How does this 3000VA Transformerless UPS work?

[intabits]

I did a teardown of this HP model R/T3000I transformerless 3000VA UPS, which seems to be a re-branded unit made by Eaton. The input and output voltages are from 200 to 240V, and it uses a 120V battery.





Of course, there are no schematics for it. The only documentation I've found are user manuals that only discuss the user interfaces, and electrician-oriented installation details. It also seems that the same model number, but of different vintages, may have completely different internals - some seem to be not transformerless.

Here's the main power board:-




And the In/Out board that connects to the input and output AC sockets, and to the power board via 3 main conductors (labelled line,neutral and load)



Being completely unfamiliar with these devices, I was simply guessing how it works as I went. My understanding is that the output of the UPS always comes from the converter, thus providing a clean conditioned output (online mode). But when a fault is detected in the converter, bypass mode is entered whereby the output power comes directly from the incoming mains. And of course, when the mains input fails, the converter is powered by the battery.

I want to do a follow-up based on a proper understanding of the unit, so I reverse-engineered the power circuitry of the main power board. But it's not quite what I was expecting, and I'm struggling to see how it works. I'm hoping someone can explain it to me. Here's my schematic:-



It looks like Q5, Q6, Q9, and Q10 form the main H-bridge that is driven by sinusoidal PWM signals, and the high frequency PWM filtered out by L1, L3 and C5, leaving pure sinewave power going to the load. Beyond that, I'm stuck:-
BUS+ looks to be powered during normal operation from LINE via K6, F2A, K5, CS2 and rectified by D22, and during power loss, from the battery via K3.

But I can't see how the negative DC bus (BUS-) gets powered.
In battery mode, maybe via K4, K5, CS2 and D23?
And in normal online mode, ??!!

Do D22, D23 and Q7, Q8 form some sort of synchronous rectifier to power the DC BUS?
Or are Q7 and Q8 part of the output inverter in some sort of H5 topology?

I keep looking for (and not finding) a more "solid" connection from BAT- and/or Neutral (via D23?) to BUS-. I've triple checked all this on the PCB, and I'm pretty sure my schematic is correct.

More questions:-

Q4 is on the large heatsink with the others, what does it do?

My guess is that Q20 is involved with charging the battery, but can't see how.

What is the function of the huge inductor L2?

How can this produce 240V output from a 120V battery?


Any insights or clues as to how this thing operates would be most appreciated. Thanks...

[NiHaoMike]

Looks like they're using L2 in a boost converter.

[intabits]

I can vaguely see how that could be going on, having an inductor, switch, diode and capacitor all in the mix there.

But I'm still looking for some connection to Neutral that allows the circuit to make use of the power and potential between Line & Neutral.
Can you provide any more detail on how you see that working? Even just the components involved would be a help.

I'm trying to simulate it, but missing that connection to neutral. The only path seems to be via Q6, but that's busy being part of the output bridge...

[The Soulman]

Being completely unfamiliar with these devices, I was simply guessing how it works as I went. My understanding is that the output of the UPS always comes from the converter, thus providing a clean conditioned output (online mode). But when a fault is detected in the converter, bypass mode is entered whereby the output power comes directly from the incoming mains. And of course, when the mains input fails, the converter is powered by the battery.

The ones I have seen use the by-pass relays in normal mode, they can advertise clean conditioned output because of the (mandatory for back-up mode) filters that the power travels thru.
The converter is much to inefficient to run 24/7 and would result in a lot of generated heat and a short overall lifespan.

I'd say q20, q6 and/or q4 are involved with charging the batteries.

[intabits]

That was my initial assumption also. I've been trying to find a definitive statement on this, or at least the information that changed my mind.

For this exact model, there is no explicit statement that the converter provides the output in normal operation, but it does describe "operate"
mode and "auto-bypass" mode. The best I can find is that "operate" mode is the normal operating mode, and it switches to auto-bypass when it detects: Output power overload, Over temperature, Fan failure, or Internal UPS failure while in operate mode (meaning??). These conditions all appear related to an operating converter.

Maybe the UPS's you described are lower power, consumer grade units? These higher power units are intended for commercial systems such as servers, and for the high cost (starting at $2K-$4K), one would expect that they always do more than just input filtering. 

This is a rebranded Eaton UPS, but I haven't found which exact Eaton model it is. At least some of the similarly rated Eaton models mention transformer AVR" (Automatic Voltage Regulation) which appears to be an active subsystem that may supply either the inverter or the output directly (in bypass). The AVR follows the input filter.
The 3000VA models produce their output from either the AVR or the inverter (but not both). However, this unit has no such AVR subsystem, just the input filtering.

Other Eaton documents describe "line-interactive" and "on-line" UPS types. Line-interactive is the type that uses AVR to correct abnormal voltages, but on-line types do continuous "double-conversion" (AC->DC->AC).

Although there's nothing definitive, all that I've seen leads me to believe that this unit is the "on-line" type, that runs the converter continuously.

Yes, Q20 and Q4 do seem separate from the converter function, and so are likely to do with battery charging. 

But I'm still lost on where the current paths are in this device for any function other than the final H-bridge output... 

[BennoG]

We used a 750VA online converter that wat $1k. So I guess the 3kVA are a lot more expensive.
I think the unit is in bypas mode in 99% of the time and when the power fails it kicks in.

If you have a variac you can simly test this by setting the variac at 230V and measure in and output voltage.
then lower it to 220V and measure again.  If in both cases the in and output are the same as the input it is the usual bypass UPS that kicks in when power is gone.
A normal server can handle a dip of 100ms (the time needed to switch the relais and start inverter)

Benno

[Doctorandus_P]

Why are there three heavy wires going to the output? (Red, Black, White? Ah: Black is AC power input.) Because it's supposed to generate a sinewave output, I should expect a symmetrical connection here. In the US, (And UK I think) it's apparently common to have a real "live" and "neutral" wire. but here in europe, plugs are not even polarized, and both are treated the same.

I have not analyzed the circuit, but I suspect they switch the battery between positive and negative halves of the sine waves, and maybe boost it a bit too, to get over the peaks of the sinewaves.

[m k]

My intuition doesn't say how the booster wave goes.

First Q20 and Q7 on.
Then loading C13 and L2.
Then Q20 and Q7 off and Q8 on.

Then positive through Q9 or Q5 and return through Q6 or Q10 to negative.

Leftover K3 and Q4 can be a charger.

[ramussons]

The basic principle of a transformerless invertor is a PWM bridge invertor where the "primary" of the transformer is the load itself, as shown in the attached schematic.
The 120 Volts from the battery is boosted to 350 volts and applied to the bridge.
The part of the layout connected to the battery galvanically will be floating.

[intabits]

Thanks for the responses!

If you have a variac you can simply test this by setting the variac at 230V and measure in and output voltage. then lower it to 220V and measure again. 

I got these things for dirt cheap, I don't even know if they're broken or operational. But they have no battery packs, so I imagine powering them on without batteries would just result in some fault condition, and not reveal very much. If I had the batteries, I would have already tried this and hopefully would have been able to answer these questions myself.

But it's a good plan, I wonder if there's a way to make it think it has a battery pack, without actually locating 10x12V SLA batteries?

Why are there three heavy wires going to the output? (Red, Black, White? Ah: Black is AC power input.) Because it's supposed to generate a sinewave output, I should expect a symmetrical connection here. In the US, (And UK I think) it's apparently common to have a real "live" and "neutral" wire. but here in europe, plugs are not even polarized, and both are treated the same.

The white is the neutral, and as far as I can tell is used as a "common" for both input and output to this power board. It connects directly to neutral on the output sockets, and to the neutral of the incoming AC, via a relay.
The black is line/live/active, and is input to the board, coming from the AC input socket, again via a relay.
The red is marked "load" and is presumably the output from the board to line on the output sockets (via relays).

These 3 wires, plus the 2 from the battery, are the only heavy wiring to the board.

I have not analyzed the circuit, but I suspect they switch the battery between positive and negative halves of the sine waves, and maybe boost it a bit too, to get over the peaks of the sinewaves.

That makes sense, but how is it done?

My intuition doesn't say how the booster wave goes.
First Q20 and Q7 on.
Then loading C13 and L2.
Then Q20 and Q7 off and Q8 on.
Then positive through Q9 or Q5 and return through Q6 or Q10 to negative.
Leftover K3 and Q4 can be a charger.

I don't think that can be right: Q20 is on a small heatsink, not on the big one with all the others. It's almost certainly part of the charging.
And C13 is just a tiny 100nF cap, likely just for suppresion. But maybe C11, it's a "Bus Link" capacitor like C3. These are low ESR, ESL and high current, with heavy terminals. (The upright flat white parts in the photo). I can't find any data on them but they look like this: https://www.taobao.com/list/item/41666261949.htm

The basic principle of a transformerless invertor is a PWM bridge invertor where the "primary" of the transformer is the load itself, as shown in the attached schematic.

Yes, that's the part of the circuit I do understand. Unfortunately, it's the only part!

The 120 Volts from the battery is boosted to 350 volts and applied to the bridge.
The part of the layout connected to the battery galvanically will be floating.

And how the boost works is also part of what I don't understand.
 

[m k]

Yes, C11, but it doesn't help if Q20 is small.

L2 is only loaded through Q7 and the transistor sextet is just like a 3-phase inverter.
But I just can't understand how L2 is boosting, since with battery connected it must boost downwards, and like somebody said, where is the return path.

So is L2 negative compared to battery negative?
So its energy is not going through D23 and positive side is not going through D22.

For negative and neutral, no suck absolute things, only potential differences.
Turn the wave going to Y-direction.

[oPossum]

My best guess...

[intabits]

Yes, C11, but it doesn't help if Q20 is small.

L2 is only loaded through Q7 and the transistor sextet is just like a 3-phase inverter.
But I just can't understand how L2 is boosting, since with battery connected it must boost downwards, and like somebody said, where is the return path.

Oh, but K3 is there, allowing Bat+ to BUS+, and Bat- can get to BUS- via K4,K5,D23 (but via D23 looks wrong to me).
And for on-line operation, yes, where is the return path?
I keep expecting to find neutral going to cathode of D23, but I've checked that area multiple times. Neutral goes only as shown in the schematic...

So is L2 negative compared to battery negative?
So its energy is not going through D23 and positive side is not going through D22.

For negative and neutral, no suck absolute things, only potential differences.
Turn the wave going to Y-direction.

I don't understand what you're saying there.

My best guess...

Yes, that does seem to be the consensus, but how is BUS- powered?
I've checked that again for the nth time and the schematic is correct.


I want to use the parts from this thing in a similar way, so it would have been helpful to understand it. And I was hoping that in my unfamiliarity with such things, I was just missing something obvious, but it seems that Eaton were practicing some non-obvious witchcraft here.

Thanks to all for your help and efforts, but let's wrap this up here, it's not so important, I was just curious.

[oPossum]

Yes, that does seem to be the consensus, but how is BUS- powered?

Via Q8 (synchronous rectifier) and D22

Q7 charges L2
Q8 discharges L2 via D22

[m k]

So is L2 negative compared to battery negative?
So its energy is not going through D23 and positive side is not going through D22.

For negative and neutral, no suck absolute things, only potential differences.
Turn the wave going to Y-direction.

I don't understand what you're saying there.

First C11 is charged through D23.
Then Q7 goes on and L2 is charged.
Then Q7 goes off and L2 is floating.
Then Q8 goes on and L2 becomes a power source.

So charged C11 and L2 are connected again, but now battery voltage is in between.
Only non solid point is between C11 and L2.
L2 can't crank battery negative up, so it itself must go down, it still has the initial potential difference and stored energy.

Charge arrow over L2 is from left to right.
When Q8 goes on arrow over L2 is still left to right.
Right of L2 is battery negative, it doesn't change.
Left side of D22 is battery positive, it doesn't change either.
So L2 start pulling BUS- down and C11 start charging more.
Finally over C11 is full voltage and battery negative is somewhere in the middle.

Then C11 is discharged to the load, I think.

[intabits]

Cool!
Crude simulation:-


Thanks!

[m k]

One more thing, after a first charge C11 and L2 wont go under battery voltage.
C11 clearly can't and L2 is disconnected before that start happening.
So L2 can recharge while C11 is discharged, though it's probably bit of a necessity also.