I have a pure sine wave home ups with a mains 1000VA transformer in it.

Transformer is 6.7/220VAC. How does it charge a 12V battery with 20A?????????

6.7V side is connected in H bridge mosfets.

Transformer Configuration Inside UPS (Ignore C1)


I want to design a charger with high current. Thats why I am asking.

Note: Charging Works even when mains voltage is as low as 100V.

  • \$\begingroup\$ Replace the transformer with a sine wave generator and simulate? It's a pretty simple configuration. \$\endgroup\$ – hekete Jul 23 at 6:01
  • 1
    \$\begingroup\$ How does simulation gonna help? All I am asking is how does battery get charged when mains voltage is close to 100VAC and transformer output 3.2 Volts??????? I heard somewhere that while charging, the bottom two mosfets do switching while upper ones are off. \$\endgroup\$ – Vikas Kumar Jul 23 at 9:26
  • \$\begingroup\$ Your question is about the charging the battery. Your schematic shows the inverter output stage. Why? \$\endgroup\$ – Transistor Jul 23 at 11:48
  • \$\begingroup\$ @Transistor: I think if you "properly abuse" the H-bridge you can use it to charge the battery - assuming lead acid batteries and you don't mind overcharging/boiling them. \$\endgroup\$ – JRE Jul 23 at 11:56
  • \$\begingroup\$ @Transistor Maybe because the same transformer is used both ways, depending if charging or working as inverter? But it is indeed confusing since "pure sine wave" was mentioned and this usually comes together with an UPS without a "short break" (when the device turns on the inverter shortly after a power failure). \$\endgroup\$ – vangelo Jul 23 at 11:56

Since I'm curious about this and nobody posted an answer, I'll post this attempt to explain, to propose a discussion.

If all MOSFETs are off, they work as a bridge rectifier, using the body diodes. But since the secondary voltage is too low to charge the batteries, if both MOSFETs at the bottom are turned on briefly and quickly at the same time, they could implement a kind of boost converter, using the secondary of the transformer.

As an attempt to explain what I mean, please check this concept. \$V_O\$ is \$20 V_{peak}\$ when the circuit is opened. The switch is controlled by a simple \$50 kHz\$ square wave (no control loop, obviously). The diodes function as the bridge rectifier:

enter image description here

The rectified voltage, for the conditions simulated here, is higher then the open circuit secondary peak:

enter image description here

  • \$\begingroup\$ Thanks. You madey day. \$\endgroup\$ – Vikas Kumar Jul 25 at 15:55

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