I'm putting together a wireless power transfer based battery charger to charge an 11.4 V, 3830 mAh battery from the output of a 15 V voltage regulator I made.

Due to a limited understanding of battery charging circuits, I'm not sure if the output of my simulation indicates that the circuit is functioning as intended to charge the battery. The voltage is very messy looking but the steady-state current looks close to where I think it should be in order to charge up the battery in 45 minutes to an hour.

Note that the simulation shows voltage and current from the 0.1 Ohm load I placed where the battery will be hooked up.

Here is the topology in LTSPICE: enter image description here

Here is the waveform. The voltage is all over the place but the average current is around 1.3 Amps. GREEN is current at the load, BLUE is voltage across the load.

enter image description here

Should the messy voltage signal across the load be a concern if the steady state current is reasonably precise to the desired value? I placed a 0.1 Ohm resistor based on Battery University's suggestion of typical charging battery internal resistance.


  • \$\begingroup\$ What's the purpose of C1? \$\endgroup\$ – winny Dec 3 '17 at 9:07
  • \$\begingroup\$ It's a resonance capacitor. It should be reducing current-ripple in the output. \$\endgroup\$ – user8357 Dec 3 '17 at 9:12
  • \$\begingroup\$ Formed by L3/L4? \$\endgroup\$ – winny Dec 3 '17 at 9:47
  • \$\begingroup\$ Also, simulating the battery as a zener diode plus series resistor comes to mind to give approximately correct V-I-characteristics instead of just an R. \$\endgroup\$ – winny Dec 3 '17 at 9:58

The chaotic, noisy, huge voltage signal (±150V from a 15V input??) is a sign that your simulation has gone off the rails, and that its results cannot be trusted.

As a first step, add a ground to the battery side of the circuit. Yes, I know it isn't actually grounded… but leaving that part of the circuit floating is likely to confuse the simulator.

If that doesn't help, try decreasing the time step.

  • \$\begingroup\$ Thank you. I omitted ground because I wanted the receiver and transmitter to be isolated. I connected a 1 Meg resistor between the transmitter and receiver grounds and that fixed the voltage to between -1 and 1 V. \$\endgroup\$ – user8357 Dec 3 '17 at 9:46

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