How does this self resonant circuit works?

Thread Starter

Ams Sma

Joined Feb 11, 2017
17
Hi
To study the wireless power transfer basics i mounted the self-resonant circuit described in the following link .

Receiver
1627408978678.png

The receiver is easy to understand. No problems here!

Transmitter
1627409107217.png1627409127092.png

You can see the original schematic in the left image and in the right I rearranged the circuit and showed the component values that I’ve used.

The indutor is center tapped
To keep the coupling coefficient constant the distance between transmitter and receiver coils is always the same.


The system works well but I don’t understand exactly how the transmitter works.

After mounting the system I made the following experiments in the transmitter:

  • I measured with the oscilloscope the transmitter resonant frequency (around 470 KHz)
  • Removing C4 doesn’t seem to affect the circuit
  • Initially I assume that L1, L2 and C2 were the tank circuit but removing C1 has a big effect in the voltage amplitude (5 times smaller)and frequency (700 KHz).
  • If C1 or C2 are removed there is a big degradation in the power transfer.
    • With no load i continue to see the voltage in the receiver using an oscilloscope (without C2 slightly the same voltage and without C1 5 times smaller)
    • If a small motor is connected in the receiver won’t work if C1 or C2 are removed.

How does this circuit works?

What is the function of C4?

Thanks
 

nsaspook

Joined Aug 27, 2009
8,667
L1 (feedback side) and L2 (driven side) form a center tapped (180 phase shift from the center tap at the end points) tuning coil with C2 as the primary capacitance for the resonant frequency. C1 is the center-tap power supply bypass to make the signal impedance between VCC and GND low to reduce resonant energy losses during energy circulation in the circuit. C3 and C4 along with R1 are voltage divider elements in a bias circuit.

It functions like a Hartley oscillator.
https://en.wikipedia.org/wiki/Hartley_oscillator
 

Thread Starter

Ams Sma

Joined Feb 11, 2017
17
L1 (feedback side) and L2 (driven side) form a center tapped (180 phase shift from the center tap at the end points) tuning coil with C2 as the primary capacitance for the resonant frequency. C1 is the center-tap power supply bypass to make the signal impedance between VCC and GND low to reduce resonant energy losses during energy circulation in the circuit. C3 and C4 along with R1 are voltage divider elements in a bias circuit.

It functions like a Hartley oscillator.
https://en.wikipedia.org/wiki/Hartley_oscillator
Thank you for your reply.

Can you tell me why i can remove C4 with no apparent change in the circuit behaviour?
 
Last edited:

Thread Starter

Ams Sma

Joined Feb 11, 2017
17
Because there is a large amount of Gate to Source capacitance in the MOSFET.
It makes sense. Acording with the IRFZ44N datasheet the gate to source capacitance has a value close to C4.

Probably C4 is needed for bipolar transistors i have to check it.

Thanks!
 
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