Why does this inverter use two bridges?

Question

I recently destroyed a 1000 W-rated 24 VDC -> 220 VAC pure sine wave inverter I bought a while back from Ebay, by connecting its output to mains power inadvertently.

Afterwards, I decided to analyze its internal circuitry. The following is a general power path, details are not represented. The transformer is rated 500 W.

I realized it uses two sets of MOSFETs. The first set/stage is the 24 V half-bridge that drives the transformer. The second is a HV DC full-bridge inverter, whose output goes to the LC filter (2-3 stages) and finally the AC output.

Why does it use two stages instead of one? I don't know what frequency it runs at, but lets assume its 25 kHz or 50 kHz. Why isn't a single full bridge used before the transformer, that generates a 25 kHz or 50 kHz carrier square wave and modulates a 60 Hz sine wave on top of it (as a PWM duty), and do away with the rectifier and 2nd set/stage of MOSFETs after the transformer and feed that then directly to the filter/output?

Is it perhaps because the first stage doesn't actually oscillate at 25/50 kHz, rather something far less to avoid losses in the transformer, so that then a 25/50 kHz oscillation is required after the transformer to keep the filter components small?

Note: The "220V DC+/-" labels in the schematic is just to indicate the high-voltage, I don't know its value exactly, probably is more like 320 V.

Also, although there are so many variants that support so many different power ratings and input voltages, they all share the same general design (the chinese ones). Here's a link to a teardown done by someone else which isn't the exact same one as mine, but follows the same principle, except uses IGBTs for the second stage instead of MOSFETs:

Teardown and Testing of a 1500W Pure Sine Inverter

I've answered the question using hints from Edin, Bobflux, & William's partial answers.

Also, although I'm mixing standards up, it's irrelevant to the question/answer, in that usually 220V-240V is used with 50Hz, and 60Hz is used with 110V-120V. So a 60Hz standard would usually use a 60kHz carrier pwm, while a 50Hz standard usually uses a 50kHz carrier.

Answer 1

Why isn't a single full bridge used before the transformer, that generates a 25kHz or 50kHz carrier square wave and modulates a 60Hz sine wave on top of it (as a pwm duty), and do away with the rectifier and 2nd set/stage of mosfets after the transformer and feed that then directly to the filter/output?

Because, if you did that, the transformer has to pass 60 Hz and that means a much larger transformer and a much higher cost. You might say... but, it's PWM 60 Hz and my answer is that it's still 60 Hz and, the transformer has to pass it.

Just because PWM is used it doesn't mean that most of the energy that needs transferring through the transformer isn't at 60 Hz. Consider this example of a PWM modulated 60 Hz waveform: -

The red trace is the input 60 Hz sinewave and the blue trace is the equivalent PWM output. The PWM frequency is 3 kHz if you are interested. I've DC biased things just to make life simpler for my micro-cap PWM macro in the simulation.

So now, I'm going to low-pass filter the output with a simple RC filter at a 300 Hz: -

It shouldn't be hard to recognize that if we stripped-off the high frequency PWM artefacts (using the LP filter) that hiding inside the clutter is still the same 60 Hz waveform at the original amplitude.

_{PWM "modulation" is not classical RF modulation. Classical RF modulation shifts a baseband voltage (DC or 60 Hz) up to a carrier frequency and, PWM does not do that (as shown above when LP filtered). If it did do that then, when applying a low-pass filter, we would not see any trace of the original 60 Hz voltage (but we do).}

So, look at the size of any 1000 VA transformer used for taking normal 60 Hz AC down to (say) 12 volts. Compare it with the size of the transformer in your product.

Answer 2

Why isn't a single full bridge used before the transformer, that generates a 25 kHz or 50 kHz carrier square wave and modulates a 60 Hz sine wave on top of it (as a PWM duty), and do away with the rectifier and 2nd set/stage of MOSFETs after the transformer and feed that then directly to the filter/output?

With the dual bridge design, the output filter is a low-pass filter - it only needs to block the 50 kHz carrier. Because of the high frequency, the filter components can be kept small.

It is not clear exactly what kind of PWM modulation of the transformer control you are suggesting. To avoid saturation, the DC current through the transformer must be kept 0. Like Andy aka said, modulating PWM so that average voltage swings at 60 Hz is pretty much identical to just feeding 60 Hz to the transformer directly.

Most reasonable is to have positive and negative pulses of varying lengths. The transformer output will be something like a 50 kHz sinewave with varying amplitude:

Simply low-pass filtering the 50 kHz wave would yield 0 volt output. Instead, you need some kind of rectification, but then you would get voltage that varies +0V to +220V, not -220V to +220V. You could get rid of the DC offset with a 60 Hz high-pass filter, but that is a lot larger than a 50 kHz low-pass filter.

I haven't found whether this kind of high-frequency PWM modulation + rectifier + high-pass filter topology is used anywhere. This document details the most common inverter topologies, and as you can see, there are many variants with different advantages and drawbacks.

There is a similar method with PWM modulation of the transformer signal, but where a second H-bridge is nevertheless needed to unfold the rectified waveform to remove the DC bias: (image source):

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This is just my interpretation of what you are proposing - maybe you have some different kind of scheme in mind? Merely saying "PWM modulation" and "output filter" can mean many things, and some PWM modulation schemes would just saturate the transformer and not work at all.

Answer 3

You are asking a good question, and some inverters do employ such technique, but that is only in the "pure sine output" inverters.
I would say the reason why they did it this way is because it's apparently easier or simpler and possibly cheaper to implement.
It also probably because they made so many copies and it is easier to just keep running things the same way.
One reason I can think of is that besides slightly more complex circuitry to track the output sine wave voltage, it requires fewer components or less material, as you would need a somewhat large inductor/capacitor combination on the output, and those non-polarized quality capacitors are both larger and more costly.
It is easier to control/maintain a DC output voltage than an AC with changing load conditions; you just set and maintain it at the peak value (~310V in this case), and then just use a simple push-pull circuit like TL494, set its pulse width to give you an effective 220-230V "modified sine wave" as these are called, but it is in effect a square wave of slightly shorter pulse width on both polarities to roughly simulate a sine wave.

The design you're suggesting will have high-frequency as its main component, and it would need to be low-pass filtered to prevent it from being radiated out over the wires. The high-voltage bridge design uses the low chopping frequency of 50-60Hz which has less of a high frequency component which requires a smaller, less-powerful filter.

One more important thing or to expand on the above mentioned: the transistors in the schematic in your question only need to be switched on and off at certain times, and that's where all the worry about their control ends, as long as the high DC voltage of 310V is stable.
However, if you make the circuit as you suggest, you need to keep tracking the output because the sine wave shape and voltage will depend on the load. This would require a more complex and more expensive circuitry.
So, in conclusion, I would say the main reason why they go with the "modified sine wave" design as the one in your drawn schematic instead of the "real" or "pure" sine wave similar to the one you suggest is both the relative ease of control, simplicity and lower price, and likely a lower weight (transport figures into the price too).

Answer 4

Why does it use two stages instead of one? I don't know what frequency it runs at, but lets assume its 25 kHz or 50 kHz. Why isn't a single full bridge used before the transformer, that generates a 25 kHz or 50 kHz carrier square wave and modulates a 60 Hz sine wave on top of it (as a PWM duty), and do away with the rectifier and 2nd set/stage of MOSFETs after the transformer and feed that then directly to the filter/output?

Suppose it feeds 50kHz PWM whose duty cycle is a 60Hz sine into the primary of the transformer.

On the transformer secondary, the output signal would be the same, scaled up in voltage, and with the DC component removed, because transformers don't pass DC. Hopefully the primary H-bridge is well balanced so there wasn't any DC to begin with.

So it would look like that:

So we lowpass filter this and we get a nice 50Hz sine.

But does it work? Only if the transformer is large enough to handle the fundamental frequency of this waveform, which is 50Hz.

It's a bit hard to see because the signal is chopped up, but this really is a 50Hz sine. So it needs exactly the same heavy chunky transformer as an oldskool 50Hz-into-the-primary inverter would need.

If a tiny switching transformer was used, it highpass the waveform and saturate quickly due to the strong 50Hz component.

Transformers require the current to not have significant frequency components below their low frequency cutoff. The inductance of a tiny switching transformer is very low, so its low frequency cutoff is much higher than 50Hz, which means these unbalanced PWM cycles would cause the "DC" current in the primary to increase from each cycle to the next, until it saturates.

Basically what you're asking is not possible.

Answer 5

There's some debate over whether a transformer "sees" the mains frequency in a pulse-width modulated input. For an intuitive understanding, it may be helpful to consider the problem in the time domain to avoid considering frequency components.

Suppose we have a high frequency PWM input to the transformer modulated with a 50 Hz sine wave. To avoid saturating the transformer, the average of the PWM input needs to be zero, so we make the PWM go between +V and -V rather than between +V and 0V.

Now, consider the first half cycle of the 50 Hz sine wave, i.e. the first 10 ms. The PWM starts at 50% duty cycle (for an average of zero), then increases to somewhere near 100% at 5 ms, then reduces back to 50% at 10 ms. For the entire half cycle, the duty cycle is above 50%, so the negative parts of the PWM input can never balance out the positive parts. This means the current into the transformer, though it goes up and down at the PWM frequency, always goes up more than it goes down and therefore overall it increases for the entire half cycle.

It is not until the negative half cycle of the sine wave that the negative part of the PWM can start outweighing the positive. The conclusion is that for the average current to be zero, we have to consider the average of the PWM for an entire 20ms cycle of the sine wave. The transformer needs to be able to handle the current rising for 10ms without saturating before the current starts falling again. If we need the transformer to be able to operate on the timescales of the 50 Hz sine wave, then the conclusion is that in some sense the transformer is seeing the 50 Hz sine wave even with the PWM modulation.

Answer 6

Both topologies can work with either a half-bridge or a full bridge, despite in this case the 2-stage topology uses a half-bridge whereas the 1-stage I describe would use a full bridge, but this is an irrelevant difference that only affects the max power the inverter can support.

Anyways, irrespective of bridge type, what it does is effectively alternate the polarity of voltage from the DC power source on the poles of the primary coil of the transformer, at high frequency. So we would draw a PWM signal plot as a voltage on the y-axis, and time for the x-axis. The positive and negative voltage swings would be equal in magnitude, thus equidistant from the horizontal time axis.

However, if it were a perfect square wave, ie 50% duty cycle, then the integral of voltage over time for both halves of the square wave cycle, would be 0 (area over and under the graph would be equal so cancel out). The integral represents the total difference in reactive current that would pass through the primary coil. This must always be zero in any case, but during each swing, the area will temporarily be non-zero, and the magnitude of this depends on frequency. So the higher the frequency, the smaller the total current difference (per swing), hence the smaller the transformer can be. A pure sine wave would have much larger areas for each half-cycle, meaning larger current swing, which requires a much larger transformer.

So far so good.

Once one modulates a sine wave over a pwm signal, no matter what the frequency, the average areas over and below the time axis, won't be equal anymore, and they would continue adding up in the same net direction, until the complete half-cycle of the modulated sine wave, because to modulate, the duty has to continuously change to reflect the instantaneous magnitude of the sine wave, so most of the time the wave will be asymmetrical in terms of +/- voltage swings, so after each period of the high-frequency carrier signal, there will remain a net current in the primary coil that continues to sum up over successive carrier periods. So one has to continue integrating over the period of the longest wavelength to get the total swing in each direction. Hence the total area change and thus the total current swing for a 60Hz sine wave modulated over a higher frequency PWM signal, will be exactly equal to the area if one had applied the 60Hz sine-wave directly, which would obviously saturate any transformer that can't already handle a pure sine wave. Yes, as the frequency increases, the area under each individual voltage swing is smaller, but the final total current swing will still accumulate until the longest half-cycle has been considered, which is in this case the 60Hz sine wave.

What adds to the confusion, is that although both topologies result in functioning inverters, and both topologies are represented and explained and actually built in various tutorials and blogs and videos on the web, a 1-stage topology still does require much larger and heavier transformers to work, yet many of these sources actually show small transformers, which is misleading and actually incorrect. Most likely they copy their designs from each other without knowing the full story so a mistake can propagate and amplify this way just like any rumor.

Moreover, with 2 stages, each stage can operate at a different pwm/carrier frequency. Hence the 1st stage can operate at the frequency that minimizes the transformer losses, while the 2nd stage can operate at the frequency that minimizes the size of the needed output filter.

Moreover, 2 stages is easier to regulate, since as the load increases, the voltage will drop, which the signal-generator will have to counter (and vice versa as the load decreases to avoid over-voltage). In 1-stage, it means the high AC voltage has to be sensed and factored in, which is difficult to do in a stable/reliable and/or isolated manner. However, with 2-stages, then the 1st stage produces a fixed frequency square wave (50% duty), and the transformer output passes through a bridge-rectifier and fills an electrolytic capacitor that hence stores high-voltage DC (the transformer is chosen so that this voltage always exceeds the peak AC voltage, to account later for voltage drop). There is no regulation necessary here and infact nothing to regulate, because you can't change a square wave in any way that would change the output voltage, and the high DC voltage on the capacitor after the rectifier-bridge, would remain constant independent of load, because current will pulse through as long as necessary (because it is effectively a true flat DC voltage after the rectifier) so maintains the voltage. So the regulation is instead done purely within/at the 2nd stage, hence maintaining isolation.

The 2nd-stage pwm modulates a 60Hz sine wave on a higher-frequency carrier, where the duty would oscillate around the 50% mark at 60Hz. What max/min duties it goes to, is a matter of regulation. It senses the output voltage, and when this drops, it increases the duty swing (all the way up to full duty swing), and when the voltage rises above the target voltage, then the duty swing is decreased as much as necessary. This is possible because with no load, the duty cannot and should not swing all the way anyways, because the source, the high-voltage DC, is already higher than the peak (and as mentioned must be to enable compensating increasing loads as well as attenuation in the filter). Also, both stages can remain fully independent/isolated from each other, ie not only use different pwm frequencies, but also independent/isolated signal generators (such as microcontrollers) and logic-level power supplies. Alternatively, the 2nd stage can keep pole "A" connected to the storage capacitor's negative pole for one 60Hz half-cycle while pwm-ing pole "B" of the output (that goes to the filter) on the positive pole of the storage capacitor, and then on the other half-cycle, it keeps pole "B" at the negative capacitor terminal while pwm-ing pole "A" on the capacitor positive pole. In this case, the duty will oscillate at 60Hz between 0 and some max value that is less than unity, and only increases towards unity as the load increases. Once it maxes out, the inverter has reached max rated power and can only give more power if the output AC voltage is allowed to drop.

How does one achieve an initial over-voltage? Well it depends on the voltage of the power source and the transformer ratio. One might choose the ratio so high that even when the batteries are near empty hence have a lower voltage, that there is still enough over-voltage at the storage capacitor that one can still provide full output rated power without any voltage-drop, but this means the over-voltage will be very high, which is risky, because if the control-loop isn't rock-solid and lightning fast, then as the load decreases (or is suddenly disconnected), then dangerous over-voltages can temporarily appear on the AC output.

Also, the Filter inductor can remain small despite 60Hz passing through it, exactly because it is blind to it and unable to react to it. The load takes care of this. Even without a load, the open terminals "react" to that. AC current still passes through the inductor, but only due to the filter capacitor, and the capacitor innately blocks low frequencies and DC. A transformer without anything connected to its secondary, would be like an inductor/filter, and an inductor/filter with another coil wound around it, would be a transformer. So they are very similar. So without a load, the transformer is like a single inductor connected in series with the power source, without any capacitor, meaning the inductor (transformer), is itself the load, and itself has to react to all frequencies present in the source. So if the source has a 60Hz component, then the inductor would have to be much larger to be able to react to this, otherwise it saturates and a short-circuit current arises. This doesn't happen with the filter inductor because it is in series with another load. This means the constraints on the transformer are tighter than on the filter inductor, which is why one may have 60Hz pass through it but not the other, and this in turn necessitates 2 stages, ie two sets of switches and pwm signals (unless you are willing to use a transformer (inductor) with a much larger core so it can store more energy, which is equivalent to being able to react to lower and lower frequencies).

Btw... if you were to connect the hypothetical secondary coil around the filter inductor to some tertiary load, then because of the coupling, it would siphon energy out of the inductor which would be compensated by whatever is powering the filter, and it is now defacto a transformer (the output voltage would depend on the turns ratio). In this case the current passing through the filter inductor can exceed its saturation current without problem, and this would happen in any transformer anyways.