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LM4050-N-Q1: How to use it as a low current negative LDO?

Part Number: LM4050-N-Q1

Hi team,

I want to use LM4050 as a negative LDO. I draw the block like below. Is it the right way to use LM4050? What do I need to pay attention about this circuit?

The -7V is the output of the isolation DCDC, and it is not very accurate and have some ripples @400kHz. 

Regards

Michael

  • Michael,

    The key to these devices is to ensure that the cathode current is met so that the voltage drop across the reference is as expected.
    Another key consideration is the capacitor connected across the device. You need to make sure that the capacitance does not make the device unstable.
    This is a fixed reference so I believe it should be stable with a wide range of caps, but you should look in the datasheet to make sure the cap you are using is valid.

    Last thing, you need to make sure that the GND can actually source current through the cathode.
    If you want to be able to source any current into the -2.5V node, the resistor from -2.5V to -7V needs to be sized such that the change in loading at the -2.5V rail does not increase the cathode current dramatically.

    Again, cathode current is key to regulation.

    Best,
    Michael
  • Michael,

    So if the current is 0-3mA load, the 1kohm resistor make it work?
    When load is 0mA, the current through LM4050 is (7-2.5)/1Kohm =4.5mA. When the load is 3mA, the current through is 1.5mA. It seems good.


    Regards
    Michael
    FAE
  • Hi Michael,

    I'll answer in place of Michael. That is correct :).

    -MZ