under volage and contactors burning

[isurunalaka]

why AC contactors burn if under voltage occur? whats the theory behind that failure?

 

[Prototyp_V1.0]

Because the magnetic force won't be enough to keep the contact (where current flows) firm.

Then there will be a combination of bad contact (high resistance) and release of the contacts, causing sparks. Both will cause the contact plates being polluted, wich in turn makes the contact even worse.

 

[isurunalaka]

why contactor coil burns?

 

[FlapJack]

My guess is that the coil burns because it is made like a solenoid. High current flows in the coil to pull the solenoid in and when the coil has a iron core it's inductance goes up and the coil draws less current.

In an under voltage condition the solenoid does not pull in and it draws high current until the coil burns up.

 

[isurunalaka]

could u pls explain more about this?

 

[betwixt]

Please answer first: Are you saying the CONTACTS burn or the COIL burns?

Brian.

 

[isurunalaka]

coil burnt.

 

[jiripolivka]

If coil burns, it is due to overcurrent, not a lower voltage. Can you measure coil current? Can you indicate coil specification by the manufacturer?
AC relays often need a higher current before the contact arm is moved to "closed" position. Then the coil current is reduced as the magnetic circuit is closed. If the arm cannot close the magnetic circuit, the coil field passes only through air and the current may stay higher, after some time burning the coil.
Make sure your relay does operate, also mechanically (lubricate if needed), and make sure the coil circuit is properly designed. Add a fuse to prevent coil burning again.

 

[mtwieg]

If coil burns, it is due to overcurrent, not a lower voltage. Can you measure coil current? Can you indicate coil specification by the manufacturer?
AC relays often need a higher current before the contact arm is moved to "closed" position. Then the coil current is reduced as the magnetic circuit is closed. If the arm cannot close the magnetic circuit, the coil field passes only through air and the current may stay higher, after some time burning the coil.

This is a plausible explanation. This is why fancy high power contactors will have supervisory circuits in the coil drivers to ensure that enough power is given to the coil to operate, or to turn off the coil current and report a fault if it is unable to do so.

 

[isurunalaka]

why coil current is reduced when magnetic circuit is closed? what is the theory?

 

[FvM]

Presumed we are talking about AC driven contactors. The coil impedance is mainly inductive at 50 Hz, closing the magnetical path (removing the air gap) reduces the current to e.g. 10 or 20 percent of the initial current.

Copied from a Telemecanique LC1 F contactor datasheet:

Average consumption at 20 °C:
- inrush 50 Hz: 550 VA; 60 Hz: 660 VA,
- sealed 50 Hz: 45 VA; 60 Hz: 55 VA, cos ϕ = 0.3.
Heat dissipation: 12…16 W.
Operating time at Uc: closing = 23…35 ms, opening = 5…15 ms.

You can assume that the 550 VA during inrush is real power to large extent.

- - - Updated - - -

According to datasheet values, you get about 160W real power, if unengaged at rated voltage. If you know the maximal voltage that doesn't yet engage the contactor, you can calculate the maximal power dissipation.

 

[isurunalaka]

could you pls tell me how to calculate

 

[chuckey]

The impedance of the coil has two parts , the DC resistance which does not change with a inductive impedance which is low impedance due to the big air gap when the solenoid is released (10s of milli H?). when the solenoid is energised and the pole piece goes in hard, the magnetic path is completed, so the inductive impedance rises a lot (2 Hs?).
In FvMs data, you have real power and apparent power, then with the coil voltage and frequency you can work out the actual value of the DC resistance and the closed inductance. With the DC resistance and the surge (apparent) power you can work out the open inductance.
Frank