Wireless charging circuit for FYP - MOSFET overheating & unsure of how to insert a USB output port

Thread Starter


Joined Mar 7, 2018
I'm currently working on the design of a wireless charging circuit to produce an output of 5V & 1A.

The transmitter circuit is supplied with 15V and generates a pulse via the use of a 555 timer (NE555P). The timer is configured to produce an output frequency of ~10kHz-13kHz.

However, the maximum current output of this timer is ~200mA (too small to feed in to a receiver coil).

To amplify this current, I have used an NPN BJT (2N2222) to drive an n-channel MOSFET (IRF540). The drain of said MOSFET is connected to an inductive load made of rounded copper wire (L ~ 0.93mH), acting as a transmitter.

The receiver coil is made identical with a 1:1 turns ratio, connected to a capacitor in parallel with a diode bridge (1N4007) to fully rectify the pulsating DC waveform received from the transmitter coil. This is then fed through to a capacitor and LM7805 5V regulator. At the moment I am connecting a load resistor (1K) in series with an LED to ensure the circuit is working (and it does).

A circuit diagram is attached (image file)

My first problem:

The IRF540 I am using has a large heat-sink connected, as it becomes quite hot very quickly. I am not sure why this is. The current measured across the output of the transmitter coil is close to 2A and 15V (I will recheck these values to confirm).

In addition, it is worth noting that this circuit does allow energy to be transferred from one circuit to another (this has shown through testing). However, I am not satisfied with the fact that the MOSFET becomes so hot so quickly. Is there a problem with its ON/OFF state? A problem with the input resistance? (I am a final year EEE student, so please forgive any lack of understanding with regards to the operation of the MOSFET in this particular instance)

My 2nd problem

The voltage regulator connected at the output produces 5V and ~4-5mA (as expected considering 1k resistor value added with LED at load, although this is not shown correctly in the circuit diagram below). I have ordered a 10ohm and 5 ohm power resistor (50W rating) to produce 0.5A and 1A, respectively (to provide 2.5W or 5W for charging). Other resistors used to control the load current with smaller power ratings become very hot very quickly (a 3.3ohm resistor I used started to become very hot, even with a 14W power rating with a 5V regulated output).

Now, if I am able to control the current using a more powerful, suitable resistor to produce the desired output current, how would I then connected this to a USB port to provide a charging mechanism? Does anybody have any experience in setting one up or trying one out? Thank you in advance to anybody who can help me. I'm not entirely sure how to incorporate this in to my current design.

Finally, please let me know if I can be any more specific in key areas ( am happy to provide any further information, if required). I am somewhat happy with the fact that my design mostly works, although if this was not a FYP then I'm sure this circuit would not be considered very good, considering it appears that it will be unable to charge anything for more than 30 seconds or so (due to overheating of components).

Circuit diagram included below

WhatsApp Image 2018-03-08 at 02.14.30.jpeg

Thread Starter


Joined Mar 7, 2018
Please bare in mind that the output of the receiver circuit in the schematic does not show 5V regulator or load resistor, although I'm sure (hopefully) that anyone reading will understand my explanation, regardless. My apologies for any inconvenience.


Joined Feb 8, 2018
What's an FYP?

How do you know what the inductance of the coil is?
Can you post a photo that shows how the coils are arranged?

The transformer coupling will be somewhere between poor and very poor. When the FET is on, substantial energy will be stored in the inductance of the primary winding. When the FET switches off, that energy has to go somewhere. Because the coupling to the secondary is poor, some fraction, probably a large fraction, of the stored energy has no place to go - actually it does have a place to go, just not the intended place. Since it is an inductance it will try to keep the current in the circuit constant. The only place the current can be forced to flow is in the FET that is turned off. The voltage will rise to whatever level is required to keep the current flowing - the inductor doesn't "care" if the voltage is 20 volts or 500 volts. In reality, some of the energy will flow through the FET as it is trying to turn off, causing high transient power dissipation. Most of the energy will cause reverse breakdown of the FET's body diode. This will probably happen at around 130-150 volts - some voltage above the rating of the FET. If there was 1 amp flowing in the coil, there is now going to be a transient power dissipation of 1 A x 130 V = 130 W. The current will continue to flow until the energy that is stored but can't couple to the secondary is dissipated. It is acceptable to allow this to happen provided the total energy is not excessive.

1N400x diodes are unsuitable for high frequency. The reverse recovery time is too long. UF400x diodes are more suitable if the current rating is adequate.