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LM2623 Output Current and Efficiency

R11
Expert 1485 points
Other Parts Discussed in Thread: LM2623, LM2621, LM2665, TPS60141, LM2700

Two questions on the LM2623:

System requirements:   Input source = 2AA batteries in series.  ---   Output = 7.5V, 5mA Max.

Q1) I ran through the design calculations and used the on line tool to calculate the inductor value.  It gave me a result around 1mH which is a LOT bigger than the 4.7uH in all the reference designs.  I assume this is because our output current is so much lower?  Is this correct?

Q2) what would the efficiency of this circuit be and what should the values of R104 and C106 be to optimize this?

I can send the schematic to your email as I don't want to post it here.

  • 0
    TI__Genius 14317 points

    Hi Rustin,

    Run Webench for the LM2621. That device is very similar to the LM2623 but with higher switch current FETs. The Webench report should provide the effeciency and other info you are looking for. In fact, I would encourage you to use the LM2621 unless you have some specific reason for choosing the LM2623 instead.

  • 0 in reply to Jeffery Pfarr
    Expert 1485 points

    Thanks Jeff.  Will try that right away.

  • 0 in reply to R11
    Prodigy 5 points

    Are there any other converters in the system, Rustin?
    If you have a buck running off the battery stack then you can add a winding on the transformer and tap off for a low current auxiliary output.
    Coupled inductors come in integer turns ratios, so if you are making a 1.875V rail then a 4:1 inductor could make 7.5V.

    1411.Auxilary Voltages at Low Cost.pdf

    The reason Webench tries to use a large inductor is that it's trying to keep inductor current continuous, which drives a larger L.
    I don't know if there is a way to change that.
    A boost running at such low current would likely operate in a PFM pulsed frequency mode which would have slightly higher output ripple voltage.

    I thought about a charge pump, for this low current, but 7.5Vout is the problem for a single device.
    LM2665 is a simple doubler, it could make 7.5V from a 3.75V source.
    We have many charge pumps that can make 3.3V from your source.
    TPS60141 makes 5V, double that with a LM2665 for 10V, then LDO down for 7.45V.

    Perhaps if we had more information about the system we could provide more possibilities.

  • 0 in reply to Ed Walker
    TI__Genius 14317 points

    A quick run of Webench with Iout at .2 A indicated that the LM2621/23 may not be good choices. However Webench does give a number of solutions like LM2700 that have efficiencies in the >81% range at 5 mA (with Iout initially specified at .2 A for the design).