JPS62166767A - Power supply circuit - Google Patents

Abstract

PURPOSE:To prevent load voltage oscillation from being generated, by connecting a condenser and a load in parallel with each other, to a reactor in series, and by breaking and making a first switch for feeding DC form a battery and a second switch for short-circuiting the reactor and the load side, in the alternately reverse directions. CONSTITUTION:A first switch 2 is made, and current is made to flow from a battery 1 to a reactor 3 and to a wiring resistance R and to the parallel circuit of a condenser 4 and a load 5. Voltage applied to the load 5 is increased higher than steady-state voltage due to transient phenomenon. When the voltage of the load 5 comes to the steady-state voltage, then the first switch 2 is turned OFF, and at the same time, a second switch 8 is turned ON. The current of the reactor 3 is increased after the first switch 2 is turned ON, but the current is reduced at the same time when the first switch 2 is turned OFF. Current for magnetic energy due to the current change flows to the reactor 3 and to the resistance R and to the condenser (4//5) and to the switch 8, and is reduced. So far as the voltage of the condenser 5 is concerned, the quantity of discharge is almost equal to the quantity of charge, and the voltage is not changed. As a result, the first switch 2 is repeated to be turned ON, and the second switch 8 is repeated to be turned OFF, and electric oscillation is prevented from being generated.

Description

DETAILED DESCRIPTION OF THE INVENTION Field of Industrial Application The present invention relates to a power supply circuit that supplies power from a suitable DC power source to a suitable load in which capacitors are connected in parallel through a circuit element having an inductance, and in particular to a power supply circuit that supplies power from a suitable DC power source to a suitable load in which capacitors are connected in parallel. The present invention relates to a power supply circuit that does not delay the time for the voltage applied to a load to reach a predetermined voltage at the start of supply, and can suppress vibrations in the voltage applied to the load.

BACKGROUND OF THE INVENTION A conventional power supply circuit of this type, as shown in FIG. 4 < Mt and a load 5 (having a conductance G) connected in series is known.

Note that in FIG. 10, the symbol R represents the resistance of the wiring. In such a circuit configuration, in order to supply power to the load 5, the timing is set as shown in FIG. 12(a). When switch 2 is closed, the current i changes from 1 → 2 → 3 → R → (4/15
) → 1, and power supply to the load 5 is started. The voltage V applied to the load 5 at this time is shown in FIG.
As shown by the solid line 6 in b), the transient response during the transition period from when the current i reaches the steady-state value is often oscillatory. Here, the response of the circuit shown in FIG. 10 is divided into three cases depending on the values of its inductance L, resistance, capacitance C, and conductance G, and becomes as follows. In the power supply circuit shown in FIG. 10, the resistance R is generally selected to be as small as possible in order to improve the efficiency of power transmission. Therefore, in the above equation, the value on the left side is smaller than the value on the right side, resulting in the oscillation of (3).

In order to suppress such vibrations in the voltage applied to the load 5,
The conventional power supply circuit has a resistor R as shown in FIG.
Connect a resistor R' sufficient to suppress the vibration of the applied voltage in series with "2", and in this state close switch 2 to start power supply, and after suppressing the vibration, short-circuit the resistor R'. things were being done.

Problems to be Solved by the Invention However, in such a power supply circuit, due to the action of the resistor R', as shown by the chain line 7 in FIG. However, the rise time t until the voltage applied to the load 5 becomes equal to a predetermined voltage, for example, the electromotive force E of the DC source 1, is delayed. Therefore, especially when there is a limit to the upper 4L time depending on the f-P of load 5, it cannot be applied. Therefore, the present invention
In particular, it is an object of the present invention to provide a power supply circuit that can suppress vibrations in the voltage applied to the load without delaying the time when the voltage applied to the load reaches a predetermined voltage at the start of power supply.

Means for Solving the Problems The means of the present invention for solving the above problems will be explained based on the drawings.

FIG. 1 is a circuit diagram showing the principle of a power supply circuit according to the present invention. This power supply circuit consists of a DC power supply 1 with an electromotive force E, a first switch 2, a circuit element such as a coil 3 with an inductance L, and a load 5 (with a conductance G) connected in parallel to a capacitor 4 with a capacitance C. A second switch 8 is provided in parallel with a load 5 in which the coil 3 and the capacitor 4 are connected in parallel. Note that in FIG. 1, the symbol R represents the resistance of the wiring.

In such a circuit configuration, in order to supply power from the DC power source 1 to the load 5, first, as shown in FIG. 2(a), time t is reached. Close the first switch 2 and turn it on. Then, in Figure 1, 1→2→3→1ku→(4,/15)
→A current flows through the circuit No. 1, and power supply to the load S starts. At this time, the voltage V applied to the load 5 rises as shown by the solid line 9 in FIG. It shoots and produces vibration.

Therefore, at the time when the voltage V applied to the load 5 becomes equal to a predetermined voltage, for example, the electromotive force E of the DC power supply 1, the first switch 2 is opened and turned off as shown in FIG. 2(n). At the same time, the second switch 8 is closed and turned on as shown in FIG. 2(b). Furthermore, during the period from time t0 to time t, the current i flowing through the coil 3, that is, the inductance L in FIG. Steady value i=E
/R=EG is a larger value. And, due to the magnetic energy caused by this current change, ′
Upstream i is 3→R→(4/15)→8→ in FIG.
It flows through the 3rd circuit. However, since the first switch 2 is turned off, the power supply from the DC power supply 1 is stopped, and the power supply to the load 5 is only the magnetic energy possessed by the inductance B, as shown in Fig. 2(d). As shown by a solid line 13, the current i decreases rapidly. At this time, the voltage of the capacitor 4 does not change much because the amount of discharge to the load 5 and the amount of charge of the inductance L due to magnetic energy are approximately equal.

In this state, as shown in FIG. 2(d).

At time t2 when the current i flowing through the inductance L becomes equal to the steady-state value EG, the second switch 8 is opened and turned off as shown in FIG.
As shown in Figure (a), the first switch 2 is closed and turned on. Then, the current i changes again from 1→2→
It flows in the circuit of 3→R→(4/15)→1. At this time,
As mentioned above, the voltage of capacitor 4 does not change much and the voltage of capacitor 4 does not change much.
The current i in the inductance L has a steady value EG shown in the solid line 11 in FIG. 2(d), so the coil 3 with the inductance L There is no energy transfer between the capacitor 4 of capacitance C and the voltage V applied to the load 5 as shown in Fig. 2 (
It does not cause vibration like C). Therefore, according to the power supply circuit of the present invention shown in FIG. 1, as shown by the solid line 9 in FIG. There will be no delay, and
Applied to the load 5: vibration of the a pressure can be suppressed.

Embodiments Hereinafter, embodiments of the present invention will be described in detail with reference to the accompanying drawings.

FIG. 3 is a circuit diagram showing a first embodiment of the power supply circuit according to the present invention. In this embodiment, a DC power source 1 with an electromotive force E
, a full-bridge inverter 15, a circuit with an inductance L, and a load 5 (with a conductance G) in which a transformer 16, diodes 17 to 20 forming a full-wave rectifier, and a capacitor 4 with a capacitance C are connected in parallel. It consists of Note that the capacitor 4 is
This is for smoothing the output voltage of a full-wave rectifier made up of diodes 17-20.

The full-bridge inverter 15 converts the direct current of the direct current power supply 1 into alternating current and supplies alternating current power to the load 5, with base currents a and b, respectively.

Transistor 21.c and d are turned on by flowing them.
22, 23, 24 and these transistors 21 to 24
are connected in antiparallel to each of transistors 1 to 21.
Flywheel diode protecting ~24 25.2G
, 27.28.

The 1-transistors 21 and 22 form one current path A of this full-bridge inverter 15, and the other transistors 23 and 24 form the other current path B. The operation in which transistors 21 and 22 are turned on simultaneously, and the transistor 23 in the other current path I3
AC power is supplied to the load 5 by alternately repeating the operation of turning on and 24 at the same time.

Note that a rest time is provided between the operations in which one current path A and the other current path B are turned on alternately, and the transistors 21 and 22 of one current path A and the other current path B are turned on. The transistors 23 and 24 are prevented from turning on at the same time.

Here, the full bridge type inverter 15 functions as the first switch 2 and the second switch 8 in the circuit shown in FIG. In other words, when transistors 21 and 22 forming one current path are turned on, for example, the fourth
As shown by solid lines in the figure, DC power supplies 1 to 21. L, 17
It supplies power to the parallel connection of capacitor 4 and load 5 through 20° 22 and returns to the DC power supply 1 through 20°22. This means that in the circuit diagram shown in Figure 1, the current i is 1→2→3→l<(
1/15)→Equivalent to the case where it flows in the 1st circuit, and the 3rd
The transistor 21 in the figure corresponds to the first switch 2 in FIG. 1, and the diode 28 in FIG. 3 corresponds to the second switch 8 in FIG. In addition, when the transistors 23 and 24 forming the other current path B are turned on, for example, as shown by the broken line in FIG. supply electricity, 19
．． L.

It returns to the DC power supply 1 via 24. This means that in the circuit diagram shown in Figure 1, the current i is 1 → 2 → 3 → R → (4/15)
→It is equivalent to the case where the current flows in circuit 1, and the transistor 23 in FIG. 3 corresponds to the first switch 2 in FIG.
The diode 2G in the figure corresponds to the second switch 8 in FIG. Note that for one of the above ffE 'dtl paths A, the transistor 22 in FIG. 3 is the first switch 2 in FIG. 1, and the diode 27 in FIG. 3 is the second switch 8 in FIG. is also possible. Regarding the other current path 13, the transistor 24 in FIG. 3 is the first switch 2 in FIG. 1, and the diode 25 in FIG. 3 is the first switch 2 in FIG.
It is also possible to use the second switch 8 shown in the figure.

Next, the operation of such a circuit configuration will be described with reference to FIG. 4, which shows the flow of current in part of the inverse conversion operation of the full-bridge inverter 15, and FIG. 5, which shows the timing thereof. Here, the reference numerals in FIG. 4 indicate the leakage inductance of the transformer 16 in FIG. 3. Also, the above transformation:! ) l The constant on the secondary side of 13 is the constant of the transformer 16
It is converted to the primary side according to the turns ratio n of .

First, as shown in FIGS. 5(a) and 5(b), the time is set. The base currents a and b are connected to transistors 21 and 22, respectively.
is input to turn on the transistors 21 and 22. As a result, current flows through one current path A of the full-bridge inverter 15 shown in FIG. 3, and the first switch 2 shown in FIG. 1 is turned on. Then, as shown by the solid line in Figure 4, 1 → 21 → L → 17 (4/15) → 20
A current flows through the →22→1 circuit (hereinafter referred to as "loop 1"), and power supply to the load 5 is started. At this time, the load current i, as shown in FIG. 5(e), rises rapidly due to the leakage inductance of the transformer 16 and the capacitance C of the capacitor 4, and the output power of the transformer 1G is 5 o'clock as shown in Figure 5 (1). Changes to the opposite (changes to sex) with respect to the previous state.

Meanwhile, at the above time. The previous shi =, Garashi. Until,
As shown in FIGS. 5(a) to 5(d), there is a rest time, and each of the 1-transistors 21 to 24 is caused by a short circuit current caused by the 1-transistors 21, 22 and 23°24 being turned on simultaneously. so as not to damage it. Therefore, at this time t-□ mustard. Until then, the power supply from the DC power source 1 to the load 5 is stopped, and power is supplied to the load 5 only by discharging the capacitor 4, so the voltage of the capacitor 4 (
(approximately equal to Vx) decreases. However, the above time 1. After that, the output voltage V of the transformer 16 becomes reverse polarity and increases, and the output voltage V' (=v/n
) becomes equal to the reduced pressure of the capacitor 4, the capacitor 4 is charged again.

Next, at the time □ when the voltage Vx applied to the load 5 becomes equal to a predetermined voltage, for example, the electromotive force E of the DC power supply 1, the base current a is stopped and one side is turned off, as shown in FIG. 5(a). One transistor 21 in current path A is turned off.

At this time, diodes 1 to 28 connected in antiparallel to transistors 1 to 24 are forward biased and turned on. As a result, the first switch 2 shown in FIG. 1 is turned off, and the second switch 8 is turned on. Then, the current that was flowing in loop ■ until now becomes L → 17 → (4/1
5)→20→22→28→L circuit (hereinafter referred to as “loop ■”)
) is d. In this circuit.

The power supply from the DC power supply 1 has been stopped, and the power supply to the load 5 is limited to the magnetic energy of the inductance L and the discharge from the capacitor 4, and as shown in FIG. 5(e), the leakage inductance L The flowing load current i, decreases rapidly.

In such a state, as shown in FIG. 5(e), at time t2 when the load current IL becomes equal to the value obtained by dividing the electromotive force E of the DC power source 1 by the resistance value of the load 5, as shown in FIG. (a
), the base current a is input to the transistor 21 to turn the transistor 21 on again. Then, the current that was flowing in loop ■ until now flows in loop ■ again. At this time, the load' current iL is the value obtained by dividing the electromotive force E of the DC power supply 1 by the resistance value of the load 5 (iL=E/R=EG)
The flow, that is, the transient phenomenon ends, and the load becomes steady at 5.
This is the value that supplies power to. Also, capacitor 4
The voltage is charged by the magnetic energy of the leakage inductance L while supplying power to the load 5 between time Li and L2.

becomes equal to the electromotive force E of the DC power supply 1 without any change.

That is, the voltage applied to the load 5 is equal to the voltage Vx+J voltage E. Then, at this time point 2, the above loop I
The voltages and currents inside are all steady values, and no transient phenomena occur. Therefore, in order to reverse the polarity of the full-bridge inverter 15, the transistors 21 and 22 of one current path A are turned off at time ta (Fig.
) and (b)), the voltage Vx applied to the load 5 does not oscillate as shown in FIG. 5(g), and power can be steadily supplied to the load.

Next, fiil from time t to t,4 is considered as a pause time, as shown in FIGS. 5(a) to 5(d).

The base currents a, b, c, and d all stop, and the transistors 21.22 in one current path A and the transistors 23.24 in the other current path B are turned off, and power is supplied from the DC power source 1 to the load 5. stops.

Thereafter, at one time t4, as shown in FIGS. 5(c) and 5(d), the base currents C and d are respectively connected to the transistors 23.
24 to turn on the transistors 23 and 24.

As a result, the full bridge type inverter 1 shown in FIG.
5. Current flows through the other flow path B of the full bridge type inverter IS.As shown in FIG. A voltage is output.Thereafter, from time ts to time t, one of the transistors 1 to 23 in the other -B flow path B is temporarily turned off, and at time t6, it is turned on again at the above-mentioned time. mustard,
The same operation as before is performed, and the vibration of the voltage VX applied to the load 5 is suppressed.

By repeating the above operations, the voltage Vx applied to the load 5 becomes a full-wave rectified waveform of the output of the full-bridge inverter 15, as shown in FIG. 5(g). When the polarity of the full-bridge inverter 15 is reversed, the power supply from the direct dε power source 1 to the load 5 is temporarily stopped, and the voltage Vx applied to the load 5 is slightly reduced.
Voltage oscillations that occur when the polarity is reversed can be suppressed.

In this first embodiment, in the first cycle in which the full-bridge inverter 15 starts operating and in the subsequent second cycle 1], the full-bridge inverter 15 is turned on and temporarily turned on. The time until it turns off (for example, from time L0 until...).

or from time t4 to t5) and its temporary off time (for example, from time L□ to t2, or from time t to 1G). That is, in the first period ['' where the operation starts, the period a, 9 (from to to 1', , ) until the operation is temporarily turned off, and the temporary off time (from shi to tz) in the second period. Make it longer than those after 11. This is due to the difference in voltage across the capacitor 4; when the full-bridge inverter 15 starts operating, the voltage across the capacitor 4 is zero;
This is because the voltage of the capacitor 11 is charged to a value substantially equal to the steady-state value E after the operation of the first period [1 is started.

FIG. 6 is a circuit diagram showing a second embodiment of the present invention. In this embodiment, the full bridge inverter 15 of the embodiment shown in FIG.
That is.

The push-pull inverter 29 reversely converts the DC of the DC power source 1 into AC to supply AC power to the load 5, and the transistors 31, 32, 33, and 34 are turned on by flowing base current e+f+g+h, respectively. ,
A diode 3 is connected in antiparallel to the transistors 31 and 33 and protects each transistor 31 and 33.
5.37, and diodes 36.38 for providing reverse breakdown voltage to the transistors 32.34. The transistors 31 and 32 form one current path of this push-pull inverter 29, and the other transistors 33 and 34 form the other current path. and the transistor 33 in the other current path are turned on alternately, and by repeating this operation, alternating current is applied to the high voltage transformer 30, and alternating current power is supplied to the load 5. Here, the push-pull type inverter 29 operates as the first switch 2 and the second switch 8 in the circuit shown in FIG.
The transistor 31 corresponds to the first switch 2, and connects the center tap 30a of the primary winding of the high voltage transformer 30 to the other end of the primary winding. 30b corresponds to the second switch 8.

Next, the operation of such a circuit configuration will be described with reference to FIG. 7, which shows the timing of base current e=h input to each transistor 31 to 34 of push-pull type inverter 29. First, as shown in Figure 7(a),
Time. Then, the base '4 current C is input to the transistor 31 on one current path to turn on the transistor 31. As a result, the first switch 2 shown in FIG. 1 is turned on.

Then, in Fig. 6, 1→30a→17→(4//
5) −+ 20430 b −+ 31 →i
(7) Current flows through the circuit, charging the capacitor 4 and starting to supply power to the load 5.

In this way, at the time □ when the voltage applied to the capacitor 4 and the load 5 becomes equal to the predetermined voltage obtained by boosting or stepping down the DC power supply 1 by the turns ratio of the high voltage transformer 30, the voltage applied to the capacitor 4 and the load 5 becomes equal to the predetermined voltage as shown in FIG. 7(a). As shown in FIG. 7(b), at the same time as the base current C is stopped and the transistor 31 is turned off, the base current f is manually applied to the 1-transistor 32 to turn on the transistor 32. As a result, the first switch 2 shown in FIG. 1 is turned off, and the second switch 8 is turned on. Then, the current is.

30-+ 17-+ (4/15) →20→3O-)
32-) Flows in the circuit from 36 to 30, and the above DC 'I' source 1
The power supply from will be stopped. Here, set the time. Gara t□
The load current flowing through transistors 1 to 31 greatly exceeds the steady value as shown in FIG. 5(e), but at time t□, the power supply from DC power supply 1 Since it stops, the load current flowing through the I transistor 32 rapidly decreases as shown in FIG. 5(c).

Next, at time 2 when the load current flowing through the transistor 32 becomes equal to the value obtained by dividing the electromotive force E of the DC power source 1 by the value of the load resistance converted to the primary side of the high voltage transformer 30, as shown in FIG. As shown in (a), the base +t current e is input to the 1-transistor 31 and the transistor 31 is turned on again. Then, the diode 36 is reverse biased, and the load current flowing through the transistor 32 and the diode 36 is commutated to the transistor 31. Therefore, as shown in FIG. 7(b), the base ff1M5f is stopped and the transistor 32 is turned off. At this time, DC power supply 1
The load current flowing from the load reaches a steady value as shown in FIG. 5(b), and the heavy pressure applied to the load 5 also reaches a predetermined heavy pressure as shown in FIG. 5(g). No transient phenomenon occurs.Therefore, until the time when the polarity of the push-pull type inverter 29 is reversed, the pressure applied to the load 5 is steadily applied to the load without causing vibration. Can be supplied.

Next, the period from time t to t4 is a rest time for reversing the polarity of the push-pull inverter 29, and as shown in FIGS. e
= h is stopped, all transistors 31 to 34 are turned off, and the power supply from the DC power supply 1 to the load 5 is stopped.6 Then, at time t, as shown in FIG. 7(C), the base current g is the transistor 3 in the other current path.
3 to turn on the transistor 33. As a result, the polarity of the push-pull type inverter 29 shown in FIG. 6 is reversed. After that, 1 full from time t, to t,
The transistor 33 on the other current path is temporarily turned off,
It is turned on again at time t, and from time tl to t.

The same operation as before is performed, and the vibration of the voltage applied to the load 5 is suppressed.

By repeating the above operations, the voltage applied to the load 5 becomes a full-wave rectified waveform of the output of the push-pull inverter 29, as shown in FIG. 5(g).

In this second embodiment, the base current f of the transistor 32 in one current path or the base current f of the transistor 34 in the other current path is indicated by a broken line in FIGS. 7(b) and 7(d). As shown, it may be manually applied in the same period as the base current C of the transistor 31 in one current path or the base current g of the transistor 33 in the other current path. This means that when the transistor 31 or 33 is on, the diode 36 or 3
8 are each reverse biased, so transistor 32
Or, even if base current f or h flows through 34, it will not turn on.

FIG. 8 is a circuit diagram showing a third embodiment of the present invention. In this embodiment, the load 5 in the embodiment shown in FIG. 3 is an X-ray tube 39, and its operation is exactly the same as the timing chart shown in FIG. The waveform of the voltage (tube voltage) applied to the X-ray tube 39 by this operation is the 9th rf! ! It becomes as shown in J. That is, in the past, as shown by the broken line 40, a large overshoot occurred at the rise of the tube voltage of the X-ray tube 39, and oscillations continued to occur every time the AC power was reversed. When the present invention is applied, as shown by the solid line 41,
Overshoot I~ does not occur when the tube voltage rises, and voltage oscillations can be suppressed even if the polarity of the full bridge inverter 15 is reversed. For example, in order to quantitatively demonstrate such effects, we compare the tube voltage pulsation by the ratio of the difference between the maximum tube voltage value and the minimum tube voltage value to the maximum tube voltage value. In the case of a current of 100 mA, it was about 28% in the conventional case, but when the present invention is applied, it can be reduced to about 10%.

Effects of the Invention As explained below, the present invention uses a switch connected in series to a DC power source 1 as the first switch 2, and a circuit element (:3) having an inductance L and a capacitor 4 in parallel. Since a second switch 8 is provided in parallel with the load 5 connected to the load 5, the first switch 2 is closed and the load 5 is connected to the
When the voltage applied to becomes equal to the predetermined value 1'a) U: and !113, the first switch 2 is opened and at the same time the second switch 8 is closed, causing the current to flow through the circuit element (3) having the inductance L. By opening the second switch 8 and simultaneously closing the first switch 2 when the current i becomes approximately equal to a predetermined value, the energy that causes an overshoot at the start of power supply to the load 5 is transferred to the load 5. It is possible to suppress vibrations in the applied voltage while supplying the voltage.

Therefore, the time for the voltage applied to the load 5 to reach the predetermined voltage at the start of power supply is not delayed from the normal rise time, and it is possible to suppress vibrations in the voltage applied to the load S. From this, the effects of the present invention are remarkable when the rise time for the applied voltage to reach a predetermined voltage is limited depending on the type of load 5.

[Brief explanation of drawings]

Fig. 1 is a circuit diagram showing the principle of the power supply circuit according to the present invention, Fig. 2 is a timing diagram showing its operation, Fig. 3 is a circuit diagram showing the first embodiment of the invention, and Fig. 4 is a circuit diagram showing the principle of the power supply circuit according to the present invention. Part of the inverse conversion operation of the full-bridge inverter shown in Figure 3'1
A circuit showing the flow of E style [A1, Fig. 5 is a timing diagram showing the operation of the circuit shown in Fig. 3, Fig. 6 is a circuit diagram showing the second embodiment of the present invention, Fig. 7 8 is a timing diagram showing its operation, FIG. 8 is a circuit diagram showing the third embodiment of the present invention, FIG. 9 is a graph showing the waveform of the voltage applied to the X-ray tube in FIG. 8, and FIG. 1 and 11 are circuit diagrams showing conventional power supply circuits, and FIG. 12 is a timing diagram showing their operation. 1... DC power supply 2... First switch 3... Coil (circuit element with inductance L) 4... Capacitor 5... Load 8... Second switch 15... Full Bridge type inverter 16...Transformers 21-24...Transistors 25-28...Diode 29...Push-pull type inverter 3o...High voltage transformer 30a...Center of the primary winding of the high voltage transformer Taps 30b, 30c...The other terminals of the primary winding of the high voltage transformer 31-34...Transistors 35-38...Diode 39...X-ray tube

Claims (4)

[Claims] (1) A DC power source, a switch, a circuit element with inductance, and a load connected in parallel are connected in series, and by closing the switch, power is supplied to the load via the circuit element with inductance. In a power supply circuit that supplies a power supply, a switch connected in series to the DC power source is a first switch, and a second switch is connected in parallel to a load in which a circuit element having the inductance and the capacitor are connected in parallel. A switch is provided, and when the first switch is closed and the voltage applied to the load becomes approximately equal to the voltage of the DC power supply or the predetermined voltage to be applied to the load, the first switch is opened and at the same time the second switch is opened. When the switch is closed and the current flowing through the circuit element having the inductance becomes approximately equal to the value to be supplied to the load, the second switch is opened and at the same time the first switch is closed. supply circuit. (2) The first and second switches are configured in a full-bridge inverter, and the first switch connects two current paths connected in series with the load in the reverse conversion operation of the full-bridge inverter. A switch for one of the current paths, and the second switch is a diode connected in antiparallel to a switch for the other current path connected in series with the first switch. A power supply circuit according to claim 1. (3) The first and second switches are configured in a push-pull type inverter, the circuit element having an inductance is a transformer having a center tap, and the second switch is configured in a push-pull type inverter, and the second switch is a transformer having a center tap. 3. The power supply circuit according to claim 1, wherein the power supply circuit is connected between a center tap of a winding and the other terminal of the primary winding. (4) The power supply circuit according to any one of claims 1 to 3, wherein the load is an X-ray tube.