# Brake Resistor Failure in Variable Speed Drive

A

#### akhil86

We are using LTi PMII+ drives for pitching of wind turbines blade. we are facing failure of Brake resistors leading to dc link over voltages.

LTi MOOG Drive PMII+

rating
Voltage: 3*400V
Current: 40A
Power rating of drive: 22kW

Chopper activation when DC Link voltage >650V
Over voltage error when DC Link voltage >=700V
Brake resistor: 300W, 20Ohms

We are using pitch drive with 7.5kW motor to pitch the blades.
Rated speed of motor 1445rpm
Rated current: 27A
eff: 0.86

max deceleration: 20deg/sec2, deceleration time: 0.4sec,
at max speed of 8 deg/sec (blades) where speed of motor is 1516 rpm of motor, taking into consideration GB ratio
max Braking duty cycle = 1.83%

We are facing brake resistor failure in this drive due to which we get over voltages on DC link and the entire system (Turbine) trip. I need help to understand what can be reason for the failure of the resistor.

I have calculated average brake power Pavg = 1072.57W. As per datasheet of Resistor we can load it 20 time *300W (rated power) = 6000W.

So required Brake resistor = 1072.57W/20 = 53W and we have used 300W Resistor.

C

#### Curt Wuollet

I'm not a wind turbine pitching expert, but If your resistor is rated at 300 watts your average power should not exceed that.

Your peak power can perhaps be 20 X your rated power. But an average power of 1kW will probably toast a 300 watt resistor.

The 20 X rating is to allow for a low duty cycle. And that's the story your faults and burnt resistor tell. In any case, I'd say you need a resistor with a higher power rating so it doesn't burn out, regardless of any math:^). I like to oversize these as well so you don't see light coming out and they don't burn open and fault your system.

Regards
cww

B

#### Bob Peterson

If the resistor is actually failing that is a sign that there is much more energy involved than what your calculations show.

There are a few things you can try. One is to force air cool the resistor. The other is to increase the rating of the resistor so it can handle the actual load.