Inverter Rebuild (technical)

[Daniel Who Wants to Know]

I am in the process of rebuilding and strengthening a 250 watt
continuous/300 for 5 minutes high frequency MSW inverter out of a Xantrex
Xpower 300 Plus Jump start/light/air compressor/inverter pack. I shorted
one of the four output MOSFETS (Fairchild IRF634A) with an inductive load
and the air compressor already threw and broke off the piston connecting rod
while airing up a tire so I decided to just junk it for parts. The battery
is the 20AH one on my UPS and I plan on putting the inverter in another
enclosure and using it as a stand-alone device. So far I have strengthened
the HV DC rail by adding 2x470µF, 2x220µF, and 1x330µF capacitors all 200V
in parallel with the original 100µF 200V which was WAYYYYY too small for the
rated power draw IMHO. The cooling fan seems to be driven via PWM based on
the load on the inverter and before I added the extra capacitors it would
spool to max speed while powering my TV during bright scenes and now just
barely runs under the same conditions and often stops when there is a dark
scene. I also am limping on only 3 MOSFETS so I am only getting half wave
AC/pulsed DC out at the moment. I next plan to use 8 ST STP22NS25Z MOSFETS
for the output stage I know they are overkill at almost 10 times the
required current each or 20 times with 2 in parallel but I want to make it
almost indestructible. I am considering freewheeling diodes to supplement
the body diodes but am not sure if they will be needed.

I plan to run capable devices such as non voltage doubling switching power
supplies directly off of the DC rail to avoid the power loss due to the Rds
of the MOSFETS and so far I have found strangely enough that my Lexmark
X7170 AIO printer will only power up with the polarity neutral positive and
my 19" TV with an SMPS input stage will only power up with neutral negative.
I am still trying to figure out why a TV or printer would use half wave
rectification instead of full.

My only obstacle now is finding an enclosure to mount it in when it is
finished. Mounting options will be difficult because the PCB doesn't have
mounting screw holes or even room for them. Right now it is temporarily in
a gutted AT computer power supply.

Just looking for some tips and advice from people who like to solder as much
as I do. As I type this I am powering a GE compact fluorescent 26 watt 100
equivalent bulb off the DC rail and it works fine as the new ones are simple
full wave rectification to DC internally. I do have an older GE 60
equivalent that lasted over 5 years and upon disassembly I found that it
doubled the voltage and as such was truly AC only. As I take things apart I
am often surprised at how all appliances are marked as AC only when DC would
work fine in many cases. Yes I am aware that DC is harder to switch without
burning the contacts and that running a pure resistive load on approx. 160V
DC will roughly double the wattage draw VS 110V AC as I use that fact to
rapid preheat my 40 watt soldering iron.

 

[Daniel Who Wants to Know]

The only danger with that much power storage on the rail is if you get a
direct short on the output, the output mosfets will probably blow before
the
inverter can shut down on overload. The low capacitance value limits the
available surge current to a level that the output mosfets can take.

If you have that much capacitance, then you need to put a low ohm resistor
in series with the output to limit surge current, if something goes wrong,
so you don't take out the entire output. Something along the line of 0.5
to
1 ohm.


If the inverter monitors voltage drop on the HV DC rail, between the high
frequency step up transformer switching pulses, to determine load current,
then the large value of capacitance on the HV rail may be throwing off the
system the inverter uses to detect load current.

It would cause a lower voltage ripple, which would make the inverter think
it's carrying a lower load, even when it's almost running at full load.

That may cause the unit to overheat, because the fan isn't running fast
enough at high levels of load. And if the overload detection system uses
the
load induced ripple in the HV side to detect an overload, it may also
render
the over load detection circuitry dysfunctional. causing the inverter to
self destruct if the output is heavily overloaded.

The output mosfets, and the rail have been reinforced, but the high
frequency switching transformer, and the mosfets driving the transformer
are
not.


You will need that kind of setup on the output to handle the surge current
from the capacitors you used, if you short the output.


Been there, done that.


The switching supply doesn't run when the TV is off. There is a little
resistor/diode supply running of the AC input that provides power for a
always on remote control/panel control monitor chip. When it detects a the
power button on the front panel, or remote being pushed, then it turns on
the switching power supply. The supply for that always on circuit, is
sometimes a single diode, and a resistive voltage divider/regulator
system.
It only gets it's power off of half the input cycle.


LOLBetter to scare strangers with.
"This place looks like a mad scientist is living here"


Been there done that.
Running stuff of of DC works pretty good in my own opinion.

Just watch out on modifying stuff in certain ways. you got to watch out
for
other systems you may be affecting unintentionally when you put stuff in,
or
take stuff out of a circuit. It is easy to blow stuff up like that. I
know,
I have blown many of thing up over my life by doing that. Of course I have
also learned a lot from fixing the stuff I blew up.

If there isn't any obvious current shunts in a power circuit that uses
load
monitoring, then they are using some other method to determine load
current.
(ie) AC ripple on the HV rail.

If you boost the capacitance, then the ripple goes down, and the inverter
thinks it's carrying less load, but the components still have the same
amount of current going through them, and they are producing the same
amount
of heat.

I have often though about using 14 small lead acid batteries (5AH) in
series
to directly drive a modified sine wave output section at the normal 170V
dc.
But over current protection would get tricky at that level of DC voltage.
Even with small batteries you would have to use a hefty DC rated fuse to
stop it from turning into an arc welder if something shorted out.

That is why I have never messed with that idea much. Shorting out a 400uF
cap with 150 to 300V on it makes a big enough <BANG>, let alone a set of
14
lead acid batteries in series that can melt wire into like a hot knife
through butter.

Ok I plugged a 650 watt round waffle iron into the half wave AC output for
325 watts as a slight overload and the fan immediately went to top speed so
there is no problem there. The original fan was a 40mm and I plan to use an
80mm instead The inverter also has a thermistor that is supposed to be hot
glued to the low voltage chopper heatsink for overheat protection. It has a
30 amp fuse on the 12 volt positive supply wire The original was a 40 amp
but I downgraded for safety and to avoid letting the smoke out. Anywho my
goal is not to increase the wattage rating it is to simply make the output
stage a little tougher and increase the surge rating slightly. The input is
center tap push-pull BTW.

Doing the math using the info in the PDF
http://www.st.com/stonline/books/pdf/docs/8486.pdf about the Rds vs. current
and guesstimating it looks like the short circuit would be less than 62A per
mosfet and they are rated for 88 surge so it seems safe.

61.4634 amps - 12 = 49.4634 amps above 12
49.4634 amps x .05 ohms Rds per amp = 2.47317 ohms
2.47317 ohms + .13 ohms Rds @ 12 amps = 2.60317 ohms @ 61.4634 amps
160 volts rail / 2.60317 ohms = 61.4635 amps
88 amps surge rated - 61.4635 = 26.5365 amps to spare.

My math could be faulty in assuming that the Rds would continue to rise at
..05 ohms per amp but it may be safe to use the ST mosfets with no current
limiting resistor other than the wiring, mosfets, and the slight ESR of the
electrolytic capacitors. Another chart in the PDF shows only 20 volts of
drop @ 60 amps so I can't say for sure. My best bet is to never plug
something in to the AC outlet with the caps already charged up I guess.

Looks like I will have to just build it and never let anyone else
touch/operate/borrow it. Let them buy their own inverter and let the
manufacturer replace it if they blow it up.

 

[Daniel Who Wants to Know]

That's good, then you don't have to worry about that


On the source on resistance chart, the DS resistance goes up 0.01 ohm for
an
increase in 3A That is 0.00333 ohms per DS amp.

For a current of 100A the DS resistance value would be. 0.42 ohm. For a
total DS voltage of 42V

We have a supply voltage of 170V

Lets make up a formula to find the current value.
To make it easy lets assume that the DS resistance goes up from 0 ohm at
0A
with a slope of 0.0033 ohm per amp

formula.
(0.0033R/A *IDS)*IDS =VDS
Solve for IDS

0.0033*IDS^2=VDS
IDS^2=VDS/0.0033
IDS = square root of (VDS/0.0033)

Plug in the values.
SqRo(170V/0.0033)=IDS
SqRo(51515)=IDS
227A=IDS.
For a single device.
But you have two devices that the voltage is across, so you will have a
slope clime of 0.0066 ohm per amp.

SqRo(170V/0.0066)=IDS
SqRo(25757)=IDS
160A =IDS

It will be a little lower than that because the slope origin is close to
.01
ohm.

Instantaneous KVA subjected to the device upon a complete short would be
close to 27.3KVA

But you have two devices in parallel on each leg. That means that you have
two circuits for 160A to flow through. That makes it 320A total.
For a total of around 55 instantaneous KVA that the device has to
dissipate.

It won't reach that level, because the capacitors, and other components
will
give before that value is reached.

You can kill an electrolytic cap by charging it, and shorting it so many
times. The contacts inside will just melt open.

To get a more realistic value, you can take into account the true slope
origin, and the internal resistance of the capacitors, but that takes a
little longer than I want to spend on it. But you get the idea.

First of all thank you for your help. I plan to play it safe and add a .82
ohm 10 watt power resistor to the list as it is only $0.55 just to be safe
even though it will drop the efficiency slightly.

I am always tackling projects like this for the fun of it and keeping myself
busy. Wal-mart here has 3 APC pnoteac150 inverters in the clearance isle so
I will see if I can get one of them for general use and portability and save
my modified one for other things.

My main issue right now is still saving up to get a Fluke 187/189 as I still
don't have an analog or digital multimeter and am finding more and more
times that I really need a good accurate one and want to rip my hair out not
having the proper tools. I would go for something cheaper but true RMS,
high accuracy, and lifetime warranty are important to me.