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SN74LVC2G125 switching oscillations

Other Parts Discussed in Thread: SN74LVC2G125, SN74LVC1G17

Hello all,

I am using the SN74LVC2G125 buffer to interface a µ-controller with Hall sensor.

The Hall sensor is supplied with +5V. The sensor signal is pulled-up to +5V through a 5 Kohms resistor, and tied to GND through a 1nF capacitor (sensor requirements). This signal comes in entry of the SN74LVC2G125.

The SN74LVC2G125 is supplied with the same +5V supply (decoupling capacitor is 100nF). As it is not used the second input (resp. enable) is tied to GND (resp. Vcc).

The SN74LVC2G125 output goes then to the µC input. The µC has internally a +5V pull-up resistor (5 Kohms), and a capacitor between input and GND (unknown value).

When the Hall sensor state is changing (rising edge or falling edge), here is my problem :

- The +5V supply has big disturbance

- The SN74LVC2G125 goes not  "directly" from high state to low state (and conversely), but there are some oscillations that can last for some hundreds of ns / some µs

I tried to solve this by adding to SN74LVC2G125 a capacitor between output and GND (I tried 33pF). The behavior is better, but there are still oscillations.

Shall I add some line resistor (500 ohms) as indicated in the datasheet on § Parameter Measurement Information, page 8, figure "load circuit" ? Or shall I just try different capacitor values until I find the right one ?

Thank you for your help,

Best regards,

Julien

  • Hi Julien ,

    The 5V supply disturbance can cause the output to oscillate . The hall sensor rate changing can state slowly (rising edge and falling edge ) can be a problem which causes oscillations , if it is not within the recommended conditions of 5ns/V . For tolerating slower inputs I would suggest using the Schmitt trigger devices like SN74LVC1G17 /14.
  • Hi Shreyas,

    Thank you for your answer.

    Reading your answer I understand my problem is caused by the fact that Hall sensor state changes too slowly.

    If I use a Schmitt trigger instead of the buffer, it will add a delay to the output signal, am I right ?

    Best regards,

    Julien

  • Hi Julien,

    Any logic device adds a delay to the output signal -- defined as propagation delay in the datasheet.

    The Schmitt trigger device will switch at a higher voltage than the standard buffer of the same family (by design), which will add some additional delay if your input signal is slow enough (ie it takes longer for your input to get to 0.7*Vcc than it does to get to 0.5*Vcc), but the two devices should have very similar propagation delays.

    The Schmitt trigger is specifically designed to operate with slow or noisy inputs, so it would be the optimal choice for your circuit.
  • Hello Emrys and Shreyas,

    I tried the Schmitt trigger, it works perfectly with Hall sensors, so thank you for your help.

    For my information, why is the SN74LVC2G125 only used with signals that have high voltage rate transitions ? is it because of the silicium implementation of the component or other reason ?

    Best regards,
    Julien
  • Hi Julien ,

    so just curious to know , did you use sn74LVC1G17 device instead of sn74lvc2g125 to get the oscillation fixed ?

    The design and schematic implementation comes down to the use case for the device and what needs to be optimized . The Schmitt trigger is optimized for slow rising and noisy inputs whereas the regular buffer is not .
  • It is because of the structure of the circuit.  A Schmitt-trigger input is more complex, and thus more expensive to build, than a standard CMOS input.  

    Most logic components have simple CMOS inputs and Schmitt-triggers are only used to 'clean up' slow or noisy inputs so the remainder of the logic components in a system can have the clean/fast edges that they need.

  • Hello,

    Yes I tried the SN74LVC1G17 instead of the LVC2G125, and now it works perfectly : only 40ns for rising and falling edges, absolutely no oscillations.

    One question about the LVC1G17 : I used a 500 ohms pull-down resistor at the trigger output, does it mean the component will consume 10mA on the +5V power supply when its output is HIGH ?

    Best regards,

    Julien

  • Julien ,

    Yes , 10mA will be the load which the power supply has to provide . The output itself will not be at 5V anymore with the 10mA load it has to drive .