VFD. XSY AT4 220v 1phase to 380v 3phase – burned-out compontent identification

[Paul Thomas 1Participant @paulthomas1]

Hi all,

I have an inverter to convert 220VAC monophase to 380VAC triphase (XSY-220T380-RY is the marking on one of the PCBs inside the box, bought as XSY AT4). It worked fine on my Schaublin 102 lathe, but when I tried running a newly acquired Schaublin 12 milling machine, something started burning in the inverter. (I later discovered that the milling machine only had 2 phases out of 3 connected through the power-on switch, so I suspect the imbalance may have caused some problems….but that's a problem for another day…..back to the inverter…).

I opened it up, and the burned component is one of 2 components on the breakout board in the bottom left of the photo. I am trying to identify it to replace it, but there are no markings. I am hoping that someone can point me in the right direction as to what it is. It is mounted beside a power relay, JQX-105F-1 024D-1HS, which is a 24VDC coil, 250VAC 30A output, form A, sealed unit that seems to still be functional. As there is normally a flyback diode mounted directly beside a relay in DC applications, I suppose that the burned-out component is a diode. However, it's not a form that I recognise. As I said, it has no markings, it is green, axial, has a fairly heavy feel to it, has some kind of matt-ish friable coating over what appears to be a metal casing.

Does anyone recognise it? If indeed it is a diode, what is the type and probable ratings so that I can find a suitable replacement?

Is it possible that the relay itself is used as some kind of AC relay, and the green component is some other kind of current limiting device, like a thermistor or something?

The first photo shows the green axial component removed from the burned board, bottom right, and placed on top of the large capacitor for the photo.

The second photo is one I copied from a posting on this site showing the component in place on a functioning board.

Thanks in advance.

[Paul Thomas 1Participant @paulthomas1]

identification of a burned out component

[John Doe 2Participant @johndoe2]

I can only see one photo, but the green component looks like a resistor to me – one that has got hot, judging by the circuit board !

Can you put a multimeter across it?

Edited By John Doe 2 on 29/12/2022 23:34:27

[noel shelleyParticipant @noelshelley55608]

Like John I think it is a resistor, carbon as I see no ridges for wire wound. check the value on a meter. It may have got hot enough to melt the solder and could be resoldered on – what else failed ? The metal cap is the end cap, and it's size indicates highish wattage. Noel.

[Anonymous]

It's a wirewound power resistor. Probably around 5 to 10W, and 5 or 10%. Often used as an inrush current limiter.

Andrew

[Grindstone CowboyParticipant @grindstonecowboy]

+1 on what Andrew said.

Rob

[Paul Thomas 1Participant @paulthomas1]

Thanks for your answers.

Sorry for the second photo. It is in the Album named "VFD", but looks like it didn't insert into the post. I'll try to get that in a follow-on post below.

The green component reads OL on the multimeter, so open circuit. There are no other components that appear to be damaged.

Any idea of what nominal value would be suitable for the resistor? I may have some old PCBs lying around that have some resistors I could recycle, otherwise it'll be a trip to the local electronics store.

Paul

[Paul Thomas 1Participant @paulthomas1]

Still trying to load the first picture…..in the meantime, here's the link to the Album entry

https://www.model-engineer.co.uk/albums/member_photo.asp?a=58742&p=917745

[Paul Thomas 1Participant @paulthomas1]

[Anonymous]

Component value codes are normally printed on the green body, which doesn't look to be damaged? The PCB that the resistor is attached to contains a relay. So almost certainly the resistor is part of a current inrush limiter. If the resistor has failed it is quite possible that the cause was a relay failure. So it would be worth checking the relay and drive circuit.

Andrew

[Robert Atkinson 2Participant @robertatkinson2]

100% what Andrew says.

Do you have a multimeter of any kind? This would be useful to check components.

On the two pole switch on the mill this is not uncommon for motor switches inside machines. If you break two out of three connections there is no current flow. It does have safety issues though as it leves all 3 motor terminals (and the motor side switch or contactor connections) live.
So have a two pole switch is not normally a problem. Note there is a possible issue with a VFD as the high frequency PWM component might find a path to ground via the capacitance of the motor and wiring. However having ANY switch between the VFD and motor is a bad idea and may have caused damage.
Unfortuntly if switching the motor was the root cause it is likely that there is more damage.

Robert G8RPI.

[SillyOldDufferModerator @sillyoldduffer]

Posted by Paul Thomas 1 on 30/12/2022 08:43:40:

…

Any idea of what nominal value would be suitable for the resistor? I may have some old PCBs lying around that have some resistors I could recycle, otherwise it'll be a trip to the local electronics store.

Paul

I attempted to guesstimate the value based on it being an inrush resistor, which I agree is likely.

Not convinced by my answer though, so please can someone else check my logic, or even better open up an AT4 and measure one!

Adverts suggest the AT4 comes in 4kW, 5.5kW and 7.5kW versions. 4kW at 240V is nearly 17A, which is a hairy load for a UK 13A socket, but maybe the lathe motor rarely or never pulls full power. I guess Paul has a 4kW unit, specified at 220V.

At 220V 4kW is 18.2A so I assumed the resistor is designed to take that current without overheating. Judging from physical size the resistor is rated about 10W.

As R = W / I²

R = 4000 / 18.2 x 18.2

R = 12 ohms.

Might be right, 12 ohms is a standard value.

Dave

[Anonymous]

Posted by SillyOldDuffer on 30/12/2022 11:30:01:

…is 18.2A so I assumed the resistor is designed to take that current without overheating…

The problem with assumptions is that they can lead one up the garden path. The resistor is there to limit the inrush current due to input filter and DC link capacitors when the unit is first switched on. It is not intended to carry full load current. Once the capacitors are charged the relay is closed and the resistor plays no further part.

Apart from value, the important resistor characteristic is its overload capability. Wirewound resistors are very good at taking short term overloads without going pop. The resistor value will be set to give a maximum inrush current and the power rating is determined by the charge characteristics. These types of resistors are designed to run hot, up to several hundred degrees!

I suspect that the value will be rather higher than 12 ohms. On a practical note these type of resistors are often only available in the E6 series, which does not include 12.

Andrew

[Robert Atkinson 2Participant @robertatkinson2]

Again +1 for Andrews comment.
If this unit was intended for use on a 13 or 16 amp domestic supply (it is not of course) you would expect it to limit the current to around 10 to 20 Amps at maximum supply. So that is 240/20 to 240/10 so 12 to 24 Ohms.
Note that even at 12 Ohmsthe resistor would have to be rated at about 3500W to carry the input current (17A) for more than a fraction of a second.

A practical example I've had recently is a Farnell G series switch mode power supply rated at 360W. These have an input circuit almost identical to a VFD and thus similar inrush. The design has a 12 Ohm inrush limiting resistor. IT is bypassed by a TRIAC (semiconductor AC switch) rather than a relay.
After over 15 years of use the power supply's AC input switch welded shut. I replaced it and a few weeks later the same happened. It turned out the TRIAC had failed bypassing the inrush resistor. New TRIAC and switch and all is good.
For the electronic engineers the failure mode of the TRIC was unusual. It was not permanently shorted. In fact it tested fine on a semiconductor tester. It appeared that it was turning on under high dV/dt conditions. This was confirmed by shorting gate to Mt2 and connecting across mains with a switch and a load of a 60W 240V halogen lamp with a 0.47 uF (X class) capacitor across it. The TRIAC would turn on at random when the supply switch was closed.

Robert G8RPI.

[SillyOldDufferModerator @sillyoldduffer]

Posted by Andrew Johnston on 30/12/2022 12:06:22:

Posted by SillyOldDuffer on 30/12/2022 11:30:01:

…is 18.2A so I assumed the resistor is designed to take that current without overheating…

The problem with assumptions is that they can lead one up the garden path. The resistor is there to limit the inrush current due to input filter and DC link capacitors when the unit is first switched on. It is not intended to carry full load current. Once the capacitors are charged the relay is closed and the resistor plays no further part.

…

Ah ha, hence if the relay fails, the resistor cooks. So a higher resistance than my sum suggests. I'd guess 150 ohms because the inrush current doesn't last long. Not betting the farm on it though.

Did a course once where 'assume' was defined as 'making an ass of u and me'

Dave

[Anonymous]

Posted by SillyOldDuffer on 30/12/2022 13:01:55

…hence if the relay fails, the resistor cooks.

Not necessarily!

If the relay fails closed there might be problems, but the resistor overheating isn't one of them. If the relay fails open the following would happen. After the capacitors are charged the current drops to a small value; just enough to run the control electronics. This current will be much lower than the design inrush current so the resistor will be fine. If the VFD is activated and starts drawing a current higher than the design inrush current from the DC link capacitors the DC link voltage will fall. That will trigger an undervoltage and the system will turn off. It's quite likely the resistor will survive.

Andrew

[Paul Thomas 1Participant @paulthomas1]

You guys are the best !

Thanks very much for your help. I'll go ahead and find a resistor and try it out. I'll come back with feedback once I've done that.

Best wishes for the new year.

Paul

[Paul Thomas 1Participant @paulthomas1]

Update.

I was able to get my hands on another VFD unit to check the value of the resistor. It turns out it is a 390 ohm.

I checked the relay by removing it from the circuit and checking it. It's fine. The capacitors aren't shorted at least.

I put it all back together, but still no joy. I suspect the mosfets may also be blown, so next step is to remove them and check.

I'll come back and give an update when I've done that.

Thanks again.

[joel baileyParticipant @joelbailey36359]

Hi Paul, do you have an update on your VFD? I think i have the same issue as your blown resistor…!

Regards

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