Application of single-ended flyback circuit in inverter power supply

　　At present, the inverter power supply powered by batteries generally consists of two stages. The front-stage DC/DC circuit converts the battery voltage into a DC voltage of about 350V, and the rear-stage DC/AC circuit converts the DC 350V voltage into an AC 220V voltage. In this type of inverter power supply, the front-stage DC/DC circuit generally has a low supply voltage (12V, 24V or 48V), a large input current, a high power tube conduction voltage drop, and a large loss, so it is difficult to improve the power supply efficiency. Its circuit forms include: single-ended flyback, single-ended forward, dual-tube forward, half-bridge and full-bridge, etc. For small and medium power (about 0.5~1kW), the single-ended flyback circuit has certain advantages, such as: simple circuit, convenient control, high efficiency, etc. This article takes 24V battery power supply and output 350V/1kW as an example to discuss the application of single-ended flyback circuit in the front-stage DC/DC circuit of the inverter power supply.

　　1 Conventional single-ended flyback circuit structure

　　The structure of a conventional single-ended flyback circuit is shown in Figure 1. The disadvantage of this circuit is that when the power tube VT is cut off, the reverse peak energy of the transformer primary is consumed by the absorption circuit composed of VD1, C1 and R1; and under the same output power, the current passing through the power tube (relative to multiple tubes in parallel) is large, the conduction voltage drop is high, and the loss is large, so the efficiency and reliability are low.

　　Figure 1 Conventional single-ended flyback circuit structure

　　2 Single-ended flyback circuit structure with multiple tubes in parallel

　　As shown in Figure 2, the characteristic of this circuit is that the main power circuit uses 4 power tubes in parallel, and the current passing through each power tube is 1/4 of that when a single tube is used (assuming that the parameters of the 4 power tubes are consistent), then the conduction voltage drop of the power tube should also be 1/4 of that when a single tube is used. According to calculations, when the output is 550W, theoretically, 4 tubes in parallel can reduce the conduction loss by about 20W compared to a single tube, and improve the efficiency by nearly 3 percentage points.

　　Figure 2 Main power circuit with 4 power tubes connected in parallel

　　3 Single-ended flyback circuit structure using energy feedback technology

　　The structure of the single-ended flyback circuit using energy feedback technology is shown in Figure 3, and its main waveform is shown in Figure 4. In this circuit, the transformer primary reverse peak absorption circuit is composed of capacitor C2, inductor L1, diodes VD1 and VD2, which can feed back most of the reverse peak energy to the input capacitor C1, reducing energy loss and improving circuit efficiency.

　　Figure 3 Primary anti-peak absorption circuit

　　Figure 4 Main waveforms of primary anti-peak absorption circuit

　　Here’s how it works:

　　(1) t0~t1 stage.

　　At t0, the power tube is cut off, the transformer primary inductance L, leakage inductance LK, capacitor C2 and power tube output capacitor C0 begin to resonate, and quickly make the C2 voltage reach U0 (N1/N2), then the secondary diode is turned on, the primary voltage is clamped to U0 (N1/N2), the primary inductance L exits resonance, and at t1, IK is 0, and the voltage on C2 and C0 reaches the maximum value, that is, the switch tube voltage US reaches the maximum value (UIN+U C2MXA).

　　(2) Stage t1~t2.

　　LK, C2, and C0 continue to resonate, and at the same time, the inductor L1 participates in the resonance. C2 and C0 feed back energy to the input capacitor C1 and replenish energy to L1. At time t2, the resonance stops and the voltage of C2 drops to U0 (N1/N2).

　　(3) t2~t3 stage.

　　Starting from time t2, inductor L1 feeds back energy to input capacitor C1.

　　The voltage of C2 is clamped at (N1/N2)U0, and the voltage on C0, that is, the switch tube, is UIN+(N1/N2)U0, both remain unchanged. At time t3, the energy in L1 is completely released.

　　(4) t3~t4 stage.

　　The switch tube is completely cut off, and the C2 voltage and C0 voltage (i.e. the switch tube voltage) continue to remain unchanged.

　　(5) t4~t5 stage.

　　At t4, the power tube is turned on, its voltage US begins to drop, C0 starts to discharge through the switch tube, and the discharge is quickly completed (all losses are in the power tube); C2 and L1 begin to resonate, that is, the energy in C2 is transferred to L1. At t5, the current in L1 reaches its maximum value, and the power tube is fully turned on.

　　(6) t5~t6 stage.

　　At t5, L1 feeds back energy to input capacitor C1 through VD1 and VD2, and charges C2 to -UIN. At t6, the energy in L1 is completely released.

　　(7) t6~t7 stage.

　　During this stage, the power tube continues to be in a fully on state.

　　The above process forms a complete working cycle. It can be seen that most of the energy in the transformer leakage inductance is fed back to the input capacitor C1 (part of the energy in C0 is consumed), so the power supply efficiency is improved.

　　4 Calculation of voltage and current stress of main components

　　From Figure 3 and principle analysis, the following calculation formula can be obtained:

　　Among them: USMAX, i.e., U C0MAX, is the maximum voltage stress borne by the power tubes VT1~VT4;

　　U INMIN is the minimum input voltage (take 21V);

　　U0 is the output voltage (take 350V);

　　N1 and N2 are the primary and secondary turns of the transformer (15 and 117 respectively);

　　△U C2 is the peak voltage caused by leakage inductance;

　　I PK is the leakage inductance, i.e. the primary peak current;

　　LK is the primary leakage inductance (take 0.4μH);

　　C2 is an external capacitor (30000pF);

　　C0 is the sum of the output capacitances of VT1~VT4 (take 4000pF);

　　I PAV is the average total current during the conduction period of the power tube;

　　η is the power efficiency (taken as 92%);

　　D MAX is the maximum duty cycle (take 0.7);

　　△Ip is the current change during the conduction period of the switch tube;

　　tONMAX is the maximum on-time of the switch (taken as 23μs);

　　L is the primary inductance of the transformer (38μH);

　　I L1MAX is the maximum current passing through L 1 (take 0.5mH);

　　PLK is the energy fed back to the input by the leakage inductance;

　　f is the power tube switching frequency (30kHz).

　　By deducing and simplifying equations (1) to (6) above, we can obtain the following equation:

　　From equations (7) to (11), we can calculate the current and voltage stresses on the power tube and inductor L1 (when the output power is 550W) and the energy fed back to the input by the anti-peak absorption circuit:

　　I PK=47A

　　USMAX=188V

　　I L1MAX=1.5A

　　P LK=13.25W

　　At the same time, it can be seen from (7) to (11) that:

　　(1) To reduce the switch current stress IPK, the duty cycle D and the transformer primary inductance L should be increased;

　　(2) If you want to reduce the switch tube voltage stress U SMAX, you should reduce the transformer primary leakage inductance LK and increase the C 2 value (the C 0 value is determined by the power tube parameters);

　　(3) If you want to reduce the maximum current I L1MAX in inductor L1, you should increase the inductance of inductor L1; (4) After using the anti-peak absorption circuit, 13.25W of energy can be saved and the power supply efficiency can be improved by about 2 percentage points.

　　From the above calculation, we can know that the rated current of the four power tubes should be at least greater than 50A. Considering the differences in the parameters of the power tubes, their on-currents are not completely equal, and a certain safety margin is generally required. Therefore, in actual application, the rated current value of each power tube should be greater than 50A, the smaller the on-resistance, the better, and the withstand voltage should be greater than 250V.

　　According to the following formula, the voltage stress U D0 and current stress I SK borne by the diode VD0 can be calculated:

　　From U DO=U 0+U INMAXN 2/N 1 (12)

　　So: U DO = 584V

　　From I PKN 1 = I SKN 2 (13)

　　So: I SK = 6A

　　Where: I SK is the secondary peak current value.

　　Generally, a certain safety margin should be left, so the rated voltage of the selected diode should be greater than 800V, and the rated current should be greater than 20A (taking into account factors such as overcurrent and short circuit).

　　5 Two-way single-ended flyback parallel circuit structure

　　If you want to increase the output power, use the parallel structure shown in Figure 5. This circuit structure can output about 1.1kW of power and can be controlled by one SG3525.

　　Figure 5 Two-way single-ended flyback parallel circuit structure

　　6 Test results

　　The inverter power supply front-stage DC/DC circuit (see Figure 5) consists of two single-ended flyback circuits connected in parallel, with an output power of about 1.1kW. The test results are shown in Table 1.

　　Table 1 Pre-stage DC/DC test results

　　The 1kVA inverter power supply composed of the above DC/DC circuit outputs AC220V50Hz sine wave. The test results are shown in Table 2. The volume of the power supply is 320×200×60mm3.

　　Table 2 1kVA inverter power supply test results

　　7 Conclusion

　　In summary, for low-voltage input inverter power supplies powered by batteries (or generators), the front-stage DC/DC implemented by single-ended flyback multi-tube parallel connection and energy feedback technology has the characteristics of simple circuit, convenient control, high efficiency, small size and high reliability compared with the front-stage DC/DC implemented in other forms.