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What is the relationship between voltage breakdown and burnt marks on the surface of the thyristor?

163.com | 06/04/2021

In most cases, the aluminum gasket or the silver gasket is melted first, followed by the silicon and molybdenum sheets. The aluminum gasket or the silver gasket will not melt in a small area, and all the effective area of ​​the gasket will melt. After the aluminum gasket or silver gasket is melted, it is possible to produce an isolation layer to open the cathode and anode, and the other is that the material of the joint between the aluminum gasket or the silver gasket and the silicon wafer may change after high temperature melting, resulting in insulating substances. , Causing the cathode and anode open circuit phenomenon. So what is the relationship between the voltage breakdown and the burnt marks on the surface of the thyristor?

1. Because the voltage parameter of the thyristor drops or the overvoltage generated by the line exceeds its rated value, its insulation strength is relatively reduced, so the arc discharge phenomenon occurs, and the temperature of the arc is very high, far greater than the melting point of the metal of the chip, so The thyristor is burnt, and because the insulation voltage between the outer edge of the chip and the surface of the chip's cathode and anode is not completely consistent, only the arc discharge is started at the point where the relative insulation voltage is lower, so the voltage breakdown appears on the surface of the chip's cathode Or there is a small black spot on the edge of the chip.

2. Due to the decrease of the current, dv/dt, leakage, turn-off time, voltage drop and other parameters of the thyristor or the circuit, the chip temperature is too high, exceeding the junction temperature, causing changes in the internal metal format of the silicon wafer, causing its insulation voltage As a result, the arc discharge phenomenon occurs. The high temperature generated by the arc melts and burns the gasket, silicon wafer, and molybdenum wafer, and also melts the metal connecting the shell and the chip. Because the chip temperature is too high, it takes a long time and it accumulates slowly, so the over-temperature area is larger, and the burn-out area is also larger.

3. Although the thyristor burned out due to di/dt and turn-on time is also a small black spot, the burned-out position is different from the real voltage breakdown, and the burn-out mechanism is the same as the above 2 It's just that the small SCR in the chip is relatively small, so the burnt traces formed are also small. In fact, the small SCR has been completely burned.

Generally speaking, as long as the thyristor is burned out, the reason is that the insulation voltage of the thyristor is relatively lowered, and then the arc discharge is started, resulting in high temperature, which melts the thyristor chip metal and even the shell metal, causing the thyristor to short-circuit and damage.

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