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LM5030: Current of Gate drivers

Part Number: LM5030

It is written "The internal gate drivers can each sink 1.5-A peak each." in the Data sheet.
But source current is not written.
How much source current of drivers?

  • Hi,

    I agree it's not clear but based on the language used in other PWM/driver data sheets, whenever it is stated the drive stage is x.x A it is assumed the source/sink capabilities are the same. A better way we sometimes specify is as ±1.5A which indicates source (+) and sink (-) but this is missing in the LM5030 data sheet. Also whenever the source and sink current are not equal, the differences will be specified in the OUT section of the parametric table or at the very least should be stated on the front page of the data sheet. Please note that since the 1.5 A is not specified in the OUT section of the data sheet, we do not test for this parameter and so these are assumed to be TYP values. Thanks for connecting through E2E.

    Regards,

    Steve M

  • Thank you for good advice.
    I agree your comment, but a little bit afraid to believe.

    Best regards,

  • You can convince yourself of the source/sink current values. If you have this converter built in the lab, you could set up the LM5030 to switch into a fixed capacitor load. Measure the rise and fall times and calculate the current by I=C*(dV/dt). Use something like 1nF load, switch from 0V to 12V in some amount of measured time. If, for example you measure the rise and fall time to be 8ns, then I=1nF(12V/8ns)=1.5A. Even though the current from the output stage of the PWM if a sharp peak of current, it will appear as a near constant current within the ns time scale where you are tasking your measurements.

    Regards,

    Steve M

  • Thank you for good advice.
    But, I have no samples of LM5030.
    Are there something data of rise and fall waves with load capacitor?
    I found data of "Rise time" and "Fall time" in OUTPUT SECTION of data sheet.
    "Rise time"="Fall time"=16ns(typ) with Cload=1nF.
    How much swing of output is?
    If output swing is 10V, I=C*(dv/dt)=0.625(A).
    Is that right?

    Best regards,

  • From the LM5030 data sheet "output section", tr=tf=16ns, Cload=1nF, VCC=10V and your calculation of 625 mA is correct. the data sheet states 1.5A peak current but often times the peak value of the current is not aligned with the Miller Plateau. The real current we care most about is the available current at the Miller Plateau which drives the transition from LOW/HIGH and vice versa. The value of peak current depends upon many factors such as PCB layout, circuit parasitics, source/sink impedance and measurement technique. the actual waveform would be helpful here but this is not included as part of our automated test coverage.

    Here is an example of the peak source/sink current for a gate driver rated at 4A. The cursor is placed near the estimated equivalent "Miller Plateau" and you can see the value of the peak current happens before the4 output voltage has a chance to rise/fall to within 10%/90% of the steady state value. When the load is a Si MOSFET which has non-linear capacitance, the effect can be so bad that the peak current value appears during the MOSFET delay time which is well before the switching transition. This is why peak current ratings should be considered with caution.

    Regards,

    Steve M

  • Thank you for confirm my calculation and offer me an example of the waves.

    I'm relieved to hear that source and sink are 0.625(A).

    Best regards,