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In half-bridge inverter, two capacitors are used to get two voltage sources, each of potentials V/2 from a single voltage source of potential V.

When either of the transistor conduct, the respective capacitor discharge through the load.

But after being discharged, how does the capacitors get charged back to potential V/2? What path does the charging current flow through? If the charging current flows through path Vs-C1-C2, then voltage of the other capacitor would also rise and go above V/2. Why doesn't it happen?

  • \$\begingroup\$ How do you sense error of V1/2 , V2/2 and V1-V2 and control Rce of IGBT \$\endgroup\$ Apr 5, 2018 at 16:07

1 Answer 1


The charging / discharging is intentionally a small fraction of the supply voltage and, providing the duty cycle is 50%, the currents from the load alternate very quickly and the average voltage at the junction of the two capacitors remains largely at Vs/2. The capacitor value has to be sufficient for this to happen of course and the duty cycle needs to be about 50%.

  • \$\begingroup\$ but what is the path of charging current of capacitor? \$\endgroup\$ Apr 5, 2018 at 16:20
  • 1
    \$\begingroup\$ The top transistor charges the lower capacitor and the bottom transistor charges the upper capacitor. \$\endgroup\$
    – Andy aka
    Apr 5, 2018 at 16:25

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