Switch electromagnet drive circuit diagram

The picture shows the driving circuit of the switching electromagnet , in which device U8 is a 74123 chip, devices 02 and 03 are solid-state relays, the coil PUSH is the suction coil of the switching electromagnet, HOLD is its sustaining coil, and the power supply VDD is 24V. Of course, when the conversion mechanism switches to the dual-fuel working state, the Y1 input from the B pin of the 74123 jumps from low level to high level. This change will cause the output terminal Q of the 74123 to output an instantaneous high-level pulse current. The maintenance time is ΔT. After passing through the inverter 7404, the potential UA at point A will appear at a low level with the same pulse width, and the input terminal of the solid-state relay 02 will also have a current of ΔT time flowing through it. At this time, the output terminal of the solid-state relay 02 will be turned on for a ΔT time. . During this period of time, a strong suction current passes through the suction coil of the switch electromagnet, generating sufficient electromagnetic force to suck into the iron core of the electromagnet. The selection of pulse width ΔT should refer to the performance parameter settings of the switching electromagnet. It must not only meet the requirements of sucking the iron core, but also not be too long to cause the coil to burn out.

ΔT is determined by selecting the values ​​of resistor R10 and capacitor C3. Here ΔT is taken as 1 second, R10=1000Ω, C3=1000μF.

In terms of the generation mechanism, the sustaining current of the switching electromagnet is the same as the operating current of the high-voltage natural gas solenoid valve: when Y1=1, there is a sustaining current, and the iron core remains in the suction state; when Y1=0, there is no There is a maintaining current, and the iron core returns to its original extended position under the action of the return spring.

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