Reducing system voltage whilst getting the same output from a PCB antenna
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:
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?
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).
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.
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.
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.
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.