Direct ACAC Step-Down Single-Phase Converter with Improved Performances

The paper presents two high-performance single-phase AC chopper circuits which perform AC-AC bidirectional direct conversion;one of them contains resistive loads and the other one is for inductive loads. It includes useful equations for chopper design. The correctfunctioning is tested by simulation. DOI: http://dx.doi.org/10.5755/j01.eee.18.10.1305


I. INTRODUCTION
The direct AC-AC converters have applications in various fields, such as AC motor drives, electronic transformers, switching AC adjustable sources, output voltage waveform restorers, etc.The AC-AC step-down converters of high frequency can replace the AC phase control made up of SCR or triacs.For frequencies of more than 20 kHz, these converters eliminate disturbing noise, have smaller filters, the output voltage is less distorted, irrespective of the load nature, the output voltage variation field is large, the efficiency is adequate and the main current can be practically sinusoidal.
In [1] and [2], the authors propose an AC chopper, with improved commutation, which leads to increased efficiency.However, this requires many switching devices and complex control circuits.The possibility of using sliding-mode control for AC-AC resonant converters is analysed in [3].[4] and [5] present AC choppers with three level converters and [6] suggests several topology versions using commercial power modules for circuit design.In [7] an evaluation methodology for AC choppers is described.
This paper presents a simple circuit of direct AC-AC single-phase buck converter, which allows eliminating downtime in control by an adequate disposition of power switching devices.Moreover, snubber circuits are no longer necessary, the work frequency can be substantially increased (over 20 kHz), the control is simple, the power flux can be bidirectional, the functioning is adequate, irrespective of the load nature, and the efficiency is high.The equivalent circuits are studied in various functioning modes and, thus, we can establish equations which are useful for the design stage.The correct functioning of the proposed circuit was

II. THE TOPOLOGY OF THE AC-AC STEP-DOWN CONVERTER
The inductance L and the capacitor C represent the input high-frequency power supply filter, which eliminates the harmonics of the current absorbed from the power supply i m .
In the positive alternations of the current i o , considering a uniform PWM technique, in the time intervals [0, DT S ], S 1 is in conduction, and in the intervals [DT S , T S ] S 2 is in conduction, while S 3 and S 4 are blocked through the entire alternation.The frequency f S =1/T S is the switching frequency used for chopping the voltage from the power supply.In the negative alternations, S 3 and S 4 are in conduction in the same time intervals, while S 1 and S 2 are blocked throughout these alternations.
The load output voltage is The circuit is simple and can be used for purely resistive loads; the shift in control between the switch pairs occurs when the voltage V, and therefore the current i o , passes through zero.There is no need for downtime when the switch pairs shift, since this occurs when voltage and currents are zero.For inductive loads, which change during functioning, it is necessary to introduce a current transducer, which detects when i o is zero.At that time, it is necessary to shift the switch pairs under control, by using downtime.When the current fundamental i o passes through zero, in a specific variation range of the duty factor D, the buck converter can function in the discontinuous conduction mode (DCM) and there is an inadequate timing of i o =0, which determines an inadequate functioning of the AC-AC converter.
ELEKTRONIKA IR ELEKTROTECHNIKA, ISSN 1392-1215, VOL. 18, NO. 10, 2012 In order to eliminate the drawbacks mentioned above, Fig. 2 shows the proposed circuit for the direct AC-AC converter.As compared to Fig. 1, another inductance is added, which allows a new disposition of the switching devices.Thus it is possible to control simultaneously the switch pairs S 1 -S 3 and S 2 -S 4 and downtime is no longer necessary, as shown in the waveforms in Fig. 3.In the positive alternations of the load current, in the time intervals DT S , i o will flow through S 1 , D 1 , L 1 , Z L , and in the intervals (1-D)T S , it will flow through S 2 , D 2 , L 1 , Z L .In the negative alternations, in the time intervals DT S , i o will flow through Z L , L 2 , D 3 , S 3 , and in the time intervals (1-D)T S , it will flow through Z L , L 2 , D 4 and S 4 .
The waveform of the current i o in Fig. 3 corresponds to an inductive load where R L and L L are the load resistance and the load inductance, respectively, f m is the main frequency of the power supply of the AC-AC converter.
Depending on the time intervals, the circuit in Fig. 2 can be in four states (also marked in Fig. 3): a) the main voltage V m is positive (has the polarity without brakes) and the current i o is positive (its direction is shown in Fig. 2); b) V m is negative (it has the polarity between brakes) and the current i o is positive; c: V m is negative and i o is negative; d: V m is positive and i o is negative.

III. THE EQUIVALENT CIRCUITS OF THE AC-AC CONVERTER
The equivalent circuits correspond to the following simplified hypotheses: the passive components are ideal, the power devices are ideal switches, the voltage V and the load current i o are sinusoidal.
Fig. 4 presents the equivalent circuits and waveforms corresponding to A and B states of the converter.The equations related to the circuits are calculated based on the assumption that, in a switching period T S , the voltage V is constant at the value 2 sin where  For the inductive load, the voltage Vo is chopped (Fig. 4c).In the intervals DT S and (1-D)T S , we consider that the output voltages remain constant respectively at the values: If the circuit is in the A state, in the intervals DT S and (1-D)T S , the voltage V L1 are: Since the average voltage on the inductance L1 is zero: In order to calculate the i L2 current, we first consider Fig. 4(b) in the switching period k-1, corresponding to the intervals (1-D)T S : In the final part of the interval (1-D)T S (Fig. 6), we have   In the DT S interval within the switching period K, according to Fig. 4(a): and in the final part of the interval (t' =DT S ) The waveform of the i L2 current is shown in Fig. 4(c) in continuous line.Under real functioning conditions, the i L2 current becomes 0 before the end of the DT S interval, even if , because of the losses on the real components of the converter.If the circuit is in the B state, we have to consider initially Fig. 4a for the calculation of the i L2 current.In the DT S intervals within the K switching period, v and v o voltages have the polarities indicated between brakes: In the following interval (1-D)T S : In the final part of this interval, t' = (1-D)T S , the i L2 current is The waveform of the i L2 current in B state is shown in Fig. 4(c) in dotted line.
For converter states C and D, when the load current becomes reverse, these equivalent circuits and equations remain valid, except for the role of inductances L 1 and L 2 , which is reversed.
For the resistive load, the voltage V o is no longer chopped and has the value: The converter will be only in states A and C. In the A state, (5) becomes Eq. ( 9) leads to the following equation for the peak current in the switching period K   12 sin The load current and the ratio of these currents in the switching period K are 2 sin The final equation can be used for the design of inductances L 1 and L 2 , with a view to imposing a specific value of the ratio (less than 1%), calculated for the lowest value of the duty factor D.

IV. SIMULATION RESULTS
The correct functioning of the proposed circuits was tested by simulation.In all circuits, the converter was powered by a voltage V m = 24 V, f m = 50 Hz, the components of the power supply filter are: L=0.4 mH, C=5µF and the switching frequency is f S =20 KHz.For the chopper in Fig. 1, used for a purely resistive load R L =5Ω, the converter inductance was L 1 =4 mH.Fig. 5 shows the waveforms of the V o voltage and of the i o current for a duty factor D=0.3, leading to an efficiency η=0.78.Fig. 6 shows the waveforms for D=0.9 and the resulting efficiency is η=0.89.The controlled switch pair was selected according to the V m voltage polarity.
For the chopper in Fig. 2, destined to inductive loads, the simulations were made for L 1 =L 2 =4mH, L L =6 mH, R L =5 Ω. Fig. 7 shows the waveforms of V o voltage and of the currents i m and i o for D=0.3 and Fig. 8 shows the same waveforms for D=0.9.In the first case, the efficiency was η=0.86 and in the second case, η=0.78.

V. CONCLUSIONS
The paper presents two simple circuits of direct AC-AC buck converter; the first contains resistive loads and a single inductance (besides the input filter); the second is for inductive loads and contains two inductances.
The switches can be controlled without downtime and there is no need for snubber circuits.Circuit simulation showed that they function correctly within a high range of values of the duty factor D, and efficiency values obtained are η=0.89for D=0.9 and η=0.78 D=0.3 in the case of resistive loads, and η=0.89 for D=0.9 and η=0.76 for D=0.3 in the case of inductive loads.
The power source voltage was evenly sampled, the control circuits are very simple and the component voltage and currents requirements are normal.The network filter, which eliminates the input current harmonics, is also very simple.
The circuits analysed are single-phase circuits, but a threephase converter can be easily obtained by joining three single-phase converters.A terminal of the load impedance is connected to the network null.

Fig. 4 .
Fig. 4. The equivalent circuits and iL2 current corresponding to A and B states: a) the equivalent circuits in the interval DTS b) the equivalent circuits in the interval (1-D)TS.c) the waveform of the current through the inductance L2 corresponding to the A, B states.