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LM74610-SQEVM: If a N-channel FET would be used, will current flow from S to D be Ok?

Nando Uitterlinden1
Prodigy 20 points
Part Number: LM74610-SQEVM
Other Parts Discussed in Thread: LM74610-Q1

If I understand correctly, when using a n-channel FET in the power line side (not in return path), the FET will be switched on by a charge pump driving the gate. If the gate would not be driven, the body diode would heat up and damage the FET right?

When expected the gate is driven, the FET goes on and will pass current with only Rds(on) as loss, right?

But looking at the direction if current, it will flow from source to drain, is that correct? This is opposite to the current flowing from drain to source when a N-channel FET is used as a normal low side ‘switch’.  We are normally speaking of Rds(on) to describe resistance from drain to source, now my question is, is the resistance equal when looking ar source to drain performance? And how fast does the charge pump drive the gate? Because in the mean time that the polarity is connected correctly to the LM74610-SQEVM but the gate is not driven yet, the body diode will warm up correct? Best regards Nando

  • 0
    TI__Guru 57737 points

    Hi Nando,

    Welcome to e2e! Thanks for reaching out to us. 

    Please see my response inline below,

    If I understand correctly, when using a n-channel FET in the power line side (not in return path), the FET will be switched on by a charge pump driving the gate. If the gate would not be driven, the body diode would heat up and damage the FET right?

    Yes, your understanding is correct. If the gate of the NFET is not driven, the FET would be turned OFF and the load current would flow through the body diode. The body diode can heat up if depending on the duration and amount of current flowing through it.

    One thing to remember is that LM74610-Q1 when connected in power path will not pull the Gate low for longer duration of time and hence the FET and body diode will be safe.

    When expected the gate is driven, the FET goes on and will pass current with only Rds(on) as loss, right?

    Yes, if the Gate is high, the FET channel is fully enhanced and the would offer low resistance = Rds(on).

    But looking at the direction if current, it will flow from source to drain, is that correct? This is opposite to the current flowing from drain to source when a N-channel FET is used as a normal low side ‘switch’.  We are normally speaking of Rds(on) to describe resistance from drain to source, now my question is, is the resistance equal when looking ar source to drain performance? And how fast does the charge pump drive the gate? Because in the mean time that the polarity is connected correctly to the LM74610-SQEVM but the gate is not driven yet, the body diode will warm up correct? 

    The Rds(on) of the FET would be equal in either direction S-D or D-S. I would recommend you to go through the section '7.3 Feature Description' of the datasheet to understand more about the charge pump and gate drive functionality of LM74610-Q1 

  • 0 in reply to Praveen GD
    Prodigy 20 points

    Thank you for your quick and detailed answer, I appreciate it very much! Thanks again. BR Nando