This case is a combination of inverters in voltage control mode (click here) and in current controlled mode interfaced to them (click here). The first inverter, Inverter 1 operates only in voltage control mode and is the main three-phase voltage supply in the system. Inverter 1 is fed by a separate three-phase supply interfaced through a three-phase diode rectifier. This three-phase supply is fed through back-to-back switches that are opened during the simulation to show the effect of intermittent supply. Inverter 1 feeds a three-phase passive resistor-inductor load. During a failure of the supply to Inverter 1, the voltage at the load bus will collapse. To prevent this, inverter 2 is connected to the load bus. Inverter 2 can operate in two modes. It can operate in current control mode under normal conditions. If Inverter 2 is supplied by a battery, it would have to draw a minimal charging current from the system to maintain battery voltage and then regulate its output current to zero while staying connected to the system. When the supply to Inverter 1 fails and it can no longer supply the load, Inverter 2 changes its mode of operation to voltage control mode and supplies the load. In this manner, the supply to the load is not interrupted. In this simulation, the supply to inverter 2 is a dc source instead of a battery for simplicity.
To begin with let us examine the operation of the circuit without details of control. The figure below shows the three-phase supply feeding the dc bus of Inverter 1. At 0.4s, the supply is interrupted by opening the switches and is restored only at 0.8s. During this time, the dc bus voltage at Inverter 1 drops to a level where Inverter 1 controls block all pulses to the devices. Typically, the dc bus voltage would continue to fall, however, this simulation does not contain a resistance small enough across the dc bus of inverter 1 to cause this. Conceptually, this does not change the results.
Next, let us examine the operation of Inverter 1. Inverter 1 operates in voltage control mode and maintains a three-phase balanced voltage of RMS value 208V at its output. At 0.4s, the output voltage begins to drop because of falling dc bus voltage. This can be seen in the next plot. When the dc bus voltage falls below 325V, Inverter 1 control switches off the inverter to prevent the output voltage becoming distorted. The output voltages which are the voltage across the filter capacitors are seen to be frozen at their values until 0.8s because they do not have bleeder resistors across them that will cause their voltage to decay. At 0.8s, when the supply is restored, the dc bus voltage rises again and Inverter 1 resumes to regulate its output voltage to 208V RMS.
Now comes the operation of Inverter 2. As stated before, Inverter 2 will under normal conditions operate under current control mode and regulate its output current to zero. Only when the load voltage falls below 100V, the mode of operation will change to voltage control mode to maintain supply to the load. Inverter 2 is switched on at 0.1s after which its output current shown below is approximately zero. At 0.4s, when load bus voltage begins to fall, its output current increases as Inverter 2 is now supplying the load. At 0.8s, when Inverter 1 resumes operation, Inverter 2 goes back to current control mode. It should be observed that the fairly large current spike seen at 0.8s is because Inverter 1 produces an output voltage which is not synchronized to the output of Inverter 2. Typically, Inverter 2 switches Inverter 1 out of the circuit when its output voltage begins to fall and switches it back on when its output voltage is regulated to normal values. However, when switching Inverter 1 back on, Inverter 2 needs to change its output to match that of Inverter 1 before switching it on or else the current spikes as seen will occur.
Below are shown the output voltages of Inverter 2. Until 0.4s, the output voltages of Inverter 2 will be approximately equal to the output voltages of Inverter 1 in order to maintain its output current to be zero. Between 0.4s and 0.8s, Inverter 2 will supply the load and therefore, its output voltages will be of magnitude 208V. After 0.8s, the output voltages of Inverter 2 will again be equal to that of Inverter 1.
Below is the plot of the voltages at the load. As can be seen from the previous plots, this voltage is merely a combination of them as the inverters maintain the load voltage together. Though complete disruption of the load voltage has been prevented for the period of 0.4s to 0.8s when the three-phase supply failed, a rigid regulation of the load voltage has not been achieved as can be seen from the dip in voltage around 0.4s and a rise at 0.8s. This is because when inverter 1 output drops at 0.4s, it takes a few cycles for Inverter 2 to detect this collapse and switch Inverter 1 off and change its mode of operation. Similarly, the improper synchronization at 0.8s results is a rush of current that causes the load voltage to rise. Depending on the load, these fluctuations may not be acceptable and the control may need to be fine tuned.
Now to examine the control in detail, let us examine specifically the time window between 0.35s and 0.55s. The plot below shows the input currents of the diode rectifier and the dc bus voltage of Inverter 1. The currents are the typical non-linear currents drawn by a diode rectifier that supplies a dc bus capacitor.
The plot below shows the output voltages of Inverter 1. At 0.4s, they can be seen to gradually fall and become distorted due to the decreasing dc bus voltage. Just before 0.46s, Inverter 1 is disconnected by Inverter 2 since the output voltage has fallen below 100V. To enable this control action, Inverter 2 monitors the output voltage of Inverter 1 as described below.
Inverter 2 monitors the magnitude of Inverter 1 phase a, b and c voltage and calculates the peak every quarter of a cycle to check for falling voltage. As can be seen, the voltages of all three-phases fall below 100V just before 0.46s. This is when Inverter 2 detects that the supply has failed.
The above online UPS will have application in the smart grid applications that will follow. Therefore, there was the need to simulate this system. A battery fed inverter can be used to back up a renewable energy based inverter when the source which may be either a PV or a wind turbine stops producing power. These will be simulated in later cases particularly in integrated systems.