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LM1084 VIN < rated VOUT

Eben Louw
Prodigy 80 points
Other Parts Discussed in Thread: LM1084, LM5060

Hello,

Can I use the LM1084 when I set the adjustable output voltage to 19V and my nominal input is 14.4V. Load current is 2A.

I'm only adding the regulator for when a user uses a larger battery, which is rated at 20V max. In this case there will be some heat generated as the power dissipation will reach ( 20V-19V = 1V * 2 A = 2W ) 

Are there any cons in using the LM1084  regulator is set to 19V and the input voltage is lower? Or are there better ways of doing this?

Thanks

  • 0
    TI__Mastermind 22610 points

    "... Can I use the LM1084 when I set the adjustable output voltage to 19V ..."

    Yes, the LM1084 will be operating in the drop-out mode.

    "... Are there any cons in using the LM1084  regulator ..." (as an active voltage clamp)

    Yes, this is not a very thermally, or electrically, efficient solution. Figure 1. Dropout Voltage in the datasheet shows an typical drop-out voltage, with 2A of output current, will be near 1V. That means that at your nominal operating point (Vin=14V and Iout=2A) your output voltage will be near 13V, and you will be dissipating about 2W.

    Using a CMOS (NMOS/PMOS) pass element rather than Bipolar (PNP) will reduce the drop-out mode losses to 'Rdson x Iout'.

    I would suggest LM5060. This is NOT a linear voltage regulator.

    The "LM5060 High-Side Protection Controller with Low Quiescent Current" offers :

        1) lower drop-out voltage (Rdson x Iout) by using external MOSFETs, and

        2) user settable Over-Voltage Protection (OVP) via external resistors.