What is the cause of the burnout of the power MOS tube?
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Date:
2017-09-18
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The working state of the MOS tube in the controller circuit: the turn-on process (the transition from cut-off to turn-on), the turn-on state, the shutdown process (the transition from lead to the cut-off), and the cut-off state.
Mos main losses also correspond to these States, switching losses (turn-on process and shutdown process), conduction loss, cutoff loss (omitted by current leakage), and avalanche energy dissipation. As long as these losses are controlled within the MOS acceptance specifications, the MOS will work properly and beyond the tolerance range, i.e., damage. And switching losses are often greater than on state losses (different MOS, this gap may be great.
Mos damage is the main reason:
The junction temperature of flow - induced sustained large current or transient large current high burn;
The source drain breakdown voltage, and gate source breakdown;
Electrostatic, electrostatic breakdown. CMOS circuits are afraid of static electricity;
Mos switch principle (brief). Mos is a voltage - driven device. As long as an appropriate voltage is supplied between the gate and the source stage, the channel between the source and drain stages is formed. The resistance of this current path is MOS internal resistance, that is, on resistance <Rds (on) >. The size of the internal resistance basically determines the maximum conduction current that the MOS chip can withstand (of course, it's related to other factors, most of all, thermal resistance). The smaller the internal resistance, the greater the current to withstand (due to small fever).
The Mos problem is far from that simple. There is an equivalent capacitance between the gate and drain stages between its gate and source stages, between the source and drain stages. Therefore, the process of charging the gate voltage is to charge the capacitor (the capacitor voltage can not be mutated), so the opening process between the MOS source stage and the drain stage is controlled by the charging process of the gate capacitor.
However, these three equivalent capacitors form a series of parallel combinations, which are independent of each other, and are independent if they are independent. One of the key capacitors is the capacitor Cgd between the gate and the drain stage, which is called the Miller capacitor. This capacitor is not constant and varies rapidly with the voltage between the gate and the drain stage. This Miller is a stumbling block to the gate and source capacitance charging capacitor, because the gate to gate source capacitance charging of Cgs reached a plateau after charging current gate must be charged to Miller capacitor Cgd, voltage between gate and source level at this time is no longer increasing, reached a plateau, this is Miller (Miller platform to platform Cgd charging), the Miller platform we first thought the trouble is Miller oscillation. (that is, the grid charges the Cgs first and then charges the Cgd when it reaches a certain platform)
Because this time the source and drain voltage change rapidly, the internal capacitance corresponding rapid charge discharge, the current pulse causes the MOS parasitic inductance has great inductance, which have the capacitance, inductance, resistance of an oscillation circuit (2 loop formation), and the stronger the current pulse frequency is higher the more volatility. So the most crucial question is how this Miller platform transition.
Excessive charge will result in intense Miller shock, but too slow charge will reduce the shock, but will extend the switch, thereby increasing switching losses. The equivalent resistance between the source stage and the drain stage of the Mos turn-on process is equivalent to a transition process from the infinite resistance to the very small internal resistance (i.e., the internal resistance is low and the MOS is only a few ohms). For example, a MOS 100A 96V maximum current, battery voltage, in the opening process, a moment (just entering the Miller platform MOS (P=V*I) heating power is the power of current has reached the maximum load, not run, all landing on the MOS tube), P=
96*100=9600w! At this time, it has the highest heating power and then the heating power decreases rapidly until the power is turned into 100*100*0.003=30w at full conduction (assuming that the MOS is 3 ohms inside). The change in the heating power during the switch is staggering.
If the opening time is slow, it means that the heating is slow from 9600w to 30W, and the MOS junction temperature will rise very much. Therefore, the slower the switch, the higher the junction temperature, easy to burn mos. In order not to burn the MOS, it can only reduce the MOS current limit or reduce the battery voltage, such as limiting the 50A or voltage to 48V by half, thus reducing the switching power loss by half. Don't burn the pipe. It is also easy to burn tube pressure control, high pressure and low pressure controller only switching loss is not the same (switching loss and battery voltage is proportional to the basic assumption, limiting the same), conduction loss entirely by MOS resistance, did not have any relationship with the battery voltage.
In fact, the entire MOS boot process is very complex. There are too many variables inside. A word that is not easy to switch to slow Miller shock, but the switching loss, pipe heating, fast switching speed of the switching loss is low (as long as it can effectively inhibit the Miller shock), but often very powerful Miller shock (if the Miller shock is very serious, may be in the Miller platform, but the burning tube) switch loss is also large. The upper arm and MOS shock is more likely to cause the lower arm MOS misconductions, forming the upper and lower arm short. So it's a test driver's wiring, wiring and wiring skills. The end is to find a balance (the general opening process is not more than 1US). The opening loss is the simplest, which is directly proportional to the through resistance.