How do AC inverters handle power factor loads? - Page 1
Forum section: electronics / beginners

author: Freesurfer
author_rank: Regular Contributor
timestamp: September 26, 2020, 09:58:50 pm
content: An inverter is simple enough to understand for a pure resistive load. All it really needs to do is to increase the PWM duty cycle in either the positive or negative direction until it reaches the reference voltage, then lower the duty cycle and let the resistive load "pull down" the voltage.However, for a power factor load, that's a little different. Imagine sticking a somewhat sized AC capacitor straight across the output of the inverter with no other loads attached. Initially that's fine for the first cycle when the inverter charges the cap from 0V to ~315V peak. But the fun begins once the sine wave begins its downward slope and the inverter has to draw current back from the capacitor to discharge it to 0V and begin the negative half of the wave. On the grid that's no issue as the capacitor will simply discharge into other loads on the grid. But on an inverter with no other loads, that does not happen.Let's say the inverter is a 230V sine wave one, and the capacitor results in a 2A RMS current. So that's 460VA of apparent power, but 0W actual power. Enough power to "worry about". Will it behave like some sort of LC tank filter passing current back or forth between the filter inductor in the inverter and the capacitor? Does the power get passed back into the DC source?I am assuming a good and properly built inverter, not some chinese junk that blows up as soon as it is challenged with something more than a 10W incandescent bulb.

author: David Hess
author_rank: Super Contributor
timestamp: September 26, 2020, 10:08:43 pm
content: Quote from: Freesurfer on September 26, 2020, 09:58:50 pmThat is exactly what happens in both cases; a current circulates between the source and load which adds to the power seen and lost in the inverter.

author: Freesurfer
author_rank: Regular Contributor
timestamp: September 27, 2020, 01:30:59 pm
content: Makes sense. The output stage of an inverter is usually some sort of LC stage where the inductor smooths out of the PWM from a H-bridge to make a sine, with the capacitor to filter away the high frequency stuff. So it makes sense that the PF load kind of backfeeds into either the inductor or capacitor in the output stage of the inverter.So if I understand it right, no new power is (in theory) supplied by the DC power source with such as load, as the LC output stage is already energised in one form or another, passing current between each other and the load. In practice there are of course some switching losses and other losses since nothing is 100% efficient or ideal. But the point is that the inverter doesn't have to burn off that backfed power as heat, either in the MOSFETs, an internal resistor or as other losses. So no energy is lost (other than the inefficiencies of the inverter)

author: chuckthevkham
author_rank: Newbie
timestamp: June 19, 2021, 08:00:59 am
content: I find it hard to understand how a grid tie inverter can feed energy into the grid ?The grid has a very low source impedance whereas an inverter has a relatively high source impedance . So how can an inverter raise the main's voltage ?The mains is a very nasty environment with all the disturbances that can be present ? For the inverter to handle the voltage sags/surges impulses and flat topping etc that are present it must be pretty bloody smart so it's not producing reactive power .If the inverter is able to increase the mains voltage then how far does this go , all the way back to the HV transformer , or just down the road a way ?

author: Circlotron
author_rank: Super Contributor
timestamp: June 19, 2021, 09:05:13 am
content: Quote from: Freesurfer on September 26, 2020, 09:58:50 pmWhen the inverter is charging the capacitor in the positive direction the upper switching device and the inductor act like a buck converter, sending energy from the DC buss into the capacitor.When the inverter is discharging the capacitor towards zero volts the lower switching device and the inductor act like a boost converter, sending the capacitor energy back to the DC buss.

author: Circlotron
author_rank: Super Contributor
timestamp: June 19, 2021, 09:12:27 am
content: Quote from: chuckthevkham on June 19, 2021, 08:00:59 amJust as drawing power from the mains doesn't necessarily decrease the mains voltage very much, pushing power back into the mains doesn't have to increase the mains voltage hardly anything at all either. If the supply impedance was zero the AC voltage wouldn't change at all. All that has to happen is for you to TRY to decrease or increase it respectively. To push power back in you need to follow the mains sinewave but try and pull it further *away* from zero than it currently is at any given moment. With a variac and an isolation transformer you can make power flow either direction simply by raising or lowering the voltage just a tiny little bit.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 19, 2021, 11:29:40 am
content: Right, the solution is actually easier.Your premise I think assumes a single switch turning on and off, the output being implicitly pulled to GND by load resistance. That is, the source impedance alternates between nearly zero (Rds(on) plus DC supply impedance) and nearly infinity (leakage). This is how you most often drive a relay coil for example, or LEDs (when not driving them from logic pins, that is).The fact that the impedance is changing, makes analysis very difficult. It is not a linear, time-invariant (LTI) system. If you put any arbitrary load on there (some RLC network, or even just some complex impedance at the driven frequency and harmonics), it's difficult to predict what will happen. Further nonlinearities are likely to get involved. For example if the load is inductive, probably when the switch turns off, the voltage flies back and is clamped either by the switch's own avalanche breakdown (a hazardous operating mode) or an opposing switch's parasitic diode. (Or if neither, it probably just blows up after that one cycle.)If we instead provide means to pull the load back down to zero, or to -V, alternating with the switch to +V, now the output is enabled at all times, and its output impedance is well defined, or much better anyway. (There is the brief period when both switches are off, where voltage can be allowed to freewheel a bit; the impedance for such short periods is determined by switch capacitances and the stray inductance between them, and so the average impedance at the driven frequency is an average between these; it's not nearly an undefined impedance, like it would be in the on-off case.)And with a constant low inverter impedance, we can use a PWM filter of the one-port-shorted type, providing termination at the output port with a shunt R+C (and perhaps a series L or R||L as well, so that low impedance loads don't screw it up by shorting out the R+C). This will be a L-input lowpass filter, and that first inductor takes the brunt of the PWM ripple as the main filter inductor so must be sized accordingly.Put another way: in the one case, we really don't have any analytical tool to bring to bear; the best we can do is shove a representative model into SPICE and simulate it from timestep to timestep. In the other case, our LTI assumptions are satisfied (or more nearly so), and we can use traditional AC steady-state analysis, and network theory and etc., to design an efficient filter for it.There are other situations in drives and controls, where it pays to keep the system linear, or only permit certain exceptions. Like, you might have a power supply that should be linear so its loop can be compensated (and that compensation is adequate for all V/I output settings), but make an exception that, under startup or fault conditions say, the compensation is bypassed to allow quick startup/shutdown. Or for a bench supply (CV/CC output mode), the compensation for each respective error amp (voltage and current) might be disabled or clamped, such that the other is able to catch the output quickly after its variable crosses the threshold, thus reducing integrator windup (which causes a big excursion in voltage and/or current when crossing between CC/CV).Tim

author: David Hess
author_rank: Super Contributor
timestamp: June 21, 2021, 02:48:21 am
content: Quote from: chuckthevkham on June 19, 2021, 08:00:59 amThe difference in impedances actually makes things easier. The grid has a very low and inflexible impedance, but the inverter can synthesize any impedance it wants at its output. A high impedance acts as a current source, and an AC current source which is synchronized to the low impedance grid will happily feed current back into the grid and the low impedance of the grid regulates the voltage as appropriate.Real grid-tie inverters do this, as well as adjusting the phase of the injected current which allows more power to be pumped back into the grid without violating voltage limits.QuoteAs I pointed out above, the grid-tie inverter may deliberately produce reactive power within the frequency limits of the grid.It *is* a very nasty environment. Better designs use a phase-locked loop to maintain synchronization with the grid despite disturbances, and they always have the option to disconnect for protection.QuoteIt has an effect all the way back to the grid source, and the effect varies with the series resistance of the grid so it has the most effect locally.