Here is the circuit for UPS 2200 va capacity, use 4 batteries 12v in series, so total is 48v, 7AH
im replacing these 12 mosfets
in the inverter circuit of UPS
The original mosfets are p65nfo6, attached datasheet
I have 75nf75 available for replacement
Attached datasheet
The inverter output goes to step up transformer
The output signal is modified sine wave, 50hz, not pure sine wave
220vac
I can see that every 3 mosfets Re connected in parallel, and that's for high current handling under load which is up to 2kw almost
What I need understand is how the switching work in this inverter?
As I see that the mosfets into two groups, each group is symmetric to other, why not all symmetric?
Can we calculate how much ampere go through mosfets under full load?, Id?
Can we estimate dead time between switching from high to low?, so we are safe when changing mosfets with little slower speed?
Attached the whole UPS circuit
What calculations we can do having these info!, of battery voltage, mosfets chosen, UPS capacity, inverter circuit, driver circuit, etc to help in choosing right replacement mosfets?

 

To repair something, you must know how it works, as you well asked. Replacing mosfets without testing them nor troubleshooting is guesswork.

 

This may help:
https://www.electricaltechnology.org/2020/09/full-bridge-inverter.html

 

To repair something, you must know how it works, as you well asked. Replacing mosfets without testing them nor troubleshooting is guesswork.

Of course I tested, they are burned already and seen by eyes that they are broken and need replace

 

You mean 2200 VA, Not 2200 KVA !!!!!
Is this your design? 2200 VA at 24 volts will be about 100 Amps !
With a 7 AH ( ) battery, you will not get more than 15 VA output. 4 MoSFeT's are enough, why 16?

 

You mean 2200 VA, Not 2200 KVA !!!!!
Is this your design? 2200 VA at 24 volts will be about 100 Amps !
With a 7 AH ( ) battery, you will not get more than 15 VA output. 4 MoSFeT's are enough, why 16?

2200 VA, yes
Sorry the machine is 4 batteries
by the way the output of this inverter goes to transformer to step up, not to output load
How did you calculated the 15va please?

 

Hi
the output stage is so called full bridge configuration. The three mosfets Q2 -Q5- Q9 , Q10- Q8- Q20 , etc work as a single big power mosfet. Found some information , how full bridge operate.
Dead time is determined by the controller U2.

First of all , check all electrolytic capacitors. The ESR meter is very helpful in that. Most of the ESR meters can determine a faulty capacitor directly on the PCB, desoldering not needed

 

You mean 2200 VA, Not 2200 KVA !!!!!
Is this your design? 2200 VA at 24 volts will be about 100 Amps !
With a 7 AH ( ) battery, you will not get more than 15 VA output. 4 MoSFeT's are enough, why 16?

Can you clarify your calculations?

 

First of all , check all electrolytic capacitors. The ESR meter is very helpful in that. Most of the ESR meters can determine a faulty capacitor directly on the PCB, desoldering not needed

I checked the capacitors with multimeter, there are two bad, I've changed

 

2200 VA, yes
Sorry the machine is 4 batteries
View attachment 336006by the way the output of this inverter goes to transformer to step up, not to output load
How did you calculated the 15va please?

Is the Battery Voltage mentioned anywhere? How do you know it as 48 Volts?
A 7 AH battery is normally discharged at [1/10 AH] amperes, 0.7 A
I erred in assuming 6 volt batteries, making it 24 volts at .7 amps and with efficiency < 100 %, rounded it off to 15 VA.
With 48 volts battery, it will be 30 VA.

That aside, the 7 AH battery may not even power up the invertor in no load state.

 

It's 4 batteries connected in series
As I've mentioned before, this output goes to step up transformer, so ta the output is 220 vac

How calculate the 2200va?, or 1320w as machine nameplate say?

 

2200 VA is 10 Amps @ 220 Volts
That is 220 * 10 / 48 = 46 Amps @ 48 Volts
With an Invertor efficiency of 80 %, it will be about 57 Amps.
I would think about using 48 Volt 600 AH Battery Bank if you plan to use the full capacity of the UPS.
Keep in mind that the recommended discharge current of the battery used in UPS (not automobiles) is 10 % of the AH value.

 

It's 4 batteries connected in series
As I've mentioned before, this output goes to step up transformer, so ta the output is 220 vac

How calculate the 2200va?, or as

2200 VA is 10 Amps @ 220 Volts
That is 220 * 10 / 48 = 46 Amps @ 48 Volts
With an Invertor efficiency of 80 %, it will be about 57 Amps.
I would think about using 48 Volt 600 AH Battery Bank if you plan to use the full capacity of the UPS.
Keep in mind that the recommended discharge current of the battery used in UPS (not automobiles) is 10 % of the AH value.

So for 3 mosfets in parallel, how you think will be the Id " drain current" for the one mosfet?

 

Approximatelly, when the UPS is powered by the 48V battery and output is 2.2kW:
Battery current is 2200W/48V=46A.
Keep in mind inverter efficiency is about 80 % , so the current is 46A/0.8= 57A.
In the full bridge three mosfets must withstand about 1/2 of that current, so 57/2=28A.
There's three mosfets in parallel, so current is shared between them : 28A/3= 9.5A

 

In the full bridge three mosfets must withstand about 1/2 of that current, so 57/2=28A.

How does the current halve?

 

It's 4 batteries connected in series
As I've mentioned before, this output goes to step up transformer, so ta the output is 220 vac

How calculate the 2200va?, or as

So for 3 mosfets in parallel, how you think will be the Id " drain current" for the one mosfet?

Is there a "Drain Current" for a MoSFeT?

 

How does the current halve?

There's 4 switches in the H-bridge. Two of them ( say S1 and S3) are open at the same time,and the other two( say S2 and S4) are closed. After a few milliseconds, their status is reversed.
Thus, the average current thru the switch is halved

 

There's 4 switches in the H-bridge. Two of them ( say S1 and S3) are open at the same time,and the other two( say S2 and S4) are closed. After a few milliseconds, their status is reversed.
Thus, the average current thru the switch is halved

And what the purpose for 3 mosfets in parallel?

 

sharing current of course. so each transistor in the pack, only handles 1/3 of the current of the group.

 

There's 4 switches in the H-bridge. Two of them ( say S1 and S3) are open at the same time,and the other two( say S2 and S4) are closed. After a few milliseconds, their status is reversed.
Thus, the average current thru the switch is halved

Oh... Did not strike me that each arm of the bridge handles only a Half Cycle Thanks.