JPH05304784A - Double voltage pulse power supply - Google Patents

Abstract

PURPOSE:To prevent a harmful residual charge from occurring by increasing the capacity of a first capacitor which is connected in parallel with a charge power supply out of first and second capacitors which are connected in series to be larger than that of the second capacitor. CONSTITUTION:When a semiconductor switching element 7 conducts current and then the charge of a first capacitor 4 is inverted after the first capacitor 4 and a second capacitor 5 are charged, they are charged initially so that the amount of charge of the first capacitor 4 is equal to that of the second capacitor 5. Namely, capacities C1 and C2 are set so that the capacity of the first capacitor 4 and that of the second capacitor 5 satisfy an equation (CIE<2>-R(t).I (t)<2>)/2=(C2E2)/2 (E; power supply voltage, C1>C2) when a synthesis resistor of the semiconductor switching element 7, a first reactor 3, and the first capacitor 4 is set to R(t) and an inversion current is set to i(t).

Description

Detailed Description of the Invention

[0001]

BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to a voltage doubler type pulse power source suitable for a high voltage pulse power source.

[0002]

2. Description of the Related Art FIG. 2 shows a conventional circuit for obtaining high-voltage pulses, which is generally called a pulse power supply. In FIG. 2, 1 is a charging power source, 2 is a reactor, 3 is a first reactor, and the first reactor 3 has a considerably smaller inductance value than the reactor 2. Reference numeral 4 is a first capacitor, 5 is a second capacitor, and their capacities are equal. 6
Is a second reactor and has an inductance value considerably smaller than that of the reactor 2. Here, the process of charging the first capacitor 4 and the second capacitor 5 will be described. The first capacitor 4 is charged with the voltage E through the reactor 2 → reactor 3 → capacitor 4, that is, along the route A to the polarity shown in FIG. In the second capacitor 5,
It is charged to the polarity shown in FIG. 2 through the reactor 2, the first reactor 3, the second capacitor 5, and the second reactor 6, that is, along the route B. Since the second reactor 6 has a much smaller inductance value than the reactor 2, the charging voltage of the second capacitor 5 is substantially the same as the charging voltage E of the first capacitor 4. A first switch 7 is, for example, a thyristor here. A second switch 8 is a magnetic switch such as a saturable reactor. Reference numeral 9 denotes a load, which corresponds to a laser discharge part or the like. When the thyristor 7 is ignited here, the charge of the first capacitor 4 is inverted through the first reactor 3, and the polarities of the first capacitor 4 and the second capacitor 5 become the same. Therefore, the combined series voltage of the voltages of the first capacitor 4 and the second capacitor 5 is approximately 2E, and therefore the name of doubled voltage is used. The magnetic switch 8 has approximately 2E
Is applied to the magnetic switch 8, which is a saturable reactor, into a saturated conductive state, and the electric charges of the first capacitor 4 and the second capacitor 5 are consumed through the magnetic switch 8 in a load 9, for example, a laser discharge unit. It

[0003]

SUMMARY OF THE INVENTION In the above description,
Although it has been described that the voltage becomes almost E when the charge of the first capacitor 4 is inverted through the switch 7, strictly speaking, the loss of the first reactor 3 and the first switch 7 during the inversion causes the first capacitor 4 to lose its voltage. Is smaller than E.
Therefore, since the capacitances of the first capacitor 4 and the second capacitor 5 are equal, the charge amount Q4 of the first capacitor 4 is smaller than the charge amount Q5 of the second capacitor 5 because the voltage is low. That is, Q4 <Q5. However, the magnetic switch 8
Is conducted and the current flowing to the load 9 is the first capacitor 4,
Since it is common to the second capacitor 5, even if the charge of the first capacitor 4 is lost, the charge remains in the second capacitor 5 having a large amount of charge and residual charge is generated. As an example of the problem caused by the residual charge, the residual discharge of the second capacitor 5 is discharged following the regular discharge in the laser discharge part of the load 9, but the voltage at that time is low due to the residual charge. There is a problem that an arc is generated, which interferes with the regular discharge for the subsequent generation of laser light.
Alternatively, if the residual charge does not have a voltage enough to generate an arc, the residual charge may be repeatedly accumulated, leading to overvoltage breakdown even when the voltage rises. Therefore, an object of the present invention is to provide a voltage doubler pulsed power supply in which this harmful residual charge is not generated.

[0004]

The object is to connect a series circuit including a first capacitor and a second capacitor connected in series, and one end of which is connected to a series connection point of the first capacitor and the second capacitor. A first reactor whose other end is connected to the positive side of the charging power source, and a first node whose anode is connected to the other end of the first reactor and whose cathode is connected to the negative side of the charging power source. Voltage, a second reactor connected in parallel with the series circuit of the first capacitor and the second capacitor, and a series circuit of a magnetic switch and a load connected in parallel with the second reactor. In the pulse power supply, this can be achieved by selecting the capacitance of the first capacitor to be larger than that of the second capacitor.

[0005]

[Function] The first cause of the above-mentioned problems is the first.
Because the charge amounts of the capacitor 4 and the second capacitor 5 are different
Therefore, the charge of the first capacitor 4 is
When inverted, the charge amount of the first capacitor 4 and the second capacitor
Charge the capacitor 5 from the beginning so that the charges are equal.
You can leave it. Inversion circuit for inversion In this case, switch 7
The combined resistance of the first reactor 3 and the first capacitor 4
If R (t) and the reversal current are i (t), then (C1 E² -R (t) -I (t)² ) / 2 = (C2 E² ) / 2 (1) However, the capacitance of the first capacitor 4 is C1, the second capacitor is
The capacity of the capacitor 5 is C2. Naturally C1
> C2. The capacitance C1 of the first capacitor 4
Make sure that the capacitance C2 of the capacitor 5 satisfies the formula (1).
Energy of the first capacitor 4 = C1.E during fluff charging² / 2 energy of the second capacitor 5 = C2.E² It is / 2. Since C1> C2, it is natural that C1
・ E² / 2> C2 · E² / 2. Next, the first con
When the charge of the capacitor 4 is reversed, R (t) • i (t)
² Energy is lost, so after the reversal, the energy of the first capacitor 4 = (C1.E² / 2) -R
(t) ・ i (t)²  Energy of the second capacitor 5 = C2.E² It becomes / 2.

From the relationship of the equation (1), since the stored charge energy of the first capacitor 4 and the stored charge energy of the second capacitor 5 are equal after the reversal, the switch 8 becomes conductive and the load 9
When the electric charges of the first capacitor 4 and the second capacitor 5 are discharged, the discharge current is common and the accumulated charge is the same, so
No electric charge remains in the capacitor 4 of FIG.

[0007]

DESCRIPTION OF THE PREFERRED EMBODIMENTS Next, an embodiment of the present invention will be described with reference to FIG. The present invention has the same original configuration as that shown in FIG. 2, but is different from the prior art in that the capacitance C1 of the first capacitor 4 and the capacitance C2 of the second capacitor 5 are different in the conventional technique.
In the present embodiment, the capacitance C1 of the first capacitor 4 is larger than the capacitance C2 of the second capacitor 5, and the difference is a value satisfying the equation (1). First of all
The first capacitor 4 is charged through the route A and the second capacitor 5 is charged through the route B as in the conventional case of FIG. Its charging energy, the energy = C1 · E ² of the first capacitor 4 / 2 the second of the energy of the capacitor 5 = C2 · E ² It becomes / 2. In the prior art, C1 · E ² / ² = C2 · E 2 / Was 2, but, in the present invention, C1> C2 So, C1 · E ² / ^2> C2 · E 2 It is / 2. Strictly speaking, (C1 · E ² / 2) -R (t) * i (t) ² = C2 · E ² It becomes / 2.

Next, the switch 7 is turned on to turn on the first capacitor.
The electric charge of the sensor 4 through the first reactor 3 switch 7
Invert, but at that time R (t) · i (t)² Lost energy
Be seen. Therefore, after reversal, the energy of the first capacitor 4 = (C1 · E² / 2)-
R (t) ・ i (t)²  Energy of the second capacitor 5 = C2.E² It is / 2. From the equation (1), the energy of the first capacitor 4 and the
The energy of the capacitor 5 of 2 becomes equal.

Then, the switch 8 is turned on and the electric charge of the first capacitor 4 and the second capacitor 5 is discharged toward the load 9, for example, the laser discharge portion. At this time, since the first capacitor 4 and the second capacitor 5 are connected in series, the discharge current is common, and the stored energy of the first capacitor 4 and the second capacitor 5 is equal, which is different from the conventional technique. There is no charge left in the.

[0010]

As described above, according to the present invention, it is possible to solve the problem that electric charge remains in one of the capacitors constituting the double voltage pulse power supply.

[Brief description of drawings]

FIG. 1 is a circuit diagram of a double voltage pulse power supply showing an embodiment of the present invention.

FIG. 2 is a circuit diagram of a conventional double voltage pulse power supply.

[Explanation of symbols]

DESCRIPTION OF SYMBOLS 1 ... Charging power source, 2 ... Reactor, 3 ... First reactor, 4 ... First capacitor, 5 ... Second capacitor, 6
... second reactor, 7 ... semiconductor switching element, 8
... saturable reactor, 9 ... load,

Claims (1)

[Claims] 1. A series circuit including a first capacitor and a second capacitor connected in series, one end of which is connected to a series connection point of the first capacitor and the second capacitor and the other end of which is a positive electrode of a charging power source. A first reactor connected to the charging side, a first switch having an anode connected to the other end of the first reactor, and a cathode connected to the negative side of the charging power source; A double voltage pulse power supply comprising a second reactor connected in parallel to a series circuit of a capacitor and a second capacitor, and a series circuit of a magnetic switch and a load connected in parallel to the second reactor, wherein
The double-voltage pulse power supply is characterized in that the capacity of the capacitor is larger than the capacity of the second capacitor.