JPH0512810Y2 - - Google Patents

Description

[Detailed explanation of the invention] This invention uses two types of AC voltage (for example, 200V and
This invention relates to a protection circuit in the event of abnormal operation in a DC power supply circuit that switches 100V) as an input and generates a DC voltage as an output.

FIG. 1 is an electrical circuit diagram of a conventional power supply circuit. 1 is an AC power supply that generates, for example, 200V or 100V, and when the voltage is 200V, switch S1 connects to contact a.
When the voltage is 100V, it is switched to contact b. The primary coils 2 and 3 of the transformer are connected between contact a and one terminal of the AC power source 1. The transformer has 4 secondary coils.
used to generate alternating current voltage for other purposes. D1 to D4 are rectifier diodes,
7 and 8 are smoothing capacitors. R ₁ and R ₂ are leakage resistances due to capacitor losses. The middle point 9 of the primary coils 2 and 3 is connected to contact b, and
It is connected to the connection point 10 of the capacitors 7 and 8. Load 12 is connected between one terminal of capacitors 7 and 8. The reason why the two capacitors 7 and 8 are used is to perform voltage double rectification when the AC power supply 1 is 100V. Switch S2 turns on at 200V. Resistor 5, fuse 6, and switch S3 constitute an inrush current reduction circuit when the power is turned on. At power on
S3 is turned off, the rush current is reduced by the resistor 5, and after the current is stabilized, S3 is turned on. Fuse 6 is disconnected if S3 is not turned on even after reaching a stable state (do not disconnect as the inrush current generation period is short).

Here, if the leakage resistances R ₁ and R ₂ are different, the voltage between the terminals of capacitors 7 and 8 may exceed the withstand voltage of the capacitor, or a large capacitor with a large withstand voltage must be used in anticipation of changes in the voltage between the terminals. . However, in such a circuit, the node 10 is connected to the midpoint of the primary coil, and the impedance of the transformer is sufficiently small compared to R ₁ and R ₂ , so the voltage at the node 10 is kept at the midpoint. It can be done. Therefore, the same voltage is generated in the capacitors 7 and 8, and since the output DC voltage is known, the voltage between the terminals will not exceed the withstand voltage.
However, in such a circuit, if the AC power supply is 200V and S3 is not turned on after reaching a stable state and fuse 6 is disconnected, the load current will flow through the midpoint of the transformer to the primary coil, causing the transformer to burn out. There are cases.

That is, when the fuse 6 is cut, the current is
The current flows into the primary coil 3 of the transformer through the coupling point 10 and the middle point 9, and merges with the current flowing from the primary coil 2 side. As a result, the primary coil 3 has the following:
A large current exceeding the allowable current flows and the transformer burns out.

FIG. 2 is an electrical circuit diagram of another conventional power supply circuit, which eliminates the drawbacks of the circuit of FIG. 1. The same parts as in FIG. 1 are given the same reference numerals. The difference between the circuit in Figure 2 and the circuit in Figure 1 is that the midpoint 9 of the transformer and the connection point 10 between the capacitors 7 and 8
Connect switch S4 between and turn on at 100V,
The two features are that it is turned off at 200V, and that resistors R ₄ and R ₅ are connected in parallel with capacitors 7 and 8 (R ₄ = R ₅ , R ₁ , R ₂ >> _{R 4} , R ₅ ). With this configuration, switch S4 is off at 200V, so
Even if the fuse 6 is cut, the transformer will not be damaged by fire. Further, since the coupling point 10 is maintained at the midpoint by the resistors R ₄ and R ₅ , the above-mentioned capacitor problem does not occur. However, in this circuit, a large voltage is constantly applied between both terminals of the resistors R ₄ and R ₅ , which consumes a large amount of power and requires the use of large resistors.

FIG. 3 is an electrical circuit diagram of a power supply circuit according to the present invention. The same parts as in FIGS. 1 and 2 are given the same reference numerals. The circuit in FIG. 3 differs from the circuit in FIG. 1 in that the parallel circuit of switch S4 and resistor 13 is
It is connected between the midpoint 9 of the transformer and the connection point 10. Resistor 13 is a transformer protection resistor. As in the case of Figures 1 and 2, when the power is turned on
S3 is off, the inrush current is reduced by resistor 5, and after reaching a stable state, S3 is turned on. Fuse 6 disconnects when S3 is not turned on.

The operation of maintaining the midpoint of the coupling point 10 of the capacitors 7 and 8 will be explained. At 100V, middle point 9 and connection point 1
0 is connected by switch S4,
As in FIG. 1, the connection point 10 remains at the midpoint.
At 200V, the difference i in leakage current between capacitors 7 and 8
flows through resistor 13 to the midpoint of the transformer. Since the impedance of the transformer is sufficiently small compared to R ₁ and R ₂ , the voltage V ₉ at the midpoint 9 is maintained at the midpoint (the voltages between the terminals of the primary coils 2 and 3 are equal).

The voltage V ₁₀ at the node 10 is V ₁₀ = (1/2) V _DC +
It becomes R ₁₃・i. If the voltage (withstanding voltage) that can be applied to the capacitor 8 is V _C , then V _C > V ₁₀ (or V _DC − V ₁₀ ) = (1/2) V _DC + R ₁₃・i (V _DC , i
must be known). Therefore, V _C =
If R ₁₃ at V ₁₀ is R _a , then R _{13 <} R _a
(Condition 1).

Next, the operation when the fuse is disconnected will be explained. At 100V, switch S4 is turned on, so there is no problem. At 200V, i is the load current. such that i is less than the allowable current of the transformer.
Select _R13 . If R ₁₃ that satisfies this condition is R _b , then R ₁₃ >R _b (condition 2). Therefore, according to conditions 1 and 2, a value of R ₁₃ that satisfies R _b <R ₁₃ <R _a may be selected.

As is clear from the above explanation, capacitor 7,
One resistor 1 holds the center point of 8 and protects the transformer.
This is achieved by 3. In addition, at 100V Switch S4
is on, so there is no power consumption by resistor _R13 , and at 200V, it is usually small, so there is almost no power consumption by resistor _R13 . R ₁₃ can be a small resistor.

[Brief explanation of drawings]

Figure 1 is an electrical circuit diagram of a conventional power supply circuit;
This figure is an electrical circuit diagram of another conventional power supply circuit, and FIG. 3 is an electrical circuit diagram of a power supply circuit according to the present invention. 1...AC power supply, S1-S4...switch, 6...
...Fuyuse, 12...Load.

Claims (1)

[Scope of utility model registration request] An alternating current power source that generates two types of alternating current voltages, a coil having both terminals connected to the alternating current power source and a terminal at a midpoint, a bridge rectifier circuit connected to both terminals of the coil, and a bridge rectifier circuit connected to both terminals of the coil. A current path connecting both terminals of the coil and the bridge rectifier circuit, comprising two capacitors connected in series to the output terminal, a series circuit of a resistor and a fuse, and a switch connected in parallel to the series circuit. and a parallel circuit of a switch and a resistor connected between the midpoint of the coil and the connection point of the capacitor.