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ALM2402F-Q1: programmable current limit (high side = power supply)

Part Number: ALM2402F-Q1

Hello,

The spec says "RLIMIT = (VS - 1.5) / ILIMIT"

This formula is very strange:

#1: The current limit would depend on the power supply voltage => not "great"

#2: Assuming a VS=10V and ILIMIT=100mA     RLIMIT = (VS - 1.5) / ILIMIT = (10-1.5)/100E-3  = 85ohm

but 85ohm * 100mA = 8.5V drop => this is a lot since this resistor supplies the high side MOSFET

Could double check the formula and comment?

Thanks

  • Frederic,

    The ALM2402F-Q1 formula you refer to, RLIMIT = (VS - 1.5) / ILIMIT, is NOT a spec but rather a calculation of a required Rlimit resistor to be used as a crude way to lower the output short-circuit sourcing current below its specified 750mA limit.  Most applications do NOT need this feature and should rely only on 750mA short-circuit limit specified in the datasheet. 

    The desire ILIMIT sourcing current may be implemented by adding Rlimit resistor in series with output stage supply voltage, VS_O(1 or 2), which has the effect of premature trioding of P-channel output transistor by increasing its Ron value  - see below.  

    There is no equivalent way to lower short-circuit sinking current because there is no separate ground pin in ALM2402F output stage.

  • Thanks Marek,

    Everything is clear now.

    I guess that "1.5" is 1.5V and is a approximate value of the VGSth of the PMOSFET.

    I was hoping to use this "feature" to limit the supply current "precisely" (significantly below 750mA and significantly before the thermal protection) in case of a short circuit on the resolver excitation but clearly, I will have to find another solution.

    Thanks,

    Frederic

  • Yes, you may think of P-channel Vgs being at 1.5V at the output current limiting point.  

    You may still limit the sourcing current quite precisely by measuring Iout and adjusting Rlimit resistor. 

    The activation of thermal protection is a function of the output current AND voltage drop inside the package.

    Therefore, Pdissipation = Iout*(VS-O - Vout)