Power Electronics Circuits
In a voltage-source PWM inverter, which of the following correctly describes the effect of the dead time on the output voltage?
- ADuring the dead time the output terminal is connected to nothing and the output voltage is always zero, so the output voltage drops by a fixed fraction regardless of the current direction
- BThe dead time only prevents shoot-through of the upper and lower switches; it does not affect the output voltage and merely adds a little loss
- CThe output voltage during the dead time is set by the current direction, causing a voltage error that depends on the direction and distorting the current at low output voltages
Show answer and explanation
Answer: C. The output voltage during the dead time is set by the current direction, causing a voltage error that depends on the direction and distorting the current at low output voltages
During the dead time both the upper and lower switches are off, so the voltage of the output terminal is determined by whichever diode carries the current. If the current flows out to the load, the lower diode conducts and the output goes to the negative side; if it flows in, the upper diode conducts and the output goes to the positive side. An error that makes the voltage slightly lower (or higher) than the command therefore arises according to the current direction, and the waveform is distorted near the zero crossings of the current. The effect is most visible at low speed, where the output voltage is small, so dead-time compensation, which corrects the command according to the current direction, is used.
- A. The voltage is not necessarily zero; whether it goes to the positive or negative side depends on the current direction.
- B. The dead time causes an error in the output voltage, with a large effect especially at low voltages.
- C. Correct. The sign of the error changes with the current direction, so the current is distorted.