Unwind di Power Electronics - 5 di 5

La topologia di un inverter trifase

La topologia di un inverter trifase è la seguente: LESSON #34Vdc si 53Ssab c56si saSo un inverter trifase si basa su topologie a semiponte, ma è utile separare l'alimentazione Vdc in 2 tensioni diverse e il nodo centrale viene utilizzato come riferimento. Il carico generico è rappresentato genericamente come un carico bilanciato con una componente resistiva, una componente induttiva e un generatore di tensione indipendente in serie. La topologia diventa quindi la seguente:Vdc2 ba Csi 5553OVdc2 ice Ib io56si sa Ra la la MlbRb lbRc te lo

La supposizione di base è che il carico sia bilanciato, quindi le resistenze sono tutte uguali, gli induttori sono tutti uguali e la somma istantanea di tutte le tensioni dei generatori di tensione è 0. I nodi più importanti del circuito sono il punto neutro dell'inverter o e il punto neutro del carico n. In generale, questi nodi non sono allo stesso potenziale, ma in casi speciali questo può

happen.

The instantaneous values of the output node of each leg can be +Vdc/2 or -Vdc/2 in respect of inverter neutral point.

The load neutral point voltage in respect of inverter neutral point is just the average of the output voltages of the 3 legs of the 3-phase inverter:

OboiNao VcoUno I3

The load voltages are the voltages between each phase and the load neutral point and can be expressed as follow:

Van Nao Ono va Io IoTbn OnoNba Nba 0vagoUno iNeoVen NÉVco_Vago

Each phase of the inverter depends by all the voltages at the output nodes of the inverter but never by a single one. This situation is called interphase interference.

Let’s see now what happens if a common mode component is added on top of inverter voltages:

VTao Ooot t tt Vorobotrobot tt VVeoVeo t t tt

Where vo(t) is a common-mode component.

Let’s see how the inverter neutral point reacts calculating the new expression of vno(t):

Vo Vo Vovaotvb.to Ooo VeoVbaOno t t t t t t tt t3 3Vovinot t t30gvaotfbot.ro

If a common

mode component is added at all the 3 phase voltages, this common mode appears equal at the load neutral point. It’s simply transmitted to the load neutral point voltage.The new load voltages becomes:TootNaoVan UnoVa NaoNo VanUnoOmo VenTbn VenTbnThe singular phase voltages remains unchanged.

The goal of the modulation techniques used in 3-phase inverter is the same as in single-phase inverters that is tosynthesize a low frequency spectral line.

The modulation techniques used in 3-phase systems are the following:

1. Square-wave modulation: when applied to 3-phase system is called six-step modulation. In this case: fs = f0.In this case low-frequency harmonics are created;
2. Pulse width modulation (PWM): in this case: fs>>f0. This modulation is called sine-triangle modulation orthird harmonic/pseudo third harmonic injection;
3. Space vector modulation (SVM): this is a more general modulation that includes both previous ones;

Let’s see square wave modulation applied to 3-phase systems.

TaoVac2   ToVdc2   DbaVdc2   ToVdc2   viVdc2   ToVdc2   io
  6         6         6         6        6         6

Vdc2 ba   Csi 5553O   Vdc2 Ibice   io56si   sa Ra la la   MlbRb lbte   Rc lo

noVdc6   tVdc6   3

VanHae3the3   tYeYe

VanHae3the3   tYe

VanHae3the3   tYeHae3

The waveform in question is a square wave, but it is much more similar to a sine wave. In order to analyze this situation, we need to examine the frequency behavior of the previous waveforms.

The frequency content of the inverter voltage (vao) is characterized by the following multiples of f0: Veo 17t llI 9 B7 215 193 15. These are the harmonics that represent the square wave.

The harmonic numbers of vno must be odd multiples of 3 times f0 because it's a square wave with a frequency that is 3 times f0: 9Uno t 213 15to3 5.313.38 7.31. The harmonics contained by vno are called triplen harmonics.

Lastly, van is defined as follows: van = vao-vno. Due to the difference, all common harmonics cancel out in the obtained waveform van. Note that this is true only if the harmonics have the same amplitude and the same initial phase, which is the case here, so they can be literally cancelled out: Van 7t 17135I 19Il.

In the van waveform, all triplen harmonics have been cancelled out, and for this reason, this waveform is different from the previous ones.

phase to the k-th harmonic of the voltage of the inverter.

The fundamental amplitude of the harmonics is: iao i

Due to the fact that the harmonic is transmitted to the load with no alterations, then the amplitude of the harmonic is: Van VenVbi i in

The fundamental amplitude of the phase-to-phase harmonic is: Tab 3

A very big problem of the square-wave modulation is that the output voltage cannot be controlled, in fact the output voltage can be only +Vdc/2 or -Vdc/2 and nothing else. Moreover there are problems in the frequency shape of the output voltages because there is a lot of low-frequency content so the load cannot see a good sinusoidal waveform.

LESSON #35

In order to study Space Vector Modulation (SVM), some mathematical tools are needed. First of all let's consider a 3-phase inverter:

LESSON #36

Vdc2 ba Csi 5553OVdc2 Ibice io56si sa Ra la la MlbRb lbteRc lo

For hypothesis the load is balanced. Let's define the following iabc: ice1i Ibabc c'e

the instantaneous values of the 3 currents.

Due to KCL at this surface the instantaneous summation of the currents is 0:io ttistt ttic o it

The instantaneous summation of the 3 components of the vector is 0.

Let’s define another vector, vabc:Vanbmabc n In

About this last vector it can be noted the following principle:Non NaoVen Vuoi3VbVb Nao tubo VcoUno Uno Uno tt tt con ttO I

The instantaneous summation of the 3 voltages is 0, so, as the previous currents vector, the instantaneous summation of the voltages is 0. Note that the last equation is not related to any Kirchhoff law but it’s a result of the topology of the 3-phase inverter.

Let’s see geometrically what does it mean that the summation of the elements of a vector is 0:b Equation of a planeXsX Xc Oat taAbc SpaceC

In a 3-space vector, if 1 element of the vector is 0 everytime, than remains only 2 grades of freedom.

Due to the previous currents and voltages equations:istio ttt ttic o itVan NaoVen Vuoi3VbUno Uno UnoVb Nao

It can be assumed that the vector of currents and of load voltages move but only on a plane. Let's write the previous plane definition in another way: IXXa Xc 0IXbt XeXatb I

The vector that contains xa, xb and xc is perpendicular to the unitary column vector. Everytime the summation of the components of a vector is 0, it can be assumed that the vector is moving on a plane perpendicular to the unitary vector. The plane is an alfa-beta plane while the direction perpendicular to the alfa-beta plane is called gamma, so to identify the direction of the vector gamma it can be used the vector: (1, 1, 1).

Alfa-beta coordinates are used to describe how the load responds to inverter action because the quantities move on the alfa-beta plane, so it can be used a change of base to analyse the quantities in alfa-beta plane. The change of base is made using Clarke transformation. Thanks to Clarke transformation it can be performed a change of basis from (a, b, c) canonical basis.

that are: 00I0 01la lb le1 i0 0 ITo the so called (alfa, beta, gamma) basis.

The Clark transformation (also defined as alfa-beta-gamma transformation) defined as follow:

XaXa ihaTappeaXpapp XeXp

Where: 1 12 2d Gi Gt0 03 3143g 2 2l l l2 2 2

The constant Gt can be different but in general it must be simply positive.

Comsidering that the gamma direction is perpendicular both to alfa and gamma direction, than the previousvectors in the new plane can be re