Three-phase Inverters

Three-phase inverters are used in high-power applications, as power can be more easily transferred by three cables than by two. They can be used with delta or wye (star) connected loads. For this article we drew upon Rashid (1995).

Three-phase full bridge inverter

As we can see in the figure below, the three-phase full bridge inverter consists of three H bridges operating together. This type of inverter is used in applications where three-phase loads can be connected independently. This implies that 6 cables are required to connect to the three-phase load. To connect a three-phase motor to this inverter, we must take those 6 cables from the inverter to the motor. This involves more expenses than simply taking 3 cables.

An inverter can be connected by means of a transformer to the load, which allows reducing the number of cables used in power transmission and reducing harmonics. In this case, the primaries can be connected independently, in delta or in star. On the other hand, the secondary ones can be connected in delta or star. If the secondary is wye (star) connected, multiples of 3 harmonics are reduced.

However, a transformer limits the type of application of the inverter, because it cannot be used in position control of synchronous motors, since it is not possible to have torque at zero frequency. A transformer needs a changing magnetic field to work.

In conclusion, in the case of an AC electric motor, it must be close to the inverter, to reduce electrical installation costs, or a transformer is put in between, which limits using the inverter only in speed control applications. As for the waveforms obtained, they will be like a single-phase inverter.

Three-phase bridge inverter

One of the disadvantages of the three-phase full-bridge inverter is the number of components, since a total of 12 switches and 12 diodes are required. The problem with this number of components is cost and reduced reliability.

As an alternative we have the three-phase bridge inverter. It consists of only 3 half-bridges, and only requires three cables to transfer power to the load. However, load waveforms are influenced by the connection type.

In the next figure, we show the three-phase bridge inverter with a load connected in delta. In this section, we will analyze the bridge waveforms. In the next section we will see the effect on the load.

In the following figure, it is shown the waveforms on the switches. As we can see, these waveforms consider that the switches operate with a 180° conduction. It is important to consider the nomenclature in the switch sequence. The sequence shown in the waveforms considers this nomenclature. It simplifies the implementation of inverter control. Compare this sequence with switching position in the preceding figure.

Delta connection

A delta-three-phase resistive load will be assumed to simplify the analysis. The following figure shows the equivalent circuit of the three-phase bridge inverter, during the 0 to $\frac{1}{3\omega}\pi$ interval.

As we can see, for that interval, we have three voltage sources connected in star. About $Rab$ we have the algebraic sum of voltages of $Va+Vb$, which gives $V_s$. The same goes for $Rbc$, where we have the sum $Vb+Vc$, but in this case the voltage is measured in the reverse direction, which gives us $-V_s$. Finally, on $Rca$ there’s no voltage drop because it is connected to two equal voltages, so the potential difference is zero. The full analysis is shown in the following figure.

Wye Connection

As we can see in the next figure, we have a three-phase bridge inverter, with the wye (star) connected load.

For the analysis, the 0 to $\frac{1}{3\omega}\pi$ interval will be considered again and a three-phase balanced resistive load, but this time the load will be wye-connected. As we can see in the following figure, we have three voltage sources connected in wye (star), but two of them in parallel. This will be done to streamline the analysis.

To determine the voltage, drop on the resistor $Ra$, $Van$, the voltage divider computing needs to be done

\[Van = \frac{\frac{Ra Rc}{Ra+Rc}}{\frac{Ra Rc}{Ra+Rc}+Rb} (Va+Vb).\]

Because the load is balanced, we’ve got that

\[Ra=Rb=Rc\]

and, on the other hand, we’got

\[Va=Vb.\]

Then we finally obtain

\[Van = \frac{1}{3} Vs.\]

For $Vbn$ and $Vcn$, we can have similar computations and obtain

\[Vbn = -\frac{2}{3} Vs\]

and

\[Vcn = \frac{1}{3} Vs.\]

If we follow the same analysis throughout the entire interval, we may get signals appearing on the next figure.

As we can see, the waveforms obtained are a better approach to a sine wave than just a square wave, but they differ from those in the delta connection.

References

Rashid, M. H. (1995). Electrónica de potencia: circuitos, dispositivos y aplicaciones (2a. Edición). México: Prentice-Hall Hispanoamérica.