#### kwired

##### Electron manager
Thanks for bringing up power factor. In my 40 years working on drives never occurred to me about the PF affecting the output current. Now that you brought that up can remember when we added PF capacitors to six 100 HP motors that ran 24/5 that would trip out maybe twice a year in a 100 degrees room years ago.None of the six motors ever tripped out after I stalling the capacitors.
PF is caused by reactance in motor windings, and is going to be very similar to running directly off the line, the non sinusoidal wave form may alter it some but will still be similar.

Supply side of the rectifier will not have a displacement power factor, may have some distortion power factor but is rather negligible in comparison to power factor on output side.

#### bwat

##### EE
MVAin as you called it is roughly the same numeric value as power in with a VFD because pf is close to unity.

MVAout is going to be more than power out because VFD output pf <1.
100% understood on all this. You're comparing VAin to POWERin, and comparing VAout to POWERout.

But what I was getting at was comparing the 'out' to the 'in'. Is VAin = VAout ? POWERin = POWERout? My initial reaction is that both of those can't be true.

#### kwired

##### Electron manager
100% understood on all this. You're comparing VAin to POWERin, and comparing VAout to POWERout.

But what I was getting at was comparing the 'out' to the 'in'. Is VAin = VAout ? POWERin = POWERout? My initial reaction is that both of those can't be true.
if power factor is 1.0 then VA and watts are the same. Therefore on the input side VA and power are nearly the same because input power factor is normally very close to 1.0. Output power factor is dependent on the load being driven.

#### petersonra

##### Senior Member
100% understood on all this. You're comparing VAin to POWERin, and comparing VAout to POWERout.

But what I was getting at was comparing the 'out' to the 'in'. Is VAin = VAout ? POWERin = POWERout? My initial reaction is that both of those can't be true.
Powerin = powerout, discounting conversion losses.

VAin = powerin, or very close anyway.

VAout value will exceed powerout value because PF < one. VA and power are related but are not the same thing.

VAin will usually be less than VAout.

Unless the VFD is off than they should be similar.

#### bwat

##### EE
if power factor is 1.0 then VA and watts are the same. Therefore on the input side VA and power are nearly the same because input power factor is normally very close to 1.0. Output power factor is dependent on the load being driven.
Yes, I understand the relationship of power, VA, and PF on the input (or output). I am talking about the comparing of in and out of the VFD. It's a very strange (but true) phenomenon of input CURRENT being less than output current, and I'm trying to wrap my mind around what's really happening in a way that I could explain it to somebody in a simple way.

Powerin = powerout, discounting conversion losses.

VAin = powerin, or very close anyway.

VAout value will exceed powerout value because PF < one. VA and power are related but are not the same thing.

VAin will usually be less than VAout.

Unless the VFD is off than they should be similar.
I think you're now getting to the root of my problem in understanding. The part I bolded (VAin < VAout) is eyebrow raising. I don't think that's possible. I don't say that with absolute certainty though, which is why I'm asking. If what you said is correct, this would mean that the VFD is creating VAs, which I don't think it can do?

What I was starting to lean toward was that the drive has the ability to convert between W and VAR, by basically shifting the phase/timing of the current, but it can't produce VA. So some of the input W would be "converted" to output VAR in a sense, but VAout has to equal VAin. So:

- VAin = POWERin = VAout

and so therefor

- POWERout < POWERIn (even with zero conversion losses!)

But I don't think that creates the situation of more measurable current on the output, so that's where I'm wondering what foundational piece I'm screwing up. Can a VFD really create VAs?

[VA = P + VAR]

#### LarryFine

##### Master Electrician Electric Contractor Richmond VA
The part I bolded (VAin < VAout) is eyebrow raising. I don't think that's possible.
I agree with you.

#### TwoBlocked

##### Senior Member
Wait a second. Watts-in is always going to be a little more than watts-out. There is always some loss regardless of what the device might be. But will VA-in always be greater than VA-out? Since watts equals VA x pf, what if the pf is poorer going out than coming in, yet there is only a little loss.? Then VA-out could be more than VA-in while the watts-in is still a little greater than the watts-out. Maybe an MG set or something?

#### TwoBlocked

##### Senior Member
From this webpage, Even at full load conditions, the input current will typically be lower than the motor current because the input still does not have any magnetizing current component in it.

#### bwat

##### EE
From this webpage, Even at full load conditions, the input current will typically be lower than the motor current because the input still does not have any magnetizing current component in it.

Thanks, but we're not really questioning whether the output current is higher than the input current. We know this to be true.

What we've been kicking around here on these last few posts is why / how exactly is this possible? It would seem that that only way that this is possible is that the output apparent power (VA) is greater than input VA. Which in a sense means that the drive is creating apparent power, which seems strange to say..

I had a little bit of an "ah ha!" moment this morning when thinking more about it. I don't have a problem saying that a perfect capacitor bank "produces VARs" without consuming any watts, which also means it's creating apparent power (VAs) in the circuit. So maybe that's the key piece to this: don't think of it as the drive producing VAs, think of it as the drive producing VARs. And if POWERin = POWERout, and it creates VARs on the output, VAout is going to be more than VAin.

How does that sound? Larry - would you buy that vowel? You seemed to be getting the same heartburn as me on this.

(and all of this is with a mythical loss-less drive of course)

I wonder if I should have split this off into a new thread..

#### TwoBlocked

##### Senior Member
Thanks, but we're not really questioning whether the output current is higher than the input current. We know this to be true.

What we've been kicking around here on these last few posts is why / how exactly is this possible? It would seem that that only way that this is possible is that the output apparent power (VA) is greater than input VA. Which in a sense means that the drive is creating apparent power, which seems strange to say..

I had a little bit of an "ah ha!" moment this morning when thinking more about it. I don't have a problem saying that a perfect capacitor bank "produces VARs" without consuming any watts, which also means it's creating apparent power (VAs) in the circuit. So maybe that's the key piece to this: don't think of it as the drive producing VAs, think of it as the drive producing VARs. And if POWERin = POWERout, and it creates VARs on the output, VAout is going to be more than VAin.

How does that sound? Larry - would you buy that vowel? You seemed to be getting the same heartburn as me on this.

(and all of this is with a mythical loss-less drive of course)

I wonder if I should have split this off into a new thread..
Reading the article, I gather what happens is the motor's magnetizing current comes from and goes back to the capacitor bank each cycle like a tuned tank circuit. This increases the VARs on the output, but not the input. Here is the entire paragraph from the webpage:

"The drive consumes input current proportional to the motor’s active torque demand, or load. The current needed for producing the magnetic field typically does not vary with speed and is provided by the drive’s main dc bus capacitors, which are charged during power up of the VFD. Under low torque conditions, the output current may seem to be much higher than the input current because the input current mirrors only the torque-producing current plus some harmonics but does not include the magnetizing current. The magnetizing current circulates between the dc bus capacitors and the motor. Even at full load conditions, the input current will typically be lower than the motor current because the input still does not have any magnetizing current component in it."

#### garbo

##### Senior Member
PF is caused by reactance in motor windings, and is going to be very similar to running directly off the line, the non sinusoidal wave form may alter it some but will still be similar.

Supply side of the rectifier will not have a displacement power factor, may have some distortion power factor but is rather negligible in comparison to power factor on output side.
Only a sparkies but between the capacitor bank that is the first thing feed from VFD input then to IGBT'S and the better line reactors that have capacitors would think the drive input power factor hopefully would be over 90% PF and not have too much distortion.

#### bwat

##### EE
Reading the article, I gather what happens is the motor's magnetizing current comes from and goes back to the capacitor bank each cycle like a tuned tank circuit. This increases the VARs on the output, but not the input. Here is the entire paragraph from the webpage:
Yesss..... that's it! That clicks now. Thinking of it having like a tank circuit component to it all, and the charging of the drive's capacitors (dc links) essentially as the prerequisite to making it all happen. Once they are ready to go, the caps supply any needed VARs in the output to match the motor inductance, and so the only thing that the drive's input needs is the real power requirement for the torque load. Hence, the input current can be less than the output current, since the output current has the real power AND the VARs bouncing back and forth between the drive's caps and the motor.

#### kwired

##### Electron manager
Thanks, but we're not really questioning whether the output current is higher than the input current. We know this to be true.

What we've been kicking around here on these last few posts is why / how exactly is this possible? It would seem that that only way that this is possible is that the output apparent power (VA) is greater than input VA. Which in a sense means that the drive is creating apparent power, which seems strange to say..

I had a little bit of an "ah ha!" moment this morning when thinking more about it. I don't have a problem saying that a perfect capacitor bank "produces VARs" without consuming any watts, which also means it's creating apparent power (VAs) in the circuit. So maybe that's the key piece to this: don't think of it as the drive producing VAs, think of it as the drive producing VARs. And if POWERin = POWERout, and it creates VARs on the output, VAout is going to be more than VAin.

How does that sound? Larry - would you buy that vowel? You seemed to be getting the same heartburn as me on this.

(and all of this is with a mythical loss-less drive of course)

I wonder if I should have split this off into a new thread..
The drive isn't creating the apparent power. The inductive load (motor) is.

If you have across the line connected motor that apparent power flows between the source and the motor. If you place power factor capacitors near the motor a majority of that apparent power is exchanged between motor and capacitor and the line to the source only sees the corrected power factor

With a VFD you have a rectifier as the front end of the drive, you then create a simulated three phase AC output with controlled DC pulses. The output is similar enough to true three phase the motor sees it as a three phase source. It still has inductance and apparent power flowing between it and the source (the drive) but that apparent power does not pass through the inverter or rectifier like it does an AC transformer to the supply circuit. At the most you only get a distortion power factor because of the high speed switching but no displacement power factor on the supply circuit.

#### LarryFine

##### Master Electrician Electric Contractor Richmond VA
So, it sounds like it's more accurate to say that the load conductors must be sized for the real plus apparent powers, just like between PF-correction equipment and a motor.

But not to say that more real power flows on the load conductors than on the line conductors. Is that correct?

#### TwoBlocked

##### Senior Member
The drive isn't creating the apparent power. The inductive load (motor) is.

...
Got a p/n for this motor that creates apparent power?

#### petersonra

##### Senior Member
You could conceivably have a VFD load that was 100% inductive. In that case the VFD input current would be close to zero but the output current would not be zero.

The is not any direct connection between input and output sections of a VFD. The input section feeds the bus capacitor. The output section takes whatever it needs from the capacitor and sends it it to motor.

It is not unusual for a VFD connected motor to actually generate power that is than thrown back to the bus capacitor. In such a case the input power is zero and the output power is negative because it is flowing from the motor back to the bus.

#### TwoBlocked

##### Senior Member
You could conceivably have a VFD load that was 100% inductive. In that case the VFD input current would be close to zero but the output current would not be zero.

The is not any direct connection between input and output sections of a VFD. The input section feeds the bus capacitor. The output section takes whatever it needs from the capacitor and sends it it to motor.

It is not unusual for a VFD connected motor to actually generate power that is than thrown back to the bus capacitor. In such a case the input power is zero and the output power is negative because it is flowing from the motor back to the bus.
"The(re) is not any direct connection between input and output sections of a VFD."

I wouldn't say that. It's not a transformer. There is a rectifier section, a capacitor section, and an inverter section. The same electrons that come in and out of the line side go out and in on the load side.

#### petersonra

##### Senior Member
"The(re) is not any direct connection between input and output sections of a VFD."

I wouldn't say that. It's not a transformer. There is a rectifier section, a capacitor section, and an inverter section. The same electrons that come in and out of the line side go out and in on the load side.
Not really true. It is like saying the same water goes into a lake as goes over the dam that makes the lake.

#### Jraef

##### Moderator, OTD
Staff member
… and I'm trying to wrap my mind around what's really happening in a way that I could explain it to somebody in a simple way.
Can you explain how Power Factor Correction Capacitors work? If so, it’s the same issue. In the VFD, the DC bus capacitors are storing and releasing the reactive power for the motor, the same as when you have PFC capacitors.

#### bwat

##### EE
Can you explain how Power Factor Correction Capacitors work? If so, it’s the same issue. In the VFD, the DC bus capacitors are storing and releasing the reactive power for the motor, the same as when you have PFC capacitors.
Yes, I was really not considering that it's the dc-link caps in the drive that make this happen with creating VARs, which increases the VA. I had in my mind that VAout couldn't be more than VAin, but that's just not true. I had no problem explaining how this works with PFC caps.

To be honest, this actually turned into a great example of how to better describe reactive power overall, and its impacts, without improperly describing it as some type of inefficiency or going into the math weeds or talking about sine waves, phase angles, or creating and collapsing fields, etc. I will use this in the future.

You can say that the 'grid' had to supply some VARs in order to first charge the caps of the VFD in the first place.
Then when the VFD runs the motor, the grid supplies the drive with real power that is then sent to the motor.
The motor needs VARs-energy for the magnetic rotation as well, and that energy comes from the previously charged VFD's caps and then bounces back and forth between the drive and the motor. It's not really produced or consumed.
We say 'produce' and 'consume' with VARs, but it's really only by convention. Realistically the energy bounces back and forth between two points.
And so in operation the grid and the drive input only sees the real power, but the conductors between the drive and motor see the current from the real power AND the VARs energy.
This shows why we try to mitigate VARs wherever possible because it causes you to increase cable sizes and creates more I^2*R losses between those to points that have the VAR-energy bouncing back and forth

Thanks all