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I was researching about inverters when I came across the Z source inverter. The Z source inverter is unique in a sense that it enables a shoot-through or a short circuit condition by closing any two switches of the same leg. This would not be possible in a traditional inverter as such a situation would cause the switches to get damaged. But a shoot-through state in a Z source inverter is possible due to the presence of the so called "LC" network. My question is this:

1) How does the LC network prevent any damage of the switch during the short circuit stage?(this is similar to a boost converter when the inductor charges during the 'on time' of the switch in a standard PWM. The closed switch and the inductor effectively creates a short circuit but there is no damage of the switch).

2) What is the use of connecting the capacitors in an "X" like fashion in between the inductors?

enter image description here

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How does the LC network prevent any damage of the switch during the short circuit stage?

I prefer this schematic for a Z source converter because it contains the all-important diode (red circle): -

enter image description here

That diode is important because without it, when T1 / T2 and/or T3 / T4 turn-on to create the shoot-through condition, energy stored in the capacitors would be lost instead of being "allowed" to produce double the supply voltage. Maybe it might be easier to see it if re-drawn: -

schematic

simulate this circuit – Schematic created using CircuitLab

In quiescent conditions node A will be at 100 volts and node B will be at 0 volts. Node C will also be at 100 volts (assuming an ideal diode)

When the switch closes (representing the shoot-through condition) the capacitors are placed in series and so node C suddenly rises to 200 volts and reverse biases the diode. This means that the applied input voltage to the inductors is twice as high. As shoot-through progresses the current ramps up as per a normal booster circuit but with potentially twice the input voltage.

What is the use of connecting the capacitors in an "X" like fashion in between the inductors?

Hopefully, with the redrawn diagram you should be able to see that now.

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  • \$\begingroup\$ Couple of questions here. 1) Assuming current flows through the inductor before it reaches the capacitor, how will node A be at 100v? Wouldn't there be a voltage drop due to the inductor? 2)"energy stored in the capacitors would be lost instead of being "allowed" to produce double the supply voltage" by "losing energy", do you mean the discharging of the capacitor? If yes, then how does the diode "prevent" this? what do you mean by "supplying" double the voltage? 3)If the voltage boosting is done by the capacitors only, what role do the inductors serve? \$\endgroup\$ – noorav Jan 20 at 17:07
  • \$\begingroup\$ In quiescent conditions, the inductor will look like 0 ohms and the capacitor will have charged up. Even in light or medium loading conditions this will be the case but, the voltage will be a few or perhaps ten volts below 100 volts. It all depends on the amount of shoot through time and regular bridge tings that have previously passed. \$\endgroup\$ – Andy aka Jan 20 at 17:34
  • \$\begingroup\$ The capacitors will try and dump energy back into the supply hence, because the combined series voltage across the two caps will be greater than the DC bus voltage, the diode will become reverse biased. \$\endgroup\$ – Andy aka Jan 20 at 17:35
  • \$\begingroup\$ The inductors are needed just like in any boost converter circuit. The capacitors providing near double the voltage make the overall job of boosting easier (if it’s needed). \$\endgroup\$ – Andy aka Jan 20 at 17:38
  • \$\begingroup\$ Ah okay. But I still didn't quite get the part about how the switch is protected from high short circuit current? Do the passive components i.e. inductor and capacitor do this job in protecting the switch? If yes, how? \$\endgroup\$ – noorav Jan 20 at 18:08

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