Is the machine crashed due to a short circuit in an IGBT of the inverter bridge arm?

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Is the machine crashed due to a short circuit in an IGBT of the inverter bridge arm? [Copy link]

A low-power inverter is made of single-tube IGBT. When using a probe to measure the drive signal and CE signal of a certain IGBT, C and G are accidentally short-circuited together, and then the inverter explodes. What causes the inverter to explode?

1. It is unlikely to be caused by the IGBT itself, because the inverter had not been turned on at that time. Even if the drive and the positive bus, that is, the collector, were short-circuited, the IGBT would not be damaged?

2. The switching power supply exploded due to a short circuit. Checking the damaged inverter, it is the part adjacent to the switching power supply and the input side. The switching power supply is a flyback type, and the drive circuit is an upper bridge. The exploded IGBT is the U-phase upper bridge or the V-phase upper bridge, it should be one of these two.

The result is that the upper bridge of the inverter bridge is completely broken down and short-circuited, the lower tube is also broken down but only a 1.6V voltage drop is left; the rectifier bridge is completely broken down but only a small voltage drop is left.

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The G pole is generally at a voltage of 15V, and the C pole may be a DC voltage of 300V or higher. Therefore, when the G pole is connected to the high voltage, the corresponding tube is directly turned on and directly breaks down the GE pole and CE pole. At the same time, the electronic components in the driving circuit will also break down and short-circuit. If the driving circuit is not isolated, all the tubes will be burned out.

乱世煮酒论天下

Vibration Test Instrument Published on 2024-8-21 00:40 The G pole is generally at a voltage of 15V, and the C pole may be a DC voltage of 300V or higher. Therefore, when the G pole is connected to the high voltage, the corresponding tube is directly turned on and...

On the upper bridge, the voltage between C and E is about 270V. For three-phase diode rectification, C is connected to G. GE breaks down first. The upper and lower limits of the drive are generally 20V. At the same time, CE will be completely turned on and the instantaneous current will be too large and then burn out. Is this what you mean?

This machine is driven by a bootstrap circuit

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Is it the IGBT of the upper tube? Then point C corresponds to the bus voltage. If C is short-circuited to point G, the machine will explode normally.

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When the G pole is at a high voltage, CE will continue to conduct, causing a direct short circuit in the power supply. The high voltage at the G pole will not be disconnected until the power is cut off. At a voltage of 270V, all components on the driver board will burn.

Without circuit isolation, failure of other driving circuits will lead to failure of other electronics, and all electronic components will burn out instantly.

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The four tubes in the bridge circuit are turned on alternately. Now one tube is turned on for a long time without being turned off, causing the two short-circuited tubes to burn out and cause a short circuit.

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Even if the inverter is not powered on, the drive circuit may already be powered.

The short circuit between C and G may damage the components in the driving circuit due to overcurrent. The impedance between the gate and emitter of the IGBT is relatively low. The short circuit may cause a large current to flow through the driving circuit instantly and cause damage.

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The switching power supply will start its own protection when it detects abnormal current,
but if the protection action is not fast enough or the fault current is too large, the switching power supply may be damaged.

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Because the short circuit of the tube caused a short circuit fault at the power supply end, the short circuit current was large and the protection circuit did not work. The CE pole was either short-circuited or burned open, and the CG pole was also short-circuited. It is possible that the driver board will burn out.