Reply to Thread

Originally Posted by hotrodrob

here's a little article i picked up in a factory training class. should explain a few fundamentals for you

Great article, reminds me of the old VCR capstan motors. Works pretty much the same way but using much lower power.

Thanks, that scratched me right where I itched. And thanks DDCDan for your input as well. I was all ready to diagram your explanation out before i saw this article. I need a visual to understand this stuff.

Originally Posted by hotrodrob

here's a little article i picked up in a factory training class. should explain a few fundamentals for you

Nice. If you had posted yours first I I wouldn't have had to type all that out! :-)

Sent from my ADR6300 using Tapatalk 2

here's a little article i picked up in a factory training class. should explain a few fundamentals for you

OK, lets see if this makes any sense:

On your diagram, label the top three IGBT's 1, 2 and 3 left-to-right. Label the bottom three 4, 5 and 6. Label the motor lead connected to IGBT 1/4 as A, 2/5 as B and 3/6 as C. The very top line is the DC positive bus, and the bottom line is negative (always, does not ever change). So far so good?

Now, let's single-phase the motor to simplify things a bit. We will disconnect C (pop the fuse out? )and ignore IGBT's 3 and 6 for now.

So, let's start out by forgetting the inverter, that is, phases A and B are wired straight to the 480V AC line. What would we observe on a scope? At the beginning of a cycle A would go positive and B would go negative at the same time , peaking at about 678 volts betwen the two (the peak value of 480V RMS), then as the cycle continued, the polarity would reverse (alternating current - right?) and A would now go negative while B goes positive. That's what the natural AC waveform does. If we go back to 3-phase, its the same thing, only B-C doing it's thing 120 degrees later, and C-A 120 after that.

OK, now lets put our inverter back in, still ignoring C and 3/6. We turn on IGBT's 1 and 5 (oh yeah, the DC bus is charged up to, you guessed it - 678 volts). As you can see, phase A goes positive (through IGBT 1) and B goes negative (through IGBT 5). That covers the first half of the cycle. Now we turn those two IGBT's off. Oops, the collapsing magnetic field from the motor generates a big inductive voltage spike in the opposite direction of the voltage we applied, that's what the diodes are there for, to snub that back-EMF. Now we turn on IGBT's 2 and 4. this reverses the voltage - A now goes negative (through IGBT 4) and B goes positive (through IGBT 2). Just like our normal AC waveform - except what I just described would really be a square wave, not sine. So, instead of turning on the IGBT pairs for a complete half cycle, we rapidly turn them on and off (PWM) to approximate a sine wave.

Now wait a few milliseconds and do the same thing with IGBT's 2/6 and 3/5 (for B-C phase), then wait a bit longer and do it with 3/4 and 1/6 (for C-A phase), and you are now generating three phase power.

As you can see, IGBT's 1 and 5 are always turned on together, as are 2/4, etc. Bad things would happen if, for instance, 1 and 4 were to come on at the same time, this would be a direct short across the DC bus, and is called shoot-through by the engineer-types. Great pains are taken to prevent it, even for an instant.

This is a little simplistic, hope it helps!

Looking at it again, i guess I have another fundamental question: on any given cycle in a single phase, do the gates work in synch, or is one on and one off at either polarity?

VFD electronics explanation

Ok, so I know this might open a can of worms on electrical/electronics theory but I need to understand this....

I just took a class on VFDs. I understand how it works but I'm hung up on something. How exactly do the IGBT's in the inverter section create both polarities of the sin wave? Here are my references: (see attached graphics)

My questions are: isn't current always flowing from + to - on the DC bus (negative current flow?) or will it flow in the other direction through the energized IGBT if the diode is checking it (my best guess at how it works)? From that, do the IGBT's conduct through the diode if the gate is not energized? If the gate is energized will current go around the diode in positive current flow? What is the significance of the orientation of the collecting and emitting ends of the IGBT's?
I know these might not be easy concepts to explain, but anything to explain the original question would help. I'm going nuts trying to understand this.

Any graphics of current flow superimposed on a diagram of the inverter section, with corresponding references to the cycle and states of the IGBT's would be lovely!