# Thread: Single Phase current draw for a 3 phase output VFD: technical discussion

1. gar
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170516-2141 EDT

mike_kilroy:

Won't Diode heating be function of Irms? Is not rms the heating amount as opposed to Iaverage?
Only part of the heating could be allocated to a linear resistance. Thus, the answer is no, and you should calculate average power from the product of the instantaneous voltage and current integrated over some long enough time period to obtain an adequate average and divide by that time period.

A good approximation for a silicon diode would be a value somewhat above 1 times average current. The number about 1 will be a function of diode junction temperature. Look at the curves in the diode datasheet I referenced above in Post #3, Figure 7.
Last edited by gar; 05-16-17 at 10:12 PM.

2. Originally Posted by mike_kilroy
Won't Diode heating be function if Irms? Is not rms the heating amount as opposed to Iaverage?
The usual simple method is to calculate it in two parts. Threshold voltage, Vto, and slope resistance, rt. But you need to know the current.
For most of the systems I've worked on/designed, an approximation of level DC current was good enough to size the semiconductors, the heatsinks, and airflow.
But, as gar has shown, the single phase diode bridge current is far from being level. You simply cant accurately apply average, RMS, or any fixed factor to go from three phase to single phase.

3. Originally Posted by steve66
But if we assume the VFD is 95% efficient (which most respectable VFD's would be), power out is still going to be pretty close to the power in. Even in a non-linear circuit....
Right there is the problem. It may be 95% efficient when 3Ø powered. We do not know that, nor should we assume that when 1Ø powered.

The principle of conservation of energy still holds. Power in equals power out... but we want a specific part of the power out, the mechanical power at the motor shaft. The difference therein determines the efficiency.

4. Originally Posted by Smart \$
Right there is the problem. It may be 95% efficient when 3Ø powered. We do not know that, nor should we assume that when 1Ø powered.
It's probably not a lot different.The input rectifier losses will be about the same just shared by fewer components. And that's kinda the issue here. How hard can you push that rectifier when it isn't firing on all six cylinders?

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What is confusing some people is thinking that Irms is LINEAR calculation. It is not of course: the square gets us the I^2 portion of extra heating due to higher peaks. THIS is why Irms indeed DOES give the EXACT average heat in the diode.

I must run Irms calculations often in my work on servo motors to pick proper size to handle the thermal load of non linear loads. There are two ways to kill a device by heat: too much heat over time - above the rating of the device - this is calculated with rms values. The other way is way too high a peak current that when rms'ed out over a long level of low current, LOOKS below the rms capability of the device, but in fact already fried it on the first high current spike; the cool down period afterwards is meaningless by then so the "rms" calculation, although correct, is of no value. We do not have this second situation in diodes in front end of vfds.

Irms is indeed 100% accurate for this discussion and defines any waveform you can throw at the diode. Saying otherwise seems to me suggesting the equivalent of "we never went to the moon - it was filmed in a room on earth."

6. gar
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170517-0832 EDT

mike_kilroy:

Your previous post is wrong. Somehow you have forgotten how to work from basic electrical concepts.

Instantaneous electrical power in a system, device, or load is the instantaneous product of the voltage across and the current into the system. This power can be positive (into) or negative (out of).

Average power is the instantaneous power averaged over some time period.

If I have an invariant resistance, then for all practical values v = i * R, and p = v * i (this power equation is not part of Ohm's law), or p can be expressed as p = i * i * R.

If i is constant (I) with respect to some particular time period, and R is constant, then average power over that time is Pave = I^2 * R. If I is changed by a factor of 10, then power changes by a factor of 100.

If instead the value of r changes with respect to voltage so that voltage remains constant at a value of V independent of current, then p = i * V. This means a change of 10 times in current only produces a 10 times change in power.

Look at the diode curve I referenced. There is about a 0.1 to 0.2 V change out of 1 V or fairly constant for about a factor of 5 to 10 change in current. Therefore, average current times a constant voltage is a good measure of average power.

.

7. gar
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170517-1050 EDT

Look at the curve for 180 C in Figure 7 at
http://www.vishay.com/docs/93516/vs-50pfrwseries.pdf .

At 1 A the voltage drop is 0.5 V, or 0.5 W. A one point resistance calculation would be 0.5 ohms.

At 50 A the drop is 1 V, or 50 W, and the power ratio from 1 A to 50 A is 100. One point resistance is 0.02 ohms.

Use I^2 as the criteria, and the power ratio is 2500, 25 times greater.

If you are calculating average power dissipation for this diode from an RMS current number how do you pick what resistance value to use?

.

8. Originally Posted by Besoeker
It's probably not a lot different.The input rectifier losses will be about the same just shared by fewer components. And that's kinda the issue here. How hard can you push that rectifier when it isn't firing on all six cylinders?
You're saying it's probably not a lot different... then you're saying it may very well be different.

It's simple physics If you put 10.5kW 3Ø in and you 10kW out, the process has a 95% efficiency (i.e. 5% loss). Then everyone (but Jraef) is saying it'll likely take significantly more 1Ø power in to get the same 10kW 3Ø power out. Logic dictates a lesser efficiency (greater loss). How can this be "not a lot different"?

9. It won't take more 1 ph input power than 3 ph
from the dc bus out loads are lower than rated (drive upsized, 10 hp on a 20 drive) so losses are likely a bit higher due to operation at a lower design point

the loading per diode section used is the same (upsized drive)
say you need 10 A drive for 3 ph and size for 17 for 1 phase use
your 17 A drive sees rated i on fewer diodes, losses should go down, eff goes up

rectifier: per unit the same, but total goes down, ie, less units
output stage: operating at a lower than rated point eff should go down a bit
but power < rated power so net diff is moot
so they offset
imo the differences are so small they can be neglected

the 1.732 is a good number
2 is a good rule of thumb

this is done all the time without issue in rural areas where 3 ph is not available

10. Well, my original point in that other thread was that 1.732 was NOT a good number to use, because the WORSE situation is the bus capacitance, and 2x is the MINIMUM size that should be used because of that. In fact on drives with no DC bus choke, I always recommend a 65% de-rate (motor load is 35% of drive rating) because the caps will run hotter, or if not a 65% de-rate, then de-rate the ambient temperature by 50%, which for many drive installations is unrealistic (i.e. drive rated for 40C can only be used at 20C, which will almost always means an A/C unit if it is enclosed). Mind you all, these are ALL "rules of thumb"; I teach these in classes where 90% of the attendees are Electricians (the basic premise of this entire forum), not EEs with the level of expertise being exhibited here.

So what should I be teaching then? Just forget mentioning input current as an issue because it is too complex to make ANY kinds of reasonable assumptions without violating some complex engineering principle? I'm OK with that, because like I said, I think of the DC bus ripple as the bigger issue anyway. The 1.732 is actually the PROBLEM I was trying to combat in selecting a VFD size, because I see a LOT of VFD salesmen use that as a guideline to sell their VFDs to electricians and customers, since it makes their offering less expensive than someone doing a more proper de-rate. But the average electrician attending my class is not stupid either, they will ask me what the input current will be. If I can't tell them "1.732x the motor current, because you are feeding the same kW load from a single phase source", then what DO I say?

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