Greeting All,

I have a Welding Inverter Machine which is Model # TECHNOLOGY 238 CE/MPGE 230V .

The welding workshop reported that the machine is damaged/not working and unfortunately they open and remove/lost the resistor without any knowledge.

And when I open the machine I saw that the resistor is missing in the circuit which cant be work without it, I plan to replace it but I am not sure what is the resistance value of it.

I will attach the part of the circuit may someone help to figure it out:

I hope Everything clear.

Thanks.
Yousif

 

Odd? looks like the two centre AC in pins of the bridge are connected?
PB5010

 

Hi Max, Assuming the resistor is for current sensing I think it is possible the the bridge with shorted output could be used to limit the voltage across the resistor to two diode forward volts drops. (About 1.3 volts) I it was AC current through the resistor it would clamp in both directions. We would need to see the etch side of the board to see if my theory is correct.

Les.

 

Not to helpful ,, but I would suggest that you get back to the people that ruined the machine, charge them for the time / loss.

 

I've just had a look at the datasheet for the PB5010 (Bridge rectifier.) and the two middle pins would normally be the AC input.
So it is not being used in the way I originally thought it might be. (The right hand pin of the bridge in the picture is the negative pin.)

Les.

 

That's why I questioned it, seemed strange?

 

Not to helpful ,, but I would suggest that you get back to the people that ruined the machine, charge them for the time / loss.

I did already and charged them for this behavior, but on another hand, I would like to try to fix it as well.

 

I've just had a look at the datasheet for the PB5010 (Bridge rectifier.) and the two middle pins would normally be the AC input.
So it is not being used in the way I originally thought it might be. (The right hand pin of the bridge in the picture is the negative pin.)

Les.

Hello Les,

The DC output of the PB5010 is to power the coil of the 2 Relay in order to pass the ac power through the relay to the high section of the board.
In Addition, the DC output that powers the relay coil goes to the 2 Cooling Fan.

I think in my opinion the resister is to limit the current that goes inside the relay which is 16A should be limited / 250v.

But I am wondering here what is the counted voltage here calculated..
250v / 5A = 50 Ohm so I go with 47 Ohm? or should I go extreme like 10A and the resistor will be 250 / 10 = 25 Ohm?

Should I count the AC voltage as 250 or less?

Please note that the 1 of the relay is burned due to overcurrent.

Regards,

 

If I want to test should I go high resistor then step down to lower? like 100 Ohm them go for less?

250 / 100 ohm = 2.5 Amp only...

 

In many instances where a large resistor is used, it is followed by a zener .
Max.