Dual-source inverter for hybrid PV–FC application

The proposed structure is shown in Fig. 1. Power can flows through both Z-impedance networks and supply the load. The proportion of power generated from any of the sources can be determined by shoot through duty cycles of two Z-source converters (Dst1 and Dst2).

Schematic diagram of the proposed dual Z-source converter

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The Main advantages of the proposed converter are as follows:

• Ability to control the power sharing between two sources by adjusting the switching mode, the amount and phase shift of the shoot through times.

• High DC-link voltage applied to the inverter, not requiring extreme values of Dst.

• Power sharing by two Impedance networks without any extra elements.

• Simple switching methods.

2.1

Circuit description

As shown in Fig. 1, instead of two inductors, Z-Source 1 contains two transformers in the structure, which the secondary voltages (Vsec1 and Vsec2) are in series with Z-Source 2, and therefore DC-link voltage of the inverter is equals to VX+ Vsec1+ Vsec2. The required inductors of Z-Source 1 are the magnetizing inductances of the transformers. Two switching methods which will be explained later, shows expect in optimum switching method. The output are three level signal and it is required that an small LC filter be placed in the output for bypassing the high frequency harmonics.

2.2

The steady state operation analysis

Due to the circuit symmetry, circuit analysis is performed for a part of the circuit. Obviously, the other part have the same behavior.

Regarding the structure of the system, Z-Source 1 works similar to the traditional Z-source converter [20] and therefore the relations for the capacitor voltages (VC1 and VC2) and the output voltage (VO1) are the same:

$$\frac{{V_{C1} }}{{V_{i} }} = \frac{{1 - D_{st1} }}{{1 - 2D_{st1} }}$$

(1)

$$\frac{{V_{O1} }}{{V_{i1} }} = \frac{1}{{1 - 2D_{st1} }}$$

(2)

where \(D_{st1}\) is the duty cycle for shoot through state of Z-Source 1. Considering the voltage drops across L1, secondary voltage of the transformer Vsec1 is positive while S1 is ON (shoot-through state for Z-Source 1) as VC1 is applied to the primary winding of the transformer (similar for Vsec2 and VC2). During this time interval (equals to Dst1T), Vsec1 is:

$$V_{sec1} = \frac{{N_{2} }}{{N_{1} }}V_{C1}$$

(3)

During non-shoot through time (S1 is off), the secondary voltage is negative (because of volt-second law for the magnetizing inductors of the transformers) and is equal to:

$$V_{sec1} = \frac{{N_{2} }}{{N_{1} }}\left( {V_{i1} - V_{O1} } \right)$$

(4)

According to the values of Dst1 and Dst2 and time position relative to each other, two operation modes is defined as shown in Fig. 2. In the following, the operation principles of the converter in both modes are described.

Possible shoot through states for \(D_{st1}\) and \(D_{st2}\). a Mode I, b Mode II

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2.2.1

Mode I

In this mode of operation, two Z-Source networks goes to shoot through state, simultaneously. But shoot through times may differ (Fig. 2a). During \((D_{st1} - D_{st2} )T\) (which is subinterval 2), diode D2 may be conducting or not, which depends on the inputs voltage level, the output current level and the duty cycles values (Dst1 and Dst2). In the next, it is supposed that all symmetric elements have same values and behavior (such as transformers turn ratio (n) or \(L_{3}\) and \(L_{4}\) or …). Figure 3 Shows the equivalent circuits for all three states of the converter in Mode I and actual current directions (red dashed lines) in all the circuit branches.

The equivalent circuits of the converter in Mode Ӏ for a \(D_{st2} T\) (subinterval 1) b \(\left( {D_{st1} - D_{st2} } \right)T\) (subinterval 2) c \((1 - D_{st1} )T\) (subinterval 3)

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According to Fig. 3a, in \(D_{st2} T\) (subinterval 1), because both Z-sources are in shoot through state, so \(V_{L3} = V_{C3} + 2V_{sec1} = V_{C3} + 2nV_{C1}\), \(V_{O2}^{1} = 0\). It is clear that \(L_{3}\) charges with the value is larger than \(V_{C1}\) (in compared to classic Z-source) and so when Z-source 2 inters in non-shoot through state it produces the value larger than the inductor voltage in classic Z-source.

When the converter is in subinterval 2 (\((D_{st1} - D_{st2} )T\)), \(V_{O2}^{2} = V_{i2} - 2V_{L3} + 2V_{sec1} = V_{i2} - 2V_{L3} + 2nV_{C1}\) and when the converter is in subinterval 3 (\(\left( {1 - D_{st1} } \right)T\)), \(V_{O2}^{3} = V_{i2} - 2V_{L3} + 2V_{sec1} = V_{i2} - 2V_{L3} + 2(V_{i1} - V_{C1} )\). Because \(V_{i1} - V_{C1} < V_{C1}\) so \(V_{O2}^{2} > V_{O2}^{3}\). In other words, \(V_{O2}\) is a three level signal which is zero when Z-source 2 is in shoot through state (subinterval 1) and have non-zero values in other states (subinterval 2 and 3). Figure 4 illustrates \(V_{O2}\) with the related shoot through duty cycles for the converter. In order to eliminate flowing the high order sinusoidal voltage harmonics, an LC filter is required in the output of the inverter.

The schematic waveform of \(V_{O2}\) for Mode I

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In the following, the values of the important variables in this Mode will be calculated. Employing volt-second law for inductor L3 yields:

$$\left( {2nV_{C1} + V_{C3} } \right)D_{st2} + \left( {V_{i2} - V_{C3} } \right)(D_{st1} - D_{st2} )T + \left( {V_{i2} - V_{C3} } \right)\left( {1 - D_{st1} } \right)T = 0$$

(5)

Simplifying (5) and implying kvl for the circuits result:

$$V_{C3} = V_{i1} + \frac{{\left( {1 - D_{st2} } \right) \cdot \left( {1 - 2D_{st1} } \right)V_{i2} }}{{2nD_{st2} \left( {1 - D_{st1} } \right)}}$$

(6)

$$V_{O2}^{2} = V_{i1} + \left( {\frac{{D_{st2} \left( {1 - 2D_{st1} } \right)}}{{2n\left( {1 - D_{st1} } \right)\left( {1 - 2D_{st2} } \right)}}} \right)V_{i2}$$

(7)

$$V_{O2}^{3} = V_{i1} + \left( {\frac{{D_{st2} \left( {1 - 2D_{st1} } \right)}}{{2n(D_{st1} D_{st2} - D_{st1} + D_{st2} )}}} \right)V_{i2}$$

(8)

where \(V_{O2}^{2}\) and \(V_{O2}^{3}\) are the DC-link voltage during subintervals 2 and 3, respectively.

In order to analyze the power sharing between two networks, the current relations of the system have to be derived in the following. Considering that in the steady state operation, the average currents of the capacitors are zero, the average values of the currents can be calculated as follows:

$$I_{L3} = I_{sec1}$$

(9)

$$I_{i1} = I_{L1} + nI_{sec1} = I_{L1} + nI_{L3}$$

(10)

$$I_{i2} = I_{L3}$$

(11)

$$I_{O1} = nI_{sec1} + I_{L1} = nI_{L3} + I_{L1}$$

(12)

According to the circuit operation in this Mode; \(I_{sec1}\) can be expressed as follow:

$$I_{sec1} = 2I_{L3} .D_{st2} + I_{Inv}^{2} .(D_{st1} - D_{st2} ) + I_{Inv}^{3} .(1 - D_{st1} )$$

(13)

where \(I_{Inv}^{2}\) and \(I_{Inv}^{3}\) are the average input values of the inverter current during subinterval 2 and 3, respectively. Employing (9) and (13) yields:

$$I_{L3} = I_{Inv}^{2} + I_{Inv}^{3} \cdot \left( {\frac{{1 - D_{st1} }}{{D_{st1} - D_{st2} }}} \right)$$

(14)

Again for \(I_{O1}\):

$$\begin{aligned} I_{O1} & = 2I_{L1} .D_{st1} + 2nI_{sec1}^{1} \cdot D_{st2} \\ & \quad + 2nI_{sec1}^{2} \cdot (D_{st1} - D_{st2} ) \\ \end{aligned}$$

(15)

In (15), \(I_{sec1}^{1}\) and \(I_{sec1}^{2}\) imply to the average current values of the transformers secondary windings in the subintervals 1 and 2, respectively. They are equal to \(2I_{L3}\) and \(I_{Inv}^{2}\), respectively. So (15) can be rewrote as follow:

$$\begin{aligned} I_{O1} & = 2I_{L1} \cdot D_{st1} + 4nI_{L3} \cdot D_{st2} \\ & \quad + 2nI_{Inv}^{2} \cdot (D_{st1} - D_{st2} ) \\ \end{aligned}$$

(16)

Considering (12), (14) and (16) shows:

$$I_{L1} = I_{Inv}^{2} + I_{Inv}^{3} \cdot \left( {\frac{{\left( {4D_{st2} - 1} \right) \cdot (1 - D_{st1} )}}{{D_{st1} - D_{st2} }}} \right)$$

(17)

Replacing (13) and (17) in (10) results:

$$I_{i1} = I_{Inv}^{2} + I_{Inv}^{3} \cdot \left( {\frac{{\left( {1 - D_{st1} } \right)\left( {2D_{st2} - D_{st1} } \right)}}{{\left( {D_{st1} - D_{st2} } \right)\left( {1 - D_{st1} - D_{st2} } \right)}}} \right)$$

(18)

Replacing (14) in (11) results:

$$I_{i2} = I_{Inv}^{2} + I_{Inv}^{3} .\frac{{1 - D_{st1} }}{{D_{st1} - D_{st2} }}$$

(19)

The values of active power produced by each sources can be calculated by multiplying the value of any source voltage value to the average value of its current by using (18) and (19). Therefore, active power of source 1 (\(P_{z1}\)) and source 2 (\(P_{z2}\)) are calculated as follow:

$$P_{z1} = V_{i1} \cdot \left( {\begin{array}{*{20}c} {I_{Inv}^{2} } \\ { + I_{Inv}^{3} \cdot \left( {\frac{{\left( {1 - D_{st1} } \right)\left( {2D_{st2} - D_{st1} } \right)}}{{\left( {D_{st1} - D_{st2} } \right)\left( {1 - D_{st1} - D_{st2} } \right)}}} \right)} \\ \end{array} } \right)$$

(20)

$$P_{z2} = V_{i2} \cdot \left( {I_{Inv}^{2} + I_{Inv}^{3} \cdot \frac{{1 - D_{st1} }}{{D_{st1} - D_{st2} }}} \right)$$

(21)

Obviously, the total output power is \(P_{z1}\) + \(P_{z2}\), so:

$$P_{out} = P_{z1} + P_{z2}$$

(22)

When the proposed topology operates in Mode I, via controlling \(D_{st1}\) and \(D_{st2}\), \(V_{O2}\) and the values of the active power flowed to the load can be controlled. Now, for evaluation the operation of the converter from power sharing viewpoint, the curve showing \(\frac{{P_{z1} }}{{P_{out} }}\) for \(V_{i1} = V_{i2}\) for different values of \(D_{st1}\) and \(D_{st2}\) is drawn.

Figure 5 shows an increase in \(D_{st1}\) or a decrease in \(D_{st2}\) causes increasing The ratio of \(P_{z1}\) to \(P_{out}\). Generally, when the system operates, the main variable which have to be controlled are the values of \(V_{O2}\) and \(P_{z1}\). This control algorithm is performed by controller. So the controller has to be solved the (7), (8) and (20) to find out the sufficient values of \(D_{st1}\) and \(D_{st2}\). That; (7) and (8) have to be calculated for achieving \(V_{O2}\) is performed by taking average from \(V_{O2}\). In other words, after filtering, the output applied to the load, is the average value of \(V_{O2}\) and so the controller has to calculate the required average value of \(V_{O2}\) by solving (7) and (8), simultaneously. Also, it can be said when \(P_{z1}\) is chose, \(P_{z2}\) can be calculated from \(P_{z2} = P_{out} - P_{z1}\).

The ratio of \(P_{z1}\) to \(P_{out}\) in Mode Ӏ for \(V_{i1} = V_{i2} , n = 1\)

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If in the converter \(D_{st1} = D_{st2}\), subinterval 2 will be eliminated, the output value will be a two level signal as in classic Z-source inverter and the non-zero level of it is calculated by (8) for \(D_{st1} = D_{st2}\). This is named Optimum Switching Method which is very useful because in the output, the LC filter isn’t required. In Fig. 5, this mode is achieved for \(D_{st1} = D_{st2}\) and it is seen that controller can simply control the \(P_{z1}\) only by one duty cycle value.

One of the advantages of the proposed structure is having high voltage gain. It means that if the curve of \(V_{O2}\) shown in Fig. 4 is compared with the output value of the classic Z-source inverter in [20], it is deduced that the proposed converter has higher gain in an special operation area. For being clear and better comparison, it is supposed that in this Mode, both input sources are in series and is applied to classic Z-source inverter. The output value which is yielded is compared to the output of proposed converter when \(D_{st1} = D_{st2} = D_{st}\). In this situation, in the proposed converter, the subinterval 1 doesn’t exist and only \(V_{o2}^{3}\) (which is named \(V_{O}\)) exists. By simplification (8), it is resulted:

$$V_{O} = \underbrace {{\left( {\frac{{2nD_{st} }}{{(1 - 2D_{st} )^{2} }}} \right)}}_{{K_{1} }}V_{i1} + \underbrace {{\frac{1}{{1 - 2D_{st} }}}}_{{K_{2} }}V_{i2}$$

(23)

The output relation for the classic Z-source inverter which its input is \(V_{i1} + V_{i2}\) is as follow:

$$\begin{aligned} V_{O} &= \frac{1}{{1 - 2D_{st} }}\left( {V_{i1} + V_{i2} } \right) \hfill \\ &= \underbrace {{\frac{1}{{1 - 2D_{st} }}}}_{{K_{1}^{'} }}V_{i1} + \underbrace {{\frac{1}{{1 - 2D_{st} }}}}_{{K_{2}^{'} }}V_{i2} \hfill \\ \end{aligned}$$

(24)

For comparison of the values of \(V_{O}\) in two converters according to (23) and (24), it can be said because the coefficients \(K_{2}\) and \(K_{2}^{'}\) are the same, so the coefficients \(K_{1}\) and \(K_{1}^{'}\) should be compared. It is possible \(n = 1\) and the coefficients are drawn for \(0 < D_{ov} < 0.5\). Figure 6 shows both curves in a plane with logarithmic vertical axis and linear horizontal axis. According to the figure if \(0.25 < D_{st}\), it can be said that \(K_{1}^{'} < K_{1}\) and so the proposed converter has the higher voltage gain than the classic Z-source inverter. In addition, if \(n > 1\), \(K_{1}\) will be more than when \(n = 1\) and so it is possible to achieve higher gain with choosing higher \(n\).

Comparision of voltage gain \(K_{1}\) and \(K_{1}^{'}\) for \(n = 1\)

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2.2.2

Mode II

In this Mode, both Z-source networks start shoot through state and finish it in different time. The most advantage of this switching method rather than Mode Ӏ is ability to control the contribution of Z-source 1 to handle the load. If the system wants to supply the load so that the Z-source 2 should have the more contribution, Mode ӀӀ will be better than Mode Ӏ. As Mode I, in one period, when Z-source 1 is in shoot through state and Z-source 2 isn’t in shoot through state, \(V_{O2}\) has its maximum value. Also, \(V_{O2}\) is a three-level signal. Figure 7 shows shoot through signals and DC-link voltage for this Mode.

DC-link voltage waveform for Mode II

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In this Mode, equivalent circuits and current directions is similar to Mode Ӏ so that subintervals 2, 3 and 4 in this Mode is equal to subintervals 1, 2 and 3 in Mode Ӏ, respectively. Only subinterval 1 is different from them and is according to Fig. 8:

The equivalent circuit of the converter in Mode ӀӀ for subinterval 1

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If the duty cycle of subinterval 2 is defined as \(D_{ov}\), Similar to the Mode I, it can be said:

$$V_{C3} = V_{i1} - \left( {\frac{{\left( {1 - D_{st2} } \right)\left( {1 - 2D_{st1} } \right)}}{{2nD_{ov} \left( {2D_{st2} + D_{st1} - 2D_{ov} - 1} \right)}}} \right)V_{i2}$$

(25)

$$V_{O2}^{3} = V_{i1} - \left( {\frac{{\left( {1 - 2D_{st1} + D_{st2} } \right)\left( {1 - 2D_{st1} } \right)}}{{n\left( {4D_{st2} + 2D_{st1} - 4D_{ov} - 1} \right)}}} \right)V_{i2}$$

(26)

$$V_{O2}^{4} = V_{i1} - \left( {\frac{{1 - 2D_{st1} }}{{2n\left( {D_{st1} - 2D_{ov} } \right)}}} \right)V_{i2}$$

(27)

where \(V_{O2}^{3}\) and \(V_{O2}^{4}\) are the DC-link voltage in subinterval 3 and 4, respectively. Similar to Mode I, it is necessary to extract the active power values produced by Z-source 1 and 2, simultaneously. The calculation and simplification of the relations governing on the system in this Mode is deduced:

$$P_{z1} = V_{i1} .\left( {I_{Inv}^{3} + I_{Inv}^{4} .\left( {\frac{{2\left( {2D_{ov} - D_{st1} } \right)\left( {1 - D_{st1} - D_{st2} + D_{ov} } \right)}}{{\left( {D_{st1} - D_{ov} } \right)\left( {1 - D_{st1} - 2D_{st2} + 2D_{ov} } \right)}}} \right)} \right)$$

(28)

$$P_{z2} = V_{i2} .\left( {I_{Inv}^{3} + I_{Inv}^{4} .\frac{{1 - D_{st1} - D_{st2} - D_{ov} }}{{D_{st1} - D_{ov} }}} \right)$$

(29)

where \(I_{Inv}^{3}\) and \(I_{Inv}^{4}\) are the average output currents in subinterval 3 and 4, respectively. As previous section, trying to study the ratio of \(P_{z1}\) to \(P_{out}\). The ratio has been drawn in Fig. 9. In this figure, it is supposed that \(D_{st1} = D_{st2} = D_{st}\), \(V_{i1} = V_{i2}\) and the curves are drawn for various \(D_{st}\) and \(D_{ov}\).

The ratio of \(P_{z1}\) to \(P_{out}\) in Mode ӀӀ for \(V_{i1} = V_{i2} , n = 1\), \(D_{st1} = D_{st2} = D_{st}\)

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Figure 9 shows an increase in the values of \(D_{st}\) or \(D_{ov}\), increases \(\frac{{P_{z1} }}{{P_{out} }}\). So with controlling the values of \(D_{st}\), \(D_{ov}\) the contribition of Z-source 1 to handle the load and so sharing the power will be controlled.