Comparison of SRF/PI- and STRF/PR-based power controllers for grid-tied distributed generation systems

Abstract

Grid-tied distributed generation (DG) system-based renewable energy sources such as wind, sun and hydrogen have recently gained a large attention due mainly to environmental issues. In this study, to provide energy for various loads, it is connected to the common direct current bus system after DG system-based fuel cell and solar cell are modeled and simulated. In order to synchronize DG system sources to utility grid, park transformation-based three-phase phase-locked loop technique is used, which is one of the most common methods in the literature. Power control method-based synchronous reference frame with proportional integral controller or stationary reference frame with proportional resonant current controller is used in the DG systems. The performances of two controllers are discussed in this study. Simulation results are obtained for various scenarios at the designed and created simulation model of DG + Grid + Load system. The system is modeled and simulated by using PSCAD/EMTDC software package.

Appendix

Grid-tied three-phase LC filter is analyzed with following equations.

$$\begin{aligned} {\frac{\mathrm{d}{I}_{\mathrm{abc}}}{\mathrm{d}{t}}} = {\frac{V_{i,\mathrm{abc}}}{L}}- {\frac{V_{c,\mathrm{abc}}}{L}}- {\frac{I_{\mathrm{abc}}}{\mathrm{RL}}} \end{aligned}$$

(21)

Dq0/abc and abc/dq0 current Park transformations are expressed as:

$$\begin{aligned} {I}_{{ dq}0}= & {} \sqrt{\frac{2}{3}}\left[ {{\begin{array}{ccc} {\cos ( {\theta } )}&{} {\cos \left( {{\theta }-\frac{2{\pi }}{3}} \right) }&{} {\cos \left( {{\theta }+\frac{2{\pi }}{3}} \right) } \\ {-\sin ( {\theta } )}&{} {-\sin \left( {{\theta }-\frac{2{\pi }}{3}} \right) }&{} {-\sin \left( {{\theta }+\frac{2{\pi }}{3}} \right) } \\ {\frac{\sqrt{2}}{2}}&{} {\frac{\sqrt{2}}{2}}&{} {\frac{\sqrt{2}}{2}} \\ \end{array} }} \right] \nonumber \\&\left[ {{\begin{array}{c} {{I}_{a} } \\ {{I}_{b} } \\ {{I}_{c} } \\ \end{array} }} \right] \end{aligned}$$

(22)

$$\begin{aligned} {I}_{\mathrm{abc}}= & {} \sqrt{\frac{2}{3}}\left[ {{\begin{array}{ccc} {\cos ( {\theta } )}&{} {-\sin ( {\theta } )}&{} {\frac{\sqrt{2}}{2}} \\ {\cos \left( {{\theta }-\frac{2{\pi }}{3}} \right) }&{} {-\sin \left( {{\theta }-\frac{2{\pi }}{3}} \right) }&{} {\frac{\sqrt{2}}{2}} \\ {\cos \left( {{\theta }+\frac{2{\pi }}{3}} \right) }&{} {-\sin \left( {{\theta }+\frac{2{\pi }}{3}} \right) }&{} {\frac{\sqrt{2}}{2}} \\ \end{array} }} \right] \nonumber \\&\left[ {{\begin{array}{c} {{I}_{d} } \\ {{I}_{q} } \\ {{I}_0 } \\ \end{array} }} \right] \end{aligned}$$

(23)

$\theta$ change with $\omega {t}$ as:

$$\begin{aligned} {I}_{d}= & {} \sqrt{\frac{2}{3}}\left[ {I}_{a} \cos ( {{\omega t}} )+{I}_{b} \cos \left( {{\omega t}-\frac{2{\pi }}{3}} \right) \right. \nonumber \\&\left. +\,\,{I}_{c} \cos \left( {{\omega t}+\frac{2{\pi }}{3}} \right) \right] \end{aligned}$$

(24)

$$\begin{aligned} {I}_{q}= & {} \sqrt{\frac{2}{3}}\left[ {I}_{a} \sin ( {{\omega t}}) +{I}_{b} \sin \left( {{\omega t}-\frac{2{\pi }}{3}} \right) \right. \nonumber \\&\left. +\,\,{I}_{c} \sin \left( {{\omega t}+\frac{2{\pi }}{3}} \right) \right] \end{aligned}$$

(25)

Taking derivative of ${I}_{{ d}}$ and ${I}_{{ q}}$, we will get (App.6) and (App.8):

$$\begin{aligned} \frac{{\mathrm{d}{I}}_{d} }{{\mathrm{d}{t}}}= & {} \sqrt{\frac{2}{3}}\left[ \frac{\mathrm{d}{{I}}_{a} }{{\mathrm{d}{t}}}\cos \left( {{\omega t}} \right) +\frac{{\mathrm{d}{I}}_{b} }{{\mathrm{d}{t}}}\cos \left( {{\omega t}-\frac{2{\pi }}{3}} \right) \right. \nonumber \\&\qquad \quad \left. +\,\,\frac{{\mathrm{d}{I}}_{c} }{{\mathrm{d}{t}}}\cos \left( {{\omega t}+\frac{2{\pi }}{3}} \right) \right] \nonumber \\&-\sqrt{\frac{2}{3}}{\omega }\left[ {I}_{a} \sin \left( {{\omega t}} \right) +{I}_{b} \sin \left( {{\omega t}-\frac{2{\pi }}{3}} \right) \right. \nonumber \\&\quad \qquad \qquad \left. +\,\,{I}_{c} \sin \left( {{\omega t}+\frac{2{\pi }}{3}} \right) \right] \end{aligned}$$

(26)

Inserting (21) and (25) into equation (26), we will get equation (27):

$$\begin{aligned} \frac{{\mathrm{d}{I}}_{d} }{{\mathrm{d}{t}}}= & {} \frac{{V}_{d} }{{L}}-\frac{{V}_{\mathrm{c},{d}} }{{L}}-\frac{{R}}{{L}}{I}_{d} +{\omega I}_{q} \end{aligned}$$

(27)

$$\begin{aligned} \frac{{\mathrm{d}{I}}_{q} }{{\mathrm{d}{t}}}= & {} -\sqrt{\frac{2}{3}}\left[ \frac{{\mathrm{d}{I}}_{a} }{{\mathrm{d}{t}}}\sin ( {{\omega t}} )+\frac{{\mathrm{d}{I}}_{b} }{{\mathrm{d}{t}}}\sin \left( {{\omega t}-\frac{2{\pi }}{3}} \right) \right. \nonumber \\&\qquad \qquad \left. +\frac{{\mathrm{d}{I}}_{c} }{{\mathrm{d}{t}}}\sin \left( {{\omega t}+\frac{2{\pi }}{3}} \right) \right] \nonumber \\&-\sqrt{\frac{2}{3}}{\omega }\left[ {I}_{a} \cos ( {{\omega t}} )+{I}_{b} \cos \left( {{\omega t}-\frac{2{\pi }}{3}} \right) \right. \nonumber \\&\quad \qquad \qquad \left. +{I}_{c} \cos \left( {{\omega t}+\frac{2{\pi }}{3}} \right) \right] \end{aligned}$$

(28)

Inserting (21) and (24) into equation (28), we will get equation (29):

$$\begin{aligned} \frac{{\mathrm{d}{I}}_{q} }{{\mathrm{d}{t}}}=\frac{{V}_{q}}{{L}}-\frac{{V}_{{ c},{q}}}{{L}}-\frac{{R}}{{L}}{I}_{q} -{\omega I}_{d} \end{aligned}$$

(29)