I'm trying to simulate the following circuit enter image description here

taken from http://www.techlib.com/electronics/battery_chargers.html.
The circuit is supposed to charge 6V battery (shown as Zener diode in the circuit). According to the author, this circuit can charge a battery with a higher voltage than what the cell generates. The solar cell is simulated with a 3V battery and a resistor. The circuit is here https://www.circuitlab.com/circuit/859v6932b322/solar-charger/
I can see that the transistors are on, but the zener diode does not turn on meaning that its voltage does not reach 6V.

Any idea why this is not working? Any suggestions for alternatives/changes?


2 Answers 2


You cannot make an analysis of a 5kHz (0.2ms between pulses) converter in steps of 1ms. Put time step to 1us (0.000001) and the stop time to 0.05 (it will last forever but you can stop-it before to see the result) and see that it works.


You forgot a diode between L1, C3.

Take 2: Not all pusblished designs work! This one is very fussy and not very efficient and the battery charger article discusses some variables. Not shown is self capacitance of inductor and ESR of caps.

The model and actual RC and RLC values for each reactive part matters and not just C or L.

Also spurious oscillations can occur.

Try adding 200 ohms in series with C2. This adds stability to the loop with phase lead/lag compensation. Then lower ESR of pile from 22 to 5 ohms.

  • \$\begingroup\$ That doesn’t make a difference. The diode’s in the circuit in the simulator. \$\endgroup\$
    – vmontazeri
    Commented Mar 28, 2018 at 23:50
  • 1
    \$\begingroup\$ Tony, is not this clearly a comment and not an answer? Or is this just a place-holder for a later answer? \$\endgroup\$
    – jonk
    Commented Mar 29, 2018 at 0:08
  • \$\begingroup\$ initially it was an answer but error later corrected on question, I don't think there is a perfect answer as this design is to fussy with parasitics and reactive imperfections. Also the Pile ESR if too high can drop source voltage while integrating coil current to 0.5A so that it loses energy. Impedance matching is crucial. \$\endgroup\$ Commented Mar 29, 2018 at 0:38

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