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TMAG5123-Q1: Value of pull-up resistor determines the output voltage

Part Number: TMAG5123-Q1
Other Parts Discussed in Thread: TMAG5123

I used the TMAG5123-Q1 in my own design as recommended in the datasheet with a 10kOhm pull-up resistor. The output voltage was significantly lower as the Vin. With Vin at 5V the Vout was about 2.2V. I replaced the 10kOhm resistor with 100Ohm and the Vout is about 4.5V 

Is that how this chip is intended to work? I did not notice any mention of such behavior or to use of different values of pull-up resistors in the datasheet. 

  • Matej,

    Thank you for reaching out.  This does not sound like expected behavior.

    The output stage of the device is an open drain configuration requiring a  pull-up resistor to set the high output state.  In either case, when the device is not active, the output should release and draw no current.  The result should be that you observe little to no voltage drop across the resistor.  It might be possible that there is a short, leakage path, or some other kind of load on this node.  

    If instead, the device is operating in the presence of a magnetic field, then the output of the device would be attempting to pull this voltage down.  The lower resistance value would require the device to sink a greater amount of current.  For instance, you mention Vcc = 5V and a resistance of 100 Ohms.  The resulting sink current would be 50 mA which exceeds the recommended 20 mA maximum operating sink current for the device.  

    As a matter of troubleshooting, I might try removing the device and looking to see if the voltage drop still occurs with the 10 kOhm resistor.  If the voltage is pulled-up correctly, I would measure the resistance to ensure that the correct resistor is installed.  After this step, I would then replace the unit with a new device to verify normal function.  If there is a voltage drop without the device installed, then it provides a starting point to chase after other possible issues on the PCB.

    Thanks,

    Scott

  • Hi,

    thanks for the quick response. I did some debugging but I couldn't find / fix the issue.

    The TMAG5123 output is connected to the https://eu.mouser.com/ProductDetail/Diodes-Incorporated/PAM2306AYPKE?qs=sGAEpiMZZMutXGli8Ay4kMsKmnQTt96q4oMlwAk8nHc%3D power supply chip. For better representation I am also attaching the images of the design:

    I changed the resistor to 10kOhm. And also replaced two new TMAG5123 chips. Always get the same ~2.3V on VOUT 

    If I remove the PAM2306 IC completely and there is no load on the TMAG5123 then I get the correct 4.9V which is also the voltage that is on VCC 

    Turning the device off with a magnet works in both cases.

    When I remove the TMAG5123 chip completely and measure the current between VOUT & VLI-ION pads I get ~1mA which seems normal to me.

    Any tips would be greatly appreciated.

    Best,

    Matej

  • Matej,

    What it looks like must be happening here is that the pull-up resistor is the primary current path for Vin1 and Vin2 of the PAM2306. Since TMAG5123 is an open drain device, it is not sourcing the current input for PAM2306, but rather, all of this current passes through the 10 kOhm resistor.  A voltage drop of 2.6 V would suggest an input current into PAM2306 of only 260 uA, and this also explains why you were able to see a higher voltage with a 100 Ohm resistor.

    Perhaps the more ideal way to setup this circuit to control the system would be to connect VIN1 and VIN2 directly to VLI-ION.  You would only use VOUT from TMAG5123 to connect to EN1 and EN2.  This way, the supply current for the rest of the system is not passing through the pull-up resistor, but would rather be sourced directly from the battery.  

    Thanks,

    Scott

  • I redesigned the circuit and now there is no more voltage drop. Thanks!