Difference between the VNLD5300 and VNLD5090

[drakejest]

I have been looking for a low-side mosfet driver IC and found these VNLD5300 which is a 300mOhm 2A driver and the VNLD5090 which is a 90mOhm 13A driver. I would like to do one last sanity check, the 5090 seems to be a much more superior IC am i right? But what strikes me is as weird for the same package can the 5090 really handle that current? they are priced very close too so i assumed that ST electronics made these 2 version intentionally with different purposes.

My application is controll and inductive coil at 1.5A both IC fit the application, would going for the 13A 5090 good for me ? What could the reason be that there is a skew with an inferior rating, because i dont think its cost, since they are priced that close.

[magic]

It can only handle 13A peak, not continuously - do the math on power dissipation. It will discharge your gate much faster than the other chip, though.

I imagine one reason for preferring a slower driver, besides price, may be less switching EMI.

[drakejest]

It can only handle 13A peak, not continuously

I assumed that the higher value amperage was the peak (25 A for the VNLD5090). Having a second look at it, i could not find the true maximum current for the device and is written as "internally limited" under the electrical characteristics

do the math on power dissipation.

can you guide me sir on how to do this ? what parameters should i use and how do i use them so that i can know if im safe on my application on open air with no heatsinks?

[magic]

For continuous current through the output stage MOSFET it's simply I²·RDS(on), like with any other MOSFET.
When driving an external FET, it depends on gate charge and switching frequency.

[drakejest]

For continuous current through the output stage MOSFET it's simply I²·RDS(on), like with any other MOSFET.

I do not quite understand how to use the formula, Isnt that a formula for power P=I²·R.  I am missing the P, R is 0.09 ohms, I² is my missing variable. So i cant solve the continue current

When driving an external FET, it depends on gate charge and switching frequency.

can we assume then 100% uptime?

[thm_w]

Rth j-amb Thermal resistance junction-ambient = 108°C/W

Use this number in the datasheet to get a rough idea of maximum power dissipation. So 150C max - 25C ambient = 125C = about 1W.
Real result will depend on how its mounted to your board, if its heatsinked, etc. and whatever headroom you want to include.

[David Hess]

The higher current chip is more expensive because it has more area but the primary difference is the current limits.  You would not want to rely on the 18 amp current limit in an application that only requires 2 amps.