JPS5819919A - Controller for dc feeding device - Google Patents

Abstract

PURPOSE:To realize economization and simplify the construction and improve the reliability, by detecting a small current in the primary side of a transformer in the AC side and calculating a DC main circuit current by load simulation. CONSTITUTION:A load current is detected as a DC voltage signal iac proportional to an AC current Iac in the AC side through a current detector 13 such as a current transformer, a rectifier 14, and a resistance 15 connected as the load of this rectifier 14. This voltage signal iac is converted to a current waveform output equivalent to a welding machine DC-side current Idc by a load simulation circuit 16. The output is converted to the effective value (or average value) of the current by a current value operating circuit 17 and is compared with a current reference IR, and a current regulator 2 is controlled by a phase controller 9. A pulse P outputted from the phase controller 9 synchronously with the turn-on signal of an electric valve is used to switch the time constant of the load simulation circuit 16.

Description

DETAILED DESCRIPTION OF THE INVENTION The present invention relates to a control device for a DC power supply, and more particularly to a control device for controlling a DC power supply device, and more particularly to a control device for controlling a DC power supply device, and more particularly to a control device for controlling a DC power supply device, and more particularly to a control device for controlling a DC power supply device. This article relates to a control device for a DC power supply device that supplies DC power to a DC load.

As an example of this type of device, the conventional direct current type S
*The circuit configuration of the current control device of the machine is shown in Figure 1. AC power is supplied to the primary side of the transformer J through a current regulator 1 consisting of a thyristor, etc. and a controllable electric valve from an AC power source 1, and outputs 9 AC power. Connect the rectifier circuit to the secondary side of transformer J,
The rectified output is supplied to the load, that is, the welding machine electrode part zK.

#The DC current flowing through the contact electrode section sK is detected by the current detector 6, and the detection signal is amplified by the amplifier 9. The load current actual value detected in this way is compared with a current reference - and the deviation thereof is supplied via an amplifier r to a phase regulator vK. This phase controller! adjust the current so that the deviation is zero, that is, the actual current value is equal to the current reference xIIk! Controls the energization phase of S1.

This conventional device had the following drawbacks.

/) Since the welding DC current is generally a large current of about KI-, I will

1) The distance between the installation location of the control device that houses the electric valve in the current control circuit and the installation location of the welding machine body that houses the transformer and rectifier circuit is often several tens of meters and one kilometer (
Since a current of several tens of meters flows through the secondary side of the transformer, it is necessary to wire a short distance.→However, with the method that detects the DC side current, the wiring distance is long, and construction work is difficult due to noise problems. It had drawbacks such as being expensive.

Therefore, an object of the present invention is to provide a control device for a DC power supply device that overcomes these drawbacks.

In order to achieve this object, the present invention calculates the DC side current from the AC side current by using a range, and converts it to the current f.
It is possible to realize a device that is % more economical than K when used as an IL actual value symbol.

Hereinafter, the present invention will be described in detail with reference to FIG. 1 showing one embodiment. The main circuit portion of FIG. 1 is functionally substantially the same as that of FIG. 1, but for convenience of explanation, some parts are shown as equivalent circuits. That is, the illustration of the transformer is omitted here, and the AC side inductance 10 including the transformer is inserted into the AC current circuit, and the DC side circuit including the load is composed of the DC side inductance it and the DC side load resistance 12. It is shown. The value of inductance // is
If the value of the load resistance 12 is R, then the DC side time constant TII'
i.

It is expressed as T-L/R.

The load current of this device ft is detected on the AC side as a DC voltage signal 1ac proportional to the AC current Ru. This voltage signal iaC is the load voltage, the welding machine DC side current 1d from the power supply/circuit/jK.
It is converted into a current waveform equivalent to a, and its output is converted into an effective current value (or average value) K by a current value calculation circuit l), and the current reference x! l, and the current regulator -
control.

A timer circuit is driven by a pulse P outputted from the phase controller 9 in synchronization with the ON signal of the electric valve, and a current value "l*4, which will be described later, is compared with the output signal PmePaK by a calculation circuit A!J-←-÷ Anzu - Seven ＾◆Compare.

From up/down counter JOK, time constant selection circuit
The time constant of the main circuit on the DC side is selected so as to match the time constant of the main circuit on the DC side, and the time constant of the load chassis loading circuit 16 is switched by the switching circuit.

The operation of the control circuit shown in FIG. 1 will be explained according to FIG. 3. - VaFi DC side output voltage, xeL DC side current 1 da
The electric current, the load stain that was damaged, the wharf 1f circuit 16
The output voltage, iaO, is the voltage proportional to 10 obtained by rectifying the alternating current 1aO, and mag indicates the stone ignition pulse when the electric valve of the current regulator 1 is turned on.

At time t1, the electric valve of the current regulator 1 is turned on with an ignition pulse Vg, and a positive current EE- is output to the DC side.

AC current 11° begins to flow (see signal taa), il
Since the inductance on the AC side of the horizontal machine device is considerably smaller than the inductance on the DC side, the current is rapidly commutated from the circulating current of the rectifier circuit to the AC side.

DC current Due to the side inductance //, regardless of the AC side, the rectifier circuit circulates in the street at 1 time 1.
~t4, the time constant T = L/RK, and therefore attenuation.

Phase controller! A pulse P is output which rises at time t, which is slightly ahead of the rise time t4 of the one-firing pulse V, and which falls at time 9, which is slightly later than that, at time t.

Timer circuit/I#iA Pulse P is generated at the rising edge of pulse ν.

At the falling edge of pulse P, a nocellus horse is output. pulse P,
, P, the comparison arithmetic circuit/1 calculates 1 time 1. At time t, a load stain immediately before the electric valve turns on is detected, and at time t, the electric valve is turned on by an ignition pulse, and the AC side current I&O is turned on. At the rising time t, the AC side current I is 1 shift 1 at 'sP%A.
．． Detects the current value l, and compares the current value I, %I.

0, for example, Xl'>1.
In this case, the up/down counter is caused to count down and the time constant is switched by the switching circuit - so that the time constant is shortened by the time constant selection circuit &. Conversely, if x>I, the time constant of the load is-surge and y circuit 14 is switched to be longer. Thus, the load shift is adjusted so that I, HX, and IIC.
By adjusting the time constant of the a-range circuit 14, the load shift t
Since the time constant of the &rage circuit/4 and the time constant of the main circuit DC side can be made to match, the current of the load switch J'a storage circuit 140 output x4 can be adjusted to the current value calculation circuit t.
Calculate the DC side current 11G from tK. In this way, the load is detected from the AC side current without directly detecting the DC side main circuit current.

If you pass it through the radar circuit, you will be able to detect it.

Fig. 1 is an enlarged diagram of the signal section in Fig. 3, and shows the case where the time constant of the DC side main circuit and the time constant of the load circuit are the same.

In other words, if the time constant is as shown in the figure, the current value X is 1 point -, and X, kq
If the time constant of the main circuit is shorter than the time constant of the main circuit on the DC side,
In other words, the time constant! In the case of , the current value X is at point a, and I8<E. In this case, the time constant selection circuit 1 is operated so that the time constant becomes longer. Reverse Kjl
If the time constant of the electric power supply/radiation circuit is longer than the time constant of the DC main circuit, the current value 1 will be 0 points, and z,')x
, so the time constant of the load shift circuit 14 is changed so that the time constant of the load shift shift circuit 14 is shortened.

By creating a minor loop such that the time constant of the load stain circuit 14 always matches the time constant of the load circuit, it is possible to calculate the DC side current only from the AC side current. becomes.

FIG. 3 shows an example in which the circuit device shown in FIG. 1 is realized using a computer. In this example,
The load signal t and the range control circuit/4C1 voice power Id are digitized through the ~Φ converter X, and this data is read by a computer (hereinafter referred to as σG) J/. ,0POJ/
calculates a current value from this data and outputs it through three four converters. On the other hand, Bars P! sPl output is OPO3/
The current value 1. shown in FIG. and X are read, calculated, and output to the time constant selection circuit V. In Fig. S, since a method is adopted in which the current value I is read in from the current value I, the output of the circuit 14, the negative balance is established by the pulse 9, and the filter of the circuit 14 is I'm trying to reset it. This situation is shown in FIG.

FIG. 6 shows a detailed configuration of the load stain, range 1 in circuit 14, and switching circuit in FIG. 5, and shows the pulse P.

Operate the switch from K and filter capacitor 4
FIG. 7 shows how the 47 is reset.

The diode axis is ideally an ideal diode (a turtle with no forward voltage drop), and can be realized with a circuit using an operational amplifier. Capacitor Hiro 7 is the inductance l of the main circuit.
Element for converting / to VTM range 1, with resistor, 3i
'd, 3/ constitute a discharge circuit for the capacitor 4I-7 and determine the time constant. Resistance 419,! The parallel switches 32 and 33 correspond to a switching circuit g, and four constants change by turning on and off the switches S- and jJ.

The one with operational amplifier is a buffer amplifier. The resistor jj and the switch are a circuit for resetting the capacitor Hiroshi at time t. Assuming that the DC main circuit has the time constant shown by the dashed line, the accurate current value 1. In order to find this, the capacitor 7 is discharged at time t4, and the load stain detection circuit is operated to match the AC side current Xao (iao) K accurately.

Although the main circuit of the nest phase has been described above, it goes without saying that the present invention can also be applied to a main circuit having a multiphase full-wave rectifier circuit in a multi-phase half-wave rectifier circuit.

Further, although only a portion of the circuit shown in FIG. S is processed by a computer, it is also possible to realize the entire circuit shown in FIG. S by a computer.

As explained above, according to the present invention, a simple current detector such as a current transformer is used to detect the small current on the side next to the transformer on the AC side, without directly detecting the large current in the DC main circuit. Since it is possible to indirectly calculate the DC main circuit current by using the load voltage control circuit, it is not economical, and at the same time, it is possible to detect the current at the control device side rather than at the load end. , construction is very easy. Furthermore, in the case of four-dack F welding, etc., the main circuit on the DC side has many moving parts, and reliability is significantly reduced due to wire breakage, etc. However, with the present invention, the reliability is significantly reduced because the current can be detected from the fixed side. improves.

[Brief explanation of the drawing]

Fig. 1 is a circuit diagram showing a conventional control device, Fig. 1 is a circuit diagram showing an embodiment of the present invention, and Fig. 2 and the reference figures are diagrams for explaining the operation of the device shown in Fig. 1. , FIG. S is a circuit diagram of a main part showing another embodiment of the present invention, FIG. 6 i is a circuit diagram showing details of a main part of the device shown in FIG. 6, and FIG. A stone with a line diagram to explain its operation. I...AC power supply, Co-current regulator, J...Transformer, Reference...rectifier circuit, S-@Inoculating electrode part,! -Phase controller, io, //...inductor ys, saw...
・Load resistance, ts...Current detector, /4-=Load electric current detection circuit, lF...Current value calculation circuit%/l
...Timer circuit%19...Comparison calculation circuit, X...
・Up/down circuit %l...time constant selection circuit, tsu...
・Switching circuit %X -... to converter, J/...I1 calculation
(OPO), 3 pieces...Hiroshi converter. Departing person's agent Ino Mata ゛ Purity 1 Kuchi Atsushi 3 Kuchiku 4 K also 6 ni Mitsu 7 Prisoner

Claims (1)

[Claims] 1. A DC power supply device that supplies a controlled DC current obtained from an AC power source through a current regulator and a rectifier circuit consisting of a controllable electric valve to a DC load having an i/ductance. The control device includes a current detection device that detects the load current on the AC side, and an off period of the current regulator according to the detection output of the current detection device and a time constant that simulates the DC load. The DC current flowing to the DC load, including the
The predistribution current is adjusted so that the DC current generated by the load switching circuit and the ζO load switching register circuit match the target value. Control *b
1. A control device for a DC power supply device, comprising: a control section; A control device of a DC power supply device that supplies a controlled DC current obtained from an AC power source through a current regulator and a rectifier circuit consisting of a controllable electric valve to a DC load having a and a current detection device that detects the load current on the AC side, and a current detection device that detects the load current on the DC load including the off period of the current regulator according to the detection output of this current detection device and the simulated time constant of the DC load. A load shield (a-litter 1 off-circuit) and this load shunt x-
A controller is used to control the current regulator so that the DC current is set by the V-V circuit to match the target value, and the simulated current value immediately before the electric valve is turned on and the AC immediately after the electric valve is turned on. 1. A control device for a DC power supply device, comprising: a switching device for comparing a detected current value and switching a time constant of a load shift circuit so that the two substantially match.