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RVast
Associate
January 17, 2020
Question

Current rating per channel of VNQ5050AK-E?

  • January 17, 2020
  • 1 reply
  • 1337 views

Can I know what is the max current rating per channel of the smart FET VNQ5050AK-E, it is quad channel IC and its total current is internally limited to 19A.

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1 reply

Andrew Lohmann
Senior
January 19, 2020

Do worse case design - that is calculate from maximum or minimum figures different ways then use the worse realistic figure.

For safe design assume 27A x 4. Table 9

That comes down to less than 7A per output over temperature. Less than 7A x4. Table 9. All four outputs in parallel gives less than 30A

Use the the thermal graph of 'C per watt to estimate the heatsink of your PCB depending on copper area. 55'C to 35'C per watt, graph 26. Look up the maximum on resistance and from that re-arrange from W = I squared R to find the maximum current.

You need to go through the datasheet first estimating what you want then checking and improving your estimate.

Table 6 tells you the maximum R on at different temperatures 50mR, 65mR and 100mR depending on Vcc and temperature for example. Use the worse case in your application.

RVast
RVastAuthor
Associate
January 21, 2020

Hello Andrew,

Thank you for your reply, it helped me in understanding a little but the Tables you are referring in your explanation should I check it in Datasheet of IC, if yes than references you have made are not matching with that of datasheet.

Thank you.

Andrew Lohmann
Senior
January 21, 2020

Yes I did make a mistake or the datasheet was a different version. I have corrected below.

 

Here is a link to the datasheet; https://my.st.com/resource/en/datasheet/vnq5050ak-e.pdf

 

Do worse case design - that is calculate from maximum or minimum figures different ways then use the worse case realistic figure.

For safety design assume 27A x 4. Table 9 --- Design the PCB power tracks to carry 110A.

That comes down to less than 7A per output over temperature. Less than 7A x4. Table 9. All four outputs in parallel gives less than 30A

Use the the thermal graph of 'C per watt to estimate the heat-sink of your PCB depending on copper area. 55'C to 35'C per watt, graph 26 Figures 26 and 29. Look up the maximum on resistance and from that re-arrange from W = I squared R to find the maximum current.

Note; Temperature rise = The device maximum working temperature - Your maximum working temperature.

  • The figures for 'C/W will change by a factor of 1.5, If the copper heat-sink is black mat copper or shiny polished copper (exposed tin for example will oxidize mat white and the cooling will improve),
  • The cooling will change by a factor of 2 depending on If the PCB is vertical or horizontal,
  • The cooling will also diminish if the copper area is underneath or on top.
  • Green Solder resist layer will reduce the efficiency of the heat-sinking (I have measured) but I have been told the opposite is true.

You need to go through the datasheet first estimating what you want then checking and improving your estimate.

Table 6 tells you the maximum R on at different temperatures 50mR, 65mR and 100mR depending on Vcc and temperature for example. Use the worse case in your application.