CN102709940B - Design method of energy storage quasi-Z source single-phase photovoltaic power generation system - Google Patents

Abstract

The invention discloses a design method of an energy storage type quasi-Z source single-phase photovoltaic power generation system, including the design of the energy storage battery voltage and capacity parameters required by the system, the selection of photovoltaic battery modules, the inductance and capacitance parameters of the quasi-Z source network Design, H-bridge inverter voltage and current level design, Z-source network diode design, quasi-Z source inverter loss calculation, etc. The design method is based on the user's voltage and power requirements, the application of the photovoltaic power generation system, and the local climate characteristics. The designed energy storage type quasi-Z source single-phase photovoltaic power generation system can meet the wide range variation of photovoltaic cell voltage 1:2. No matter how the voltage changes, the system will output the voltage required by the load, and it is suitable for single-stage power conversion to complete the up/down Voltage, inverter and energy storage requirements. The invention provides a simple, effective and fast method for the design and realization of an energy storage type quasi-Z source single-phase photovoltaic power generation system.

Description

A Design Method for Energy Storage Type Quasi-Z Source Single-phase Photovoltaic Power Generation System

technical field

The invention relates to the technical field of photovoltaic power generation, in particular to a design method for an energy storage type quasi-Z source single-phase photovoltaic power generation system.

Background technique

Photovoltaic power generation is an ideal sustainable energy source, and a power converter/inverter is essential for its development and utilization. However, generally speaking, the output voltage of photovoltaic cells can vary by a factor of two, and using a traditional single-stage inverter structure will result in a doubling of the design capacity of the inverter. If a two-stage structure is adopted, the introduction of a DC/DC converter will increase costs and reduce efficiency. To this end, researchers began to study new technologies, using Z-source and quasi-Z source inverters to overcome these problems, because it realizes the traditional two-stage inverter composed of DC/DC and inverter in the form of single-stage power conversion. The conversion function does not increase the capacity of the inverter, and is compatible with traditional systems in the field of photovoltaic power generation.

On the other hand, the power of photovoltaic power generation is highly dependent on light and temperature. Due to the erratic light and temperature, the output voltage and power of photovoltaic cells vary in a wide range. Direct grid connection or independent power supply will have a negative impact on the grid or load. Therefore, in addition to the application of energy storage batteries in traditional independent photovoltaic power generation systems, energy storage battery technology has also been adopted in grid-connected solar power generation systems in recent years to buffer energy and stabilize grid-connected power. The commonly used method is to connect the energy storage battery to the DC side through a bidirectional DC/DC converter to realize the grid-connected power stabilization function, but an additional DC/DC converter is added, which increases the cost of the system. In order to overcome this problem, the invention patent [Application No. 201010234868.X, single-stage buck-boost energy storage type photovoltaic grid-connected power generation control system] connects the energy storage battery in parallel with the constant capacitor, realizing the single-stage power conversion to complete the step-up/step-down , inverter and energy storage. In this system, the energy storage battery is directly connected in parallel with the capacitor without adding additional equipment, which is economical and practical. However, so far, there is no literature on how to design various parameters in the system, such as energy storage battery voltage and capacity, photovoltaic battery voltage level, modulation index, capacitance, inductance, power switching device loss analysis, etc. For an energy storage photovoltaic grid-connected power generation control system, the determination of these parameters is very important.

Contents of the invention

In order to solve the above problems, the present invention discloses a design method of an energy storage type quasi-Z source single-phase photovoltaic power generation system. The energy storage type quasi-Z source single-phase photovoltaic power generation system includes: an energy storage battery, an H bridge inverter Transformer, quasi-Z source network diode, first electrolytic capacitor, second electrolytic capacitor, first inductance, second inductance, LC filter, photovoltaic cell, power grid and partial load; the LC filter includes output filter inductance and Composed of an output filter capacitor; and, the negative pole of the second electrolytic capacitor is connected to the anode of the quasi-Z source network diode, and the positive pole of the second electrolytic capacitor is connected to the positive pole of the H-bridge inverter; the The cathode of the quasi-Z source network diode is connected to the positive pole of the first electrolytic capacitor and the second inductor at the same time; the other end of the second inductor is connected to the positive pole of the H-bridge inverter; the first electrolytic capacitor The negative pole of the H-bridge inverter is connected to the negative pole; one end of the first inductor is connected to the positive pole of the photovoltaic cell; the other end of the first inductor is connected to the negative pole of the second electrolytic capacitor; The output of the H-bridge inverter is connected to the power grid after passing through the LC filter, or supplies power to local loads; the energy storage battery is connected across the two ends of the first electrolytic capacitor, and the positive pole of the energy storage battery is connected to the second A positive electrode of an electrolytic capacitor; based on the user's voltage and power requirements, the purpose of the photovoltaic power generation system, and the local climate characteristics, the method includes the design of the energy storage battery voltage and capacity parameters required by the system, the selection of photovoltaic battery modules, and the quasi-Z source network inductance , Capacitor parameter design, H-bridge inverter voltage and current level design, Z-source network diode design, quasi-Z source inverter loss calculation, etc.

Further, as a preference, the H-bridge inverter voltage and current level design includes the following steps:

Step 1, calculate the voltage amplitude v _al output by the single-phase inverter according to the load voltage or the grid voltage at the grid connection point;

Step 2, set the operating voltage range of photovoltaic cells to be 1:2, and the maximum photovoltaic cell voltage is V _in =v _al , and the minimum is V _in =v _al /2, correspondingly, when the photovoltaic cell voltage is maximum, the inverter Modulation index M=1, direct duty cycle D=0, peak value of DC bus voltage is V _PN =v _al , when photovoltaic cell voltage is minimum, direct duty cycle D=1/3, inverter modulation index M=2/ 3. The peak value of DC bus voltage is V _PN =1.5*v _al .

Further, as a preference, the design of the capacitance and inductance parameters includes:

Step 3, calculate the voltage of the capacitor _C1 and the voltage of the parallel battery V _C1 =V _B =v _al ;

Step 4, calculate the maximum peak value of the DC bus voltage as V _PN =1.5×v _al ;

Step 5, calculate the maximum voltage of capacitor _C2 as V _C2 =0.5×v _al ;

Step 6, the maximum direct duty cycle D=1/3.

Further, as a preference, the selection of the photovoltaic cell module includes:

Step 7, according to the load or grid-connected power P _o , considering that the maximum power occurs at the maximum voltage, the photovoltaic cell current is

Step 8, the discharge of the energy storage battery occurs when the photovoltaic power is insufficient. According to the above design, the minimum operating voltage of the photovoltaic battery is v _al /2. When the minimum light intensity allowed to work is set, the current provided by the photovoltaic battery at the minimum power is the maximum power. One part of k; when P _o power is to be output, the battery needs to provide power as P _B =P _o -0.5×v _al ×iL ₁ /k, then the battery current is I _B =P _B /V _B ; then the inductance L The current of ₂ is i _L2 =i _L1 +i _B .

Further, as a preference, the design of the energy storage battery voltage and capacity parameters includes:

Step 9, determine the capacity of the battery, apply statistical data, according to the local climate characteristics, combined with the purpose of the designed system, determine the capacity of the battery; In addition, the battery is charged for x hours on average with the power of P _x every day for use when there is no sunlight at night, and the battery discharge depth is y% each time, then the battery capacity is required to be [P _x ×x/v _al ]/( 1-y%)(Ah).

Further, as a preference, the selection of the photovoltaic cell module includes:

Step 10, determine the photovoltaic cell module and its quantity, select the photovoltaic cell module, and determine the maximum operating voltage v _pv and the maximum operating current i _pv in the maximum power point of the module according to the local climate characteristics, then the photovoltaic cell quantity can be calculated as Take an integer to get n, Take an integer to get m, and the number of photovoltaic cell modules is m*n.

Further, as a preference, the design of the capacitance and inductance parameters includes:

Step 11, capacitor _C2 is designed, and the system adopts a constant through zero vector modulation method, then the pulsating frequency of the capacitor voltage on the Quasi-Z network is 2f _s ; during steady state, the initial value and final value of the capacitor voltage in one cycle are equal; with Take straight-through as an example, the ripple ΔV _C2 of the Quasi-Z network capacitor C ₂ voltage is in, -i _L1 ＝i _C2 , f _s is the carrier frequency, so we have If the ripple of the given capacitor voltage is ΔV _C2 ≤ αV _C2 , then there is In order to suppress double-frequency voltage ripple, the required capacitor is In the formula, ε is the ratio of double-frequency voltage ripple to V _PN , and f is the load or grid voltage frequency, since the capacitance C is greater than And the capacitor C ₁ is connected in parallel with the battery, so the capacitor C ₂ is designed as $C_2=\frac{P_o}{4\mathrm{\πf}{V}_{\mathrm{PN}}^2\mathrm{\ϵ}};$

Step 12, capacitor _C1 is designed, and the system adopts a constant through zero vector modulation method, then the pulsating frequency of the capacitor voltage on the Quasi-Z network is 2f _s ; in steady state, the initial value and final value of the capacitor voltage in one cycle are equal, with Take straight-through as an example, the ripple ΔV _C1 of the voltage of the Quasi-Z network capacitor C ₁ is in, i _C1 ＝i _B -i _L2 , f _s is the carrier frequency, so we have If the ripple of the given capacitor voltage is ΔV _C1 ≤ αV _C1 , then there is $C_1\&Greater\; Equal;(i_B-i_{L2})\frac{\mathrm{D.}}{2f_{\mathrm{the\; s}}\mathrm{\α}{V}_{C1}};$

Step 13, the design of the inductor L ₂ , in a steady state, the initial value and final value of the inductor current in one cycle are equal. Taking straight-through as an example, the ripple Δi _L2 of the Quasi-Z network inductor current is in, v _L2 =V _C1 , so we have If the ripple of the given inductor current is Δi _L2 ≤ bi _L2 , then there is

Step 14, the design of the inductor L1, in the steady state, the initial value and the final value of the inductor current in one cycle are equal, taking the straight-through as an example, the ripple Δi _L1 of the inductor current in the Quasi-Z network is in, V _in +V _C2 ＝v _L1 , so we have If the ripple of the given inductor current is Δi _L1 ≤ bi _L1 , then there is

Step 15, the parameters of the diodes in the Quasi-Z network, the diodes in the Quasi-Z network are turned off under reverse pressure in the through state, and turned on in the non-through state. The diode is designed according to the current flowing through it in the straight-through state; in the straight-through state, the diode withstands the reverse voltage V _PN , and the current through the diode in the non-through state is i _D ≤i _L1 +i _C2 =i _L1 +i _L2 -i _d , in the traditional zero vector time, when i _d =0, the maximum current flowing through the diode is i _D =i _L1 +i _L2 ;

Step 16, based on the parameters of the inverter power device, according to the load or grid-connected power P _o , and the load or grid-connected grid voltage, calculate the effective value of the load current The peak value V _PN of the DC bus voltage calculated in step 4 and the effective value of the load current calculated above are used as the selected voltage and current parameters of the inverter power device.

Further, as a preference, the quasi-Z source inverter loss calculation includes:

Step 17, energy storage type quasi-Z source single-phase photovoltaic inverter loss estimation, there are 4 IGBTs and their anti-parallel diodes, a Z-source network diode, the loss can be calculated separately according to the device;

1) Loss estimation of 4 IGBTs and their anti-parallel diodes. This part of the loss can be divided into loss in the traditional sense and loss caused by direct connection. The loss in the traditional sense includes switching loss and conduction loss, and the switching loss is

$P_{\mathrm{SW}}=\frac{4}{\mathrm{\π}}{f}_{\mathrm{the\; s}}\·({\mathrm{E.}}_{\mathrm{ON},I}+{\mathrm{E.}}_{\mathrm{OFF},I}+{\mathrm{E.}}_{\mathrm{OFF},\mathrm{D.}})\·\frac{V_{\mathrm{PN}}}{V_{\mathrm{ref}}}\&Center\; Dot;\frac{i_L}{i_{\mathrm{ref}}},$ In the formula, E _{ON, I} , E _{off, I} , E _{OFF, D} are the turn-on loss, turn-off loss and diode reverse recovery loss energy of the power device at the voltage V _ref and current i _ref respectively, which can be obtained from the device manual ; i _L is the peak value of the load current, i _L =1.414*i _a , f _s is the switching frequency; for the conduction loss of each IGBT in the traditional sense, it is calculated according to the following formula

${\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; cv</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; I</span>}}<span\; class="notranslate">\; =</span>\frac{{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; CE</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}}{\mathrm{<span\; class="notranslate">\; 2</span>}\mathrm{<span\; class="notranslate">\; \&pi;</span>}}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; +</span>\frac{\mathrm{<span\; class="notranslate">\; \&pi;M</span>}}{\mathrm{<span\; class="notranslate">\; 4</span>}}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; +</span>\frac{{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; CE</span>}}{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{\mathrm{<span\; class="notranslate">\; 2</span>}}}{\mathrm{<span\; class="notranslate">\; 2</span>}\mathrm{<span\; class="notranslate">\; \&pi;</span>}}<span\; class="notranslate">\; (</span>\frac{\mathrm{<span\; class="notranslate">\; \&pi;</span>}}{\mathrm{<span\; class="notranslate">\; 4</span>}}<span\; class="notranslate">\; +</span>\frac{\mathrm{<span\; class="notranslate">\; 2</span>}\mathrm{<span\; class="notranslate">\; m</span>}}{\mathrm{<span\; class="notranslate">\; 3</span>}}<span\; class="notranslate">\; -</span>\frac{\mathrm{<span\; class="notranslate">\; \&pi;D</span>}}{\mathrm{<span\; class="notranslate">\; 4</span>}}<span\; class="notranslate">\; )</span>$

$P_{\mathrm{cv},\mathrm{D.}}=\frac{V_{f,0}\&Center\; Dot;i_L}{2\mathrm{\&pi;}}(1-\frac{\mathrm{\&pi;M}}{4}-\mathrm{D.})+\frac{r_f\&Center\; Dot;i_L^2}{2\mathrm{\&pi;}}(\frac{\mathrm{\&pi;}}{4}-\frac{2m}{3}-\frac{\mathrm{\&pi;D}}{4}),$ In the formula, V _CE0 , V _F,0 are the saturation voltage drop and conduction voltage drop of IGBT and diode; r _CE , r _F are the on-resistance of IGBT and diode; M and D are modulation index and direct duty cycle respectively , then the conduction loss of the 4 IGBTs and their diodes is P _CV =4*(P _{CV, I} +P _{CV, D} ), for the loss caused by the straight-through, it is only necessary to calculate the switching loss and conduction loss from the IGBT, namely The switching loss is $P_{\mathrm{SW},\mathrm{SH}}=4f_{\mathrm{the\; s}}\&Center\; Dot;({\mathrm{E.}}_{\mathrm{ON},I}+{\mathrm{E.}}_{\mathrm{OFF},I})\&CenterDot;\frac{V_{\mathrm{PN}}}{V_{\mathrm{ref}}}\&CenterDot;\frac{i_{L1}}{i_{\mathrm{ref}}},$ The conduction loss is $P_{\mathrm{cv},\mathrm{SH}}=4*(V_{\mathrm{CE},0}{i}_{L1}\mathrm{D.}+r_{\mathrm{CE}}{i}_{L1}^2\mathrm{D.}+\frac{r_{\mathrm{CE}}{i}_L^2\mathrm{D.}}{8})$

2) The loss of the Z-source network diode, the conduction loss of the Z-Source network diode of each module is $P_{\mathrm{cv},\mathrm{D.}}=V_{\mathrm{DF},0}[{2i}_{L1}(1-\mathrm{D.})-\frac{{\mathrm{Mi}}_L}{2})+r_{\mathrm{DF}}\frac{1-\mathrm{D.}}{2}[{8i}_{L1}^2+i_L^2]-r_{\mathrm{DF}}\frac{8(1-\mathrm{D.})}{\mathrm{\&pi;}}{i}_L{i}_{L1},$ In the formula, V _{DF, 0} is the conduction voltage drop of the Z-source network diode; r _DF is the conduction resistance of the Z-source network diode, and its reverse recovery loss is In the formula, E _rec is the reverse recovery loss energy of Z-source network diode at I _FM and _VR ; the total loss is P _SW + P _CV + P _{SW, SH} + P _{CV, SH} + P _{CV, D} + P _{CV, DR} , by estimating the loss, you can compare alternative devices, as a condition for optimizing device selection.

Through the method of the invention, the energy storage type quasi-Z source single-phase photovoltaic power generation system can be designed simply, effectively and quickly.

Description of drawings

A more complete and better understanding of the invention, and many of its attendant advantages, will readily be learned by reference to the following detailed description when considered in conjunction with the accompanying drawings, but the accompanying drawings illustrated herein are intended to provide a further understanding of the invention and constitute A part of the present invention, the exemplary embodiment of the present invention and its description are used to explain the present invention, and do not constitute an improper limitation of the present invention, wherein:

Figure 1 is a schematic structural diagram of an energy storage type quasi-Z source single-phase photovoltaic power generation inverter;

Figure 2 shows the design method of this energy storage type quasi-Z source single-phase photovoltaic power generation system.

Detailed ways

Embodiments of the present invention will be described below with reference to FIGS. 1-2.

In order to make the above objects, features and advantages more comprehensible, the present invention will be further described in detail below in conjunction with the accompanying drawings and specific embodiments.

An embodiment of a design method for an energy storage type quasi-Z source single-phase photovoltaic power generation control system.

As shown in Figure 1, the energy storage type quasi-Z source single-phase photovoltaic power generation inverter involved in the present invention includes: an energy storage battery, an H-bridge inverter, a quasi-Z source network diode, and a first electrolytic capacitor C1 , a second electrolytic capacitor C2, a first inductor L1, a second inductor L2, an LC filter, a photovoltaic cell, a power grid, and a local load; the LC filter includes an output filter inductor L _f and an output filter capacitor C _f ; and, The negative pole of the second electrolytic capacitor C2 is connected to the anode of the quasi-Z source network diode, and the positive pole of the second electrolytic capacitor C2 is connected to the positive pole of the H-bridge inverter; the quasi-Z source network The cathode of the diode is connected to the anode of the first electrolytic capacitor C1 and the second inductance L2 at the same time; the other end of the second inductance L2 is connected to the anode of the H-bridge inverter; the anode of the first electrolytic capacitor C1 The negative pole is connected to the negative pole of the H-bridge inverter; one end of the first inductor L1 is connected to the positive pole of the photovoltaic cell; the other end of the first inductor L1 is connected to the negative pole of the second electrolytic capacitor C2 ; the output of the H-bridge inverter is incorporated into the grid after passing through the LC filter, or supplies local loads; the energy storage battery is connected across the two ends of the first electrolytic capacitor C1, and the energy storage battery The positive pole is connected to the positive pole of the first electrolytic capacitor C1.

As shown in Figure 2, a design method for an energy storage type quasi-Z source single-phase photovoltaic power generation system, including:

S1. Know the user's voltage and power requirements, the purpose of the photovoltaic power generation system, and the local climate characteristics;

S2. Design of energy storage battery voltage and capacity parameters;

S3. Photovoltaic battery module selection;

S4. Design of quasi-Z source network inductance and capacitance parameters;

S5, H-bridge inverter voltage and current level design;

S6, Z-source network diode design;

S7. Loss calculation of the quasi-Z source inverter.

Example

The purpose is to design an energy storage type quasi-Z source single-phase photovoltaic power generation control system, the load/grid phase voltage is 120V, and the power is 1700W. Then it can be determined according to the steps:

Step 1, calculate the single-phase voltage amplitude v _al =120×1.414=170V;

Step 2, set the operating voltage range of the photovoltaic cell to 1:2, and the maximum photovoltaic cell voltage is 170V, and the minimum is 85V. Correspondingly, when the photovoltaic cell voltage is at its maximum, the modulation index of the inverter is M=1, the through-duty ratio D=0, and the peak value of the DC bus voltage is

V _PN =170

When the photovoltaic cell voltage is at its minimum, V _in =85V, the direct duty ratio D=1/3, the inverter modulation index M=2/3, and the peak value of the DC bus voltage is

V _PN =85*3=255V

Step 3, design capacitor _C1 voltage and parallel battery voltage

V _C1 =V _B = _val =170 V

Step 4, calculate the maximum peak value of the DC bus voltage as

V _PN =170+85=255V

Step 5, calculate the maximum value of the capacitor _C2 voltage as

V _C2 =85 V

Step 6, the maximum through-duty ratio D=1/3;

Step 7, according to the load or grid-connected power P _O , considering that the maximum power occurs at the maximum voltage, the current of the photovoltaic cell module is

${\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; =</span>\frac{{\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; o</span>}}}{{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; in</span>}}}<span\; class="notranslate">\; =</span>\frac{{\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; o</span>}}}{{\mathrm{<span\; class="notranslate">\; v</span>}}_{\mathrm{<span\; class="notranslate">\; al</span>}}}<span\; class="notranslate">\; =</span>\frac{\mathrm{<span\; class="notranslate">\; 1700</span>}}{\mathrm{<span\; class="notranslate">\; 170</span>}}<span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 10</span>}\mathrm{<span\; class="notranslate">\; A</span>}$

Step 8. The discharge of the energy storage battery occurs when the photovoltaic power is insufficient. According to the above design, the minimum operating voltage of the photovoltaic battery is 85V. When the minimum photovoltaic power is set, the current provided is one-third of the maximum power (for example, the maximum light is 1000W/ m ² , the minimum hour is 200W/m ² ). To output 1700W power, the battery needs to provide power of

P _B =1700-85×10/3=1416.7W

Then the battery current is

i _B =P _B /V _B =8.3A

Then the current in inductor _L2

i _L2 =i _L1 +i _B =8.3+3.3=11.6 A

Step 9, determine the capacity of the storage battery

Apply statistical data, according to the local climate characteristics, combined with the purpose of the designed system, determine the capacity of the storage battery. For example, it is required that the designed photovoltaic system not only supply power to the local load/grid under the sunshine conditions of a certain place, but also charge the battery with an average power of 1000W for 4 hours a day for use when there is no sunshine at night, and the battery is discharged every time The depth is y%. A battery capacity of

[1000*4/170]/(1-y%)=23.5/(1-y%)Ah

Step 10, determine the photovoltaic cell module and its quantity

Suppose a photovoltaic panel is at S=1000W/m ² , T=25°, the maximum power point voltage v _pv =42.4V, and the maximum power point current i _pv =5A. As the battery board temperature rises, its maximum power point voltage drops, considering the 1:2 working range, the lowest is v _pv =21.2V.

Then, the number of photovoltaic cells can be calculated as

$\mathrm{<span\; class="notranslate">\; no</span>}<span\; class="notranslate">\; =</span>\frac{{\mathrm{<span\; class="notranslate">\; v</span>}}_{\mathrm{<span\; class="notranslate">\; al</span>}}}{{\mathrm{<span\; class="notranslate">\; v</span>}}_{\mathrm{<span\; class="notranslate">\; PV</span>}}}<span\; class="notranslate">\; =</span>\frac{\mathrm{<span\; class="notranslate">\; 170</span>}}{\mathrm{<span\; class="notranslate">\; 42.4</span>}}<span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 4</span>}<span\; class="notranslate">\; ,</span>$ $\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; =</span>\frac{\mathrm{<span\; class="notranslate">\; 10</span>}}{\mathrm{<span\; class="notranslate">\; 5</span>}}<span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 2</span>}$

The number of photovoltaic cell modules is m×n=8.

Step 11, Capacitor C ₂ Design

The system adopts the constant through zero vector modulation method, so the pulsation frequency of the capacitor voltage on the Quasi-Z network is 2f _s .

In steady state, the initial value and final value of the capacitor voltage in one cycle are equal. Taking straight-through as an example, the ripple ΔV _C2 of the voltage of the Quasi-Z network capacitor C ₂ is

${\mathrm{<span\; class="notranslate">\; \&Delta;V</span>}}_{\mathrm{<span\; class="notranslate">\; C</span>}\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; C</span>}\mathrm{<span\; class="notranslate">\; 2</span>}}\frac{\mathrm{<span\; class="notranslate">\; \&Delta;t</span>}}{{\mathrm{<span\; class="notranslate">\; C</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}}$

in, -i _L1 =i _C2 , f _s is the carrier frequency, so we have

${\mathrm{<span\; class="notranslate">\; C</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}\frac{\mathrm{<span\; class="notranslate">\; D.</span>}}{\mathrm{<span\; class="notranslate">\; 2</span>}{\mathrm{<span\; class="notranslate">\; f</span>}}_{\mathrm{<span\; class="notranslate">\; the\; s</span>}}\mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; C</span>}\mathrm{<span\; class="notranslate">\; 2</span>}}}$

If the ripple of the given capacitor voltage is

ΔV _C2 ≤αV _C2

then there is

${\mathrm{<span\; class="notranslate">\; C</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; \&Greater\; Equal;</span>{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}\frac{\mathrm{<span\; class="notranslate">\; D.</span>}}{{\mathrm{<span\; class="notranslate">\; 2</span>}\mathrm{<span\; class="notranslate">\; f</span>}}_{\mathrm{<span\; class="notranslate">\; the\; s</span>}}\mathrm{<span\; class="notranslate">\; \&alpha;</span>}{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; C</span>}\mathrm{<span\; class="notranslate">\; 2</span>}}}$

If set α=1%, carrier frequency f _s =10kHz, then

${\mathrm{<span\; class="notranslate">\; C</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; \&Greater\; Equal;</span>\mathrm{<span\; class="notranslate">\; 10</span>}<span\; class="notranslate">\; \&times;</span>\frac{\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; /</span>\mathrm{<span\; class="notranslate">\; 3</span>}}{\mathrm{<span\; class="notranslate">\; 2</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; 10000</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; 0.01</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; 85</span>}}<span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 196</span>}\mathrm{<span\; class="notranslate">\; \&mu;F</span>}$

For single-phase systems, there is twice the frequency of pulsation. For the system in Figure 1, capacitor _C1 is connected in parallel with the energy storage battery, and then connected in series with capacitor _C2 . To suppress double-frequency ripple, the total capacitance required is

$\mathrm{<span\; class="notranslate">\; C</span>}<span\; class="notranslate">\; =</span>\frac{{\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; o</span>}}}{\mathrm{<span\; class="notranslate">\; 4</span>}\mathrm{<span\; class="notranslate">\; \&pi;f</span>}{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; PN</span>}}\mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; PN</span>}}}<span\; class="notranslate">\; =</span>\frac{{\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}}{\mathrm{<span\; class="notranslate">\; 4</span>}\mathrm{<span\; class="notranslate">\; \&pi;f</span>}{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; PN</span>}}^{\mathrm{<span\; class="notranslate">\; 2</span>}}\mathrm{<span\; class="notranslate">\; \&epsiv;</span>}}$

In the formula, ε is the ratio of double-frequency voltage ripple to V _PN , ΔV _PN = εV _PN , and f is the load or grid voltage frequency. Set the double-frequency voltage pulse ratio as ε=1%, f=50Hz, V _PN =255V, P _PV =1700W, then

$\mathrm{<span\; class="notranslate">\; C</span>}<span\; class="notranslate">\; =</span>\frac{\mathrm{<span\; class="notranslate">\; 1700</span>}}{\mathrm{<span\; class="notranslate">\; 4</span>}\mathrm{<span\; class="notranslate">\; \&pi;</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; 50</span>}<span\; class="notranslate">\; \&times;</span>{\mathrm{<span\; class="notranslate">\; 255</span>}}^{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; \%</span>}<span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 4.16</span>}\mathrm{<span\; class="notranslate">\; f</span>}$

Since the capacitor _C1 is connected in parallel with the energy storage battery, it can be regarded as an infinite capacitor during design. Therefore, the capacitance _C2 is 4.16mF.

Step 12, Capacitor C ₁ design

The system adopts the constant through zero vector modulation method, so the pulsation frequency of the capacitor voltage on the Quasi-Z network is 2f _s .

In steady state, the initial value and final value of the capacitor voltage in one cycle are equal. Taking straight-through as an example, the ripple ΔV _C1 of the voltage of the Quasi-Z network capacitor C ₁ is

$\mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; C</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; C</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}\frac{\mathrm{<span\; class="notranslate">\; \&Delta;t</span>}}{\mathrm{<span\; class="notranslate">\; C</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}$

in, i _C1 ＝i _B -i _L2 , f _s is the carrier frequency, so we have

${\mathrm{<span\; class="notranslate">\; C</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; =</span><span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; B</span>}}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span>\frac{\mathrm{<span\; class="notranslate">\; D.</span>}}{\mathrm{<span\; class="notranslate">\; 2</span>}{\mathrm{<span\; class="notranslate">\; f</span>}}_{\mathrm{<span\; class="notranslate">\; the\; s</span>}}\mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; C</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}}$

If the ripple of the given capacitor voltage is

ΔV _C1 ≤ αV _C1 ,

then there is

${\mathrm{<span\; class="notranslate">\; C</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; \&Greater\; Equal;</span><span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; B</span>}}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span>\frac{\mathrm{<span\; class="notranslate">\; D.</span>}}{{\mathrm{<span\; class="notranslate">\; 2</span>}\mathrm{<span\; class="notranslate">\; f</span>}}_{\mathrm{<span\; class="notranslate">\; the\; s</span>}}\mathrm{<span\; class="notranslate">\; \&alpha;</span>}{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; C</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}}$

Set other parameters as above, and I _B -I _L2 =-I _L1 =-10A, then

${\mathrm{<span\; class="notranslate">\; C</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; 10</span>}<span\; class="notranslate">\; \&times;</span>\frac{\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; /</span>\mathrm{<span\; class="notranslate">\; 3</span>}}{\mathrm{<span\; class="notranslate">\; 2</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; 10000</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; 0.01</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; 170</span>}}<span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 98</span>}\mathrm{<span\; class="notranslate">\; \&mu;F</span>}$

Step 13, design of inductance _L2

In steady state, the initial value and final value of the inductor current in one cycle are equal. Taking straight-through as an example, the ripple Δi _L2 of the Quasi-Z network inductor current is

$\mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; v</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 2</span>}}\frac{\mathrm{<span\; class="notranslate">\; \&Delta;t</span>}}{{\mathrm{<span\; class="notranslate">\; L</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}}$

in, v _L2 =V _C1 , so we have

${\mathrm{<span\; class="notranslate">\; L</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; C</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}\frac{\mathrm{<span\; class="notranslate">\; D.</span>}}{{\mathrm{<span\; class="notranslate">\; 2</span>}\mathrm{<span\; class="notranslate">\; f</span>}}_{\mathrm{<span\; class="notranslate">\; the\; s</span>}}\mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 2</span>}}}$

If the ripple of the given inductor current is

Δi _L2 ≤ bi _L2

then there is

${\mathrm{<span\; class="notranslate">\; L</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; \&Greater\; Equal;</span>{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; C</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}\frac{\mathrm{<span\; class="notranslate">\; D.</span>}}{\mathrm{<span\; class="notranslate">\; 2</span>}{\mathrm{<span\; class="notranslate">\; f</span>}}_{\mathrm{<span\; class="notranslate">\; the\; s</span>}}\mathrm{<span\; class="notranslate">\; b</span>}{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 2</span>}}}$

Set b=20%, i _L2 =11.6A then

${\mathrm{<span\; class="notranslate">\; L</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; \&Greater\; Equal;</span>\mathrm{<span\; class="notranslate">\; 170</span>}<span\; class="notranslate">\; \&times;</span>\frac{\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; /</span>\mathrm{<span\; class="notranslate">\; 3</span>}}{\mathrm{<span\; class="notranslate">\; 2</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; 10000</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; 0.2</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; 11.6</span>}}<span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1.22</span>}\mathrm{<span\; class="notranslate">\; mH</span>}$

Step 14, design of inductor _L1

In steady state, the initial value and final value of the inductor current in one cycle are equal. Taking straight-through as an example, the ripple Δi _L1 of the Quasi-Z network inductor current is

${\mathrm{<span\; class="notranslate">\; \&Delta;i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; v</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}\frac{\mathrm{<span\; class="notranslate">\; \&Delta;t</span>}}{{\mathrm{<span\; class="notranslate">\; L</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}}$

in, V _in +V _C2 ＝v _L1 , so we have

${\mathrm{<span\; class="notranslate">\; L</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; =</span><span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; in</span>}}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; C</span>}\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span>\frac{\mathrm{<span\; class="notranslate">\; D.</span>}}{\mathrm{<span\; class="notranslate">\; 2</span>}{\mathrm{<span\; class="notranslate">\; f</span>}}_{\mathrm{<span\; class="notranslate">\; the\; s</span>}}\mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}}$

If the ripple of the given inductor current is

Δi _L1 ≤ bi _L1

then there is

${\mathrm{<span\; class="notranslate">\; L</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; \&Greater\; Equal;</span><span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; in</span>}}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; C</span>}\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span>\frac{\mathrm{<span\; class="notranslate">\; D.</span>}}{\mathrm{<span\; class="notranslate">\; 2</span>}{\mathrm{<span\; class="notranslate">\; f</span>}}_{\mathrm{<span\; class="notranslate">\; the\; s</span>}}\mathrm{<span\; class="notranslate">\; b</span>}{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}}$

V _in =85V, current i _L1 =10A, then

${\mathrm{<span\; class="notranslate">\; L</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; \&Greater\; Equal;</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 85</span>}<span\; class="notranslate">\; +</span>\mathrm{<span\; class="notranslate">\; 85</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; \&times;</span>\frac{\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; /</span>\mathrm{<span\; class="notranslate">\; 3</span>}}{\mathrm{<span\; class="notranslate">\; 2</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; 10000</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; 0.2</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; 10</span>}}<span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1.42</span>}\mathrm{<span\; class="notranslate">\; mH</span>}$

Step 15, Parameters of the diodes in the Quasi-Z network

The diodes in the Quasi-Z network are turned off under reverse pressure in the through state and turned on in the non-through state. Therefore, the diode can be designed according to the voltage across it in the straight-through state and the current flowing through it in the non-straight-through state.

In the straight-through state, the maximum reverse voltage that the diode can withstand is V _PN =255V.

The current through the diode in the non-through state is

i _D ≤i _L1 +i _C2 =i _L1 +i _L2 -i _d

In the traditional zero vector time, when i _d =0, the maximum current flowing through the diode is

i _D =i _L1 +i _L2 =20 A

Step 16, parameters of inverter power devices

According to the load or grid-connected power P _o , and the load or grid-connected grid voltage, calculate the effective value of the load current

${\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; a</span>}}<span\; class="notranslate">\; =</span>\frac{\sqrt{\mathrm{<span\; class="notranslate">\; 2</span>}}{\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; o</span>}}}{{\mathrm{<span\; class="notranslate">\; v</span>}}_{\mathrm{<span\; class="notranslate">\; al</span>}}}<span\; class="notranslate">\; =</span>\frac{\sqrt{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; 1700</span>}}{\mathrm{<span\; class="notranslate">\; 170</span>}}<span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 14.1</span>}\mathrm{<span\; class="notranslate">\; A</span>}$

The peak value V _PN of the DC bus voltage calculated in step 4 and the effective value of the load current calculated above are used as the selected voltage and current parameters of the inverter power device.

Step 17, energy storage type quasi-Z source photovoltaic inverter loss estimation

For the circuit shown in Figure 1, there are 4 IGBTs (and their anti-parallel diodes), a Z-source network diode, and their losses can be calculated separately according to the device.

1) Loss estimation for 4 IGBTs (and their anti-parallel diodes)

This part of the loss can be divided into the loss in the traditional sense and the loss caused by the pass-through. The loss in the traditional sense includes switching loss and conduction loss, and the switching loss is

${\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; SW</span>}}<span\; class="notranslate">\; =</span>\frac{\mathrm{<span\; class="notranslate">\; 4</span>}}{\mathrm{<span\; class="notranslate">\; \&pi;</span>}}{\mathrm{<span\; class="notranslate">\; f</span>}}_{\mathrm{<span\; class="notranslate">\; the\; s</span>}}<span\; class="notranslate">\; \&Center\; Dot;</span><span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; ON</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; I</span>}}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; OFF</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; I</span>}}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; OFF</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; D.</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; \&Center\; Dot;</span>\frac{{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; PN</span>}}}{{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; ref</span>}}}<span\; class="notranslate">\; \&Center\; Dot;</span>\frac{{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}}{{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; ref</span>}}}$

In the formula, E _{ON, I} , E _{off, I} , E _{OFF, D} are the turn-on loss, turn-off loss and diode reverse recovery loss energy of the power device at the voltage V _ref and current i _ref respectively; i _L is the load current Peak value, i _L =i _a *1.414A, f _s is the switching frequency. According to the voltage and current parameters calculated in steps 4 and 16, select the IGBT module SGH30N60RUFD as the inverter power switch, and calculate its loss. According to the data provided by the device, when V _ref = 300V, i _ref = 30A, the switching loss energy during turn-on is E _{ON, I} = 0.919mJ/P, and the switching loss energy during turn-off is E _{off, I} = 0.814mJ /P, reverse recovery switching loss energy is E _{OFF, D} =0.067mJ/P. i _L =14.1*1.414=19.94A, V _PN =255V, f _s =10kHz, then calculate P _SW =12.9W.

For the conduction loss of each IGBT in the traditional sense, it is calculated according to the following formula

${\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; cv</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; I</span>}}<span\; class="notranslate">\; =</span>\frac{{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; CE</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}}{\mathrm{<span\; class="notranslate">\; 2</span>}\mathrm{<span\; class="notranslate">\; \&pi;</span>}}<span\; class="notranslate">\; \&Center\; Dot;</span>{<span\; class="notranslate">\; \&Integral;</span>}_{\mathrm{<span\; class="notranslate">\; 0</span>}}^{\mathrm{<span\; class="notranslate">\; \&pi;</span>}}\mathrm{<span\; class="notranslate">\; sin</span>}\mathrm{<span\; class="notranslate">\; \&omega;t</span>}<span\; class="notranslate">\; \&CenterDot;</span><span\; class="notranslate">\; (</span>\frac{\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; +</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; t</span>}<span\; class="notranslate">\; )</span>}{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; -</span>\frac{\mathrm{<span\; class="notranslate">\; D.</span>}}{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; \&CenterDot;</span>\mathrm{<span\; class="notranslate">\; d\&omega;t</span>}<span\; class="notranslate">\; +</span>\frac{{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; CE</span>}}{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{\mathrm{<span\; class="notranslate">\; 2</span>}}}{\mathrm{<span\; class="notranslate">\; 2</span>}\mathrm{<span\; class="notranslate">\; \&pi;</span>}}<span\; class="notranslate">\; \&CenterDot;</span>{<span\; class="notranslate">\; \&Integral;</span>}_{\mathrm{<span\; class="notranslate">\; 0</span>}}^{\mathrm{<span\; class="notranslate">\; \&pi;</span>}}{\mathrm{<span\; class="notranslate">\; sin</span>}}^{\mathrm{<span\; class="notranslate">\; 2</span>}}\mathrm{<span\; class="notranslate">\; \&omega;t</span>}<span\; class="notranslate">\; \&CenterDot;</span><span\; class="notranslate">\; (</span>\frac{\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; +</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; t</span>}<span\; class="notranslate">\; )</span>}{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; -</span>\frac{\mathrm{<span\; class="notranslate">\; D.</span>}}{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; \&CenterDot;</span>\mathrm{<span\; class="notranslate">\; d\&omega;t</span>}$

$<span\; class="notranslate">\; =</span>\frac{{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; CE</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}}{\mathrm{<span\; class="notranslate">\; 2</span>}\mathrm{<span\; class="notranslate">\; \&pi;</span>}}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; +</span>\frac{\mathrm{<span\; class="notranslate">\; \&pi;M</span>}}{\mathrm{<span\; class="notranslate">\; 4</span>}}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; +</span>\frac{{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; CE</span>}}{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{\mathrm{<span\; class="notranslate">\; 2</span>}}}{\mathrm{<span\; class="notranslate">\; 2</span>}\mathrm{<span\; class="notranslate">\; \&pi;</span>}}<span\; class="notranslate">\; (</span>\frac{\mathrm{<span\; class="notranslate">\; \&pi;</span>}}{\mathrm{<span\; class="notranslate">\; 4</span>}}<span\; class="notranslate">\; +</span>\frac{\mathrm{<span\; class="notranslate">\; 2</span>}\mathrm{<span\; class="notranslate">\; m</span>}}{\mathrm{<span\; class="notranslate">\; 3</span>}}<span\; class="notranslate">\; -</span>\frac{\mathrm{<span\; class="notranslate">\; \&pi;D</span>}}{\mathrm{<span\; class="notranslate">\; 4</span>}}<span\; class="notranslate">\; )</span>$

${\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; cv</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; D.</span>}}<span\; class="notranslate">\; =</span>\frac{{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; f</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; 0</span>}}{<span\; class="notranslate">\; \&CenterDot;</span>\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}}{\mathrm{<span\; class="notranslate">\; 2</span>}\mathrm{<span\; class="notranslate">\; \&pi;</span>}}<span\; class="notranslate">\; \&CenterDot;</span>{<span\; class="notranslate">\; \&Integral;</span>}_{\mathrm{<span\; class="notranslate">\; 0</span>}}^{\mathrm{<span\; class="notranslate">\; \&pi;</span>}}\mathrm{<span\; class="notranslate">\; sin</span>}\mathrm{<span\; class="notranslate">\; \&omega;t</span>}<span\; class="notranslate">\; \&CenterDot;</span><span\; class="notranslate">\; (</span>\frac{\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; t</span>}<span\; class="notranslate">\; )</span>}{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; -</span>\frac{\mathrm{<span\; class="notranslate">\; D.</span>}}{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; \&CenterDot;</span>\mathrm{<span\; class="notranslate">\; d\&omega;t</span>}<span\; class="notranslate">\; +</span>\frac{{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; f</span>}}{<span\; class="notranslate">\; \&CenterDot;</span>\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{\mathrm{<span\; class="notranslate">\; 2</span>}}}{\mathrm{<span\; class="notranslate">\; 2</span>}\mathrm{<span\; class="notranslate">\; \&pi;</span>}}<span\; class="notranslate">\; \&CenterDot;</span>{<span\; class="notranslate">\; \&Integral;</span>}_{\mathrm{<span\; class="notranslate">\; 0</span>}}^{\mathrm{<span\; class="notranslate">\; \&pi;</span>}}{\mathrm{<span\; class="notranslate">\; sin</span>}}^{\mathrm{<span\; class="notranslate">\; 2</span>}}\mathrm{<span\; class="notranslate">\; \&omega;t</span>}<span\; class="notranslate">\; \&CenterDot;</span><span\; class="notranslate">\; (</span>\frac{\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; t</span>}<span\; class="notranslate">\; )</span>}{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; -</span>\frac{\mathrm{<span\; class="notranslate">\; D.</span>}}{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; \&Center\; Dot;</span>\mathrm{<span\; class="notranslate">\; d\&omega;t</span>}$

$<span\; class="notranslate">\; =</span>\frac{{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; f</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; \&Center\; Dot;</span>{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}}{\mathrm{<span\; class="notranslate">\; 2</span>}\mathrm{<span\; class="notranslate">\; \&pi;</span>}}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; -</span>\frac{\mathrm{<span\; class="notranslate">\; \&pi;M</span>}}{\mathrm{<span\; class="notranslate">\; 4</span>}}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; +</span>\frac{{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; f</span>}}<span\; class="notranslate">\; \&Center\; Dot;</span>{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{\mathrm{<span\; class="notranslate">\; 2</span>}}}{\mathrm{<span\; class="notranslate">\; 2</span>}\mathrm{<span\; class="notranslate">\; \&pi;</span>}}<span\; class="notranslate">\; (</span>\frac{\mathrm{<span\; class="notranslate">\; \&pi;</span>}}{\mathrm{<span\; class="notranslate">\; 4</span>}}<span\; class="notranslate">\; -</span>\frac{\mathrm{<span\; class="notranslate">\; 2</span>}\mathrm{<span\; class="notranslate">\; m</span>}}{\mathrm{<span\; class="notranslate">\; 3</span>}}<span\; class="notranslate">\; -</span>\frac{\mathrm{<span\; class="notranslate">\; \&pi;D</span>}}{\mathrm{<span\; class="notranslate">\; 4</span>}}<span\; class="notranslate">\; )</span>$

In the formula, V _CE0 , V _F,0 are the saturation voltage drop and conduction voltage drop of IGBT and diode; r _CE , r _F are the on-resistance of IGBT and diode respectively; M and D are the modulation index and direct duty Compare. Since V _CE0 =2.2V, V _{F, 0} =1.3V, r _CE =0.02Ω, r _F =0.01Ω, M=2/3, D=1/3, the conduction loss of 4 IGBTs and their diodes for

P _CV =4*(P _CV，I +P _CV，D )=45.8W

For the loss caused by the shoot-through, it is only necessary to calculate the switching loss and conduction loss from the IGBT, that is, the switching loss is

${\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; SW</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; SH</span>}}<span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 4</span>}{\mathrm{<span\; class="notranslate">\; f</span>}}_{\mathrm{<span\; class="notranslate">\; the\; s</span>}}<span\; class="notranslate">\; \&Center\; Dot;</span><span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; ON</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; I</span>}}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; OFF</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; I</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; \&Center\; Dot;</span>\frac{{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; PN</span>}}}{{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; ref</span>}}}<span\; class="notranslate">\; \&Center\; Dot;</span>\frac{{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}}{{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; ref</span>}}}<span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 19.6</span>}\mathrm{<span\; class="notranslate">\; W</span>}$

The conduction loss is

${\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; cv</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; SH</span>}}<span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 4</span>}<span\; class="notranslate">\; *</span><span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; CE</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; CE</span>}}{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}^{\mathrm{<span\; class="notranslate">\; 2</span>}}\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; +</span>\frac{{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; CE</span>}}{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{\mathrm{<span\; class="notranslate">\; 2</span>}}\mathrm{<span\; class="notranslate">\; D.</span>}}{\mathrm{<span\; class="notranslate">\; 8</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 6.2</span>}\mathrm{<span\; class="notranslate">\; W</span>}$

2) Loss of Z-source network diode

The Z-Source network diode conduction loss of each module is

${\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; cv</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; D.</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; DF</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; [</span>{\mathrm{<span\; class="notranslate">\; 2</span>}\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; -</span>\frac{{\mathrm{<span\; class="notranslate">\; Mi</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}}{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; DF</span>}}\frac{\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; D.</span>}}{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; [</span>{\mathrm{<span\; class="notranslate">\; 8</span>}\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}^{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; DF</span>}}\frac{\mathrm{<span\; class="notranslate">\; 8</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; )</span>}{\mathrm{<span\; class="notranslate">\; \&pi;</span>}}{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}$

Its reverse recovery loss is

${\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; cv</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; DR</span>}}<span\; class="notranslate">\; =</span>\frac{\mathrm{<span\; class="notranslate">\; 2</span>}{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; rec</span>}}{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; PN</span>}}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; \&pi;i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; i</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}<span\; class="notranslate">\; )</span>}{\mathrm{<span\; class="notranslate">\; \&pi;</span>}{\mathrm{<span\; class="notranslate">\; I</span>}}_{\mathrm{<span\; class="notranslate">\; FM</span>}}{\mathrm{<span\; class="notranslate">\; V</span>}}_{\mathrm{<span\; class="notranslate">\; R</span>}}}\mathrm{<span\; class="notranslate">\; 2</span>}{\mathrm{<span\; class="notranslate">\; f</span>}}_{\mathrm{<span\; class="notranslate">\; the\; s</span>}}$

According to the diode parameters calculated in step 15, select APT40DQ60B as the Z-source network diode, when I _FM =30A, V _R =600V, the reverse recovery loss energy E _rec =0.06mJ/P, V _PN =255V, V _DF0 = 1.7V. Then P _{CV, D} = 14.7W, P _{CV, DR} = 0.124W.

The total loss is P _SW + P _CV + P _{SW, SH} + P _{CV, SH} + P _{CV, D} + P _{CV, DR} = 118.2W.

By estimating losses, alternative devices can be compared as one of the criteria for optimal device selection.

It can be seen from the above embodiments that the main parameters of the energy storage type quasi-Z source photovoltaic power generation system can be effectively designed according to the present invention. In the designed system, when the voltage of the photovoltaic cell is low, the circuit boosts the voltage to meet the load requirements; when the voltage of the photovoltaic cell is high, it can meet the requirements without boosting the voltage; at night, the energy storage battery can directly provide energy to the load. load. The whole system is implemented with a single-stage power circuit, which has the simplest structure and low cost. And the system is suitable for independent photovoltaic power generation, also suitable for grid-connected photovoltaic power generation.

As mentioned above, although the Example of this invention was demonstrated in detail, it is obvious to those skilled in the art that many modifications can be made as long as the inventive point and effect of this invention are not substantially deviated. Therefore, all such modified examples are also included in the protection scope of the present invention.

Claims (7)

1. A design method for an energy storage type quasi-Z source single-phase photovoltaic power generation system, the energy storage type quasi-Z source single phase photovoltaic power generation system includes: energy storage battery, H bridge inverter, quasi-Z source network diode , a first electrolytic capacitor, a second electrolytic capacitor, a first inductor, a second inductor, an LC filter, a photovoltaic cell, a power grid, and a local load; the LC filter includes an output filter inductor and an output filter capacitor; and, the The negative pole of the second electrolytic capacitor is connected to the anode of the quasi-Z source network diode, and the positive pole of the second electrolytic capacitor is connected to the positive pole of the H-bridge inverter; the cathode of the quasi-Z source network diode is simultaneously It is connected to the positive pole of the first electrolytic capacitor and the second inductor; the other end of the second inductor is connected to the positive pole of the H-bridge inverter; the negative pole of the first electrolytic capacitor is connected to the H-bridge inverter connected to the negative pole of the device; one end of the first inductor is connected to the positive pole of the photovoltaic cell; the other end of the first inductor is connected to the negative pole of the second electrolytic capacitor; the output of the H-bridge inverter passes through The LC filter is connected to the power grid, or supplies local loads; the energy storage battery is connected across the two ends of the first electrolytic capacitor, and the positive pole of the energy storage battery is connected to the positive pole of the first electrolytic capacitor; It is characterized in that, based on the user's voltage and power requirements, the purpose of the photovoltaic power generation system, and the local climate characteristics, the method includes the design of the energy storage battery voltage and capacity parameters required by the system, the selection of photovoltaic battery modules, and the quasi-Z source network inductance and capacitance. Parameter design, H-bridge inverter voltage and current level design, Z-source network diode design, quasi-Z source inverter loss calculation; Among them, the H-bridge inverter voltage and current level design includes the following steps: Step 1, calculate the voltage amplitude v _al output by the single-phase inverter according to the load voltage or the grid voltage at the grid connection point; Step 2, set the operating voltage range of the photovoltaic cell to be 1:2, and the maximum photovoltaic cell voltage is V _in =v _al , and the minimum is V _in =v _al /2. Correspondingly, when the photovoltaic cell voltage is maximum, the inverter Modulation index M=1, direct duty cycle D=0, peak value of DC bus voltage is V _PN =v _al , when photovoltaic cell voltage is minimum, direct duty cycle D=1/3, inverter modulation index M=2/ 3. The peak value of DC bus voltage is V _PN =1.5*v _al . 2. a kind of design method of energy storage type quasi-Z source single-phase photovoltaic power generation system as claimed in claim 1, is characterized in that, described capacitance, inductance parameter design comprise: Step 3, calculate the voltage of the first electrolytic capacitor C1 and the voltage of the parallel energy storage battery V _C1 =V _B = _val ; Step 4, calculate the maximum peak value of the DC bus voltage as V _PN =1.5×v _al ; Step 5, calculating the maximum voltage value of the second electrolytic capacitor C ₂ as V _C2 =0.5×v _al ; Step 6, the maximum direct duty cycle D=1/3. 3. a kind of design method of energy storage type quasi-Z source single-phase photovoltaic power generation system as claimed in claim 1, is characterized in that, described photovoltaic cell module selection comprises: Step 7, according to the load or grid-connected power P _o , considering that the maximum power occurs at the maximum grid voltage, the photovoltaic cell current is Step 8, the discharge of the energy storage battery occurs when the photovoltaic power is insufficient. According to the above design, the minimum operating voltage of the photovoltaic battery is v _al /2. When the minimum light intensity allowed to work is set, the current provided by the photovoltaic battery at the minimum power is the maximum power. One part of k; when P _o power is to be output, the energy storage battery needs to provide power as P _B =P _o -0.5×v _al ×i _L1 /k, then the energy storage battery current is I _B =P _B /V _B ; Then the current of the second inductor L ₂ is i _L2 =i _L1 +i _B . 4. A kind of design method of energy storage type quasi-Z source single-phase photovoltaic power generation system as claimed in claim 1, is characterized in that, described energy storage battery voltage and capacity parameter design comprise: Step 9. Determine the capacity of the energy storage battery, apply statistical data, and determine the capacity of the energy storage battery according to the local climate characteristics and the purpose of the designed system; In addition to the local load/grid, the battery is charged for x hours on average with the power of P _x every day for use when there is no sunshine at night, and the discharge depth of the energy storage battery is y% each time, then the capacity of the energy storage battery is required to be [P _x ×x/v _al ]/(1-y%)(Ah). 5. The design method of a kind of energy storage type quasi-Z source single-phase photovoltaic power generation system as claimed in claim 1, is characterized in that, described photovoltaic cell module selection comprises: Step 10, determine the photovoltaic cell module and its quantity, select the photovoltaic cell module, and determine the maximum operating voltage _vpv and The maximum operating current i _pv , then the number of photovoltaic cells can be calculated as Take an integer to get n, Take an integer to get m, and the number of photovoltaic cell modules is m*n. 6. A kind of design method of energy storage type quasi-Z source single-phase photovoltaic power generation system as claimed in claim 1, it is characterized in that, described capacitor inductance parameter design comprises: Step 11, the second electrolytic capacitor _C2 is designed, and the system adopts a constant direct-through zero-vector modulation method, then the pulse frequency of the second electrolytic capacitor voltage on the quasi-Z source network is 2f _s ; during steady state, the second electrolytic capacitor within one cycle The initial value and final value of the voltage are equal; in the case of straight-through, the ripple ΔV _C2 of the voltage of the second electrolytic capacitor C ₂ of the quasi-Z source network is in, -i _L1 ＝i _C2 , f _s is the carrier frequency, so we have If the ripple voltage of the second electrolytic capacitor is given as ΔV _C2 ≤ αV _C2 , then In order to suppress double-frequency voltage ripple, the second electrolytic capacitor required is In the formula, ε is the ratio of double-frequency voltage ripple to V _PN , f is the load or grid voltage frequency, and the second electrolytic capacitor C required is greater than And the first electrolytic capacitor _C1 is connected in parallel with the energy storage battery, so the second electrolytic capacitor _C2 is designed as $C_2=\frac{P_o}{4\mathrm{\&pi;f}{V}_{\mathrm{PN}}^2\mathrm{\&epsiv;}};$ Step 12, the first electrolytic capacitor _C1 is designed, and the system adopts a constant through zero vector modulation method, then the pulsation frequency of the first electrolytic capacitor voltage on the quasi-Z source network is 2f _s ; during steady state, the first electrolytic capacitor within one cycle The initial value and final value of the voltage are equal, and the ripple ΔV _C1 of the voltage of the first electrolytic capacitor C ₁ of the quasi-Z source network is in, i _C1 ＝i _B -i _L2 , f _s is the carrier frequency, so we have If the ripple of the given first electrolytic capacitor voltage is ΔV _C1 ≤ αV _C1 , then there is $C_1\&Greater\; Equal;(i_B-i_{L2})\frac{\mathrm{D.}}{2f_{\mathrm{the\; s}}\mathrm{\&alpha;}{V}_{C1}};$ Step 13, the design of the second inductance _L2 , in the steady state, the initial value and the final value of the second inductance current in one cycle are equal, and the ripple Δi _L2 of the second inductance current in the quasi-Z source network is $\mathrm{\&Delta;}{i}_{L2}=v_{L2}\frac{\mathrm{\&Delta;t}}{L_2},$ in, $\mathrm{\&Delta;t}=\frac{\mathrm{D.}}{2f_{\mathrm{the\; s}}},$ v _L2 =V _C1 , so we have $L_2=V_{C1}\frac{\mathrm{D.}}{2f_{\mathrm{the\; s}}\mathrm{\&Delta;}{i}_{L2}},$ If the ripple of the given second inductor current is Δi _L2 ≤ bi _L2 , then Step 14, the design of the first inductance _L1 , in the steady state, the initial value and the final value of the first inductance current in one cycle are equal, and in the straight-through state, the ripple ΔiL1 of the first inductance current in the quasi-Z source network is in, V _in +V _C2 ＝v _L1 , so we have If the ripple of the given first inductor current is Δi _L1 ≤ bi _L1 , then $L_1\&Greater\; Equal;(V_{\mathrm{in}}+V_{C2})\frac{\mathrm{D.}}{2f_{\mathrm{the\; s}}b{i}_{L1}};$ Step 15, the parameters of the quasi-Z source network diode, the quasi-Z source network diode is turned off under the back pressure in the straight-through state, and is turned on in the non-through state. In the direct state, the current flowing through it is used to design the diode; in the straight-through state, the quasi-Z source network diode withstands the back pressure as VPN, and in the non-straight-through state, the current passing through the quasi-Z source network diode is i _D ≤ i _L1 +i _C2 = i _L1 +i _L2 -i _d , in the traditional zero vector time, when i _d =0, the maximum current flowing through the quasi-Z source network diode is i _D =i _L1 +i _L2 ; Step 16, based on the parameters of the inverter power device, according to the load or grid-connected power P _o , and the load or grid-connected grid voltage, calculate the effective value of the load current The peak value V _PN of the DC bus voltage calculated in step 4 and the effective value of the load current calculated above are used as the selected voltage and current parameters of the inverter power device. 7. A kind of design method of energy storage type quasi-Z source single-phase photovoltaic power generation system as claimed in claim 1, is characterized in that, described quasi-Z source inverter loss calculation comprises: Step 17, energy storage type quasi-Z source single-phase photovoltaic inverter loss estimation, there are 4 IGBTs and their anti-parallel diodes, a quasi-Z source network diode, the loss is calculated separately according to the device; 1) Loss estimation of 4 IGBTs and their anti-parallel diodes. This part of the loss can be divided into loss in the traditional sense and loss caused by direct connection. The loss in the traditional sense includes switching loss and conduction loss, and the switching loss is $P_{\mathrm{SW}}=\frac{4}{\mathrm{\&pi;}}{f}_{\mathrm{the\; s}}\&Center\; Dot;({\mathrm{E.}}_{\mathrm{ON},I}+{\mathrm{E.}}_{\mathrm{OFF},I}+{\mathrm{E.}}_{\mathrm{OFF},\mathrm{D.}})\&CenterDot;\frac{V_{\mathrm{PN}}}{V_{\mathrm{ref}}}\&Center\; Dot;\frac{i_L}{i_{\mathrm{ref}}},$ In the formula, E _ON,I , E _off,I , E _OFF,D are the turn-on loss, turn-off loss and diode reverse recovery loss energy of the power device at the voltage V _ref and current i _ref respectively, which can be obtained from the device manual ; i _L is the peak value of the load current, i _L =1.414*i _a , f _s is the switching frequency; for the conduction loss of each IGBT in the traditional sense, it is calculated according to the following formula $P_{\mathrm{cv},I}=\frac{V_{\mathrm{CE},0}{i}_L}{2\mathrm{\&pi;}}(1+\frac{\mathrm{\&pi;M}}{4}-\mathrm{D.})+\frac{r_{\mathrm{CE}}{i}_L^2}{2\mathrm{\&pi;}}(\frac{\mathrm{\&pi;}}{4}+\frac{2m}{3}-\frac{\mathrm{\&pi;D}}{4})$ $P_{\mathrm{cv},\mathrm{D.}}=\frac{V_{f,0}{i}_L}{2\mathrm{\&pi;}}(1+\frac{\mathrm{\&pi;M}}{4}-\mathrm{D.})+\frac{r_f{i}_L^2}{2\mathrm{\&pi;}}(\frac{\mathrm{\&pi;}}{4}-\frac{2m}{3}-\frac{\mathrm{\&pi;D}}{4}),$ In the formula, V _CE0 , V _F,0 are the saturation voltage drop and conduction voltage drop of IGBT and diode; rCE, rF are the on-resistance of IGBT and diode; M and D are modulation index and direct duty cycle respectively, then The conduction loss of 4 IGBTs and their diodes is P _CV =4*(P _CV,I +P _CV,D ), for the loss caused by the straight-through, it is only necessary to calculate the switching loss and conduction loss from the IGBT, that is, the switching loss for $P_{\mathrm{SW},\mathrm{SH}}=\frac{4}{\mathrm{\&pi;}}{f}_{\mathrm{the\; s}}\&CenterDot;({\mathrm{E.}}_{\mathrm{ON},I}+{\mathrm{E.}}_{\mathrm{OFF},I}+{\mathrm{E.}}_{\mathrm{OFF},\mathrm{D.}})\&CenterDot;\frac{V_{\mathrm{PN}}}{V_{\mathrm{ref}}}\&CenterDot;\frac{i_L}{i_{\mathrm{ref}}},$ The conduction loss is $P_{\mathrm{cv},\mathrm{SH}}=4*(V_{\mathrm{CE},0}{i}_{L1}\mathrm{D.}+r_{\mathrm{CE}}{i}_{L1}^2\mathrm{D.}+\frac{r_{\mathrm{CE}}{i}_L^2\mathrm{D.}}{8})$ 2) The loss of the quasi-Z source network diode, the conduction loss of the quasi-Z source network diode of each module is $P_{\mathrm{cv},\mathrm{D.}}=V_{\mathrm{DF},0}[{2i}_{L1}(1-\mathrm{D.})-\frac{{\mathrm{Mi}}_L}{2})+r_{\mathrm{DF}}\frac{1-\mathrm{D.}}{2}[{8i}_{L1}^2+i_L^2]-r_{\mathrm{DF}}\frac{8(1-\mathrm{D.})}{\mathrm{\&pi;}}{i}_L{i}_{L1},$ In the formula, V _{DF, 0} is the conduction voltage drop of the quasi-Z source network diode; r _DF is the conduction resistance of the quasi-Z source network diode, and its reverse recovery loss is In the formula, E _rec is the reverse recovery loss energy of the quasi-Z source network diode at the time of I _FM and _VR ; the total loss is P _SW +P _CV +P _SW,SH +P _CV,SH +P _CV,D + P _CV,DR , by estimating the loss, alternative devices can be compared as a condition for optimizing device selection.