6.9. P–Q Control of a Three-Phase Voltage Source Inverter

Search in book...
Toggle Font Controls
Create new playlist

Name your new playlist

Playlist description (optional)
Sign In

Email address

Password

Forgot Password?

or

Continue with Facebook

Continue with Google
Sign Up

Full Name

Email address

Confirm Email Address

Password

or

Continue with Facebook

Continue with Google

6.9. P–Q Control of a Three-Phase Voltage Source Inverter

Today a power electronics inverter is used efficiently to control the flow of the active and reactive power between a dc power source and the utility grid acting as a power STATCOM which also finds applications in the interfacing of RES. Fig. 6.135 presents the block diagram of a three-phase VSI, connected to a utility grid (PCC), where

{\vec{V}}_{conv, 1}

${\vec{V}}_{conv, 1}$

is the fundamental component of the inverter output phase voltage that operates at a voltage amplitude of

{\hat{V}}_{conv, 1}

${\hat{V}}_{conv, 1}$

and with a phase displacement δ with respect to the utility grid phase voltage

{\vec{V}}_{g}

${\vec{V}}_{g}$

. The basic and minimum requirement of the VSI is to control the flow of the active and reactive power between the renewable source (i.e., the inverter) and the utility grid. The choice of the control technique of an inverter, which is connected to a utility grid, depends on the utility connection specifications, which include power quality and the control of the power flow from and to the utility grid (P–Q control). As mentioned before in this chapter, the inverter has the ability to operate bidirectionally, thus when is feeding power from its dc input to the utility grid operates as an inverter and when is absorbing power from the utility grid operates as a rectifier. For this reason the inverter in this section will be called converter. Since the voltage of the utility grid is not defined by the converter and can be assumed to be constant, the value and the quality of power delivered or received from the utility grid depend on the generated current at the output of the converter. Assuming that

{\vec{V}}_{conv, 1}

${\vec{V}}_{conv, 1}$

is synchronized with

{\vec{V}}_{g}

${\vec{V}}_{g}$

, the phase difference between

{\vec{V}}_{conv, 1}

${\vec{V}}_{conv, 1}$

and

{\vec{V}}_{g}

${\vec{V}}_{g}$

to be δ° and ωL >> R, the equivalent single-phase of Fig. 6.135 can be represented by the equivalent circuit of Fig. 6.136(a).

Figure 6.135 Block diagram of a three-phase voltage source converter connected to a point of common coupling of the utility grid.
(a) Three-phase block diagram; (b) single-phase block diagram.

Figure 6.136 Single-phase interconnection equivalent circuit and operation vector diagrams.
(a) Equivalent interconnection circuit of the system when ωL >> R; (b) vector diagram of the system when the inverter phase voltage ${\tilde{V}}_{i, 1} > {\tilde{V}}_{g}$ ${\tilde{V}}_{i, 1} > {\tilde{V}}_{g}$ and leading with respect to the utility phase voltage.

Using Fig. 6.136(a), the following results are obtained:

${\vec{V}}_{conv, 1} - {\vec{V}}_{L, 1} - {\vec{V}}_{g} = 0 or {\vec{V}}_{conv, 1} = j X_{L, 1} {\vec{I}}_{conv, 1} + {\vec{V}}_{g}$ ${\vec{V}}_{conv, 1} - {\vec{V}}_{L, 1} - {\vec{V}}_{g} = 0 or {\vec{V}}_{conv, 1} = j X_{L, 1} {\vec{I}}_{conv, 1} + {\vec{V}}_{g}$

(6.212)

${\vec{I}}_{conv, 1} = \frac{{\vec{V}}_{conv, 1} - {\vec{V}}_{g}}{j X_{L, 1}}$ ${\vec{I}}_{conv, 1} = \frac{{\vec{V}}_{conv, 1} - {\vec{V}}_{g}}{j X_{L, 1}}$

(6.213)

Also, assuming that the output voltage of the converter is pure sinusoidal and knowing, that the conjugate of a + jb = a − jb, then the per-phase apparent power provided by the utility at the PCC (i.e., utility side) is given by:

$\begin{matrix} {\vec{S}}_{g} = {\vec{V}}_{g} {\vec{I}}_{conv, 1}^{*} = {\vec{V}}_{g} {[\frac{{\vec{V}}_{conv, 1} - {\vec{V}}_{g}}{j X_{L, 1}}]}^{*} = {\tilde{V}}_{g} ∠ 0 ° {[\frac{{\tilde{V}}_{conv, 1} ∠ + δ ° - {\tilde{V}}_{g} ∠ 0 °}{j X_{L, 1}}]}^{*} \\ = {\tilde{V}}_{g} {[\frac{({\tilde{V}}_{conv, 1} \cos δ + j \sin δ) - {\tilde{V}}_{g}}{j X_{L, 1}}]}^{*} \\ = {\tilde{V}}_{g} {[\frac{j ({\tilde{V}}_{conv, 1} (\cos δ + j \sin δ) - {\tilde{V}}_{g})}{{(j)}^{2} X_{L, 1}}]}^{*} \\ = {\tilde{V}}_{g} {[\frac{j {\tilde{V}}_{conv, 1} \cos δ - {\tilde{V}}_{conv, 1} \sin δ - j {\tilde{V}}_{g}}{- X_{L, 1}}]}^{*} \\ = {\tilde{V}}_{g} {[\frac{{\tilde{V}}_{conv, 1} \sin δ - j ({\tilde{V}}_{conv, 1} \cos δ - {\tilde{V}}_{g})}{X_{L, 1}}]}^{*} \\ = {\tilde{V}}_{g} [\frac{{\tilde{V}}_{conv, 1} \sin δ + j ({\tilde{V}}_{conv, 1} \cos δ - {\tilde{V}}_{g})}{X_{L, 1}}] \\ = \frac{{\tilde{V}}_{g} {\tilde{V}}_{conv, 1} \sin δ}{X_{L, 1}} + j (\frac{{\tilde{V}}_{g} {\tilde{V}}_{conv, 1} \cos δ - {\tilde{V}}_{g}^{2}}{X_{L, 1}}) \\ = P_{g} + j Q_{g} \end{matrix}$ $\begin{matrix} {\vec{S}}_{g} = {\vec{V}}_{g} {\vec{I}}_{conv, 1}^{*} = {\vec{V}}_{g} {[\frac{{\vec{V}}_{conv, 1} - {\vec{V}}_{g}}{j X_{L, 1}}]}^{*} = {\tilde{V}}_{g} ∠ 0 ° {[\frac{{\tilde{V}}_{conv, 1} ∠ + δ ° - {\tilde{V}}_{g} ∠ 0 °}{j X_{L, 1}}]}^{*} \\ = {\tilde{V}}_{g} {[\frac{({\tilde{V}}_{conv, 1} \cos δ + j \sin δ) - {\tilde{V}}_{g}}{j X_{L, 1}}]}^{*} \\ = {\tilde{V}}_{g} {[\frac{j ({\tilde{V}}_{conv, 1} (\cos δ + j \sin δ) - {\tilde{V}}_{g})}{{(j)}^{2} X_{L, 1}}]}^{*} \\ = {\tilde{V}}_{g} {[\frac{j {\tilde{V}}_{conv, 1} \cos δ - {\tilde{V}}_{conv, 1} \sin δ - j {\tilde{V}}_{g}}{- X_{L, 1}}]}^{*} \\ = {\tilde{V}}_{g} {[\frac{{\tilde{V}}_{conv, 1} \sin δ - j ({\tilde{V}}_{conv, 1} \cos δ - {\tilde{V}}_{g})}{X_{L, 1}}]}^{*} \\ = {\tilde{V}}_{g} [\frac{{\tilde{V}}_{conv, 1} \sin δ + j ({\tilde{V}}_{conv, 1} \cos δ - {\tilde{V}}_{g})}{X_{L, 1}}] \\ = \frac{{\tilde{V}}_{g} {\tilde{V}}_{conv, 1} \sin δ}{X_{L, 1}} + j (\frac{{\tilde{V}}_{g} {\tilde{V}}_{conv, 1} \cos δ - {\tilde{V}}_{g}^{2}}{X_{L, 1}}) \\ = P_{g} + j Q_{g} \end{matrix}$

(6.214)

where

$\begin{matrix} {\vec{I}}_{conv, 1} = \frac{{\vec{V}}_{conv, 1} - {\vec{V}}_{g}}{j X_{L, 1}} = \frac{{\tilde{V}}_{conv, 1} (\cos δ + j \sin δ) - {\tilde{V}}_{g}}{j X_{L, 1}} \\ = \frac{{\tilde{V}}_{conv, 1} \cos δ + j {\tilde{V}}_{conv, 1} \sin δ - {\tilde{V}}_{g}}{j X_{L, 1}} \\ = \frac{j {\tilde{V}}_{conv, 1} \cos δ - {\tilde{V}}_{conv, 1} \sin δ - j {\tilde{V}}_{g}}{- X_{L, 1}} \\ = \frac{{\tilde{V}}_{conv, 1} \sin δ + j ({\tilde{V}}_{g} - {\tilde{V}}_{conv, 1} \cos δ)}{X_{L, 1}} \\ = \frac{{\tilde{V}}_{conv, 1} \sin δ}{X_{L, 1}} + j \frac{{\tilde{V}}_{g} - {\tilde{V}}_{conv, 1} \cos δ}{X_{L, 1}} = I_{p} + j I_{q} \end{matrix}$ $\begin{matrix} {\vec{I}}_{conv, 1} = \frac{{\vec{V}}_{conv, 1} - {\vec{V}}_{g}}{j X_{L, 1}} = \frac{{\tilde{V}}_{conv, 1} (\cos δ + j \sin δ) - {\tilde{V}}_{g}}{j X_{L, 1}} \\ = \frac{{\tilde{V}}_{conv, 1} \cos δ + j {\tilde{V}}_{conv, 1} \sin δ - {\tilde{V}}_{g}}{j X_{L, 1}} \\ = \frac{j {\tilde{V}}_{conv, 1} \cos δ - {\tilde{V}}_{conv, 1} \sin δ - j {\tilde{V}}_{g}}{- X_{L, 1}} \\ = \frac{{\tilde{V}}_{conv, 1} \sin δ + j ({\tilde{V}}_{g} - {\tilde{V}}_{conv, 1} \cos δ)}{X_{L, 1}} \\ = \frac{{\tilde{V}}_{conv, 1} \sin δ}{X_{L, 1}} + j \frac{{\tilde{V}}_{g} - {\tilde{V}}_{conv, 1} \cos δ}{X_{L, 1}} = I_{p} + j I_{q} \end{matrix}$

(6.215)

$I_{p} = real part of {\vec{I}}_{conv, 1} = \frac{{\tilde{V}}_{conv, 1} \sin δ}{X_{L, 1}}$ $I_{p} = real part of {\vec{I}}_{conv, 1} = \frac{{\tilde{V}}_{conv, 1} \sin δ}{X_{L, 1}}$

(6.216)

$I_{q} = imaginary part of {\vec{I}}_{conv, 1} = \frac{{\tilde{V}}_{g} - {\tilde{V}}_{conv, 1} \cos δ}{X_{L, 1}}$ $I_{q} = imaginary part of {\vec{I}}_{conv, 1} = \frac{{\tilde{V}}_{g} - {\tilde{V}}_{conv, 1} \cos δ}{X_{L, 1}}$

(6.217)

{\tilde{I}}_{conv, 1} = rms value of the converter output line current fundamental component = \sqrt{I_{p}^{2} + I_{q}^{2}}

${\tilde{I}}_{conv, 1} = rms value of the converter output line current fundamental component = \sqrt{I_{p}^{2} + I_{q}^{2}}$

{\tilde{V}}_{conv, 1} = rms value of the converter output phase voltage fundamental component

${\tilde{V}}_{conv, 1} = rms value of the converter output phase voltage fundamental component$

{\tilde{V}}_{g} = amplitude of the utility grid phase voltage

${\tilde{V}}_{g} = amplitude of the utility grid phase voltage$

v_L,1 = inductor voltage fundamental component

{\vec{I}}_{conv, 1}^{*} = conjugate of {\vec{I}}_{conv, 1}

${\vec{I}}_{conv, 1}^{*} = conjugate of {\vec{I}}_{conv, 1}$

X_L,1 = reactance of the interconnection inductor at fundamental frequency

δ = phase difference between utility grid and converter phase voltages

φ = phase displacement between the utility voltage and inverter output current

$P_{g} = per-phase active power at PCC = \frac{{\tilde{V}}_{g} {\tilde{V}}_{conv, 1} \sin δ}{X_{L, 1}}$ $P_{g} = per-phase active power at PCC = \frac{{\tilde{V}}_{g} {\tilde{V}}_{conv, 1} \sin δ}{X_{L, 1}}$

(6.218)

$Q_{g} = per-phase reactive power at PCC = \frac{{\tilde{V}}_{g} {\tilde{V}}_{conv, 1} \cos δ - {\tilde{V}}_{g}^{2}}{X_{L, 1}}$ $Q_{g} = per-phase reactive power at PCC = \frac{{\tilde{V}}_{g} {\tilde{V}}_{conv, 1} \cos δ - {\tilde{V}}_{g}^{2}}{X_{L, 1}}$

(6.219)

Eqs. (6.218) and (6.219) indicate that the active and reactive power at PCC depend on the converter output voltage

{\tilde{V}}_{conv, 1}

${\tilde{V}}_{conv, 1}$

and the phase difference δ. According to Eq. (6.128), the active power at PCC can be controlled by varying the converter output voltage

{\tilde{V}}_{conv, 1}

${\tilde{V}}_{conv, 1}$

(through the amplitude modulation factor, m_a) and the phase angle δ (through the phase displacement between the reference signal v_r of the converter and the utility grid). Moreover, according to Eq. (6.127) at the same time, the reactive power at PCC can be controlled only by varying the inverter output voltage

{\tilde{V}}_{conv, 1}

${\tilde{V}}_{conv, 1}$

. Treating the utility grid as a load to the converter when δ > 0 (i.e.,

{\vec{V}}_{conv, 1}

${\vec{V}}_{conv, 1}$

is leading to

{\vec{V}}_{g}

${\vec{V}}_{g}$

), then at PCC P_g > 0, which means that the utility is absorbing active power from the converter. When δ < 0 (i.e.,

{\vec{V}}_{conv, 1}

${\vec{V}}_{conv, 1}$

is lagging to

{\vec{V}}_{g}

${\vec{V}}_{g}$

), then P_g < 0, which means that the utility is feeding active power to the converter. When

{\tilde{V}}_{conv, 1} > {\tilde{V}}_{g}

${\tilde{V}}_{conv, 1} > {\tilde{V}}_{g}$

, then Q_g > 0, which means that the utility is absorbing reactive power from the converter. When

{\tilde{V}}_{conv, 1} < {\tilde{V}}_{g}

${\tilde{V}}_{conv, 1} < {\tilde{V}}_{g}$

, then Q_g < 0, which means that the utility is feeding reactive power to the converter. Table 6.21 summarizes the power exchange operation modes between the converter and the utility grid with reference the PCC.

Fig. 6.137 presents the P_g and Q_g variation with respect to δ and keeping

{\tilde{V}}_{conv, 1}, {\tilde{V}}_{g} and X_{L, 1}

${\tilde{V}}_{conv, 1}, {\tilde{V}}_{g} and X_{L, 1}$

constant. As can be seen in Fig. 6.137 there is a critical value of angle δ_critical = 90° at which maximum active power is achieved. If δ is increased beyond this point the system becomes unstable. To avoid such a problem the interconnected supplies must operate at small phase angles (δ ≤ 20°).

Fig. 6.138 presents the P_g–Q_g plane operation modes of the inverter-utility transmission system. The range of the phase angle φ is found from the above given figures and equations.

Following the same steps as before the active and reactive power provided by the inverter to the utility can be found as follows:

$\begin{matrix} {\vec{S}}_{conv} = {\vec{V}}_{conv, 1} {\vec{I}}_{conv, 1}^{*} = {\vec{V}}_{conv, 1} {[\frac{{\vec{V}}_{conv, 1} - {\vec{V}}_{g}}{j X_{L, 1}}]}^{*} = {\tilde{V}}_{i, 1} ∠ δ ° {[\frac{{\tilde{V}}_{conv, 1} ∠ δ ° - {\tilde{V}}_{g} ∠ 0 °}{j X_{L, 1}}]}^{*} \\ = {\tilde{V}}_{conv, 1} (\cos δ + j \sin δ) {[\frac{({\tilde{V}}_{conv, 1} \cos δ + j \sin δ) - {\tilde{V}}_{g}}{j X_{L, 1}}]}^{*} \\ = - \frac{{\tilde{V}}_{conv, 1} {\tilde{V}}_{g} \sin δ}{X_{L, 1}} + j (\frac{{\tilde{V}}_{conv, 1}^{2} - {\tilde{V}}_{conv, 1} {\tilde{V}}_{g} \cos δ}{X_{L, 1}}) \\ = P_{conv} + j Q_{conv} \end{matrix}$ $\begin{matrix} {\vec{S}}_{conv} = {\vec{V}}_{conv, 1} {\vec{I}}_{conv, 1}^{*} = {\vec{V}}_{conv, 1} {[\frac{{\vec{V}}_{conv, 1} - {\vec{V}}_{g}}{j X_{L, 1}}]}^{*} = {\tilde{V}}_{i, 1} ∠ δ ° {[\frac{{\tilde{V}}_{conv, 1} ∠ δ ° - {\tilde{V}}_{g} ∠ 0 °}{j X_{L, 1}}]}^{*} \\ = {\tilde{V}}_{conv, 1} (\cos δ + j \sin δ) {[\frac{({\tilde{V}}_{conv, 1} \cos δ + j \sin δ) - {\tilde{V}}_{g}}{j X_{L, 1}}]}^{*} \\ = - \frac{{\tilde{V}}_{conv, 1} {\tilde{V}}_{g} \sin δ}{X_{L, 1}} + j (\frac{{\tilde{V}}_{conv, 1}^{2} - {\tilde{V}}_{conv, 1} {\tilde{V}}_{g} \cos δ}{X_{L, 1}}) \\ = P_{conv} + j Q_{conv} \end{matrix}$

(6.220)

Fig. 6.139 presents the block diagram of a STATCOM.

Figs. 6.140–6.142 show the waveforms of the STATCOM for three different operating conditions.

In Chapter 11 the application of the VSI as an active filter will be examined.

Table 6.21

Power exchange modes between utility grid and converter

Operation condition	Phasor diagrams	Active and reactive power flow
${\tilde{V}}_{conv, 1} > {\tilde{V}}_{g}$ ${\tilde{V}}_{conv, 1} > {\tilde{V}}_{g}$ and δ = 0° Current flows from the converter into the grid.		P_g = 0 W Q_g > 0 • No active power flow. • Grid absorbs reactive power from the converter. • Grid acts as an inductor and the converter as a capacitor.
${\tilde{V}}_{conv, 1} < {\tilde{V}}_{g}$ ${\tilde{V}}_{conv, 1} < {\tilde{V}}_{g}$ and δ = 0° Current flows from the grid into the converter.		P_g = 0 W Q_g < 0 • No active power flow. • Grid feeds reactive power to the converter. • Grid acts as a capacitor and the converter as an inductor.
${\tilde{V}}_{conv, 1} = {\tilde{V}}_{g}$ ${\tilde{V}}_{conv, 1} = {\tilde{V}}_{g}$ and δ = 0° No load mode. No current flow.		P_g = 0 W Q_g = 0 VAR No power flow.
${\tilde{V}}_{conv, 1} = {\tilde{V}}_{g}$ ${\tilde{V}}_{conv, 1} = {\tilde{V}}_{g}$ and δ > 0°		P_g > 0 Q_g < 0 • Grid absorbs active power from the converter and feeds reactive to the converter. • Grid acts as a capacitor and the converter as an inductor.
${\tilde{V}}_{conv, 1} = {\tilde{V}}_{g}$ ${\tilde{V}}_{conv, 1} = {\tilde{V}}_{g}$ and δ < 0°		P_g < 0 Q_g < 0 • Grid feeds active and reactive power to the converter. • Grid acts as a capacitor and the converter as an inductor.
${\tilde{V}}_{conv, 1} > {\tilde{V}}_{g}$ ${\tilde{V}}_{conv, 1} > {\tilde{V}}_{g}$ and δ < 0°		P_g < 0 Q_g > 0 • Grid feeds active power to the converter and absorbs reactive. • Grid acts as an inductor and the converter as a capacitor.
Table Continued

Operation condition	Phasor diagrams	Active and reactive power flow
${\tilde{V}}_{conv, 1} < {\tilde{V}}_{g}$ ${\tilde{V}}_{conv, 1} < {\tilde{V}}_{g}$ and δ < 0°		P_g < 0 Q_g < 0 • Grid feeds active and reactive power to the converter. • Grid acts as a capacitor and the converter as an inductor.
${\tilde{V}}_{conv, 1} > {\tilde{V}}_{g}$ ${\tilde{V}}_{conv, 1} > {\tilde{V}}_{g}$ and δ > 0°		P_g > 0 Q_g > 0 • Grid absorbs active and reactive power from the converter. • Grid acts as an inductor and the converter as a capacitor.
${\tilde{V}}_{conv, 1} < {\tilde{V}}_{g}$ ${\tilde{V}}_{conv, 1} < {\tilde{V}}_{g}$ and δ > 0°		P_g > 0 Q_g < 0 • Grid absorbs active and feeds reactive power to the converter. • Grid acts as a capacitor and the converter as an inductor.

Figure 6.137 P_g and Q_g variation with respect to δ.

Figure 6.138 P_g–Q_g plane operation modes at the utility side (i.e., at point of common coupling) $({\vec{I}}_{d} and {\vec{I}}_{q} are the real and imaginary compoents of {\vec{I}}_{conv, 1} respectively)$ $({\vec{I}}_{d} and {\vec{I}}_{q} are the real and imaginary compoents of {\vec{I}}_{conv, 1} respectively)$ .

Figure 6.139 Block diagram of synchronous static compensator.

Figure 6.140 Static compensator waveforms when δ = 0°, ${\tilde{V}}_{g} = 176 V$ ${\tilde{V}}_{g} = 176 V$ , and ${\tilde{V}}_{conv, 1} = 220 V$ ${\tilde{V}}_{conv, 1} = 220 V$ . (The utility acts as an inductor absorbing reactive power from the converter.)
(a) Switches S₁ and S₄ gating signals; (b) utility phase voltage, inverter output phase voltage, and current; (c) dc capacitor current.

Figure 6.141 Static compensator waveforms when δ = 0°, ${\tilde{V}}_{g} = 286 V$ ${\tilde{V}}_{g} = 286 V$ , and ${\tilde{V}}_{conv, 1} = 220 V$ ${\tilde{V}}_{conv, 1} = 220 V$ . (The utility acts as a capacitor feeding reactive power to the converter.)
(a) Switches S₁ and S₄ gating signals; (b) utility phase voltage, inverter output phase voltage, and current; (c) converter dc capacitor current.

Figure 6.142 Static compensator waveforms when δ = 10°, ${\tilde{V}}_{g} = 176 V$ ${\tilde{V}}_{g} = 176 V$ , and ${\tilde{V}}_{conv, 1} = 220 V$ ${\tilde{V}}_{conv, 1} = 220 V$ . (The utility acts as an inductor absorbing reactive power from the converter. Moreover, the utility is absorbing active power from the converter.)
(a) Switches S₁ and S₄ gating signals; (b) utility phase voltage, inverter output phase voltage, and current; (c) Converter capacitor current.

Example 6.13

A three-phase VSI is connected in parallel to a utility grid to be used as a STATCOM. The converter and the utility voltages are respectively given:

{\tilde{V}}_{conv, 1} = 2000 ∠ - 5 ° V and {\tilde{V}}_{g} = 1900 ∠ 0 ° V

${\tilde{V}}_{conv, 1} = 2000 ∠ - 5 ° V and {\tilde{V}}_{g} = 1900 ∠ 0 ° V$

If the interconnection impedance is X_L = 7 Ω, calculate:

a) The active and reactive power supplied or received by each source.

b) The value of the converter output current.

c) If the

{\tilde{V}}_{conv, 1} = 1900 ∠ - 5 ° V and {\tilde{V}}_{g} = 2000 ∠ 0 ° V

${\tilde{V}}_{conv, 1} = 1900 ∠ - 5 ° V and {\tilde{V}}_{g} = 2000 ∠ 0 ° V$

, find the operation mode of the converter.

Solution

a) The values of the active and reactive power at the PCC are given by:

$P_{g} = \frac{{\tilde{V}}_{g} {\tilde{V}}_{conv, 1} \sin δ}{X_{L, 1}} = \frac{1900 \cdot 2000 \sin (- 5 °)}{7} = - 47.313 kW$ $P_{g} = \frac{{\tilde{V}}_{g} {\tilde{V}}_{conv, 1} \sin δ}{X_{L, 1}} = \frac{1900 \cdot 2000 \sin (- 5 °)}{7} = - 47.313 kW$

$Q_{g} = \frac{{\tilde{V}}_{g} {\tilde{V}}_{conv, 1} \cos δ - {\tilde{V}}_{g}^{2}}{X_{L, 1}} = \frac{1900 \cdot 2000 \cos (- 5 °) - {(1900)}^{2}}{7} = 25.1 kVAR$ $Q_{g} = \frac{{\tilde{V}}_{g} {\tilde{V}}_{conv, 1} \cos δ - {\tilde{V}}_{g}^{2}}{X_{L, 1}} = \frac{1900 \cdot 2000 \cos (- 5 °) - {(1900)}^{2}}{7} = 25.1 kVAR$

Using Table 6.21 the meaning of the above results is that the grid is feeding active power to the converter and at the same time is absorbing reactive power from the converter (i.e., the utility is acting as a capacitor and the converter as an inductor). According to Fig. 6.138 the system operates in the third quadrant of the P_g–Q_g plane and, consequently, the phase angle φ must be between −90° and −180°.

b) According to Eq. (6.215) the rms values of the fundamental component of the line current and its components fed into the utility are given by:

$I_{p} = real part of {\vec{I}}_{conv, 1} = \frac{{\tilde{V}}_{conv, 1} \sin δ}{X_{L, 1}} = \frac{2000 \sin (- 5)}{7} = - 24.9 A$ $I_{p} = real part of {\vec{I}}_{conv, 1} = \frac{{\tilde{V}}_{conv, 1} \sin δ}{X_{L, 1}} = \frac{2000 \sin (- 5)}{7} = - 24.9 A$

$I_{q} = imaginary part of {\vec{I}}_{conv, 1} = \frac{{\tilde{V}}_{g} - {\tilde{V}}_{conv, 1} \cos δ}{X_{L, 1}} = \frac{1900 - 2000 \cos (- 5)}{7} = - 13.2 A$ $I_{q} = imaginary part of {\vec{I}}_{conv, 1} = \frac{{\tilde{V}}_{g} - {\tilde{V}}_{conv, 1} \cos δ}{X_{L, 1}} = \frac{1900 - 2000 \cos (- 5)}{7} = - 13.2 A$

${\tilde{I}}_{conv, 1} = \sqrt{{(- 24.9)}^{2} + {(- 13.2)}^{2}} = 28.2 A$ ${\tilde{I}}_{conv, 1} = \sqrt{{(- 24.9)}^{2} + {(- 13.2)}^{2}} = 28.2 A$

Since the system operates in the third quadrant of the P_g–Q_g plane, the phase angle φ is given by:

$φ = - 180 ° + \tan^{- 1} (\frac{13.2}{24.9}) = - 180 ° + 27.9 ° = - 152 °$ $φ = - 180 ° + \tan^{- 1} (\frac{13.2}{24.9}) = - 180 ° + 27.9 ° = - 152 °$

The vector diagram of the system when

{\tilde{V}}_{conv, 1} > {\tilde{V}}_{g}

${\tilde{V}}_{conv, 1} > {\tilde{V}}_{g}$

and converter phase voltage is lagging with respect to the utility phase voltage by 5° will be as given in Fig. 6.143.

c) The active and reactive power components are given by:

$P_{g} = \frac{{\tilde{V}}_{g} {\tilde{V}}_{conv, 1} \sin δ}{X_{L, 1}} = \frac{2000 \times 1900 \sin (- 5 °)}{7} = - 47.313 kW$ $P_{g} = \frac{{\tilde{V}}_{g} {\tilde{V}}_{conv, 1} \sin δ}{X_{L, 1}} = \frac{2000 \times 1900 \sin (- 5 °)}{7} = - 47.313 kW$

Figure 6.143 Static compensator vector diagram when ${\tilde{V}}_{conv, 1} = 2000 ∠ - 5 ° V and {\tilde{V}}_{g} = 1900 ∠ 0 ° V$ ${\tilde{V}}_{conv, 1} = 2000 ∠ - 5 ° V and {\tilde{V}}_{g} = 1900 ∠ 0 ° V$ .

$Q_{g} = \frac{{\tilde{V}}_{g} {\tilde{V}}_{conv, 1} \cos δ - {\tilde{V}}_{g}^{2}}{X_{L, 1}} = \frac{2000 \times 1900 \cos (- 5 °) - {(2000)}^{2}}{7} = - 30.637 kVAR$ $Q_{g} = \frac{{\tilde{V}}_{g} {\tilde{V}}_{conv, 1} \cos δ - {\tilde{V}}_{g}^{2}}{X_{L, 1}} = \frac{2000 \times 1900 \cos (- 5 °) - {(2000)}^{2}}{7} = - 30.637 kVAR$

As it was expected since δ has the same value as before the active power remains as in part (a). However, since the utility voltage becomes higher than the inverter voltage, the reactive power will have negative value meaning that the utility will feed reactive power to the converter (i.e., the utility is acting as an inductor and the converter as a capacitor). According to Fig. 6.138 for this mode of operation the system operates in the fourth quadrant of the P_g–Q_g plane.

According to Eq. (6.215) the rms values of the fundamental component of the line current and its components fed into the utility are given by:

$I_{p} = real part of {\vec{I}}_{conv, 1} = \frac{{\tilde{V}}_{conv, 1} \sin δ}{X_{L, 1}} = \frac{1900 \sin (- 5)}{7} = - 23.7 A$ $I_{p} = real part of {\vec{I}}_{conv, 1} = \frac{{\tilde{V}}_{conv, 1} \sin δ}{X_{L, 1}} = \frac{1900 \sin (- 5)}{7} = - 23.7 A$

$I_{q} = imaginary part of {\vec{I}}_{conv, 1} = \frac{{\tilde{V}}_{g} - {\tilde{V}}_{conv, 1} \cos δ}{X_{L, 1}} = \frac{2000 - 1900 \cos (- 5)}{7} = 15.3 A$ $I_{q} = imaginary part of {\vec{I}}_{conv, 1} = \frac{{\tilde{V}}_{g} - {\tilde{V}}_{conv, 1} \cos δ}{X_{L, 1}} = \frac{2000 - 1900 \cos (- 5)}{7} = 15.3 A$

${\tilde{I}}_{conv, 1} = \sqrt{{(- 23.7)}^{2} + {(15.3)}^{2}} = 28.2 A$ ${\tilde{I}}_{conv, 1} = \sqrt{{(- 23.7)}^{2} + {(15.3)}^{2}} = 28.2 A$

Since the system operates in the fourth quadrant of the P_g–Q_g plane, the phase angle φ must be in the range of 90° to 180° and is given by:

$φ = 180 ° - \tan^{- 1} (\frac{15.3}{23.7}) = 180 ° - 32.92 ° = 147 °$ $φ = 180 ° - \tan^{- 1} (\frac{15.3}{23.7}) = 180 ° - 32.92 ° = 147 °$

The vector diagram of the system when

{\tilde{V}}_{conv, 1} < {\tilde{V}}_{g}

${\tilde{V}}_{conv, 1} < {\tilde{V}}_{g}$

and converter phase voltage is lagging with respect to the utility phase voltage by 5° will be as given in Fig. 6.144.

Figure 6.144 Static compensator vector diagram when ${\tilde{V}}_{conv, 1} = 1900 ∠ - 5 ° V and {\tilde{V}}_{g} = 2000 ∠ 0 ° V$ ${\tilde{V}}_{conv, 1} = 1900 ∠ - 5 ° V and {\tilde{V}}_{g} = 2000 ∠ 0 ° V$ .

6.9.1. P–Q Control Based on the Decoupling Control of d–q Current Components

As mentioned before, implementing a control technique in the synchronous rotating frame with d–q dc components eliminates the steady-state errors and exhibits fast transient response. In this section the P–Q control of a three-phase balanced converter, which is based on the decoupling control of d–q current components, will be analyzed.

From Fig. 6.135(a), the following equations hold:

$\begin{matrix} v_{conva} (t) = R i_{a} (t) + L \frac{d i_{a} (t)}{dt} + v_{ga} (t) \\ v_{convb} (t) = R i_{b} (t) + L \frac{d i_{b} (t)}{dt} + v_{gb} (t) \\ v_{convc} (t) = R i_{c} (t) + L \frac{d i_{c} (t)}{dt} + v_{gc} (t) \end{matrix}$ $\begin{matrix} v_{conva} (t) = R i_{a} (t) + L \frac{d i_{a} (t)}{dt} + v_{ga} (t) \\ v_{convb} (t) = R i_{b} (t) + L \frac{d i_{b} (t)}{dt} + v_{gb} (t) \\ v_{convc} (t) = R i_{c} (t) + L \frac{d i_{c} (t)}{dt} + v_{gc} (t) \end{matrix}$

(6.221)

where

v_conva, v_convb, v_convc = converter phase voltages with respect to the utility neutral

v_ga, v_gb, v_gc = grid phase voltages

i_a, i_b, i_c = converter line currents

L = interconnection inductance

R = interconnection resistance

Applying the Clarke transformation to Eq. (6.221), the following equations are obtained:

$\begin{matrix} v_{convα} (t) = R i_{α} (t) + L \frac{d i_{α} (t)}{dt} + v_{gα} (t) \\ v_{convβ} (t) = R i_{β} (t) + L \frac{d i_{β} (t)}{dt} + v_{gβ} (t) \end{matrix}$ $\begin{matrix} v_{convα} (t) = R i_{α} (t) + L \frac{d i_{α} (t)}{dt} + v_{gα} (t) \\ v_{convβ} (t) = R i_{β} (t) + L \frac{d i_{β} (t)}{dt} + v_{gβ} (t) \end{matrix}$

(6.222)

or in matrix form

$[\begin{matrix} v_{convα} \\ v_{convβ} \end{matrix}] = R [\begin{matrix} i_{α} \\ i_{β} \end{matrix}] + L [\begin{matrix} \frac{d i_{α}}{dt} \\ \frac{d i_{β}}{dt} \end{matrix}] + [\begin{matrix} v_{gα} \\ v_{gβ} \end{matrix}]$ $[\begin{matrix} v_{convα} \\ v_{convβ} \end{matrix}] = R [\begin{matrix} i_{α} \\ i_{β} \end{matrix}] + L [\begin{matrix} \frac{d i_{α}}{dt} \\ \frac{d i_{β}}{dt} \end{matrix}] + [\begin{matrix} v_{gα} \\ v_{gβ} \end{matrix}]$

(6.223)

or in vector form

${\vec{V}}_{convαβ} = R {\vec{I}}_{αβ} + L \frac{d {\vec{I}}_{αβ}}{dt} + {\vec{V}}_{gαβ}$ ${\vec{V}}_{convαβ} = R {\vec{I}}_{αβ} + L \frac{d {\vec{I}}_{αβ}}{dt} + {\vec{V}}_{gαβ}$

(6.224)

Since the variables v_convα and v_convβ are time variant, they have to be transformed to dc quantities using the Park transformation as follows:

$\begin{matrix} [\begin{matrix} v_{convd} \\ v_{convq} \end{matrix}] = [\begin{matrix} \cos ω_{e} t & \sin ω_{e} t \\ - \sin ω_{e} t & \cos ω_{e} t \end{matrix}] [\begin{matrix} v_{iα} \\ v_{iβ} \end{matrix}] \\ = [\begin{matrix} \cos ω_{e} t & \sin ω_{e} t \\ - \sin ω_{e} t & \cos ω_{e} t \end{matrix}] [\begin{matrix} R i_{α} \\ R i_{β} \end{matrix}] + [\begin{matrix} \cos ω_{e} t & \sin ω_{e} t \\ - \sin ω_{e} t & \cos ω_{e} t \end{matrix}] [\begin{matrix} L \frac{d i_{α}}{dt} \\ L \frac{d i_{β}}{dt} \end{matrix}] + [\begin{matrix} \cos ω_{e} t & \sin ω_{e} t \\ - \sin ω_{e} t & \cos ω_{e} t \end{matrix}] [\begin{matrix} v_{gα} \\ v_{gβ} \end{matrix}] \end{matrix}$ $\begin{matrix} [\begin{matrix} v_{convd} \\ v_{convq} \end{matrix}] = [\begin{matrix} \cos ω_{e} t & \sin ω_{e} t \\ - \sin ω_{e} t & \cos ω_{e} t \end{matrix}] [\begin{matrix} v_{iα} \\ v_{iβ} \end{matrix}] \\ = [\begin{matrix} \cos ω_{e} t & \sin ω_{e} t \\ - \sin ω_{e} t & \cos ω_{e} t \end{matrix}] [\begin{matrix} R i_{α} \\ R i_{β} \end{matrix}] + [\begin{matrix} \cos ω_{e} t & \sin ω_{e} t \\ - \sin ω_{e} t & \cos ω_{e} t \end{matrix}] [\begin{matrix} L \frac{d i_{α}}{dt} \\ L \frac{d i_{β}}{dt} \end{matrix}] + [\begin{matrix} \cos ω_{e} t & \sin ω_{e} t \\ - \sin ω_{e} t & \cos ω_{e} t \end{matrix}] [\begin{matrix} v_{gα} \\ v_{gβ} \end{matrix}] \end{matrix}$

(6.225)

Using Eq. (6.225) and transforming the variables i_α and i_β into dc components and knowing that

$[\begin{matrix} \cos ω_{e} t & \sin ω_{e} t \\ - \sin ω_{e} t & \cos ω_{e} t \end{matrix}] [\begin{matrix} \cos ω_{e} t & \sin ω_{e} t \\ - \sin ω_{e} t & \cos ω_{e} t \end{matrix}] = 1$ $[\begin{matrix} \cos ω_{e} t & \sin ω_{e} t \\ - \sin ω_{e} t & \cos ω_{e} t \end{matrix}] [\begin{matrix} \cos ω_{e} t & \sin ω_{e} t \\ - \sin ω_{e} t & \cos ω_{e} t \end{matrix}] = 1$

the following equation is obtained:

$[\begin{matrix} v_{convd} \\ v_{convq} \end{matrix}] = R [\begin{matrix} i_{d} \\ i_{q} \end{matrix}] + [\begin{matrix} \cos ω_{e} t & \sin ω_{e} t \\ - \sin ω_{e} t & \cos ω_{e} t \end{matrix}] \cdot L \frac{d}{dt} {[\begin{matrix} \cos ω_{e} t & \sin ω_{e} t \\ - \sin ω_{e} t & \cos ω_{e} t \end{matrix}]}^{- 1} [\begin{matrix} i_{d} \\ i_{q} \end{matrix}] + [\begin{matrix} v_{gd} \\ v_{gq} \end{matrix}]$ $[\begin{matrix} v_{convd} \\ v_{convq} \end{matrix}] = R [\begin{matrix} i_{d} \\ i_{q} \end{matrix}] + [\begin{matrix} \cos ω_{e} t & \sin ω_{e} t \\ - \sin ω_{e} t & \cos ω_{e} t \end{matrix}] \cdot L \frac{d}{dt} {[\begin{matrix} \cos ω_{e} t & \sin ω_{e} t \\ - \sin ω_{e} t & \cos ω_{e} t \end{matrix}]}^{- 1} [\begin{matrix} i_{d} \\ i_{q} \end{matrix}] + [\begin{matrix} v_{gd} \\ v_{gq} \end{matrix}]$

(6.226)

$[\begin{matrix} v_{convd} \\ v_{convq} \end{matrix}] = R [\begin{matrix} i_{d} \\ i_{q} \end{matrix}] + L \frac{d}{dt} [\begin{matrix} i_{d} \\ i_{q} \end{matrix}] + ω_{e} L [\begin{matrix} - i_{q} \\ i_{d} \end{matrix}] + [\begin{matrix} v_{gd} \\ v_{gq} \end{matrix}]$ $[\begin{matrix} v_{convd} \\ v_{convq} \end{matrix}] = R [\begin{matrix} i_{d} \\ i_{q} \end{matrix}] + L \frac{d}{dt} [\begin{matrix} i_{d} \\ i_{q} \end{matrix}] + ω_{e} L [\begin{matrix} - i_{q} \\ i_{d} \end{matrix}] + [\begin{matrix} v_{gd} \\ v_{gq} \end{matrix}]$

(6.227)

Therefore, assuming that the quantities i_d and i_q are constant, which means that

L \frac{d}{dt} (\begin{matrix} i_{d} \\ i_{q} \end{matrix}) = 0

$L \frac{d}{dt} (\begin{matrix} i_{d} \\ i_{q} \end{matrix}) = 0$

, and using Eq. (6.227), the d–q components of the inverter output voltages in the synchronous rotating frame are given by:

$\begin{matrix} v_{convd} = R i_{d} - ω_{e} L i_{q} + v_{gd} \\ v_{convq} = R i_{q} + ω_{e} L i_{d} + v_{gq} \end{matrix}$ $\begin{matrix} v_{convd} = R i_{d} - ω_{e} L i_{q} + v_{gd} \\ v_{convq} = R i_{q} + ω_{e} L i_{d} + v_{gq} \end{matrix}$

(6.228)

Fig. 6.145 presents the block diagram of a P–Q controller which is based on the SVPWM control technique.

As shown in Fig. 6.145, to synchronize the utility grid with the converter a phase locked loop (PLL) is used which detects the grid phase angle θ_e = ω_et and provides this information to the Park transformation blocks.

For a three-phase balanced system the active and reactive power of the converter expressed with the voltages and currents of the a–b–c reference frame are given by the following equations:

$\begin{matrix} P_{conv} = converter active power \\ = \frac{1}{T} \int_{t - T}^{t} (v_{conva} i_{a} + v_{convb} i_{b} + v_{convc} i_{c}) dt \end{matrix}$ $\begin{matrix} P_{conv} = converter active power \\ = \frac{1}{T} \int_{t - T}^{t} (v_{conva} i_{a} + v_{convb} i_{b} + v_{convc} i_{c}) dt \end{matrix}$

(6.229)

$\begin{matrix} Q_{conv} = converter reactive power \\ = \frac{1}{T \sqrt{3}} \int_{t - T}^{T} (v_{convbc} i_{a} + v_{convca} i_{b} + v_{convab} i_{c}) dt \end{matrix}$ $\begin{matrix} Q_{conv} = converter reactive power \\ = \frac{1}{T \sqrt{3}} \int_{t - T}^{T} (v_{convbc} i_{a} + v_{convca} i_{b} + v_{convab} i_{c}) dt \end{matrix}$

(6.230)

where v_convab, v_convbc, and v_convca are the converter line-to-line voltages.

Figure 6.145 Block diagram of P–Q controller based on space vector pulse width modulation control technique (∗, reference signals).

Transforming the voltages and the currents of Eqs. (6.229) and (6.230) into α–β reference frame yields:

$P_{conv} = \frac{1}{T} \int_{t - T}^{t} (v_{convα} i_{convα} + v_{convβ} i_{convβ}) dt$ $P_{conv} = \frac{1}{T} \int_{t - T}^{t} (v_{convα} i_{convα} + v_{convβ} i_{convβ}) dt$

(6.231)

$Q_{conv} = \frac{1}{T} \int_{t - T}^{t} (v_{β} i_{α} - v_{iα} i_{β}) dt$ $Q_{conv} = \frac{1}{T} \int_{t - T}^{t} (v_{β} i_{α} - v_{iα} i_{β}) dt$

(6.232)

The active and reactive power expressed in voltages and currents of the d–q reference frame must equal the active and reactive power expressed in voltages and currents of the a–b–c reference frame. Therefore, substituting Eq. (6.200) into Eqs. (6.231) and (6.232), the following results are obtained:

$P_{conv} = \frac{3}{2} (v_{d} i_{d} + v_{q} i_{q})$ $P_{conv} = \frac{3}{2} (v_{d} i_{d} + v_{q} i_{q})$

(6.233)

$Q_{conv} = \frac{3}{2} (v_{q} i_{d} - v_{d} i_{q})$ $Q_{conv} = \frac{3}{2} (v_{q} i_{d} - v_{d} i_{q})$

(6.234)

where the factor 3/2 is a scaling factor due to the choice of the scaling factor 2/3 used in the transformation Eq. (6.200).

Next, using Eqs. (6.233) and (6.234) the reference currents in the d–q synchronous rotating frame with respect to the reference values

P_{conv}^{*}

$P_{conv}^{*}$

and

Q_{conv}^{*}

$Q_{conv}^{*}$

can be found and are given by the following equations:

$i_{d}^{*} = \frac{2}{3} (\frac{P_{conv}^{*} v_{gd} + Q_{conv}^{*} v_{gq}}{v_{gd}^{2} + v_{gq}^{2}})$ $i_{d}^{*} = \frac{2}{3} (\frac{P_{conv}^{*} v_{gd} + Q_{conv}^{*} v_{gq}}{v_{gd}^{2} + v_{gq}^{2}})$

(6.235)

$i_{q}^{*} = \frac{2}{3} (\frac{P_{conv}^{*} v_{gq} - Q_{conv}^{*} v_{gd}}{v_{gd}^{2} + v_{gq}^{2}})$ $i_{q}^{*} = \frac{2}{3} (\frac{P_{conv}^{*} v_{gq} - Q_{conv}^{*} v_{gd}}{v_{gd}^{2} + v_{gq}^{2}})$

(6.236)

When the converter is synchronized to the utility grid (through the PLL system), the d-axis of the d–q frame is aligned with the vector of the grid voltage

{\vec{V}}_{g}

${\vec{V}}_{g}$

. With this alignment, the component v_q = 0 and, consequently, control decoupling is achieved and Eqs. (6.233)–(6.236) become:

$P_{conv} = \frac{3}{2} v_{d} i_{d}$ $P_{conv} = \frac{3}{2} v_{d} i_{d}$

(6.237)

$Q_{conv} = - \frac{3}{2} v_{d} i_{q}$ $Q_{conv} = - \frac{3}{2} v_{d} i_{q}$

(6.238)

$i_{d}^{*} = \frac{2}{3} \frac{Q_{conv}^{*}}{v_{gq}}$ $i_{d}^{*} = \frac{2}{3} \frac{Q_{conv}^{*}}{v_{gq}}$

(6.239)

$i_{q}^{*} = \frac{2}{3} \frac{P_{conv}^{*}}{v_{gq}}$ $i_{q}^{*} = \frac{2}{3} \frac{P_{conv}^{*}}{v_{gq}}$

(6.240)

Therefore, assuming that the utility grid is balanced and stable, the active and reactive power delivered by the converter to the utility grid is determined by the reference currents

i_{d}^{*} and i_{q}^{*}

$i_{d}^{*} and i_{q}^{*}$

, respectively. If the reference current

i_{q}^{*} = 0

$i_{q}^{*} = 0$

, the inverter voltages and currents are in phase and the converter operates at unity power factor.

Example 6.14

Transform the following three-phase time variant voltages to d–q reference frame.

v_a(t) = 380 cos ωt

v_b(t) = 380 cos(ωt − 120°)

v_c(t) = 380 cos(ωt − 240°)

Solution

The transformation from a–b–c reference frame to α–β is given by:

$\begin{matrix} [\begin{matrix} v_{α} \\ v_{β} \end{matrix}] = \frac{2}{3} [\begin{matrix} 1 & - \frac{1}{2} & - \frac{1}{2} \\ 0 & \frac{\sqrt{3}}{2} & - \frac{\sqrt{3}}{2} \end{matrix}] [\begin{matrix} v_{an} \\ v_{bn} \\ v_{cn} \end{matrix}] = \frac{2}{3} [\begin{matrix} 1 & - \frac{1}{2} & - \frac{1}{2} \\ 0 & \frac{\sqrt{3}}{2} & - \frac{\sqrt{3}}{2} \end{matrix}] [\begin{matrix} 380 \cos ωt \\ 380 \cos (ωt - 120 °) \\ 380 \cos (ωt - 240 °) \end{matrix}] \\ = \frac{2}{3} [\begin{matrix} \frac{3}{2} (380) \cos ωt \\ \frac{3}{2} (380) \sin ωt \end{matrix}] \end{matrix}$ $\begin{matrix} [\begin{matrix} v_{α} \\ v_{β} \end{matrix}] = \frac{2}{3} [\begin{matrix} 1 & - \frac{1}{2} & - \frac{1}{2} \\ 0 & \frac{\sqrt{3}}{2} & - \frac{\sqrt{3}}{2} \end{matrix}] [\begin{matrix} v_{an} \\ v_{bn} \\ v_{cn} \end{matrix}] = \frac{2}{3} [\begin{matrix} 1 & - \frac{1}{2} & - \frac{1}{2} \\ 0 & \frac{\sqrt{3}}{2} & - \frac{\sqrt{3}}{2} \end{matrix}] [\begin{matrix} 380 \cos ωt \\ 380 \cos (ωt - 120 °) \\ 380 \cos (ωt - 240 °) \end{matrix}] \\ = \frac{2}{3} [\begin{matrix} \frac{3}{2} (380) \cos ωt \\ \frac{3}{2} (380) \sin ωt \end{matrix}] \end{matrix}$

Finally, the transformation from time varying α–β stationary frame to synchronous time invariant d–q frame is given by:

$[\begin{matrix} v_{d} \\ v_{q} \end{matrix}] = [\begin{matrix} \cos ωt & \sin ωt \\ - \sin ωt & \cos ωt \end{matrix}] [\begin{matrix} v_{α} \\ v_{β} \end{matrix}] = [\begin{matrix} \cos ωt & \sin ωt \\ - \sin ωt & \cos ωt \end{matrix}] \cdot \frac{2}{3} [\begin{matrix} \frac{3}{2} (380) \cos ωt \\ \frac{3}{2} (380) \sin ωt \end{matrix}] = [\begin{matrix} 380 \\ 0 \end{matrix}]$ $[\begin{matrix} v_{d} \\ v_{q} \end{matrix}] = [\begin{matrix} \cos ωt & \sin ωt \\ - \sin ωt & \cos ωt \end{matrix}] [\begin{matrix} v_{α} \\ v_{β} \end{matrix}] = [\begin{matrix} \cos ωt & \sin ωt \\ - \sin ωt & \cos ωt \end{matrix}] \cdot \frac{2}{3} [\begin{matrix} \frac{3}{2} (380) \cos ωt \\ \frac{3}{2} (380) \sin ωt \end{matrix}] = [\begin{matrix} 380 \\ 0 \end{matrix}]$

Therefore: v_d = 380 V and v_q = 0 V.

..................Content has been hidden....................

You can't read the all page of ebook, please click here login for view all page.

Table of Contents for 6.9. P–Q Control of a Three-Phase Voltage Source Inverter

Create new playlist

Sign In

Sign Up

6.9. P–Q Control of a Three-Phase Voltage Source Inverter

Solution

6.9.1. P–Q Control Based on the Decoupling Control of d–q Current Components

Solution

Table of Contents for
6.9. P–Q Control of a Three-Phase Voltage Source Inverter