A practical high-power current square wave inverter circuit
Abstract: This paper introduces an engine current square wave inverter circuit for electromechanical control of mechanical components of high-power electronic experimental machine. Based on the general capacitor commutation current inverter circuit, the circuit realizes the current square wave waveform of "two-level" and "three-level" only by introducing a pair of conversion thyristors. And the circuit is simple and practical, and has achieved good results in practical application
key words: high power current type inverter three electric square wave 1 Introduction in the exploration of underground resources, such as groundwater exploration, oil exploration, etc., the traditional exploration method often adopts the so-called "seismic method", that is, using explosives to artificially create a "micro earthquake" in the exploration area, and analyzing the nature and distribution of underground resources by measuring "seismic waves". This method costs a lot of manpower and material resources, and its application range is limited. Therefore, it has been gradually replaced by "electrical exploration" in many occasions. The so-called electrical exploration is to inject a certain frequency of "current square wave" into the underground in the exploration area, and analyze the underground resources by measuring the geological electric field response. The key to underground exploration by electrical exploration is to have a source of alternating current square wave. It is characterized by high DC voltage, high power and steep waveform requirements of inverter current. The inverter current can work at both positive and negative alternating "two-level" and positive, zero and negative "three-level", but the frequency requirement is not high (generally 0 ~ 20Hz adjustable). As the exploration is in the field, the power supply device must be very reliable. The power circuit introduced in this paper adopts AC-DC-AC frequency conversion mode. The inverter part is based on the capacitor commutation thyristor current inverter circuit, and only a pair of conversion thyristors are added, so that the inverter current wave can also replace non-ferrous metals for precision structural parts such as clocks, copiers, cameras, etc; Shape function requirements, simple and practical. The output power of the power supply device can reach 300kW (DC 1500v/200a), the front and back edges of the current square wave are less than 1ms, and the frequency of 0 ~ 20Hz is adjustable. It has achieved good practical results in the exploration of new oil fields such as Tuha and Korla. 2. The main circuit diagram 1 of the power supply system shows that the main electrical route of the power supply system is three-phase bridge controllable rectification (vt11 ~ vt16), which outputs 0 ~ 15 to avoid the generation of dangerous 00v adjustable DC voltage, so as to meet the needs of different power outputs. The DC side is connected in series with large reactance LD to ensure current mode output. Vt21 ~ vt24, vd21 ~ vd24 and C2 constitute capacitance commutation current inverter circuit. Vd02, R2, vt25, vt26 and C1 realize the conversion of current waveform "three-level" working mode. Due to the high DC voltage of the power supply and the low requirements of the inverter frequency, and considering the high requirements of the working reliability and the steep requirements of the current square wave waveform, the switching elements of the inverter circuit adopt fast thyristors. 3. The working principle and analysis of the circuit are shown in Figure 1. If the inverter part of the circuit keeps vt26 on and vt25 off, and periodically triggers two pairs of thyristors vt21, vt24, vt22 and vt23 at the same time at an interval of 180 °, it is a basic capacitor commutation current inverter, and the frequency adjustable "two-level" current square wave can be obtained on the load. If the switching elements are controlled according to the law shown in Figure 2, the "three-level" current square wave with adjustable frequency and width can be obtained on the load. For example, at time T0, vt21, vt24 and vt26 are simultaneously triggered to turn on, and the negative
load current IL is positive. At this time, C1 is charged with left positive and right negative voltage through vd02 and R2. When vt25 is triggered to turn on at time T1, the voltage on C1 is reversely applied to vt26 to force it to turn off. C1 discharges and then reverses the charge. The charge ends and the load is cut off. At time T2, vt22, vt23 and vt26 are simultaneously triggered to turn on, reversing the load current. At the same time, the voltage on C1 is reversely applied to vt25, forcing vt25 to turn off, C1 discharges and charges in a forward direction. At time T3, vt25 is triggered to turn on, forcing vt26 to turn off and load to cut off. By analogy, a "three-level" current square wave with positive, negative and zero can be obtained on the load. It is not difficult to see that changing the time interval between t1~t2 and t3~t4 can change the current square wave width. Changing the time interval between t0~t2 and t2~t4 can change the AC current square wave on the load, mainly the detection frequency of various materials
the transition of current from one "level" to another requires a certain process. Taking the current conversion from "positive level" to "zero level" as an example, when vt25 triggers conduction, C1 applies back voltage to vt26 to make it turn off quickly, and the current of vt26 is transferred to C1. At first, C1 discharges at a constant current. When it discharges to vd02 positive bias, vd02
is turned on and flows through current I. At this time, the equivalent circuit is shown in Figure 3, which should meet the following requirements: ID LD id- i
the transition of current from one "level" to another "level" requires a certain process. Taking the current conversion from "positive level" to "zero level" as an example, when vt25 triggers conduction, C1 applies back voltage to vt26 to make it turn off quickly, and the current of vt26 is transferred to C1. At first, C1 discharges at a constant current. When vd02 is discharged to a positive bias, vd02 is turned on and flows through current I. At this time, the equivalent circuit is shown in Figure 3, which should meet:
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