Inverter circuit

Abstract

An inverter circuit (100) is provided with: a series circuit which is connected in parallel to a DC power supply (Vin) and to which a control switch (Q1) and a synchronous rectification switch (Q2) are connected in series; a reactor (L), one end of which is connected to a connection point between the control switch (Q1) and the synchronous rectification switch (Q2); an output capacitor (Co) connected between the power supply line of the DC power supply (Vin) and the other end of the reactor (L); and a control circuit that controls the ON time (tonQ1) of the control switch (Q1) and the ON time (tonQ2) of the synchronous rectification switch (Q2) such that the reactor (L) generates a reverse current (Ir) over the entire range of the instantaneous value (VO) of the AC output voltage (vo).

Description

Technical Field

[0001] The present invention relates to an inverter circuit capable of converting a DC input voltage into an AC output voltage of a predetermined frequency. Background Art

[0002] In a conventional critical inverter circuit that performs synchronous rectification, if the ratio of the DC input voltage Vin to the instantaneous value VO of the AC output voltage vo is not greater than a specified value (Vin ≥ 2VO), the zero voltage switching (Zero Voltage Switching) (ZVS) of the control switch cannot be performed. In addition, in the critical inverter circuit, the switching frequency fs of the control switch and the synchronous rectification switch changes greatly according to the change of the instantaneous value VO of the AC output voltage vo, so high efficiency cannot be obtained.

[0003] Prior art literature

[0004] Patent Literature

[0005] Patent Document 1: Japanese Patent Application Laid-Open No. 62-178176 Summary of the invention

[0006] Problems to be solved by the invention

[0007] One aspect of the present invention provides a high-efficiency inverter circuit that can perform zero voltage switching (ZVS) of controlled switching within a wide range of the instantaneous value VO of the AC output voltage vo, and has a small change in the switching frequency fs corresponding to the change in the instantaneous value VO of the AC output voltage vo.

[0008] Means for solving problems

[0009] In order to solve the above-mentioned problems, an inverter circuit involved in one embodiment of the present invention comprises: a series circuit connected in parallel with a DC power supply and having a control switch and a synchronous rectification switch connected in series; an inductor, one end of which is connected to a connection point between the control switch and the synchronous rectification switch; an output capacitor connected between a power line of the DC power supply and the other end of the inductor, and outputting the AC output voltage at both ends; and a control circuit, which performs the following control, namely, after the control switch is turned on for a first conduction time, the control switch is turned off and the synchronous rectification switch is turned on, and after the synchronous rectification switch is turned on for a second conduction time, the synchronous rectification switch is turned off and the control switch is turned on, so that the inductor generates a reverse current in the full range of the instantaneous value of the AC output voltage. BRIEF DESCRIPTION OF THE DRAWINGS

[0010] Figure 1 This is a diagram showing a step-down chopper circuit (a conceptual diagram for explaining a conventional inverter circuit).

[0011] Figure 2 This is a circuit diagram when the synchronous rectification switch Q2 of the step-down chopper circuit is turned off.

[0012] Figure 3 This is a diagram showing the minimum reverse current with respect to the instantaneous value of the AC output voltage when specific conditions (parameters) are set in the step-down chopper circuit.

[0013] Figure 4 1 is a diagram showing an envelope curve (solid line) of a switching current and an AC output current (dashed line) with respect to a phase (part of a half cycle) of an AC output voltage of a step-down chopper circuit.

[0014] Figure 5 It is a diagram showing the relationship between the phase (part of a half cycle) of the AC output voltage and the switching frequency.

[0015] Figure 6 1 is a diagram showing the relationship between the instantaneous value of the AC output voltage and the reverse current in the inverter circuit adopting the first embodiment.

[0016] Figure 7 1 is a diagram showing the relationship between the instantaneous value of the AC output voltage and the reverse current in the inverter circuit adopting the second embodiment.

[0017] Figure 8 1 is a diagram showing an envelope curve (solid line) of a switching current and an AC output current (dashed line) with respect to a phase (half-cycle portion) of an AC output voltage in an inverter circuit adopting method 1. FIG.

[0018] Fig. 9 1 is a diagram showing an envelope curve (solid line) of a switching current and an AC output current (dashed line) with respect to a phase (half-cycle portion) of an AC output voltage in an inverter circuit adopting method 2.

[0019] Fig.10 1 is a diagram showing the relationship between the phase (half cycle portion) of the AC output voltage and the switching frequency in the inverter circuit adopting the first embodiment.

[0020] Fig.11 1 is a diagram showing the relationship between the phase (half cycle portion) of the AC output voltage and the switching frequency in the inverter circuit adopting the second embodiment.

[0021] Fig.12 1 is a diagram showing an envelope curve (solid line) of a switching current with respect to a phase (half-cycle portion) of an AC output voltage in an inverter circuit adopting method 3 and an AC output current (dashed line).

[0022] Fig.131 is a diagram showing the relationship between the phase (half cycle portion) of the AC output voltage and the switching frequency in the inverter circuit adopting the third embodiment.

[0023] Fig.14 This is a timing chart showing the waveform of the switching current flowing through the first switch Q1 and the second switch Q2 when the first switch Q1 and the second switch Q2 are switched on and off as in Patterns 1 to 3.

[0024] Fig.15 This is a circuit diagram of an inverter circuit according to the first embodiment.

[0025] Fig.16 It is a circuit diagram of an inverter circuit according to the second embodiment.

[0026] Fig.17 This is a circuit diagram of an inverter circuit according to a third embodiment.

[0027] Fig.18 It is a circuit diagram of an inverter circuit according to a fourth embodiment. DETAILED DESCRIPTION

[0028] Hereinafter, an inverter circuit according to an embodiment of the present invention will be described in detail with reference to the drawings.

[0029] First, refer to Figure 1 The basic operation of the conventional inverter circuit is described in detail. Figure 1 In FIG. 1 , Vin represents a DC voltage, Q1 represents a control switch (first switch), Q2 represents a synchronous rectification switch (second switch), C / 2 represents a capacitor, L represents a reactor, and Co represents an output capacitor. The step-down chopper circuit can output the positive half cycle of the AC output voltage vo to the output capacitor Co. The capacitor C / 2 includes the parasitic capacitance of each of the switches Q1 and Q2 or the capacitance of an external capacitor.

[0030] Figure 2 is Figure 1 The circuit network when the first switch Q1 is turned on and the second switch Q2 is turned off in the step-down chopper circuit. Figure 1 As for the two capacitors C / 2 shown, it is assumed that the charging and discharging amounts are the same and are combined into one capacitor C connected in parallel with the first switch Q1.

[0031] The switching frequency fs of switches Q1 and Q2 (for example, several hundred kHz) is sufficiently larger than the frequency fo of the AC output voltage vo (for example, 50 Hz). Therefore, the value of the AC output voltage vo in each switching cycle of switches Q1 and Q2 can be regarded as constant (DC).

[0032] like Figure 2 The current i(t) flowing through the circuit as shown by the arrow in FIG. 1 is expressed by the voltage v(t) of the capacitor C using the equation 1. Here, t represents time.

[0033] [Mathematical formula 1]

[0034]

[0035] If the reverse excitation current of the reactor L is set to Ir, the current i(0) and voltage v(t) at time t=0, that is, under the initial conditions, are expressed by Equation 2.

[0036] [Mathematical formula 2]

[0037]

[0038] also, Figure 2 The circuit network equation is expressed by Equation 3.

[0039] [Mathematical formula 3]

[0040]

[0041] According to equation 1, equation 2, and equation 3, the following equation 4 is obtained.

[0042] [Formula 4]

[0043]

[0044] The condition for obtaining the reverse excitation current Ir (hereinafter referred to as reverse current) with respect to the output voltage vo is obtained from equation 4. At least the voltage v(t) must be 0 V or less, so the condition as equation 5 is obtained.

[0045] [Mathematical formula 5]

[0046]

[0047] Formula 5 can be rewritten as Formula 6.

[0048] [Mathematical formula 6]

[0049]

[0050] In order to obtain the minimum reverse current Ir for the first equation of equation 6 to be satisfied, the first equation of equation 6 may be satisfied when the cosine function is minimum, that is, when it is -1 (when the following equation 7 is satisfied).

[0051] [Mathematical formula 7]

[0052]

[0053] Therefore, the reverse current Ir only needs to satisfy the following formula 8.

[0054] [Mathematical formula 8]

[0055]

[0056] For example, under the condition shown in the following equation 9, the minimum reverse current Ir with respect to the instantaneous value VO of the output voltage vo becomes as follows: Figure 3 Like that.

[0057] [Mathematical formula 9]

[0058] L=1l00uH

[0059] C=200pF

[0060] Vin=400Vdc...(Equation 9)

[0061] Here, as shown in Formula 10, the output power Po is set to 500 W, the effective voltage Vrms of the AC output voltage vo is set to AC 200 V, the instantaneous value of the AC output current io is represented as IO, and the angular frequency of the AC output current is represented as ωo (=2πfo).

[0062] [Formula 10]

[0063] Po=500W

[0064] Vrms = 200 V rms

[0065]

[0066] When the instantaneous value of the AC output voltage vo is represented by VO, it is not necessary to reverse the current for zero voltage switching in the range of 2VO≥Vin (=400V). When the instantaneous value of the AC output voltage VO is in the range of 200V to 400V, the reverse current Ir is zero.

[0067] In AC output, if the parameters of equations 9 and 10 are used, the envelope of the critical switch current becomes as follows: Figure 4 Like the solid line. Figure 4 Only the phase of a portion of the half cycle of the AC output voltage vo is shown, and the current is reversed only in the interval of the phase that satisfies Vin≥2VO. Figure 4 In FIG. 1 , the dotted line represents the AC output current (500 W / 200 Vrms=2.5 Arms).

[0068] Figure 5 Shown with Figure 4 The corresponding relationship between the phase of the AC output voltage vo and the switching frequency fs of the switches Q1 and Q2. Figure 5As shown in Figure 1, the maximum and minimum values ​​of the switching frequency fs have a difference of about 10 times. Especially when the phase of the AC output voltage vo is small, although the switching current is Figure 4 As shown in the figure, the switching frequency fs is as small as Figure 5 As shown in , it is difficult to reduce the switching loss.

[0069] exist Figure 4 , 5 The calculation formula used in is as follows.

[0070] (1) When Vin<2VO

[0071] Under this condition, the control switch Q1 cannot perform zero voltage switching, but performs hard switching. As shown in equation 11, the reverse current Ir is zero.

[0072] [Mathematical formula 11]

[0073] Ir=0...(Formula 11)

[0074] The on time tonQ1 of the first switch Q1 is expressed by Equation 12.

[0075] [Mathematical formula 12]

[0076]

[0077] The on-time tonQ2 of the second switch Q2 is expressed by equation 13.

[0078] [Mathematical formula 13]

[0079]

[0080] Furthermore, a dead time (dead time) tdead from when the second switch Q2 is turned off to when the first switch Q1 is turned on is expressed by equation 14 based on equations 7 and 11.

[0081] [Formula 14]

[0082]

[0083] The dead time from the first switch Q1 turning off to the second switch Q2 turning on can also be omitted based on the principle of synchronous rectification. Figure 4 The switching frequency fs in the phase interval of Vin<2VO.

[0084] (2) Vin ≥ 2VO

[0085] Under this condition, the first switch Q1 can perform zero voltage switching.

[0086] The reverse current Ir is expressed by equation 15.

[0087] [Mathematical formula 15]

[0088]

[0089] The on time tonQ1 of the first switch Q1 is expressed by Equation 16.

[0090] [Formula 16]

[0091]

[0092] The on time tonQ2 of the second switch Q2 is expressed by equation 17.

[0093] [Mathematical formula 17]

[0094]

[0095] Furthermore, the dead time tdead from when the second switch Q2 is turned off to when the first switch Q1 is turned on is expressed by Equation 18 based on Equation 7 and Equation 15.

[0096] [Mathematical formula 18]

[0097]

[0098] The dead time from the first switch Q1 turning off to the second switch Q2 turning on can also be omitted based on the principle of synchronous rectification. Therefore, the inverse of the sum of equations 16, 17, and 18 becomes Figure 4 The switching frequency fs in the phase interval of Vin≥2VO.

[0099] As described above, in the conventional step-down chopper circuit (conventional inverter circuit), whether zero voltage switching (ZVS) can be performed is determined by the phase angle of the AC output voltage. Especially when digital control is performed, the calculation formula is complicated and the amount of calculation is large.

[0100] In the phase interval where the instantaneous value VO of the AC output voltage vo is Vin<2VO, the voltage v(t) of the capacitor C does not drop to 0V, and the first switch Q1 performs hard switching. In addition, the switching current in this interval is also large. Therefore, the switching loss increases, making it difficult to improve the efficiency of the inverter circuit.

[0101] Therefore, in the present embodiment, a low-loss critical inverter circuit is provided that can reliably perform zero-voltage switching at all phase angles of the AC output voltage. As specific embodiments for this purpose, the following two are first listed.

[0102] (Method 1) The on-period of the second switch Q2 is set so as to draw a reverse current envelope that increases and decreases according to the phase of the AC output voltage vo, thereby reducing the difference in the switching frequency fs and obtaining high efficiency.

[0103] (Method 2) The on-period of the second switch Q2 is set so that the reverse current envelope becomes constant regardless of the phase of the AC output voltage vo, thereby reducing the difference in the switching frequency fs and obtaining high efficiency.

[0104] In method 1, for Figure 3 The curve showing the minimum reverse current Ir relative to the instantaneous value VO of the AC output voltage vo is set to be not less than the curve, such as Figure 6 The linear function is represented by the following equation 19.

[0105] [Mathematical formula 19]

[0106]

[0107] Equation 19 is simpler than Equation 15, and the amount of calculation required for digital control is small. In method 1, the amount of reverse current Ir is determined according to the change in the instantaneous value VO of the AC output voltage vo. Specifically, the smaller the instantaneous value VO, the larger the amount of reverse current Ir.

[0108] The on-time tonQ1 of the first switch Q1 in mode 1 is expressed by equation 20.

[0109] [Mathematical formula 20]

[0110]

[0111] The on-time tonQ2 of the second switch Q2 in mode 1 is expressed by equation 21.

[0112] [Mathematical formula 21]

[0113]

[0114] Furthermore, the dead time tdead from when the second switch Q2 is turned off to when the first switch Q1 is turned on is expressed by Equation 22 based on Equation 7 and Equation 19.

[0115] [Mathematical formula 22]

[0116]

[0117] The dead time from the first switch Q1 turning off to the second switch Q2 turning on can also be omitted based on the principle of synchronous rectification. Therefore, the reciprocal of the sum of equations 20, 21, and 22 becomes the switching frequency fs in method 1.

[0118] In method 2, for Figure 3 The reverse current Ir curve of the instantaneous value VO of the AC output voltage vo is set to be not less than the curve, such as Figure 7 The straight line is represented by the following equation 23.

[0119] [Mathematical formula 23]

[0120]

[0121] In Expression 23 of the second embodiment, the reverse current Ir becomes a constant value, so the calculation of the value of the reverse current Ir can be further simplified compared to Expression 19 of the first embodiment.

[0122] The on-time tonQ1 of the first switch Q1 in mode 2 is expressed by equation 24.

[0123] [Mathematical formula 24]

[0124]

[0125] The on-time tonQ2 of the second switch Q2 in mode 2 is expressed by (Formula 25).

[0126] [Mathematical formula 25]

[0127]

[0128] Furthermore, the dead time tdead from when the second switch Q2 is turned off to when the first switch Q1 is turned on is expressed by Equation 26 based on Equation 7 and Equation 23.

[0129] [Mathematical formula 26]

[0130]

[0131] The dead time from the first switch Q1 turning off to the second switch Q2 turning on can also be omitted based on the principle of synchronous rectification. Therefore, the reciprocal of the sum of equations 24, 25, and 26 becomes the switching frequency fs in method 2.

[0132] Figure 6 , Figure 7 The linear function and the straight line both represent Figure 3 The reverse current amount contained in the zero voltage switching (ZVS) region, therefore, according to equations 19 and 23, the reverse current amount that satisfies zero voltage switching is obtained.

[0133] Figure 3 The curve of is the boundary line that divides the non-zero voltage switching region and the zero voltage switching region. In the method 1 and the method 2 that provide a reverse current Ir larger than the curve, in order to maintain Figure 4 The AC output current shown by the dotted line also needs to increase the peak current by the amount corresponding to the increase in the reverse current Ir. That is, it is necessary to increase the on-time tonQ1 of the first switch Q1. Along with this, the on-time tonQ2 of the second switch Q2 is also increased, thereby reducing the switching frequency fs. The effect of reducing the switching loss brought about by the reduction in the switching frequency fs exceeds the effect of increasing the switching loss brought about by the increase in the peak current.

[0134] Figure 8 The current envelope (solid line) and the AC output current (dashed line) with respect to the phase of the AC output voltage vo in the inverter circuit adopting the first embodiment are shown. Fig. 9 The current envelope (solid line) and the AC output current (dashed line) with respect to the phase of the AC output voltage vo in the inverter circuit adopting the second embodiment are shown. Fig.10 The relationship between the phase of the AC output voltage vo and the switching frequency fs in the inverter circuit adopting the method 1 is shown. Fig.11 The relationship between the phase of the AC output voltage and the switching frequency fs in the inverter circuit adopting the second embodiment is shown.

[0135] Will Figure 6 (Method 1) and Figure 7 (Method 2) Compare, Figure 7 The reverse current of the AC output voltage of (method 2) is greater in the entire phase range. Therefore, in order to maintain the AC output current of the dotted line, Fig. 9 (Method 2) The peak value ratio of the current envelope Figure 8 The peak value of the current envelope of (method 1) is slightly larger. However, regarding the peak value of the switching frequency fs, Fig.11 (Method 2) Fig.10 (Method 1) has decreased. In addition, Fig.10 (Method 1) and Fig.11 (Method 2) Figure 5 In comparison, both can sufficiently reduce the peak value of the switching frequency fs.

[0136] The following describes the distinction between method 1 and method 2.

[0137] In method 1, the amount of reverse current Ir, the on-time of switches Q1 and Q2, and the like are calculated while sampling the instantaneous value VO of the AC output voltage vo every time switches Q1 and Q2 are switched.

[0138] Mode 2 is suitable for the case where switches Q1 and Q2 are switched at a higher frequency, or when a low-grade microcomputer or digital signal processor (DSP) is used to digitally control the inverter circuit. As in Mode 2, if the reverse current Ir is set to a fixed value greater than the constant value of Formula 23 throughout all phases of the AC output voltage vo, even if there is a deviation in the components used in the inverter circuit, the peak value of the switching frequency fs and the current envelope (switching current) will not change to the extent that the efficiency is affected, and zero voltage switching can be reliably performed. That is, in Mode 2, compared with Mode 1, it is not necessary to set a strict value for the reverse current Ir.

[0139] Next, method 3 is described. In method 3, the reverse current Ir is set to a fixed value equal to or greater than the constant value of equation 23 plus a harmonic component of the frequency fo of the AC output voltage vo, thereby providing an inverter circuit in which the switching frequency fs hardly changes.

[0140] In method 3, the reverse current Ir is set as shown in Expression 27.

[0141] [Mathematical formula 27]

[0142]

[0143] Here, β is a harmonic component of three times the AC output voltage vo alternating at an angular frequency ωo=2πfo. β may include not only harmonic components of three times the AC output voltage vo, but also harmonic components of odd numbers such as 5 times and 7 times.

[0144] The on-time tonQ1 of the first switch Q1 in mode 3 is expressed by (Formula 28).

[0145] [Mathematical formula 28]

[0146]

[0147] The on-time tonQ2 of the second switch Q2 in mode 3 is expressed by (Formula 29).

[0148] [Mathematical formula 29]

[0149]

[0150] Since the other equations are the same as those in Example 2, they are omitted.

[0151] Fig.12The relationship between the phase of the AC output voltage vo and the envelope of the switching current (solid line) and the AC output current (dashed line) is shown when the third harmonic component is added to the frequency fo of the AC output voltage vo as β in equation 27. In this way, in mode 3, the peak value of the switching current does not increase much, so high efficiency can be expected. Fig.13 27, and the third harmonic component of the frequency fo of the AC output voltage vo is added as β in the equation 27. Fig.13 It can be seen that the switching frequency fs hardly changes. In the inverter circuit of mode 3, it is easy to take measures against noise such as radio wave interference (electromagnetic interference (EMI)). In addition, by adding not only the harmonic components of 3 times the AC output voltage vo to β, but also the harmonic components of odd numbers such as 5 times and 7 times, the switching frequency fs becomes flatter relative to the phase, and the switching frequency fs is further reduced, thereby further reducing the switching loss.

[0152] Fig.14 It is a timing diagram showing the waveform of the switching current flowing through the first switch Q1 and the second switch Q2 when the control switch (first switch) Q1 and the synchronous rectification switch (second switch) Q2 are turned on and off as in Mode 1, Mode 2, and Mode 3.

[0153] Specifically, switches Q1 and Q2 are controlled as follows. First, after the second switch Q2 is turned on, when the current reaches the set reverse current Ir after the on-time tonQ2 of the second switch Q2, the second switch Q2 is turned off. Then, after the dead time tdead, the first switch Q1 is turned on again, and after the on-time tonQ1 of the first switch Q1, the first switch Q1 is turned off, and the second switch Q2 is turned on again. In addition, the dead time tdead is extremely short compared to the on-time tonQ1 of the first switch Q1 and the on-time tonQ2 of the second switch Q2, so Fig.14 Omitted.

[0154] exist Figure 15-18 Implementation methods of inverter circuits that can realize Mode 1, Mode 2, and Mode 3 are shown in FIG.

[0155] (First Embodiment)

[0156] Reference Fig.15 The inverter circuit 100 according to the first embodiment will be described.

[0157] The inverter circuit 100 outputs only the positive polarity of the AC output voltage vo.

[0158] In the inverter circuit 100, a series circuit of a control switch (first switch) Q1 and a synchronous rectification switch (second switch) Q2 is connected in parallel to a DC voltage (DC power supply) Vin. In the first embodiment, the first switch Q1 and the second switch Q2 use N-channel MOSFETs. A first capacitor C1 is connected in parallel to the first switch Q1. A second capacitor C2 is connected in parallel to the second switch Q2. Capacitors C1 and C2 may also be parasitic capacitors of switches Q1 and Q2.

[0159] The connection point between the first switch Q1 and the second switch Q2 is connected to one end of the reactor L. The AC output voltage vo is outputted via the output capacitor Co connected between the other end of the reactor L and the negative electrode (power line) of the DC voltage Vin. The on and off of the first switch Q1 and the second switch Q2 are controlled by the control circuit 200.

[0160] The control circuit 200 includes an error amplifier 4, comparators 3, 5, and RS flip-flop circuits 6, 7. The control circuit 200 actually performs digital control based on a computer program, so the circuit diagram of the control circuit 200 is a conceptual diagram of digital control ( Figure 16 to Figure 18 The error amplifier 4 amplifies the error voltage between the instantaneous value VO of the AC output voltage vo and the reference voltage Vref, and outputs it to the inverting input terminal of the comparator 5. The current sensor 1 detects the instantaneous value IO of the current io flowing through the power line on the anode side of the inverter circuit 100.

[0161] When the error voltage from the error amplifier 4 is lower than the voltage based on the instantaneous value IO of the current io detected by the current sensor 1, the comparator 5 outputs a high level to the reset terminal R of the RS flip-flop circuit 7. At this time, a low level is output from the output terminal Q of the RS flip-flop circuit 7, so the first switch Q1 is turned off. A high level is output from the inverting output terminal of the RS flip-flop circuit 7 to the set terminal S of the RS flip-flop circuit 6, so the second switch Q2 is turned on. In this way, the control circuit 200 turns on and off the first switch Q1 and the second switch Q2 alternately, thereby enabling control so that the instantaneous value VO of the AC output voltage vo and the instantaneous value IO of the current io flowing through the power line on the anode side of the inverter circuit 100 become predetermined values.

[0162] By turning on the first switch Q1, current flows in the loop of Vin→Q1→L→Co→Vin, and the reactor L is excited. Then, if the first switch Q1 is turned off and the second switch Q2 is turned on, current flows in the loop of L→Co→Q2→L, and the excitation of the reactor L is reset. If the current flowing through the second switch Q2 becomes zero, a reverse current flows in the loop of Co→L→Q2→Co, and the reactor L starts to be excited in the reverse direction. This is Fig.14 The switching current flowing through the second switch Q2 becomes a reverse current flowing in the negative direction, and the value at which the absolute value of the reverse current becomes the maximum is referred to as the “reverse current Ir”.

[0163] If the reverse current Ir detected by the current sensor 2 for reverse current detection connected to the power supply line on the negative side of the inverter circuit 100 reaches the set value Ir_set of the reverse current Ir, the comparator 3 outputs a high level to the reset terminal R of the RS flip-flop circuit 6, and outputs a low level from the output terminal Q of the RS flip-flop circuit 6, so the second switch Q2 is turned off. A high level is output from the inverting output terminal of the RS flip-flop circuit 6 to the set terminal S of the RS flip-flop circuit 7, so the first switch Q1 is turned on. Here, by setting the set value Ir_set of the reverse current Ir to the value of the reverse current Ir of any one of the methods 1 (Formula 19), the method 2 (Formula 23), and the method 3 (Formula 27) or a value corresponding thereto, it can correspond to any one of the methods 1, 2, and 3. In Formulas 19, 23, and 27, the value of the capacitor C=C1+C2.

[0164] The current sensor 1 and the current sensor 2 may be integrated into one current sensor. When integrated into one current sensor, the instantaneous value IO of the current and the reverse current Ir are in opposite directions, and therefore they need to be monitored separately.

[0165] To simplify the description, a dead time tdead from when the second switch Q2 is turned off to when the first switch Q1 is turned on is omitted.

[0166] In this way, with respect to the inverter circuit 100 involved in the first embodiment, by setting the set value Ir_set of the reverse current of the comparator 3 to the value of the reverse current Ir of any one of Formula 19 (Method 1), Formula 23 (Method 2), and Formula 27 (Method 3) or a value corresponding thereto, operation is performed in any one of Method 1, Method 2, and Method 3.

[0167] Therefore, according to the inverter circuit 100 according to the first embodiment, zero voltage switching can be reliably performed in all phases of the AC output voltage vo, that is, in the entire range of the instantaneous value VO of the AC output voltage vo.

[0168] (Second Embodiment)

[0169] Reference Fig.16 An inverter circuit 100A according to the second embodiment will be described.

[0170] The inverter circuit 100A also outputs only the positive polarity of the AC output voltage vo, similarly to the inverter circuit 100 according to the first embodiment.

[0171] In the inverter circuit 100A, Fig.15 The inverter circuit 100 according to the first embodiment shown in the figure omits the current sensor 2 for detecting the reverse current, and the control circuit 200A calculates the time to reach the reverse current Ir to perform zero voltage switching. The inverter circuit 100A is the same as the inverter circuit 100 according to the first embodiment in terms of other configurations.

[0172] In order to calculate the time to reach the reverse current Ir, it is necessary to find the on-time tonQ1 of the first switch Q1 and the on-time tonQ2 of the second switch Q2. As shown in Formula 20 (method 1), Formula 24 (method 2), and Formula 28 (method 3), the on-time tonQ1 of the first switch Q1 is calculated based on the DC input voltage Vin, the instantaneous value VO of the AC output voltage vo, and the instantaneous value IO of the AC output current io. As shown in Formula 21 (method 1), Formula 25 (method 2), and Formula 29 (method 3), the on-time tonQ2 of the second switch Q2 is calculated based on the DC input voltage Vin, the instantaneous value VO of the AC output voltage vo, and the on-time tonQ1 of the first switch Q1.

[0173] Therefore, in the second embodiment, the control circuit 200A includes a tonQ1 calculator 21 for calculating the on-time tonQ1 of the first switch Q1 and a tonQ2 calculator 22 for calculating the on-time tonQ2 of the second switch Q2.

[0174] The tonQ1 calculator 21 receives the instantaneous value IO of the current io flowing through the power line on the anode side of the inverter circuit 100 detected by the input current sensor 1, the instantaneous value VO of the AC output voltage vo of the inverter circuit 100, and the DC input voltage Vin, and calculates and outputs the on-time tonQ1 of the first switch Q1 according to any one of Formula 20 (Method 1), Formula 24 (Method 2), and Formula 28 (Method 3).

[0175] The tonQ2 calculator 22 receives the instantaneous value VO of the AC output voltage vo of the inverter circuit 100 and the DC input voltage Vin, calculates (Vin-VO) / VO as the coefficient of Formula 21 (method 1) and Formula 25 (method 2) and outputs it, and then multiplies the output of the tonQ2 calculator 22 by the output of the tonQ1 calculator 21 through the multiplier 23, thereby calculating and outputting the conduction time tonQ2 of the second switch Q2.

[0176] Furthermore, the calculated on-time tonQ1 of the first switch Q1 and the calculated on-time tonQ2 of the second switch Q2 are added together by the adder 24 , thereby calculating the switching period T.

[0177] The switching cycle T calculated by the adder 24 is input to the sawtooth wave generating circuit 25. The sawtooth wave generating circuit 25 outputs a sawtooth wave whose height changes according to the value of the switching cycle T. The sawtooth wave output by the sawtooth wave generating circuit 25 and the on-time tonQ1 of the first switch Q1 calculated by the tonQ1 calculator 21 are input to the comparator 26. According to the output of the comparator 26, the on-off of the first switch Q is switched (switched), and at the same time, according to the signal that inverts the output of the comparator 26 through the inverter circuit 27, the on-off of the second switch Q2 is switched.

[0178] The on-time tonQ1 calculated according to any one of Formula 20 (method 1), Formula 24 (method 2), and Formula 28 (method 3) depends on the reverse current Ir, so the reverse current Ir is also calculated at the same time.

[0179] To simplify the description, a dead time tdead from when the second switch Q2 is turned off to when the first switch Q1 is turned on is omitted.

[0180] Thus, in the inverter circuit 100B according to the second embodiment, the on-time tonQ1 calculated by any one of the equation 20 (method 1), equation 24 (method 2), and equation 28 (method 3) is set by the tonQ1 calculator 21, and the on-time tonQ2 calculated by the equation 21 (methods 1, 2, and 3) is set by the tonQ2 calculator 22. Thus, the operation is performed by any one of the methods 1, 2, and 3.

[0181] Therefore, in the inverter circuit 100A according to the second embodiment, zero voltage switching can be reliably performed in all phases of the AC output voltage vo, that is, in the entire range of the instantaneous value VO of the AC output voltage vo.

[0182] (Third Embodiment)

[0183] Reference Fig.17An inverter circuit 100B according to the third embodiment will be described.

[0184] The inverter circuit 100A according to the second embodiment outputs the AC output voltage vo only in the positive polarity, whereas the inverter circuit 100B according to the third embodiment outputs the AC output voltage vo in both positive and negative polarities. This is different.

[0185] In the inverter circuit 100B involved in the third embodiment, a series circuit composed of a first switch Q1 and a second switch Q2, and a series circuit composed of a third switch Q3 and a fourth switch are connected in parallel to a DC input voltage Vin. In the third embodiment, switches Q1-Q4 use N-channel MOSFETs. A first capacitor C1 is connected in parallel to the first switch Q1. A second capacitor C2 is connected in parallel to the second switch Q2. A third capacitor C3 is connected in parallel to the third switch Q3. A fourth capacitor C4 is connected in parallel to the fourth switch Q4. In addition, capacitors C1-C4 may also be parasitic capacitors of switches Q1-Q4.

[0186] One end of the first inductor L1 is connected to the connection point of the first switch Q1 and the second switch Q2. One end of the second inductor L2 is connected to the connection point of the third switch Q3 and the fourth switch Q4. The AC output voltage vo is output via the output capacitor Co connected between the other end of the first inductor L1 and the other end of the second inductor L2. The value of the inductance of the first inductor L1 and the second inductor L2 is L1=L2=L / 2. The on and off of the switches Q1-Q4 are controlled by the control circuit 200B.

[0187] The control circuit 200B controls the first switch Q1 and the third switch Q3 to be switched at the same time, and the second switch Q2 and the fourth switch Q4 to be switched at the same time, that is, controls the diagonal switches to be switched at the same time as a full-bridge inverter circuit.

[0188] When the AC output voltage vo is in a positive half cycle, the first switch Q1 and the third switch Q3 become control switches, and the second switch Q2 and the fourth switch Q4 become synchronous rectification switches. In addition, when the AC output voltage vo is in a negative half cycle, the second switch Q2 and the fourth switch Q4 become control switches, and the first switch Q1 and the third switch Q3 become synchronous rectification switches.

[0189] The control circuit 200B includes a polarity determination unit 30 , a full-wave rectification circuit 31 , polarity switching units 32 , 33 , a tonQ1 calculator 21 , a tonQ2 calculator 22 , a multiplier 23 , an adder 24 , a sawtooth wave generation circuit 25 , a comparator 26 , and an inverter circuit 27 .

[0190] Fig.17The structures and operations of the tonQ1 calculator 21, tonQ2 calculator 22, multiplier 23, adder 24, sawtooth wave generating circuit 25, comparator 26, and inverter circuit 27 of the control circuit 200B shown in FIG. Fig.16 These are the same as those of the second embodiment shown, and therefore, their description is omitted here.

[0191] The polarity determination unit 30 determines the positive or negative polarity of the AC output voltage vo, and outputs the positive or negative polarity to the polarity switching units 32 and 33 .

[0192] The full-wave rectifier circuit 31 performs full-wave rectification on the AC output voltage vo. That is, when the AC output voltage vo is negative, the instantaneous value VO of the AC output voltage vo reversed to positive polarity is input to the tonQ1 calculator 21.

[0193] When the instantaneous value IO of the current io flowing through the anode side of the inverter circuit 100 detected by the current sensor 1 is positive, the polarity switching unit 32 inputs the original value to the tonQ1 calculator 21, and when it is negative, reverses the polarity and inputs it as a positive value to the tonQ1 calculator 21.

[0194] When the polarity of the AC output voltage vo determined by the polarity determination unit 30 is positive, the polarity switching unit 33 switches the first switch Q1 and the third switch Q3 to control switches, and switches the second switch Q2 and the fourth switch Q4 to synchronous rectification switches. When the polarity of the AC output voltage vo determined by the polarity determination unit 30 is negative, the second switch Q2 and the fourth switch Q4 are switched to control switches, and the first switch Q1 and the third switch Q3 are switched to synchronous rectification switches.

[0195] That is, when the polarity of the AC output voltage vo determined by the polarity determination unit 30 is positive, the polarity switching unit 33 switches the first switch Q1 and the third switch Q3 on and off by the output of the comparator 26, and switches the second switch Q2 and the fourth switch Q4 on and off by the output of the inverter circuit 27. On the other hand, when the polarity of the AC output voltage vo determined by the polarity determination unit 30 is negative, the polarity switching unit 33 switches the second switch Q2 and the fourth switch Q4 on and off by the output of the comparator 26, and switches the first switch Q1 and the third switch Q3 on and off by the output of the inverter circuit 27.

[0196] More specifically, in the half cycle in which the AC output voltage vo is positive, first, the first switch Q1 and the third switch Q3 as control switches are turned on, and the second switch Q2 and the fourth switch Q4 as synchronous rectification switches are turned off, thereby causing the current to flow in the positive direction in the loop of Vin→Q1→L1→Co→L2→Q3→Vin, and the reactors L1 and L2 are excited. Next, the first switch Q1 and the third switch Q3 as control switches are turned off, and the second switch Q2 and the fourth switch Q4 as synchronous rectification switches are turned on, thereby resetting the excitation of the reactors L1 and L2 in the loop of L1→Co→L2→Q4→Vin→Q2→L1.

[0197] Furthermore, in the half cycle in which the AC output voltage vo is negative, first, the second switch Q2 and the fourth switch Q4 as control switches are turned on, and the first switch Q1 and the third switch Q3 as synchronous rectification switches are turned off, whereby the current flows in the negative direction in the loop of Vin→Q4→L2→Co→L1→Q2→Vin, and the reactors L1 and L2 are excited. Next, the second switch Q2 and the fourth switch Q4 as control switches are turned off, and the first switch Q1 and the third switch Q3 as synchronous rectification switches are turned on, whereby the excitation of the reactors L1 and L2 is reset in the loop of L2→Co→L1→Q1→Vin→Q3→L2.

[0198] The value of the capacitor C may be set to C=C2+C4 in the positive half cycle of the AC output voltage vo, and may be set to C=C1+C3 in the negative half cycle. The value of the inductor L may be L=L1+L2.

[0199] In this way, in the inverter circuit 100B involved in the third embodiment, the on-time tonQ1 calculated by any one of Formula 20 (method 1), Formula 24 (method 2), and Formula 28 (method 3) is set by the tonQ1 calculator 21, and the on-time tonQ2 calculated by Formula 21 (methods 1, 2, and 3) is set by the tonQ2 calculator 22, thereby being able to set the value of the reverse current Ir to any one of Formula 19 (method 1), Formula 23 (method 2), and Formula 27 (method 3), so that operation can be performed by any one of Methods 1, 2, and 3.

[0200] Therefore, in the inverter circuit 100B according to the third embodiment, zero voltage switching can be reliably performed in all phases of the AC output voltage vo, that is, in the entire range of the instantaneous value VO of the AC output voltage vo.

[0201] (Fourth Embodiment)

[0202] Reference Fig.18An inverter circuit 100C according to a fourth embodiment will be described.

[0203] Fig.18 The inverter circuit 100C according to the fourth embodiment shown is a totem pole type inverter circuit capable of performing zero voltage switching in the entire range of AC output voltage.

[0204] In the inverter circuit 100C involved in the fourth embodiment, a series circuit composed of a first switch Q1 and a second switch Q2, and a series circuit composed of a third switch Q3 and a fourth switch are connected in parallel to a DC input voltage Vin. In the fourth embodiment, switches Q1-Q4 use N-channel MOSFETs. A first capacitor C1 is connected in parallel to the first switch Q1. A second capacitor C2 is connected in parallel to the second switch Q2. A third capacitor C3 is connected in parallel to the third switch Q3. A fourth capacitor C4 is connected in parallel to the fourth switch Q4. In addition, capacitors C1-C4 may also be parasitic capacitors of switches Q1-Q4.

[0205] One end of the reactor L is connected to the connection point between the first switch Q1 and the second switch Q2. The AC output voltage vo is outputted via the output capacitor Co connected between the other end of the reactor L and the connection point between the third switch Q3 and the fourth switch Q4. The on / off of the switches Q1-Q4 is controlled by the control circuit 200C.

[0206] In the half cycle when the AC output voltage vo is positive, the first switch Q1 is a control switch, the second switch Q2 is a synchronous rectification switch, the third switch Q3 is turned on, and the fourth switch Q4 is turned off. In the half cycle when the AC output voltage vo is negative, the second switch Q2 is a control switch, the first switch Q1 is a synchronous rectification switch, the fourth switch Q4 is turned on, and the third switch Q3 is turned off. That is, the inverter circuit 100C according to the fourth embodiment is a totem pole type inverter circuit.

[0207] The control circuit 200C includes a polarity determination unit 30 , a full-wave rectification circuit 31 , polarity switching units 32 , 33 a , and 33 b , a tonQ1 calculator 21 , a tonQ2 calculator 22 , a multiplier 23 , an adder 24 , a sawtooth wave generation circuit 25 , a comparator 26 , and an inverter circuit 27 .

[0208] Fig.18 The structure and operation of the tonQ1 calculator 21, tonQ2 calculator 22, multiplier 23, adder 24, sawtooth wave generating circuit 25, comparator 26, inverter circuit 27, polarity determination unit 30, full-wave rectifier circuit 31, and polarity switching unit 32 of the control circuit 200C shown in FIG. Fig.17These are the same as those of the third embodiment shown, and therefore, their description is omitted here.

[0209] When the polarity of the AC output voltage vo determined by the polarity determination unit 30 is positive, the polarity switching unit 33a switches the first switch Q1 to a control switch and the second switch Q2 to a synchronous rectification switch. When the polarity of the AC output voltage vo determined by the polarity determination unit 30 is negative, the second switch Q2 is switched to a control switch and the first switch Q1 is switched to a synchronous rectification switch.

[0210] That is, when the polarity of the AC output voltage vo determined by the polarity determination unit 30 is positive, the polarity switching unit 33a switches the first switch Q1 on and off by the output of the comparator 26, and switches the second switch Q2 on and off by the output of the inverter circuit 27. In addition, when the polarity of the AC output voltage vo determined by the polarity determination unit 30 is negative, the polarity switching unit 33a switches the second switch Q2 on and off by the output of the comparator 26, and switches the first switch Q1 on and off by the output of the inverter circuit 27.

[0211] The polarity switching unit 33b turns on the third switch Q3 and turns off the fourth switch Q4 when the polarity of the AC output voltage vo determined by the polarity determination unit 30 is positive, and turns on the fourth switch Q4 and turns off the third switch Q3 when the polarity of the AC output voltage vo determined by the polarity determination unit 30 is negative.

[0212] More specifically, in the half cycle in which the AC output voltage vo is positive, when the third switch Q3 is on and the fourth switch Q4 is off, the first switch Q1 as the control switch becomes on, and the second switch Q2 as the synchronous rectification switch becomes off, thereby causing a current in the positive direction to flow through the loop of Vin→Q1→L→Co→Q3→Vin, and thereby the reactor L is excited. Next, the first switch Q1 as the control switch becomes off, and the second switch Q2 as the synchronous rectification switch becomes on, thereby causing a current to flow through the loop of L→Co→Q3→Q2→L, and thereby the excitation of the reactor L is reset. If the current flowing through the second switch Q2 becomes zero, this time a reverse current flows through the loop of Co→L→Q2→Q3→Co, and thereby the reactor L begins to be excited in the reverse direction. This is in Fig.14 The current flowing through the second switch Q2 becomes a reverse current flowing in the negative direction.

[0213] Furthermore, in the half cycle in which the AC output voltage vo is negative, when the fourth switch Q4 is on and the third switch Q3 is off, the second switch Q2 as the control switch becomes on, and the first switch Q1 as the synchronous rectification switch becomes off, thereby causing a current in the negative direction to flow through the loop of Vin→Q4→Co→L→Q2→Vin, thereby causing the reactor L to be excited. Next, the second switch Q2 as the control switch becomes off, and the first switch Q1 as the synchronous rectification switch becomes on, thereby causing a current to flow through the loop of L→Q1→Q4→Co→L, thereby resetting the excitation of the reactor L. If the current flowing through the first switch Q1 becomes zero, a reverse current flows through the loop of Co→Q4→Q1→L→Co, thereby causing the reactor L to start being excited in the reverse direction. This is equivalent to Fig.14 Q1 and Q2 are swapped to reverse the polarity of the current.

[0214] In addition, in the half cycle when the AC output voltage vo is positive, the value of the capacitor C is set to C=C1+C2+C4, and in the half cycle when the AC output voltage vo is negative, the value of the capacitor C is set to C=C1+C2+C3, thereby enabling research to be conducted using the calculation formula derived so far.

[0215] As described above, in the inverter circuit 100C involved in the fourth embodiment, the on-time tonQ1 calculated by any one of Formula 20 (method 1), Formula 24 (method 2), and Formula 28 (method 3) is set by the tonQ1 calculator 21, and the on-time tonQ2 calculated by Formula 21 (methods 1, 2, and 3) is set by the tonQ2 calculator 22, thereby being able to set the value of the reverse current Ir to any one of Formula 19 (method 1), Formula 23 (method 2), and Formula 27 (method 3), so that the operation can be performed by any one of Methods 1, 2, and 3.

[0216] Therefore, in the inverter circuit 100C according to the fourth embodiment, zero voltage switching can be reliably performed in all phases of the AC output voltage vo, that is, in the entire range of the instantaneous value VO of the AC output voltage vo.

[0217] Description of Reference Numerals

[0218] 100, 100A, 100B, 100C: inverter circuit, 200, 200A, 200C, 200C: control circuit, Co: output capacitor, fs: switching frequency, Ir: reverse current, L: inductor, Q1: first switch (control switch), Q2: second switch (synchronous rectification switch), Q3: third switch, Q4: fourth switch, tonQ1: on-time of control switch, tonQ2: on-time of synchronous rectification switch, Vin: DC input voltage (DC power supply), vo: AC output voltage.

Claims

1. An inverter circuit capable of converting a DC input voltage of a DC power supply into an AC output voltage of a predetermined frequency, the inverter circuit comprising: A series circuit connected in parallel with the DC power supply and having a control switch and a synchronous rectification switch connected in series; A reactor, one end of which is connected to a connection point between the control switch and the synchronous rectification switch; an output capacitor connected between a power line of the DC power supply and the other end of the reactor, and outputting the AC output voltage at both ends; as well as The control circuit performs the following control, namely, after the control switch is turned on for a first conduction time, the control switch is turned off and the synchronous rectification switch is turned on, and after the synchronous rectification switch is turned on for a second conduction time, the synchronous rectification switch is turned off and the control switch is turned on, so that the inductor generates a reverse current in the full range of the instantaneous value of the AC output voltage.

2. The inverter circuit according to claim 1, wherein: The current value of the reverse current is obtained based on the inductance value of the reactor, the capacitance value of the capacitor connected in parallel with the control switch, the detected value of the DC input voltage, and the detected value of the instantaneous value of the AC output voltage.

3. The inverter circuit according to claim 1, wherein: The current value of the reverse current is a fixed value obtained based on the inductance value of the reactor, the capacitance value of the capacitor connected in parallel with the control switch, and the detected value of the DC input voltage.

4. The inverter circuit according to claim 1, wherein: The current value of the reverse current is a value obtained by adding a harmonic component of the frequency of the AC output voltage to a fixed value obtained based on the inductance value of the reactor, the capacitance value of the capacitor connected in parallel to the control switch, and the detected value of the DC input voltage.

5. The inverter circuit according to any one of claims 1 to 4, wherein: The control circuit controls the first on-time and the second on-time based on the detected value of the DC input voltage, the detected value of the instantaneous value of the AC output voltage, and the detected value of the AC output current so that the reverse current becomes a predetermined current value.

6. The inverter circuit according to any one of claims 1 to 4, wherein: The control circuit performs control so that a dead time in which both the control switch and the synchronous rectification switch are turned off is provided during a period from when the synchronous rectification switch is turned off to when the control switch is turned on.

7. The inverter circuit according to any one of claims 2 to 4, wherein: The capacitor connected in parallel with the control switch is a parasitic capacitance of the control switch.

8. An inverter circuit capable of converting a DC input voltage of a DC power supply into an AC output voltage of a predetermined frequency, the inverter circuit comprising: A first series circuit is connected in parallel with the DC power supply and is connected in series with a first switch and a second switch; A second series circuit is connected in parallel with the DC power supply and is connected in series with a third switch and a fourth switch; a reactor, one end of which is connected to a connection point between the first switch and the second switch; an output capacitor connected between a connection point between the third switch and the fourth switch and the other end of the reactor, and outputting the AC output voltage at both ends; as well as The control circuit performs the following control in the half cycle when the AC output voltage is positive, that is, the third switch is turned on and the fourth switch is turned off, and at the same time, after the first switch as the control switch is turned on for a first conduction time, the first switch is turned off and the second switch as the synchronous rectification switch is turned on, and after the second switch is turned on for a second conduction time, the second switch is turned off and the first switch is turned on, and in the half cycle when the AC output voltage is negative, the control circuit performs the following control, that is, the third switch is turned off and the fourth switch is turned on, and at the same time, after the second switch as the control switch is turned on for a first conduction time, the second switch is turned off and the first switch as the synchronous rectification switch is turned on, and after the first switch is turned on for a second conduction time, the first switch is turned off and the second switch is turned on, so that the inductor generates a reverse current in the full range of the instantaneous value of the AC output voltage.

9. The inverter circuit according to claim 8, wherein: The current value of the reverse current is obtained based on the inductance value of the reactor, the capacitance value of the capacitor connected in parallel with the control switch, the detected value of the DC input voltage, and the detected value of the instantaneous value of the AC output voltage.

10. The inverter circuit according to claim 8, wherein: The current value of the reverse current is a fixed value obtained based on the inductance value of the reactor, the capacitance value of the capacitor connected in parallel with the control switch, and the detected value of the DC input voltage.

11. The inverter circuit according to claim 8, wherein: The current value of the reverse current is a value obtained by adding a harmonic component of the frequency of the AC output voltage to a fixed value obtained based on the inductance value of the reactor, the capacitance value of the capacitor connected in parallel to the control switch, and the detected value of the DC input voltage.

12. The inverter circuit according to any one of claims 8 to 11, wherein: The control circuit controls the first on-time and the second on-time based on the detected value of the DC input voltage, the detected value of the instantaneous value of the AC output voltage, and the detected value of the AC output current so that the reverse current becomes a predetermined current value.

13. The inverter circuit according to any one of claims 8 to 11, wherein: The control circuit performs control so that a dead time in which both the control switch and the synchronous rectification switch are turned off is provided during a period from when the synchronous rectification switch is turned off to when the control switch is turned on.

14. The inverter circuit according to any one of claims 9 to 11, wherein: The capacitor connected in parallel with the control switch is a parasitic capacitance of the control switch.

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