I have seen in some posts on the Internet that if the MOS tube is turned on, it may burn out if it is working in the current amplification area. This is because the ID current in the linear area is large, and the RDS is also large, and the power consumption is high. To.
But I looked at the MOS application manual, and the RDS mentioned above is all mΩ after the turn-on. Is this a large resistance? Isn't this contradicting the above introduction?
In addition, how should the power consumption of MOS be calculated?
Kendall 發表於 November 23, 2020
If the drain-source resistance really reaches the order of milliohms, it has entered the on-state of the switch mode (or type D), and it is no longer a linear amplification state.
Emmanuel 發表於 November 23, 2020
RDS in the application manual refers to the resistance after saturation, not in the linear region.
Stella 發表於 November 23, 2020
There is a diagram of Vgs/Id in the MOS tube manual. Above it is determined that Vds=a value, such as 30V. According to the curve, you can calculate different resistances under different Vgs.
The resistance of a few milliohms in the manual is measured above Vgs>10V. For example, when Vgs=4v, the resistance is 25 ohms for 3315, and Vgs=10v is 0.082 ohms.
