# VSD for induction motor maximum current

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1. May 28, 2017

### tim9000

Hi,
I'm trying to ascertain when you're driving an asynchronous motor with VVVF or whatever else a VSD uses, when do you draw the most current?
For instance, say you're talking about a train/tram: I would have thought intuitively that since power = torque / angular speed, that starting up, to move the big mass the motor and system would draw the most power (torque very high, speed very low). But I heard a rumour that the vehicles actually draw the most power when motoring at full speed.
Now, obviously when you're starting up you lower the voltage and the frequency, so is there some sort of algorithm that VSDs use as a standard to draw the most power/current when operating at full speed....some industry profiles I should know about?
Thanks very much!

P.S. I'm not sure what conservation of energy says about keeping a mass moving (overcoming friction etc) at full speed, vs starting a mass to move from stationary.

2. May 29, 2017

### Asymptotic

I'm assuming a typo here.
Horsepower= (torque * RPM)/5252, where torque is in lb-ft.
An alternative formulation is Watts = 2π*torque *RPS, where torque is in newton meters, and RPS is revolutions per second.

It depends upon the characteristics of the driven load.

Let's set that aside for now, and consider a scalar (as opposed to vector) "Volts/Hertz" (V/Hz) inverter drive rated 6 amp continuous current, and a 460 volt/60 Hz/3 PH/1760 RPM motor. These ratings define the basic Volts/Hertz relationship, specifically, 460/60, or 7.67 V/Hz. If commanded speed is 1 Hz (equal to 1/60, or 1.67% full speed = (1/60)*1760, or 29.3 RPM) 7.67 volts is fed to the motor windings; at 2 Hz (3.34% full speed, 58.8 RPM) 15.3 volts is placed across the motor windings, and so on.

3 phase power is 1.732*volts*amps*power factor. To avoid confusion about the core principles, we'll lie a bit, and assume PF = 1.0.

Power is 80 watts at 1 Hz (7.67 volts) and 6 amp continuous current drive rating. If it takes 60 watts to overcome static friction of the load, then the motor shaft will rotate at 29 RPM, and current will drop to about 4.5 amps. Bump speed command to 2 Hz, and (providing load power requirement remains at 60 watts) the motor shaft accelerates to 59 RPM, voltage rises to 15.3 volts, and (because power = volts * amps * 1.732) current falls to 2.3 amps.

80 watts may not be enough power to overcome static load friction. In that case, it depends on drive design. Most drives can source anywhere from 150% to 300% of rated continuous current for brief periods before faulting. Additionally, V/Hz drives are also often designed with some sort of "Voltage Boost" parameter to temporarily violate the V/Hz relationship when under x amount of speed command, and pour on additional voltage. Both these things in combination are usually enough to get the load moving, although it may not be possible to run successfully at very low speed if current demand remains above the continuous rating.

OK, so high current demand will occur at very low speed until the load begins to move.

Depending upon load characteristics, it may also be a limiting factor at higher speeds.
For instance, consider pump and fan affinity laws. Flow rate is proportional to input shaft speed, pressure is proportional to shaft speed squared, and power is proportional to shaft speed cubed.