

An answer that explains how this works (in simple terms for a non-EE) would allow me to understand my own question and perhaps refine the design further.

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2 Answers 2

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First of all, it isn't the resistors that are limiting the current out of the gates — the AC impedance of the capacitors at 5 MHz is more than enough for that.

A more important function of the resistors is to isolate the resonant tank circuit from the low output impedance of the gates, allowing the Q of the circuit to be higher than it would be without them.

If anything, I would be tempted to raise the values of the resistors, to maybe 150Ω or 220Ω, to provide a better match to the antenna.

Otherwise, if you want to redesign the antenna to work at a lower voltage and higher current, you're going to have to reduce the inductance and raise the capacitances by the same ratio in order to keep the same resonant frequency.

This type of antenna only interacts directly with the magnetic field of an EM wave, so its "power" is directly related to the magnetic field it can produce, which is a function of the current through the inductor and the number of turns in it (Ampere-turns).

Since inductance is proportional to N2, halving the number of turns allows roughly 4× the current to flow for a given voltage, which gives a net 2× improvement in the overall Ampere-turns value. This is an extreme example that shows the principle — in practice, you'll probably want to make a smaller adjustment to your present design, and you do still need to pay attention to the amount of current the gates can supply.

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  • \$\begingroup\$ Interesting. On this circuit if resistance is infinite or zero, I see the Q being infinite so there should be a resistance value that would create lowest Q . \$\endgroup\$
    – Andy aka
    Jan 18, 2014 at 18:53
  • \$\begingroup\$ @Andyaka: Well, yes, but keep in mind that the resonant frequency also shifts since you're effectively shorting out one of the capacitors. \$\endgroup\$
    – Dave Tweed
    Jan 18, 2014 at 19:59
  • \$\begingroup\$ +1. | Possibly useful due to OP's experience to note that turns/2 needs C x 4 (approx) to get back to resonance. Also that resonance is vital. | De-Qing is presumably aimed at accommodating spread of component values to allow no SOT. Tank caps are probably ceramic. Using eg polystyrene caps would give less spread and as long as inductor construction was as consistent as possible would allow higher Q and usefully more range (assuming data rate is not vast). . \$\endgroup\$
    – Russell McMahon
    Jan 18, 2014 at 23:20
  • \$\begingroup\$ @DaveTweed thanks, this is very helpful. You suggest raising resistance for match, but in the second schematic it is much lower in the tank. Why the difference? \$\endgroup\$
    – David
    Jan 19, 2014 at 18:44
  • \$\begingroup\$ The resistor in the second schematic isn't for matching. I suspect it's there merely to reduce the Q of the tank in order to make it less sensitive to component variations. \$\endgroup\$
    – Dave Tweed
    Jan 19, 2014 at 19:03
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The antenna appears to be a magnetic field transmitter tuned by the capacitors. The resistors in your design reduce the Q of the circuit. This has the effect of widening the bandwidth of the "antenna" AND reducing the maximum amplitude. The reason is that it is easier to tune the antenna to the transmitter frequency.

Transmission range won't be great because you won't be emitting a full and proper EM wave. It will work at 3.3volts at a reduced distance but reducing the resistors and a possible retune of the capacitors should be enough to recover that range.

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  • \$\begingroup\$ +!. Note that R is in tank proper in 2nd cct (and not in 1st.) \$\endgroup\$
    – Russell McMahon
    Jan 18, 2014 at 23:14
  • \$\begingroup\$ A small loop antenna certainly does emit a full and proper EM wave (albeit not as efficiently as say, a resonant dipole). Is that not what this is? \$\endgroup\$
    – Phil Frost
    Jan 19, 2014 at 14:35
  • \$\begingroup\$ @PhilFrost because it's multiturn it won't be effective in the E field compared to H field. Also given its size versus operating frequency this will also make it virtually a H field device. \$\endgroup\$
    – Andy aka
    Jan 19, 2014 at 14:59
  • \$\begingroup\$ @RussellMcMahon interesting observation, what difference is there if the R is in the tank vs. not in it? \$\endgroup\$
    – David
    Jan 19, 2014 at 17:12

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