Question about hall sensor and current output limitation?

Discussion in 'General Electronics Chat' started by Jon K, Nov 27, 2009.

  1. Jon K

    Thread Starter New Member

    Nov 27, 2009
    3
    0
    Hey all - using this sensor for a project here.

    [​IMG]

    I am going to be using a 5v supply for the sensor, a chassis ground, and the output is going to an engine control unit. Ignore the pull-up resistor, that image is just built into the datasheet, I am going to use it sinking to ground as is default.

    It says max current output 25mA. On the package it said "An external resistor must be fitted to assure 25mA max current" - ok, so where should I hook the resistor? I did the maths to determine a 200 ohm resister for 25mA max. Is this correct or should I get further from the "max", and where in the circuit should the resistor fit?

    Thank you!
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    OK, see how the Hall-effect sensor is connected in the datasheet? Complete with the resistor?

    That's how it needs to be connected.

    They couldn't include the resistor in the design of the Hall-effect sensor, because that would limit the design to operation in a narrow range of supply voltages. You have to calculate the resistor depending upon your supply voltage.

    Now you have to consider your load. It may have a capacitive input, which current from the resistor will have to charge, and the Hall-effect sensor's open-collector output will have to discharge.

    To get the fastest charge and discharge times, divide the maximum sink current by two.
    25mA/2 = 12.5mA.
    Then calculate the resistor value:
    5v / 12.5mA = 5/0.0125 = 400 Ohms.
    400 Ohms is not a standard value of resistance. Here is a table of standard resistances:
    http://www.logwell.com/tech/components/resistor_values.html
    Usually, hobbyists can find E12 and E24 values (the yellow and green columns) locally.
    390 = 12.82mA
    430 = 11.63mA
    470 = 10.64mA
    Alternatively, you could use two 200 Ohm resistors in series to get exactly 400 Ohms.
    However, a single 430 Ohm or 390 Ohm resistor would be adequate for most purposes.

    It isn't shown in the schematic, but I would suggest adding a 0.1uF (100nF) capacitor across the Vcc and GND terminals of the Hall-effect sensor. This will help a great deal to keep the output signal "clean".

    If your Hall-effect sensor will be located far from the ECU, you would be better off to have the pull-up resistor located right next to the ECU. This will help to reduce electrical noise on the wire.
     
  3. Jon K

    Thread Starter New Member

    Nov 27, 2009
    3
    0
    I follow everything except why the current should be halved. I will probably skip putting a capacitor across it because I think the ECU does a lot of filtering - the 5v source is regulated out of the ECU and the grounds are a common sensor ground. Thank you for your help!
     
  4. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    It is best to operate components at 1/2 of their maximum ratings, or you risk rapid failures.

    I suggested the capacitor across the Vcc and GND leads of the Hall-effect sensor for a good reason; it's to help ensure the reliability of the sensor. If you don't include the capacitor as I suggested, you will likely have problems.
     
Loading...