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PI control of grid connected inverter (grid connected mode)
2022-07-02 09:37:00 【Quikk】
Grid connected inverter PI control
1. Inverter topology and mathematical model
The following figure shows the basic structure model of the inverter .

According to the model, write the mathematical model of the inverter as follows :
{ U a − L d i a d t − i a R − e a = 0 U b − L d i b d t − i b R − e b = 0 U c − L d i c d t − i c R − e c = 0 \begin{cases}{} U_a-L\frac{di_a}{dt}-i_aR-e_a=0\\ U_b-L\frac{di_b}{dt}-i_bR-e_b=0\\ U_c-L\frac{di_c}{dt}-i_cR-e_c=0 \end{cases} ⎩⎪⎨⎪⎧Ua−Ldtdia−iaR−ea=0Ub−Ldtdib−ibR−eb=0Uc−Ldtdic−icR−ec=0
2. Common transformations
2.1 abc- α β \alpha\beta αβ Transformation and its inverse transformation
The specific principle is not deduced here , Please refer to other materials .
[ U α U β ] = m × [ 1 − 1 2 − 1 2 0 3 2 − 3 2 ] [ U a U b U c ] \left[ \begin {matrix} U_\alpha\\ U_\beta \end{matrix}\right] =m \times \left[ \begin {matrix} 1 & -\frac{1}{2} & -\frac{1}{2}\\ 0 & \frac{\sqrt{3}}{2} & -\frac{\sqrt{3}}{2} \end{matrix} \right] \left[ \begin {matrix} U_a\\ U_b\\ U_c \end{matrix} \right] [UαUβ]=m×[10−2123−21−23]⎣⎡UaUbUc⎦⎤
m The value of is related to the system requirements ;
{ m = 2 3 work rate phase etc. change in m = 2 3 picture value phase etc. change in \begin {cases}{} m =\sqrt{\frac{2}{3}} & Equal power conversion \\ m =\frac{2}{3} & Equal amplitude transformation \end{cases} { m=32m=32 work rate phase etc. change in picture value phase etc. change in
In control , We often choose amplitude invariant algorithm for control .
When m = 2 3 m=\frac{2}{3} m=32 Correspondingly, there is inverse transformation :
[ U a U b U c ] = m × [ 1 0 − 1 2 3 2 − 1 2 − 3 2 ] [ U α U β ] \left[ \begin {matrix} U_a\\ U_b\\ U_c \end{matrix}\right]= m\times \left[ \begin {matrix} 1 & 0 \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{1}{2} & -\frac{\sqrt{3}}{2} \end{matrix} \right] \left[ \begin {matrix} U_\alpha\\ U_\beta\\ \end{matrix} \right] ⎣⎡UaUbUc⎦⎤=m×⎣⎢⎡1−21−21023−23⎦⎥⎤[UαUβ]
2.2 α β \alpha\beta αβ-dq Axis transformation
[ U d U q ] = [ c o s φ s i n φ − s i n φ c o s φ ] [ U α U β ] \left[ \begin {matrix} U_d\\ U_q\\ \end{matrix} \right]= \left[ \begin {matrix} cos\varphi & sin\varphi \\ -sin\varphi & cos\varphi \end{matrix} \right] \left[ \begin {matrix} U_\alpha\\ U_\beta\\ \end{matrix} \right] [UdUq]=[cosφ−sinφsinφcosφ][UαUβ]
The corresponding inverse transformation is
[ U α U β ] = [ c o s φ − s i n φ s i n φ c o s φ ] [ U d U q ] \left[ \begin {matrix} U_\alpha\\ U_\beta\\ \end{matrix} \right]= \left[ \begin {matrix} cos\varphi & -sin\varphi \\ sin\varphi & cos\varphi \end{matrix} \right] \left[ \begin {matrix} U_d\\ U_q\\ \end{matrix} \right] [UαUβ]=[cosφsinφ−sinφcosφ][UdUq]
Here we need to pay attention to :
1. In a general way , In the control of the inverter, the grid voltage is oriented to d Axis , Therefore, the angle in the transformation formula is the phase angle of the power grid .
2. according to dq And α β \alpha\beta αβ The equal positions of the coordinate axes are different , There are different forms of transformation , Specific reference to . Like European electric four dq Transformation
2.3 abc-dq Transformation
[ U d U q ] = 2 3 [ c o s φ c o s ( φ − 2 3 π ) c o s ( φ + 2 3 π ) − s i n φ − s i n ( φ − 2 3 π ) − s i n ( φ + 2 3 π ) ] [ U a U b U c ] \left[ \begin {matrix} U_d\\ U_q\\ \end{matrix} \right] =\frac{2}{3} \left[ \begin {matrix} cos\varphi & cos(\varphi-\frac{2}{3}\pi) & cos(\varphi+\frac{2}{3}\pi) \\ -sin\varphi & -sin(\varphi-\frac{2}{3}\pi) & -sin(\varphi+\frac{2}{3}\pi) \end{matrix} \right] \left[ \begin {matrix} U_a\\ U_b\\ U_c \end{matrix} \right] [UdUq]=32[cosφ−sinφcos(φ−32π)−sin(φ−32π)cos(φ+32π)−sin(φ+32π)]⎣⎡UaUbUc⎦⎤
3.dq Equation of parallel inverter in coordinate system
Will type (1) Conduct dq Change can get :
[ U d U q ] = L d d t [ i a i b i c ] × T a b c _ d q + [ i d i q ] R + [ e d e q ] \left[ \begin {matrix} U_d\\ U_q\\ \end{matrix} \right] =L\frac{d}{dt} \left[ \begin {matrix} i_a \\ i_b\\ i_c \end{matrix} \right]\times T_{abc\_dq} + \left[ \begin {matrix} i_d\\ i_q\\ \end{matrix} \right]R + \left[ \begin {matrix} e_d\\ e_q\\ \end{matrix} \right] [UdUq]=Ldtd⎣⎡iaibic⎦⎤×Tabc_dq+[idiq]R+[edeq]
among :
d [ i a i b i c ] × T a b c _ d q d t = d [ i a i b i c ] d t × T a b c _ d q + d T a b c _ d q d t × [ i a i b i c ] \frac{d\left[ \begin {matrix} i_a \\ i_b\\ i_c\end{matrix}\right]\times T_{abc\_dq}}{dt}= \frac{d\left[ \begin {matrix} i_a \\ i_b\\ i_c \end{matrix} \right]}{dt}\times T_{abc\_dq} + \frac{dT_{abc\_dq}}{dt}\times \left[ \begin {matrix} i_a \\ i_b\\ i_c \end{matrix} \right] dtd⎣⎡iaibic⎦⎤×Tabc_dq=dtd⎣⎡iaibic⎦⎤×Tabc_dq+dtdTabc_dq×⎣⎡iaibic⎦⎤
among φ = ω t \varphi=\omega t φ=ωt
d T a b c _ d q d t = ω [ − s i n φ − s i n ( φ − 2 3 π ) − s i n ( φ + 2 3 π ) − c o s φ − c o s ( φ − 2 3 π ) − c o s ( φ + 2 3 π ) ] \frac{dT_{abc\_dq}}{dt}=\omega \left[ \begin {matrix} -sin\varphi & -sin(\varphi-\frac{2}{3}\pi) & -sin(\varphi+\frac{2}{3}\pi) \\ -cos\varphi & -cos(\varphi-\frac{2}{3}\pi) & -cos(\varphi+\frac{2}{3}\pi) \end{matrix} \right] dtdTabc_dq=ω[−sinφ−cosφ−sin(φ−32π)−cos(φ−32π)−sin(φ+32π)−cos(φ+32π)]
therefore :
d T a b c _ d q d t × [ i a i b i c ] = ω × [ i q − i d ] \frac{dT_{abc\_dq}}{dt}\times \left[ \begin {matrix} i_a \\ i_b\\ i_c \end{matrix} \right]= \omega \times \left[ \begin {matrix} i_q \\ -i_d\\ \end{matrix} \right] dtdTabc_dq×⎣⎡iaibic⎦⎤=ω×[iq−id]
Substitute into the original formula to get :
d [ i a i b i c ] d t × T a b c _ d q = d [ i d i q ] d t − ω × [ i q − i d ] \frac{d\left[ \begin {matrix} i_a \\ i_b\\ i_c \end{matrix} \right]}{dt}\times T_{abc\_dq}= \frac{d\left[ \begin {matrix} i_d \\ i_q \\ \end{matrix} \right]}{dt}-\omega \times \left[ \begin {matrix} i_q \\ -i_d\\ \end{matrix} \right] dtd⎣⎡iaibic⎦⎤×Tabc_dq=dtd[idiq]−ω×[iq−id]
Then the mathematical model of the grid connected inverter is :
[ U d U q ] = L d d t [ i d i q ] − ω L × [ i q − i d ] + [ i d i q ] R + [ e d e q ] \left[ \begin {matrix} U_d\\ U_q\\ \end{matrix} \right] =L\frac{d}{dt}\left[ \begin {matrix} i_d \\ i_q\\ \end{matrix} \right] -\omega L \times \left[ \begin {matrix} i_q \\ -i_d\\ \end{matrix} \right] + \left[ \begin {matrix} i_d\\ i_q\\ \end{matrix} \right]R + \left[ \begin {matrix} e_d\\ e_q\\ \end{matrix} \right] [UdUq]=Ldtd[idiq]−ωL×[iq−id]+[idiq]R+[edeq]
namely :
{ U d = L d i d d t − ω L i q + R i d + e d U q = L d i q d t + ω L i d + R i q + e q \left\{ \begin{matrix} U_d = L\frac{di_d}{dt}-\omega Li_q+Ri_d+e_d\\ U_q = L\frac{di_q}{dt}+\omega Li_d+Ri_q+e_q \end{matrix} \right. { Ud=Ldtdid−ωLiq+Rid+edUq=Ldtdiq+ωLid+Riq+eq
Laplace transform can get :
{ U d = ( L s + R ) i d − ω L i q + e d U q = ( L s + R ) i q + ω L i d + e q \left\{ \begin{matrix} U_d = (Ls+R)i_d-\omega Li_q+e_d\\ U_q = (Ls+R)i_q+\omega Li_d+e_q \end{matrix} \right. { Ud=(Ls+R)id−ωLiq+edUq=(Ls+R)iq+ωLid+eq
4. Closed loop control
According to the inverter model , The input to be built is Id* Output is Ud Of PI Control link . Decoupling control of inverter can be realized .( First modification : Decoupling is actually introduced here to eliminate the influence of disturbance on the system , such as Grid voltage 、 Current feedforward wait , After elimination, the system will become simple to pass PI Link to correct the system .)
introduce PI Links are available :
{ U d = ( K p + K i s ) ( i d ∗ − i d ) − ω L i q + e d U q = ( K p + K i s ) ( i q ∗ − i q ) + ω L i d + e q \left\{ \begin{matrix} U_d = (Kp+\frac{Ki}{s})(i_d^*-i_d)-\omega Li_q+e_d\\ U_q = (Kp+\frac{Ki}{s})(i_q^*-i_q)+\omega Li_d+e_q \end{matrix} \right. { Ud=(Kp+sKi)(id∗−id)−ωLiq+edUq=(Kp+sKi)(iq∗−iq)+ωLid+eq
The following figure shows the overall control block diagram of the inverter system :
According to this equation, we can build PI Current closed-loop control block diagram :

Some current models are bidirectional rectifier models in current control , That is, the current is defined as the flow of the power grid to the DC side . At this time, the circuit equation will change , The control block diagram will also happen . At this time, the control model is :
{ U d = − ( K p + K i s ) ( i d ∗ − i d ) + ω L i q + e d U q = − ( K p + K i s ) ( i q ∗ − i q ) − ω L i d + e q \left\{ \begin{matrix} U_d = -(Kp+\frac{Ki}{s})(i_d^*-i_d)+\omega Li_q+e_d\\ U_q = -(Kp+\frac{Ki}{s})(i_q^*-i_q)-\omega Li_d+e_q \end{matrix} \right. { Ud=−(Kp+sKi)(id∗−id)+ωLiq+edUq=−(Kp+sKi)(iq∗−iq)−ωLid+eq
Build the corresponding circuit according to the model .
5. Simulated main circuit
According to the relevant parameters , Build the simulation main circuit :

Simulation results :
Current amplitude given waveform :

Output current waveform :
Output power waveform

Model links
Model links , I need to take it myself : Grid connected inverter PI control
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