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CSD87350Q5D: What's the side effect to input <4.5V to VGS for high side FET

Part Number: CSD87350Q5D

Hi Team,

For CSD87350, TI said this is optimized for 5V. Is this mean the VDD of driver IC are 5V?

For some application, the VDD are 5V. But there are a voltage drop caused by bootstrap diode. 

 If the voltage of VGS lower than 4.5V, what's the side effect for it? 

Below is an example that apply 5V on VDRV but will got <=4.5V VGS for high side FET.

Thanks

Shaq

  • Hi Shaq,
    Thanks for the questions. We spec and test rds(on) at VGS = 4.5V. We cannot guarantee rds(on) when VGS < 4.5V. In figures 20 & 21 in the datasheet, you will find the typical rds(on) vs. VGS curves for both FETs. As you can see the knee of the curve is well below 4V. Although we cannot guarantee rds(on) @ VGS = 4V, the converter should work and you can use the curves to estimate the increased conduction loss in the FETs. I would caution against operating anywhere close to the knee of the curve (VGS in the 3V to 3.5V range) as rds(on) increases exponentially with decreasing VGS. There will also be variation in the threshold of the FETs that can make this worse. Lastly, most modern gate drivers use a transistor - usually a PFET - for the bootstrap instead of an actual diode to minimize the voltage drop and maximize the gate drive voltage provided to the high side FET.