I have already asked this question here but this may be a better place.

I have built the following circuit to power a wireless LED but my calculations and measurements do not give the same values. I am measuring a voltage nearly 6 times larger than I would expect in my receiver circuit.

Circuit Diagram

My Circuit

I measured the AC frequency of the circuit to be \$\frac{\omega}{2\pi} = f = 350 \ kHz.\$ Assuming the magnetic field is given by

\begin{align} B &= \frac{N \mu_0 I}{2R}, \quad I = I_0\cos(\omega t) \approx 0.052A \cos(\omega t), \end{align}

and I use Faraday's law of induction, the induced EMF in the receiver circuit should be given by

\begin{align} \mathcal{E} &= - \frac{d\Phi}{dt} = - A \frac{dB}{dt} \\ &= A \frac{N \mu_0 I_0}{2R} \omega sin(\omega t), \quad A = \pi R^2 \\ & = \frac{N \pi R I_0 \mu_0}{2} \omega \sin(\omega t) \\ & = \frac{(30)\pi (0.05m)(0.052 A) (4\pi \times 10^{-7}H/m)}{2} (2\pi \times 350 \times 10^3 Hz) \sin(\omega t) \\ & \approx 0.339 \sin(\omega t) \text{ Volts } \end{align}

When I read the voltage in the receiver, I am actually getting a much larger value of over \$1.9 V\$.

Voltage Measurement

From searching, I believe this has something to do with resonant frequencies, but I do not understand this. If someone can show me why my calculation is wrong and provide the right one, I would appreciate it.

  • \$\begingroup\$ pardon me if this is obvious since I'm not experienced in magnetic calculations - but why is the magnetic field divided by 2R ? \$\endgroup\$ Commented Oct 21, 2022 at 19:55
  • \$\begingroup\$ @user253751 You can calculate this using the Biot Savart Law: hyperphysics.phy-astr.gsu.edu/hbase/magnetic/curloo.html. You multiply by N if there are N loops. \$\endgroup\$
    – Dayton
    Commented Oct 21, 2022 at 19:56
  • \$\begingroup\$ That's just calculating the field at the center (where it's weakest!), not the total field \$\endgroup\$ Commented Oct 21, 2022 at 19:57
  • \$\begingroup\$ @user253751 I know! So the voltage should be smaller if anything, not larger... \$\endgroup\$
    – Dayton
    Commented Oct 21, 2022 at 19:57
  • \$\begingroup\$ I think you should calculate a voltage that's too low since you're calculating a field that's too weak, and weaker magnetic field produces less voltage. \$\endgroup\$ Commented Oct 21, 2022 at 19:58

1 Answer 1


You measure AC RMS voltage with a device which is specified for max 1 kHz.

The device can be a frequency counter for megahertzes, but the voltage and current measurement frequency range is 40...1000Hz.

In addition you have non-linear load that has not been taken into the account. The current is surely quite different to different directions.

Finally: you are right, resonances can occur and cause difficult-to-predict effects.

An oscilloscope would be extremely useful if you wanted to see what's going on. One voltage, 1.935 volts, is quite thin piece of information of the waveforms.

You shouldn't expect sinusoidal signal from your transistor oscillator, not even in case the load is linear (now it's a diode). Single frequency calculations are worthless if there's heavy distortion.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.