Variable Frequency Drive

i'm measuring using a true RMS ammeter of course.

As for the VFD manual, do you mean that i will be able to find the formulas that explain the reason why there is a difference between the output and the input ampere ratings ?

if that is what you meant, then i would be grateful if you could possibly provide me with a VFD manual that you have used or read before.

thank you.

No formulas as such, the manual gives the input current and corresponding output current at normal maximum rated load. Output current scales depending on VFD output (eg- motor) load. For simplicity just scale the marked VFD input vs output current at different loads or corresponding to different input or output currents as a linear relationship.

Why do you want to know the output current anyway?

Are you still working on the motor project?

Measure it.

Not everything in the world can be put into a formula.

You are a student. Learn the practical way to solve problems as well as the theoretical way, if there is one.

Power consumption is highly dependent on LOAD.

I have a broad spectrum of info here...

https://www.joliettech.com/information/faq-variable-speed-drives-frequently-asked-questions/

http://www.eng-tips.com/faqs.cfm?fid=1062

"VFDs do not draw sinusoidal current from a power line, they draw pulsating currents, which can appear to many measuring instruments as DC currents. This non-linear amp draw can be mitigated somewhat by using phase shifted input bridges, which have become common in high horsepower VFD equipment, but nonetheless, it remains non-sinusoidal. This often causes errors in external current measuring equipment."

"Due to the non-sinusoidal currents, the input voltage can be distorted by currents flowing at frequencies other than the power line 60 hz. The equal harmonics tend to cancel each other, but the odd harmonics do not. Consequently, one can find significant currents flowing at the 3rd and 5th harmonics especially (180 hz and 300 hz) and these currents can, if unmitigated, not only cause power line and equipment problems, but can result in erroneous power and current measurements."

"Since the power being consumed by a VFD will always equal the power output of a VFD plus the internal drive losses, this means that if the output voltage is less than the input, then the amperes will be higher than the input. (Power = Volts times Amps)"

http://www.emainc.net/newsletter/possible-reasons-for-different-ampere-readings-between-the-input-and-output-of-a-vfd-variable-frequency-drive/

http://support.automationdirect.com/faq/faq_group.php?product_id=40

Broadly speaking, output voltage varies proportional to output frequency. So if frequency is say 25Hz (on a nominal 50Hz system) voltage is halved and current is 2x, ignoring inverter efficiency.

Power In - Power Losses = Power Out

Single Phase

Power = E x I x PF => Voltage (E) x Current (I) x Power Factor (PF)/Efficiency

Two Phase - Four Wire

Power = E x I x PF x 2

Three Phase

Power = E x I x PF x 1.73

There can't be a formula for that, the input current will vary from drive to drive depending on its internal power consumption, due to the intricacy of its control circuitry, it depends also on the switching frequency (whether fix or selectable); if you're using the built-in power supply to energize external control circuitry like sensors, indicators or meters AND how well or efficiently this internals were designed.

Even the selection of components would make a difference for the same design.

Regarding your second question; current at the input and output will vary, but not the power ratio between them.

It can draw less current than it supplies if the output voltage is lower than the input voltage. Gotcha!

Exactly, that's what I said in #6.

Looking in this way, if a motor runs same Load on two different type of starters i-e D.O.L and VFD

a. Why when motor is connected to VFD, Current at the Input of VFD is differnt from the Current at the output of VFD (motor terminals)?

b.Why for the same Load,Current consumed in case of D.O.L is different from the current consumed when the motor is connected to VFD ?

Case-a: Differance in Current Value at the Input and Out Put of VFD

At Input VFD takes Voltage and Current in the form of sine wave. VFD when supplies power to the motor,only current is in the form of Sine wave. Volatge is not in the form of Sine wave. Also magnitude of Volatge is changed to maintained V/F ratio. VFD applies its own algorithms for voltage and current(Vector Control, Scaler Control etc ). So value of current now will be changed. It will differ from the value of current as it was at the Input of VFD. This is one of the reason

Case-b:Differance in Current in case of Two different Starter

2. Consider a 18.5 KW, 400 V, 50 Hz, 35.5 A, P.F 0.83, 1465 RPM motor. It is connected to a Constant torque Load (Belt Conveyor) and consumes 12 KW when connected to a D.O.L starter.

12000 = 1.732 * 400 * I * 0.83 (Electrical Power Consumed)

I = 20.86 A

Mechanical Power delivered = 12000 * 0.9062 =10.874 kW

Load Torque = 70.88 N.m

When motor is connected to D.O.L, It consumes 20.86 A and Load torque of Belt Conveyor is 70.88 N.m

In case of D.O.L we have Constant Volatge in magnitude, waveform and have constant frequency

When the Same Belt conveyor runs on the Same motor with same load (12 kW) and same Load torque (70.88 N.m), using VFD, many things will change

1. Value of Volatge will be adjusted according frequency to keep V/F ratio to maintained torque.

2. Wave form of Voltage will now not be Sinosidel at the terminal of the motor as previously it was.

3. Change in Voltage causes change in current

4. Even motor runs at 50 Hz, it is not necessary that V/F ratio would be 8. It will be decided by the algorithm of VFD

5. Though V/F is 8 at 50 Hz, then at 50 Hz you don't have Sinosidel waveform of Voltage.

6.Also account for losses in VFD

Thanks & Kind Regards

It's all much MUCH simpler to explain than all that, but at the same time, not so simple to just read data sheets for it and understand. So I get it that you are confused, you are not alone.

When you are measuring* the current going from the VFD to the motor, that current is at the motor's displacement power factor. If at the same time you measure the current on the INPUT side of the VFD, that displacement power factor has been corrected by the VFD. Sot if you MEASURE the current on the input and compare that to the output current as per the VFD display*, the reason why they will not agree is simply because of the power factor of the motor. (* measuring the output is futile with even "true RMS" meters unless you spent upward of $1000US for it at least.)

This however may not be so clear if you just read the specification data from the VFD manufacturer. Yes, the input amps will be different from the output amps. But the problem then becomes how the VFD mfr is wording their data.

Some do it by looking at the output amp capacity of the VFD, because that is limited by the thermal capacity of the transistors, then providing the input amps AT THAT OUTPUT CAPACITY. So for example, I'm looking at a spec sheet for an Allen Bradley 10HP 480V drive right now that says the output current rating is 14A, and the input current rating is 12.5A. They are saying that when the VFD is putting out 14A, which is how the drive is rated, the maximum input current will be 12.5A. That follows the issue I stated regarding the power factor. 14A of motor circuit current going to a fully loaded motor with a .8 power factor is ACTUALLY going to mean that 17.5A of current is flowing through the transistors at that moment; 11.2A of motor power, 2.8A of reactive power. But that reactive power is being stored and supplied by the DC bus caps, so it does not have to come from the line side components.

Others unfortunately sometimes state the maximum input amps capacity based upon the thermal capacity of the input rectifier components, independently from what the output devices might be doing at that moment, and will also be based on a RANGE of input voltages, so they put the worst case scenario on their nameplate. It can be confusing, I'm warning you. You must be very careful about what you read if it's important to you. As an example, I am looking at the same 10HP VFD from ABB, which is rated for 15.6A output, but they state that the maximum input current rating is 24.3A. That's a two-fold issue: 1) that drive is listing the current ratings for 380-480V at the same 10HP, so the 15.6A represents the current for a 380V 10HP motor and 2) the 24.3A input current rating has to do with the rated capacity of the diode bridge rectifier components, not what the diode bridge will be pulling when the motor is drawing 14A, as the Allen Bradley spec is saying. That drive should NEVER be capable of drawing 24.3A from the line side, because that would equate to a severe overload on the output side, in excess of what the output transistors could safely handle. Why do they do it that way? No idea.

Does that matter? YES, if you are in North America! Our codes 9NEC and CEC) have special provisions for selecting conductors feeding VFDs, requiring that the conductors must be sized at 125% of the VFDs published maximum Input Amps. So in the above example, the Allen Bradley drive is shown as 12.5A so the conductors must be sized for 15.6A, meaning #12 wire for us. But the ABB drive is rated for 24.3A input, so even though they are feeding the SAME motor, you would need to size the conductors based on 30.4A, which might result in you needing to use #8 conductors, 2 sizes larger.

It sounds as though the OP has either not yet understood how power is measured/calculated, or he has not reached that point in his studies.....which would be strange to my mind...... High school should have brought him that much at least!!

You haven't been to high school in a while have you. They don't teach anything like electronics today and as far back as the early 70's. At least not in the Midwest, I friend and I knew more about electronics then the teacher because his brother was a hobbyist in electronics. Past the simplest of circuits our teacher had no Idea other then the lesson plan he was given, and even then he did not understand what was happening. He could only explain what the book told him and not what was really happening. It was amazing he got past low voltage DC circuits without setting his hair on fire.

So then, everything has been dumbituded down in Hameerika to the point that they no longer teach a one year requirement in General Physics? Shame on these lazy bastages, who are neither teachers, nor good role models. Greater shame on the politicians who let things devolve to this point so Little Johnny's feelings won't be hurt.

Worse shame on us all when our country completely shuts down for lack of understanding and wisdom!