sinewave inverters reactive power handling

[2dye4]

How do designs that transform dc-ac sinewave handle the return energy of a reactive load. I know the basic building block of an inverter is an H bridge with a filter after, using PWM to implement a sine wave wave output. But if you connect a large reactive load the power flow must be bi-directional. So energy must be transferred from a lower voltage to the dc bus voltage.

How is this done???
thanks

 

[dpc]

Not sure what type of reactive loads you are thinking about, but in general, real power only flows into the load, not out. The fact that the load is reactive just means the current is not in phase with the voltage. Inverters can handle this.

 

[CJCPE]

Whatever the inverter topology is, I believe that the antiparallel diodes allow current to flow to the DC bus during the portion of each cycle where current is flowing in opposition to the applied voltage. The bus capacitors receive energy from the load during that portion of each cycle and deliver energy to the load during the rest of each cycle. As long as the net energy flow in from the inverter to the load, there is no problem.

 

[davidbeach]

As dpc said, everything flows to the load, no return energy. The current is just not in phase with the voltage, but watts and vars both flow to the load.

 

[2dye4]

Have to disagree with the above about power flow.
Driving a reactive load entails energy transferred to the
reactive device during part of the cycle and then energy returned to the source during the other part of the cycle.

In a sine wave inverter fed from a DC bus power must be withdrawn from the load and returned to the bus despite various voltage levels.
In short 4 quadrant operation.
But how to do this is my question

CJCPE
In order for current to flow back to the antiparallel diodes the load voltage must have risen to the dc bus value.

 

[Skogsgurra]

Yes, there is more to your question than most guys think. I have had a hard time visualizing the current/voltage/time relationships. It does need some diagramming to become clear - I will search my books for such diagrams.

BBL



Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[CJCPE]

It is a lot easier to state in general terms what is going on than it is to explain it. Perhaps these will help:

H-Bridge-Inv-1.jpg

H-Bridge-Inv-2.jpg

H-Bridge-Inv-3.jpg

 

[2dye4]

CJCPE
That is close to the output stage i envision, except the
thyristors are FET or IGBT and they are driven with a PWM wave to get a clean sine wave. Your drawing appears to me to be a square wave inverter.
The reason for wanting a sine wave is that it causes no problems with any load designed for the grid as that is what the grid is (sine wave). This type of inverter is gaining usefullness in alternative power applications and anywhere sensitive equipment is being driven by back up.

I have heard that an additional full stage is often added in parallel in a flyback arrangement to achieve power return. I hope for something better though

 

[CJCPE]

The type of switching device doesn't change the handling of the return energy. PWM doesn't change it except that the diode vs. switch conduction periods are divided into small increments. I may have a diagram for that. In a sense, the diodes do constitute an additional full stage. You can consider them to be a full wave rectifier bridge in parallel with the inverter bridge.

The addition of a filter may make it desireable to change the configuration, but filters are available as an add-on accessory, so a different inverter configuration can't be essential.

There are a lot of multilevel inverter configurations. It may be that some of them require a different return energy handling method.

It may be that inverters that are designed for an output voltage that is always much below the input voltage may gain some advantage from a different return energy configuration. The fact that motor drive inverters have an output voltage operation range that is very wide demonstrates that the inverse parallel diode configuration is suitable for high input V / low output V operation.

 

[2dye4]

I am having trouble understanding the operation say in a region where the dc bus voltage is 170 V and the output voltage crosses zero but the current direction is still the same polarity as the prior cycle. Now this current will go through the anti-parallel diodes and the voltage across the load will be -170 Vdc because the inductive load is discharging back into the Dc bus. This it seems
would create square pulses at the Bus voltage until the inductive stored energy was absorbed then abruptly the voltage would fall to the level dictated by the PWM switching now occuring in the opposite legs. Some waveforms with sin output would help me understand.
Thanks for all the help so far

 

[Skogsgurra]

Still searching for that book!

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[bjenks]

The reactive currents are converted to heat in the inverter/output transistors, coil, and capacitors. Look at any UPS or Inverter and output power factor is one of the most important specs of the device. Most manufacture of UPS systems have a PF bandwidth they are able to handle. It is almost impossible to find this spec as they are assuming you are feeding a switch mode power supply that usually won't have a PF of less than .67. The real concern is not low PF loads, but high PF loads. Reason is that they rate these items in kVA and assuming you will have a non-linear power supply from the old days. So if you purchase a 10kVA UPS, it's inverter might only be rated for 7kW. This is a sneaky way of thinking you have a larger UPS than you really do as most power supplies are now power factor corrected. Thus if you purchase a 12kVA UPS for a 120V x 100A load and you think it will hold it because the UPS is rated for 12kVA you will probably overload it. This is because the inverter is rated at 10kW and the PF of the load is .98

I hope this sheds some light on the subject.