PF correction by inverters

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Consulting Electrical Engineer - Photovoltaic Systems
How do they do it?
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An inverter is generating the voltage and current sine waves. Reactive power control as part of the solar PV inverter can allow monitoring of the grid connection and the inverter can product active and reactive power, resulting in PF correction.
 
An inverter is generating the voltage and current sine waves. Reactive power control as part of the solar PV inverter can allow monitoring of the grid connection and the inverter can product active and reactive power, resulting in PF correction.
Actually, I know that. What I am asking is how the inverter produces reactive power, i.e., how it increases the phase angle between the voltage and current waveforms, if that is what it is doing.
 
Reduce active power by adjusting the phase angle relationship between voltage and current to be the opposite of the grid side connection?
 
Actually, I know that. What I am asking is how the inverter produces reactive power, i.e., how it increases the phase angle between the voltage and current waveforms, if that is what it is doing.
My limited understanding of a grid-tied inverter is that it has a DC bus at a voltage at or above the AC peak voltage, and it does high frequency (relative to the 60 Hz grid frequency) PWM switching between the bus and the output to produce a signal whose average voltage over multiple PWM cycles (still a small fraction of a 60 Hz cycle) is matched to the grid voltage. Filtering between the PWM switching and the output would provide smoothing.

Then if the current while the PWM is on is constant, the resulting filtered current waveform would be in phase with the grid voltage waveform. So I think the question is how does the inverter independently control the current waveform so that it can introduce a phase shift?

Cheers, Wayne
 
Not for sure, but this might pertain to it. I'd guess the VFD has active filtering AND fluctuates the voltage.

I remember controlling a shipboard-based generator (so running in island mode)... if we wanted the system more leading, we would increase the voltage. I think the reason is that higher voltage lowers current for constant power loads, so it reduces the lagging reactive power (but I have not worked it out).

I think it would also depend on the type of loads (constant current, constant power, or constant impedance. , other types?)

If you think about it, you could change the time that the MOSFET's (IGBT's) fire in each cycle, and that would change the power factor (of a rotating machine)... right? Maybe? Like variable timing on a car engine.
 

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Yes, that indeed is my question.
An inverter producing real power only, will produce a voltage waveform that matches the grid waveform, except for some epsilon of an increase in amplitude to account for voltage loss through the line impedance to the point of interconnection. An epsilon of voltage difference that directly determines the current. The inverters use a feedback loop to determine what the grid impedance is, in order to produce the voltage necessary for the desired power output.

To produce reactive power, it will need to produce a waveform that differs not only in amplitude, but also a difference in phase. The vector difference in the voltage waveforms between the inverter's voltage and the grid voltage, will produce current that is out of sync with the grid voltage.

For instance, consider a perfect 120V grid voltage at a phase angle of zero by definition, and the line impedance to the point of interconnection is 0.4 Ohms of resistance and negligible reactance. We'd like our inverter to provide 1.2 kW of real power and 0.9 kVAr to support an inductive load. We ask the question, what amplitude and phase angle, must the inverter produce?

With V & I as vectors, we set up the following equation:
V_inv = V_grid + I*R

V_grid = <120V, 0>
V_inv = <Vx, Vy>
I = <Ix, Iy>

Based on given kW & kVAr values:
Ix = +10 Amps
Iy = +7.5 Amps
If it supports a capacitive load, Iy would be negative.

This allows us to calculate Vx & Vy:
Vx = V_grid + Ix*R
Vy = Iy*R
Vx = 124V
Vy = 3V

Find the corresponding magnitude:
|V_inv| = sqrt(Vx^2 + Vy^2)
|V_inv| = 124.0363 Volts

And the phase angle:
angle(V_inv) = arctan(Vy/Vx)
angle(V_inv) = 1.386 degrees

The inverter voltage would lead the grid voltage in phase, in order to support an inductive load where current lags voltage inside the load. Opposite for a capacitive load.
 
The PWM bridge doesn't match the grid voltage. Instead it produces an output that is different from the grid voltage so that current flows from inverter to grid.

In principle the output of the PWM bridge can be driven to generate any waveform and any difference relative to the grid waveform.

So the output current could be shifted relative to the grid voltage, or the output current could intentionally have harmonic content.

This same concept is used for active rectifiers that consume real power from the grid, but produce VARs or harmonics to compensate for other loads.
 
I am an extreme novice in the field of power electronics, but I have in fact constructed a 4 kW pure sine wave inverter. I know how my inverter works. Sinusoidal pulse width modulation. Wouldn't adjusting the phase angle just be a matter of adjusting the timing of the waveform generator? Or maybe I don't understand the question exactly.
 
Actually, I know that. What I am asking is how the inverter produces reactive power, i.e., how it increases the phase angle between the voltage and current waveforms, if that is what it is doing.
A grid tied inverter does not directly generate the voltage waveform, it is a current source. It monitors the grid waveform and injects ac current. It can adjust the phase angle, i.e the timing of how it injects the sinusoidal current in relation to the grid voltage signal. If it advances the current it produces slightly in relation to the grid voltage waveform, it can bring a lagging power factor closer to 1. That's how I look at it, fwiw.
 
To produce reactive power, it will need to produce a waveform that differs not only in amplitude, but also a difference in phase. The vector difference in the voltage waveforms between the inverter's voltage and the grid voltage, will produce current that is out of sync with the grid voltage.
Your post is a nice description of the above, which I would summarize as saying that if a device can produce an arbitrary sinewave voltage at its terminals, and it is connected to an idealized voltage source V in series with an impedance Z, to push out a current I it needs to produce a voltage V + IZ. And the above holds for arbitrary complex phasors V, I, and Z.

But under the hood, how does a commercial grid-following inverter do this? I don't believe it's measuring the impedance Z and doing the calculation. Rather, there must be some feedback loop that takes in, at a given moment, the voltage at its terminals, the current, and the desired output current, and uses those inputs to decide how to control some switching. Can you comment on that?

Thanks, Wayne
 
The inverter has to match the service voltage in magnitude, frequency, and phase, so it must be able to time (phase) shift the current to correct the PF.
 
Your post is a nice description of the above, which I would summarize as saying that if a device can produce an arbitrary sinewave voltage at its terminals, and it is connected to an idealized voltage source V in series with an impedance Z, to push out a current I it needs to produce a voltage V + IZ. And the above holds for arbitrary complex phasors V, I, and Z.

But under the hood, how does a commercial grid-following inverter do this? I don't believe it's measuring the impedance Z and doing the calculation. Rather, there must be some feedback loop that takes in, at a given moment, the voltage at its terminals, the current, and the desired output current, and uses those inputs to decide how to control some switching. Can you comment on that?

Thanks, Wayne
Indeed these devices do this because they measure (and calculate) the voltage and current VERY fast, basically instantly. This does gives you an 'instant' Z, I, and V. Modern solid-state measuring can sample the voltage and current hundreds, thousands, or some millions (more like an oscilloscope) of times per cycle (which is already 1/60 seconds here in the States). This is done with a lot of different electric device... CT's, PT's, ADC's, uC's, iC's, PFM:)'s, etc... Note too, this is also how a sub-cycle trip signal can be generated.

That is where the magic lies... not only can it be measured fast, it can be controlled fast too. Know how sound-cancelling headphones work, right? Active filtering is analogous to this except with electricity instead of sound. I think (somebody please correct me if I am wrong) these headphones were invented after active filtering was already being used...? Not sure.
 
The inverter has to match the service voltage in magnitude, frequency, and phase, so it must be able to time (phase) shift the current to correct the PF.
Yet must be mismatched by enough to cause it to transmit energy from the inverter to the grid. Too much mismatch will destroy things though.
 
Not really. PV grid tied inverters are current sources.
Well, they behave as current sources by putting out a voltage at their terminals equal to V + IZ where V is the no load terminal voltage (no load either way), and Z is the source impedance, to achieve a current I. Or maybe it's more accurate to say that the inverter controls the PWM pattern of connecting the DC bus to the output stage to result in a current and terminal voltage that obeys the above relationship, and the control algorithm likely is targeting a particular real power output (typically maximum possible given the constraints of available DC power and circuitry limits) along with a particular phase shift between I (actual output current) and V (no load terminal voltage).

[Still looking for a simple description of that control algorithm, something like "based on these particular instantaneous measurements and this target signal, do or do not connect the DC bus to the output in this specific fashion."]

Anyway, for the linear case, current sources and voltage sources are equivalent per Norton's / Thevin's theorem.

Cheers, Wayne
 
If the inverter voltage is below the grid voltage will it push anything into the grid? Doesn't have to be much different to do what it needs to do.
I didn't say it could be lower, just equal. Thought experiment: connect an inverter to the service with superconducting conductors, so Vd=0. The voltage will be the same at both ends of the conductors and the inverter will still export power.
 
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