Incoming current on a VSD is lower than the motor current

[ER]

Hi community, I noticed on a drive installation that the incoming current (line side of VSD) is lower than the motor current. How this could be?

[BJacqz · Accepted solution]

Hello, the incoming current of a drive is basically totally different shape than the output current.

output current to motor is made from a PWM voltage signal, this voltage signal allows to create a nearly sinewave signal for current. the motor current is then very close to a sinewave. 

Regarding input signals, and due to the input bridge constitution of the drive, the input current request will have different shape than a sinewave.

The value of this current will be directly linked to the line supply impedance.

It means that a high power transfomer has low impedance compared to a small one.

In this case it is possible  to add some line impedance (choke) and this will reduce the value of input current.

[ReinhardK]

Motor current 300A and a Input current on VSD side of only 150A - yes, such a behavior sounds a bit wired in the first step.
But we need to consider, that a VSD runs a motor with variable frequency and voltage.
Full speed 50Hz/400V, at half speed 25Hz/200V.

The power is defined as  P = V * I * 1,73 * cosϕ * η; 
When considering, that power must be equal on in and output, the VSD input current will be half of its output
current at half motor speed (when output voltage is half)  

P (in)= 400V * 150A  = P (out) = 200V * 300A  

This leads to the effect, that the VSD line side current is a linear function of the motor speed.

• Zero speed - near to zero current (only the losses are covered) 
• Nominal speed - nominal current (at nominal load)

Therefore we also find no inrush current effect at starting compared with DOL or Y/D start. 

[TC]

It is a normal observation because at rated load, power factor of VSD (~0.95) is higher than motor (~0.8). At low/no load, motor power factor is even lower (~0.3), whereas VSD power factor still hold quite well (~0.7).

[Anonymous user]

To explain it very simply:
The power (P) must be the same on both sides. (motor and mains side)
On the motor side I change the output voltage, on the mains side I must change the current because the voltage is fixed!
This means on motor side:

0 revolutions 0 volts
half speed; half voltage (200V)
full speed; full voltage (400V)
Since on the input side (mains side) the voltage is fixed, the current must change if the power remains the same!

Simplified example : P=UxI
Motor side 10A                                                                     Mains side
0 speed -> 0V = P= UxI = 0Vx10A = 0W                              I=P/U =0W/400V = 0A
50 % speed -> 200V = P= UxI = 200Vx10A = 2000W          I=P/U =2000W/400V = 50A
100 % speed -> 400V = P= UxI = 400Vx10A = 4000W        I=P/U =4000W/400V = 10A