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LMR16006: Practical maximum output current

David Dean
Expert 2325 points
Part Number: LMR16006


What is the best way to interpret both the rated output current and the switch current limit to determine how much current I can get out of the device in practice?  I have done some testing with similar TI regulators that have a 600mA "output current" and ~1200mA switch current, and I am able to get around 1000mA from these regulators without much trouble.  The 600mA figure is not actually listed anywhere within the electrical specifications, while the switch limit is, so I wonder in what situations each parameter is significant.  Perhaps 600mA is the max over the full voltage range, but higher outputs are available within a range that does not violate the switch limit?  

  • 0
    TI__Mastermind 25590 points
    David,

    The 600mA figure is the output current for which this part is optimized. Higher output current means the efficiency will start to drop off and you may start to run into thermal issues. This is the output current which is equal to the average inductor current.

    The switch current limit refers to the peak inductor current. The inductor current ramps up and down in a triangular waveform. 1.2A peak current limit means the output current will be limited to 1.2A - deltaIL/2. This max IOUT increases as deltaIL decreases (higher SW frequency, higher inductance, duty cycle farther from 50%).

    -Sam
  • 0 in reply to Samuel Jaffe
    Expert 2325 points
    Thanks Sam. My application has a variable input voltage but it tends to be pulled down when Iout increases, so if the current went above 600mA VIN would be fairly close to Vout and the duty cycle would be very high. This is the case where the maximum output current (ignoring thermal issues) would be at it's highest (close to the switch limit) - correct?

    Supposing I'm not too worried about efficiency, are there any guidelines to designing around an output current up to 1A, whether that's thermal or anything else? Or is the official TI position "DON'T DO THAT"?
  • 0 in reply to David Dean
    TI__Mastermind 25590 points

    To your first statement, generally yes. Your ripple is related to duty cycle (see red line) so VIN close to VOUT will give you a smaller ripple than a higher VIN (up to VIN/2 = VOUT where duty cycle = 50% and this curve changes direction).

    Using a 600mA part at 1A means the part will heat up a lot. If you must use this part then I'd recommend allowing a lot of copper connected to the VIN and GND pins. Also have as much copper on the other pins as possible. The heat will come out of the pins so copper is good to get that heat away. 1oz copper is okay. 2oz is better. Lots of thermal vias and other layers of unobstructed copper. Heatsink if you can get away with it. Keep it in a cool environment. Allow for airflow.

    Or get a part rated for 1A+. You're right, the official TI position is, "We don't recommend this" but it's up to you to test/validate your own designs so I can't stop you :)

    -Sam