Anyway, these heatsinks are too small to dissipate 500W (e.g. if it had 90% efficiency)

The Aluminium case of the inverter is the heatsink.

Power losses also occur in the transformers, not only in the transistors. The fan removes heat generated inside the unit.

Timescope

 

The Aluminium case of the inverter is the heatsink.

Power losses also occur in the transformers, not only in the transistors. The fan removes heat generated inside the unit.

Timescope

You are right about the heatsink and the copper bars. Anyway, 500W on the case, this one will get hot, like an electric baseboard heater.

So are you saying that this inverter is good for 5000W continuous power?

It is not.

It wasn't me who invented the 10V, it's in the spec sheet.
Look at the DC input cable gauge, looks like 3 wires in parallel for + and - for each 8 subcircuits, that would be 555A/3 = 185A continuous for 3 of these wires. That's too much.

This converter doesn't have the correct input wire gauge + they expect current to be shared equally between all subcircuits (which it will not). Neither are all FETs switching at exactly the same moment nor are transformers giving out the EXACT same voltage.

I actually built DC/AC and DC/DC converters with up to 160A input and up to 300A output current and they looked very different from this unit.

I would almost like to get one in my hands to make a REAL 5000W test with it.

 

So are you saying that this inverter is good for 5000W continuous power?

No, it obviously is not suitable for continuous operation at 5kW but it may be able to serve the intended purpose if it is repaired.

The OP asked for help so let's see how we can provide that help for example, how to remove the burnt transistor from the board, what components to check (the gate driver circuits may be damaged), a soft start circuit for the motors etc.

Timescope

 

I would probably just remove the two FETs from the one subcircuit visibly damaged. Then measure resistance from drain to source of each FET. Verify if the gate signal trace is shorted to ground or Vdd. Follow the gate signal trace to the driver, see if it is visibly damaged.

Measure input resistance from + to -. Maybe you can already measure a short there.

In the spec sheet it said 0.8A current without load so I would start with a 1A or 1.5A fast fuse in series with the the power supply when powering it up again.

 

One method I use to remove transistors from double sided boards is to apply solder to the leads to bridge them together then heat the three leads simultaneously. I then use a solder sucker to remove the solder from the holes. This works best with 60/40 solder as it has a lower melting point than lead free solder.

The gate drivers are usually two complementary pairs of emitter followers, one for each drive phase.

I have no experience with variable frequency drives. Could it be used to provide the soft start function ?

Timescope

 

No, no problem. These are probably lot and manufacturing date numbers. Doesn't matter.



Post a picture of the resistor. A SMD resistor "100" would be 10 Ohms.
Testing resistors "in-circuit" doesn't mean much, unless you know the circuit.



Good luck with that. It's not exactly a small PCB. Some testing has to be done with the proper equipment, like oscilloscope, a benchtop power supply with current limiting etc.

Look also for burnt traces, check if there are solder residues between MOSFET pins etc.

Hey, I have another question. I am still making my way through this "junk" inverter. Not really trying to fix it I just am learning first, would love to know why it failed. Anyways. I have two pictures. The first one is of the same board from each inverter that "should" be exactly the same. I have pointed red arrows at something that on one the component is labeled "RS1M". On the other the round component has no markings. I looked up RS1M to see it is a diode but in the places where diodes are they have a white letter "D" on the PCB and they also have markings on them. This spot that both components are in have white letter "U" on the PCB. I though the round component looked of a diode but not sure nor am I sure its right size.

The other photo is of a fuse which I think says its 10A but the board says 15A...is that correct?

I'm learning. Thanks.

 

You are right about the heatsink and the copper bars. Anyway, 500W on the case, this one will get hot, like an electric baseboard heater.

So are you saying that this inverter is good for 5000W continuous power?

It is not.

It wasn't me who invented the 10V, it's in the spec sheet.
Look at the DC input cable gauge, looks like 3 wires in parallel for + and - for each 8 subcircuits, that would be 555A/3 = 185A continuous for 3 of these wires. That's too much.

This converter doesn't have the correct input wire gauge + they expect current to be shared equally between all subcircuits (which it will not). Neither are all FETs switching at exactly the same moment nor are transformers giving out the EXACT same voltage.

I actually built DC/AC and DC/DC converters with up to 160A input and up to 300A output current and they looked very different from this unit.

I would almost like to get one in my hands to make a REAL 5000W test with it.

If you don't live too far so that shipping isn't too bad I will gladly ship this to you and you can have at it....I would ask for complete honesty and of course I'll pay for any costs if you think they are worth it and as long as it works

 

I would probably just remove the two FETs from the one subcircuit visibly damaged. Then measure resistance from drain to source of each FET. Verify if the gate signal trace is shorted to ground or Vdd. Follow the gate signal trace to the driver, see if it is visibly damaged.

Measure input resistance from + to -. Maybe you can already measure a short there.

In the spec sheet it said 0.8A current without load so I would start with a 1A or 1.5A fast fuse in series with the the power supply when powering it up again.

I have replaced the FET. I am new to electronics...actually I am a chemical engineer and now think I would have had more fun in EE....anyways could you elaborate a little on how to accomplish what you said above minus the FET replacement. Thanks.

 

I have replaced the FET. I am new to electronics...actually I am a chemical engineer and now think I would have had more fun in EE....anyways could you elaborate a little on how to accomplish what you said above minus the FET replacement. Thanks.

I actually said "remove" not "replace".

Looking at the printed side of the FET the leftmost terminal, where does it go to? Follow that trace to the driver. Could be a complementary pair of transistor, possible TO-92 package or similar. Could also be an IC.

With a multimeter, measure continuity (the diode symbol on your meter) from drain to source of all other FETs (although it looked like they are all in parallel. What do you measure?

At the input, measure with an ohmmeter (multimeter) from the plus to minus. What do you measure?

If you are sure that nothing is directly shorted you can try to power it up with a fast acting fuse of 1 to 1,5A at the input. Startup current could be higher though... Best would be a power supply with adjustable current limit.

I live in Canada, but this is not a job for a few hours. I'm afraid I won't have time to solve this quickly for you.

 

Didn't you want to post 2 pictures?

 

Ha yes I posted them just now...totally messed that post up. Anyways I will get started on your recommendations and report back soon. Thanks.

 

I actually said "remove" not "replace".

Looking at the printed side of the FET the leftmost terminal, where does it go to? Follow that trace to the driver. Could be a complementary pair of transistor, possible TO-92 package or similar. Could also be an IC.

With a multimeter, measure continuity (the diode symbol on your meter) from drain to source of all other FETs (although it looked like they are all in parallel. What do you measure?

At the input, measure with an ohmmeter (multimeter) from the plus to minus. What do you measure?

If you are sure that nothing is directly shorted you can try to power it up with a fast acting fuse of 1 to 1,5A at the input. Startup current could be higher though... Best would be a power supply with adjustable current limit.

I live in Canada, but this is not a job for a few hours. I'm afraid I won't have time to solve this quickly for you.

I was trying to see where the leftmost terminal went and trace it out but am confused. I took a few photos in order explain my confusion. If they are difficult to read please let me know and I will explain a different way but in essence I cant see that it goes anywhere. Will attach the photos this time

 

Ok, I forgot to mention the gate resistors. Where are the traces before the gate resistors going to?

 

Hey, I have another question. I am still making my way through this "junk" inverter. Not really trying to fix it I just am learning first, would love to know why it failed. Anyways. I have two pictures. The first one is of the same board from each inverter that "should" be exactly the same. I have pointed red arrows at something that on one the component is labeled "RS1M". On the other the round component has no markings. I looked up RS1M to see it is a diode but in the places where diodes are they have a white letter "D" on the PCB and they also have markings on them. This spot that both components are in have white letter "U" on the PCB. I though the round component looked of a diode but not sure nor am I sure its right size.

The other photo is of a fuse which I think says its 10A but the board says 15A...is that correct?

I'm learning. Thanks.

This can be modifications they made during production. Or they just didn't have the right diode package at hand. Or this is a unit that failed after coming out of production and they repaired it.. or... I wouldn't worry too much about it.

 

I actually said "remove" not "replace".

Looking at the printed side of the FET the leftmost terminal, where does it go to? Follow that trace to the driver. Could be a complementary pair of transistor, possible TO-92 package or similar. Could also be an IC.

With a multimeter, measure continuity (the diode symbol on your meter) from drain to source of all other FETs (although it looked like they are all in parallel. What do you measure?

At the input, measure with an ohmmeter (multimeter) from the plus to minus. What do you measure?

If you are sure that nothing is directly shorted you can try to power it up with a fast acting fuse of 1 to 1,5A at the input. Startup current could be higher though... Best would be a power supply with adjustable current limit.

I live in Canada, but this is not a job for a few hours. I'm afraid I won't have time to solve this quickly for you.

Ok, I forgot to mention the gate resistors. Where are the traces before the gate resistors going to?
_____________________

The first photo shows where the traces start, the second photo shows where the traces go. It is the same on the opposite side and the traces both end at a connector which connects to the other board in the inverter.
For photo #3 My thoughts are the odd numbered gate R are supposed to be connected to the lower trace. None of these holes (in photo 3) are filled with anything and only the even numbered gate R are connected to the top trace.
Hope this makes sense.

 

Ok, I forgot to mention the gate resistors. Where are the traces before the gate resistors going to?

_______________________

Couldn't attach #4 on that last post so here it is.

 

I actually said "remove" not "replace".

Looking at the printed side of the FET the leftmost terminal, where does it go to? Follow that trace to the driver. Could be a complementary pair of transistor, possible TO-92 package or similar. Could also be an IC.

With a multimeter, measure continuity (the diode symbol on your meter) from drain to source of all other FETs (although it looked like they are all in parallel. What do you measure?

At the input, measure with an ohmmeter (multimeter) from the plus to minus. What do you measure?

If you are sure that nothing is directly shorted you can try to power it up with a fast acting fuse of 1 to 1,5A at the input. Startup current could be higher though... Best would be a power supply with adjustable current limit.

I live in Canada, but this is not a job for a few hours. I'm afraid I won't have time to solve this quickly for you.

Still working on my instructions....I have finished from where the traces start to where they end before doing something else. I drew in red where they start after they are connect on the second board...photo #4 from last post.

 

The holes are called vias. They connect different layers of a PCB. Depending on their size and the amount of solder that was applied they can be open or filled. It doesn't matter.

What is printed on T9 to T12?

 

T9 has J3Y and T12 has 2TY.

 

I actually said "remove" not "replace".

Looking at the printed side of the FET the leftmost terminal, where does it go to? Follow that trace to the driver. Could be a complementary pair of transistor, possible TO-92 package or similar. Could also be an IC.

With a multimeter, measure continuity (the diode symbol on your meter) from drain to source of all other FETs (although it looked like they are all in parallel. What do you measure?

At the input, measure with an ohmmeter (multimeter) from the plus to minus. What do you measure?

If you are sure that nothing is directly shorted you can try to power it up with a fast acting fuse of 1 to 1,5A at the input. Startup current could be higher though... Best would be a power supply with adjustable current limit.

I live in Canada, but this is not a job for a few hours. I'm afraid I won't have time to solve this quickly for you.

So I measured a spare FET to see how its done. Worked good. Then I tried to measure one on the board and nothing. Do they have to be removed from the board to properly test? Also, I attached my multimeter to the positive terms and negative terms and tested the Ohms and it was zero.