Increase input current in series resonant circuit

Question

I have an antenna/coil for an RFID application that I am resonating with a series capacitor. However, the current remains extremely low and does not agree with the calculated/theoretical values for its impedance. I am expecting high levels of current. See my setup below:

Vo (Vpp)	Vc(Vpp)	Io (mArms)	Theoretical Io (Arms)
29.2	29	70.7	2.06

*Operating frequency=125 kHz. Current was calculated using the voltage difference in Rs.

Do you have any idea of what I am missing to consider? I have tried reducing the parasitic resistance of the coil further (using Litz wire) and got similar results. I have done some simulations and they show non consistent values of current that depend on the inductance value (?). I have designed lower inductance coils (50uH) with even worse results. Obviously, I have increased the input voltage, but this is impractical since I am looking for >10A input current. Still, the input impedance of the resonating circuit shouldn't be that large. Finally, I have also made a frequency sweep to find the optimal operating point. Any help or advice will be appreciated!

Answer 1

Source

Those are mine theoretical values.

$$V_{rms}=\frac{29.2V}{2\sqrt{2}}=10.32V$$

Peak to peak voltege is divided by 2, so we get a peak voltage then it's divided by sqrt(2) to get RMS value.

$$I_{rms}=\frac{10.32V}{\sqrt{574.2^2+5^2}\Omega}\approx 18mA$$

The equivalent capacitance is approx. 1.387nF when you combine both Cs and Cr. So I have inserted the LC values in the online calculator and you see the absolute impedance value is 574.2 ohm at 125kHz. Since there is no resistors yet, this value is purely reactive. I have add series resistance of 5 ohm and calculated the absolute impedance with this added resistance. So the current shall be something around 18mA.

Now, why do you measure 70mA? Possibly due to coil stray capacitance which you didn't take into account.

Answer 2

Well, the RMS of the input current is given by:

$$\overline{\text{I}}_\text{in}=\frac{1}{\sqrt{2}}\cdot\frac{\text{amplitude input voltage}}{\left|\text{input impedance}\right|}=\frac{1}{\sqrt{2}}\cdot\frac{\frac{29.2}{2}}{\left|\underline{\text{Z}}_\text{in}\right|}\tag1$$

Where:

$$\underline{\text{Z}}_\text{in}=1+\frac{1}{\text{j}\cdot2\pi\cdot125\cdot1000\cdot853\cdot10^{-9}}+\text{j}\cdot2\pi\cdot125\cdot1000\cdot1.9\cdot10^{-3}+$$ $$4+\frac{1}{\text{j}\cdot2\pi\cdot125\cdot1000\cdot1.39\cdot10^{-9}}=5+\left(475\pi-\frac{341756000}{118567 \pi }\right)\text{j}\tag2$$

So, we get:

$$\overline{\text{I}}_\text{in}=\frac{73}{5 \sqrt{2 \left(\left(\frac{341756000}{118567 \pi }-475 \pi \right)^2+25\right)}}\approx0.0179611\space\text{A}\tag3$$

---

Finding the point where resonance occurs, we get:

$$\hat{\text{f}}=\frac{1}{2\pi}\sqrt{\frac{\text{C}_\text{res}+\text{C}_\text{s}}{\text{C}_\text{res}\text{C}_\text{s}\text{L}}}\tag4$$

I found that using Mathematica:

In[1]:=x = (1/Sqrt[
      2])*(((U)/
       2))/(R1 + (1/(I*2*Pi*f*C1)) + ((1/(I*2*Pi*f*C2)) + 
        R2 + (I*2*Pi*f*L)));
y = FullSimplify[
  Sqrt[(ComplexExpand[Re[x]])^2 + (ComplexExpand[Im[x]])^2], 
  Assumptions -> 
   U > 0 && R1 > 0 && R2 > 0 && C1 > 0 && C2 > 0 && L > 0 && 
    f > 0]; FullSimplify[
 Solve[{D[y, f] == 0, 
   U > 0 && R1 > 0 && R2 > 0 && C1 > 0 && C2 > 0 && L > 0 && f > 0}, 
  f]]

Out[1]={{f -> ConditionalExpression[Sqrt[(C1 + C2)/(C1 C2 L)]/(2 \[Pi]), 
    R2 > 0 && R1 > 0 && C2 > 0 && C1 > 0 && L > 0 && U > 0]}}

At that point we get:

$$\overline{\text{I}}_\text{in}\left(\hat{\text{f}}\right)=\frac{\text{V}_\text{in}}{2\sqrt{2}\left(\text{R}_1+\text{R}_2\right)}\tag5$$

Where \$\text{V}_\text{in}\$ is the amplitude of the input voltage and \$\text{R}_1\$ is the first resistor seen from the left side and \$\text{R}_2\$ is the second resistor.

Using your values we get:

• $$\hat{\text{f}}=\frac{500000 \sqrt{\frac{854390}{2252773}}}{\pi }\approx98014.3\space\text{Hz}\tag6$$
• $$\overline{\text{I}}_\text{in}\left(\hat{\text{f}}\right)=\frac{73}{25 \sqrt{2}}=2.06475\space\text{A}\tag7$$