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Does the LM35 output require a pull down resistor?

James Nadir
Expert 1120 points
Other Parts Discussed in Thread: LM35

The LM 35 datasheet states that the output impedance is about 0.5 ohm but it can only source current (10ma) and that its current sinking capability is limited to 1 uA. Does this mean that while the sensor can easily drive the input to a ADC high it needs a pull down resistor to help it go low? If so are there any guidelines for sizing the pull down resistor?

If the ADC is a delta-sigma design should I also add an external capacitor between the LM35 and the ADC to minimize voltage noise? If so, are there any guidelines for sizing the capacitor?

In both cases, my sample rate is around 10KHz.

thank you

jim

 

  • 0
    TI__Expert 7220 points
    Hi James,

    The LM35 only needs a pull down resistor if the anticipated temperature will be less than 2C. If your system does not go down to 2C then no pull down resistor is necessary.

    We usually recommend an external capacitor between the output of the LM35 and the input of the ADC. This is external capacitor is used to quickly transfer charge to the sample and hold capacitor inside the ADC during sampling. Since all ADC's have internally different topologies we cannot recommend a one size fit all capacitor. You will need to do some experiments with the ADC you are using, if the ADC codes swing up or down then a higher value capacitor may be needed. Note that the LM35 can drive a 50pF load alone and any higher loads will need a R and C combination as seen in section 8.1.1 in the datasheet.

    -Michael Wong
  • 0 in reply to Michael Wong
    Expert 1120 points

    Hi Michael

    Thank you very much for your response. My system operates from 10C to 45C and falls into the "no external resistor required" category. What I don't understand is how the sensor can follow a dropping temperature if it is only able to sink 1ua?  Do I need to wait for a period of time to allow the 1ua to discharge the local capacitance? And would an external 25pf load make the situation worse?

    Thank you so very much for your support

    jim

  • 0 in reply to James Nadir
    TI__Expert 7220 points
    Hi James,

    The extra charge when the LM35 drops in temperature is sinked through the output pin at 1uA. The voltage rate of change is defined as I = C*dV/dt. The C term is the external capacitor between the ADC and the output pin and I is the sink current. We usually do not need to wait for the capacitor to discharge before taking a sample from the ADC.

    -Michael Wong
  • 0 in reply to Michael Wong
    Expert 1120 points
    thank you again. I understand. Just out of curiosity is there a reason that the designers of the LM35 prefer a high source current but a low sink current ?
  • 0 in reply to James Nadir
    TI__Expert 4905 points
    Hi James - it's mainly due to die area and circuit simplicity. The LM35 was the first of it's kind on a bipolar process. One cool thing though is you can tie LM35 outputs together and the hottest temp will always win.

    Take care,