I have a resonant parallel circuit, uing an electric capacitor, rectifier, oscillator and 2 coils.

I want to charge a cell phone wirelessly for a project at my school, so I researched this.

Generally, a cell phone is charged with 3.0 - 4.0 volts. So, how do I calculate how many ohms should my resistance be and the amps of my electric current? The formula is: V = IR, then

4 = IR, but I have two variables, then, how do I get it?

The diagram is this:

enter image description here

  • \$\begingroup\$ this instructable might help with your project: instructables.com/id/Wireless-Mobile-Charger You are charging by induction meaning your transmitter will convert a voltage to a magnetic field to be received/converted back to a voltage by another coil, rectified/regulated, and sent to the load. Not quite sure specifically what you are asking about resistance and current. As long as your circuit is capable of delivering enough current to the load to charge as quickly as you want, you'll be fine. \$\endgroup\$ May 31, 2018 at 22:38
  • \$\begingroup\$ that isn't resonance i think \$\endgroup\$
    – ESCM
    May 31, 2018 at 23:22
  • \$\begingroup\$ by "electrical intensity", do you mean current? \$\endgroup\$
    – Hearth
    May 31, 2018 at 23:25
  • \$\begingroup\$ Yes the current \$\endgroup\$
    – ESCM
    May 31, 2018 at 23:29
  • \$\begingroup\$ Have you read any information about how the Qi (Chi) wireless charging technology works? You could brouse wirelesspowerconsortium.com to start. Adafruit has a nice tutorial and there are even other posts on this platform. You don't strictly need to use their protocols to implement similar methods, but there are some good reasons that they made the protocol complicated. \$\endgroup\$ Feb 23, 2021 at 5:52

2 Answers 2


I am not familiar with resonant circuits but offer a couple of pointers.

Generally, a cell phone is charged with 3.0 - 4.0 volts.

While the battery may be 3 to 4 V, generally a phone is charged using an external 5 V supply. The charge controller is built into the phone. You need to supply 5 V.

So, how do I calculate how many ohms should my resistance be and the amps of my electric current?

You can get an estimate of the amps required from the charger supplied with the phone - typically 1 to 2 A. If you assume 2 A worst case then you can make a test load using a resistor of \$ R = \frac {V}{I} = \frac {5}{2} = 2.5 \ \Omega \$.

enter image description here

Figure 1. A USB power meter will allow monitoring of your charging load. Source: A random image search.

You can also calculate the power being drawn from \$ P = VI = 5 \cdot 1.64 = 8.2 \ \text W \$ for the device in Figure 1.

You will have some losses in the receiver, rectifier and voltage regulator so you will probably require 20 to 50% extra on the transmitter end just for these - and a lot more for the inductive losses.

  • \$\begingroup\$ And without that usb power meter? \$\endgroup\$
    – ESCM
    May 31, 2018 at 23:21
  • \$\begingroup\$ Cut a USB cable and put an ammeter into the red +5 V line. You can assume the voltage remains at 5 V. Expect the current to decrease at some point in the charge cycle as the battery charge level approaches full. This might be somewhere around the 80% level. \$\endgroup\$
    – Transistor
    Jun 1, 2018 at 7:09

Based on advice given to the OP on this question, it is paramount that to get success in this project, a more detailed block diagram is needed: -

enter image description here

You can't ignore the fact that you will need a proper bridge rectifier (schottky type) AND a smoothing capacitor AND a somewhat non-standard voltage regulator to be successful.

The bridge and the capacitor are easy but the voltage regulator is more difficult and success hinges on the performance of this part. As others have said, to charge a phone via its socket you will likely need to output a regulated 5 volts.

Then you have to consider how much current is needed by the phone to do basic charging. Don't raise expectations of several hundreds of mA unless you are going to develop a voltage regulator that is a buck switching type. This is because the receive tuned circuit will operate more effectively (better energy transfer) when producing a higher DC voltage on the smoothing capacitor.

This boils down to the Q of the tuned circuit and the potentially paltery amount of energy that can be gleaned through a weak magnetic coupling if you do not manage things correctly. You need to nearly always maintain as high a Q as you can to efficiently extract energy from the prevailing magnetic field.

It's almost like a solar charger application that uses MPPT control to maximize the set point to acheive maximum power transfer. Electric car battery chargers use MPPT control to obtain the best possible power transfer for instance.

But full MPPT control is probably not needed here BUT a buck regulator almost certainly is and that buck regulator needs to be able to run from a DC voltage nearly as low as 5 volts to possibly over 100 volts.

When the coils are far apart, the interaction between them is very slight and Q will be high. That's great for this type of application because you need a high Q to be able to capture the last few milli joules that are available at bigger distances.

When the coils are very close, the detuning effect on each other is significant and a form of regulation takes place automatically. That is also fine but, there can be a middle ground where the peak voltages produced by the receiver coil is many tens of volts so, you need to have a buck regulator that works across a broad range of input voltages.

So, concentrate on finding a decent buck regulator circuit is my advice. If you use decent schottky diodes, a decent smoothing capacitor and a decent voltage regulator you are giving your project the best chance of success. It then boils down to laws of physics and how you wind your coils and tune them.


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.