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LM385-2.5: application question

Part Number: LM385-2.5

1. Can you describe how use LM385, what's the input voltage range?

2. How to choose the pull up voltage and resistor? Can we change P5V to 3.3V and get the stable 2.5Vref out?

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  • Hi Colin,

    this is a shunt regulator. To maintain regulation a minimum current of 20µA must be able to flow into the LM385-2.5.

    Let's have an example: If the resistor is chosen to be 75R, then a current of (3.3V - 2.5V) / 75R = 10.7mA will flow through this resistor. If the load current is 0A, then all the 10.7mA is flowing into the LM385-2.5. If, on the other hand, the load current is 10mA, then the difference 10.7mA - 10mA = 700µA is flowing into the LM385-2.5.

    The current through the LM385-2.5 must never be outside the range of 20µA...20mA. Otherwise the LM385-2.5 will stop working properly.

    Kai
  • Hi Colin,

    Kai has described the design process well for how to use the LM385-2.5.
    Keep in mind that the pull up voltage has to be greater than the 2.5-V of the device. There is no limitation on how high the pull up voltage can because an appropriately sized resistor can be chooses to give the proper current.

    -Marcoo Z