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Is it possible to use two LED with a 3-pin non-latching Hall effect sensor?

Discussion in 'Electronics Homework Help' started by Hastaga, Jan 10, 2018.

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  1. Hastaga

    Hastaga

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    Jan 10, 2018
    The assignment requires me to build a circuit that uses a 3-pin Hall effect sensor which detects THE POLARITY OF A MAGNET.
    As in, if it's south pole, LED A will light up. If it's north pole, LED B will light up. If no magnet's nearby, neither would light up.

    I have tested the sensor I got from my lecturer; it lights up only when a side of the magnet gets close, and goes off as I remove the magnet.

    I had searched through youtube and online examples and had only seen two LEDs being used with a 4-pin hall effect sensor. Moreover, in the case of using a 4-pin Hall effect sensor, one LED would always light up whether a magnet is present or not.

    Please help, I really need some pointers on this one.

    Thank you for your attention
     
  2. Bluejets

    Bluejets

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    Oct 5, 2014
    Most hall effect sw are open collector.
     
  3. Hastaga

    Hastaga

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    Jan 10, 2018
    Still thinking. My understanding of a NPN transistor is that by supplying enough voltage at 'base', emitter and collector remains "open" or connect with each other, thus constructs the "ON" and "OFF" states. But an "ON" state #2 eh....
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    What is the output signal of the sensor for the states
    - no magnet present?
    - south pole near magnet?
    - north pole near magnet?
    Alternatively link to the datasheet? We need to know the output signal under these conditions to devise a way of indication as required.

    Note that we will not present a ready made solution, rather we will guide you in finding your own solution. It is our understanding that this will help you so much more togain an understanding.

    Harald
     
  5. Hastaga

    Hastaga

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    Jan 10, 2018
    @Harald Kapp

    Thank you for your reply.
    The part number according to my lecturer is drv5053 (although the numbers and letters on it look different, it says 720JF or 72OJF + ALOA ), since mine's long body, i assume it is the 3-pin TO-92.

    http://www.ti.com/lit/ds/symlink/drv5053.pdf

    this is the datasheet i found online.

    (if for whatever reason the part is not proper to the task, I still have two 44e/938 from a different source. I could not find the datasheet of that however, not even with google.)

    As for the output signal of the sensor for the states, the following are direct quotes from my assignment sheet:
    "The circuit will have 2 LEDs, 1 will indicate the (+) magnetic field, south pole of the magnet and the other LED will indicate the north pole or (-) magnetic field. Both LEDs will be off when no magnetic field is present"

    My understanding to my assignment is like this after referencing the below youtube video:



    There are some differences however. Firstly, the person in the video was using a 4-pin (instead of 3-pin) hall effect sensor. Second, one of the LEDs would remain on in the video even without a magnetic source nearby.

    My earlier test with the hall effect sensor in question with a single LED connected to the output and ground showed that the LED would light up when i put a magnet nearby and goes out when i withdraw the magnet from the sensor.

    Thank you for your attention.
     
  6. Alec_t

    Alec_t

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    According to the datasheet, the drv50553 output as a function of the sensed magnetic field is like this :-
    HallOutput.PNG
    This should give you a clue as to how you can distinguish the magnetic field direction.
    Note also that the output voltage range is 0V-1.8V for this field.
     
  7. Hastaga

    Hastaga

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    Jan 10, 2018
    I am not very good at this -_- do I need a large magnet? And 2 AA batteries as power supply should be enough, right?
     
  8. Alec_t

    Alec_t

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    Jul 7, 2015
    The size of the magnet is irrelevant: it's the magnetic field strength which is relevant. Any small modern magnet should do.
    You will need to think of a circuit to condition the sensor output to enable the LEDs to be driven. Depending on the circuit it will possibly require a supply voltage greater than the ~3V that 2 AA cells would provide.
     
  9. Hastaga

    Hastaga

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    Jan 10, 2018
    Please correct me if I am wrong:

    so what I am supposed to do here is to design a circuit that with a flick of the direction/polarity of the magnet, without doing anything else, a different LED would light up? And the key to achieve that is to 'condition the sensor output'? (i think i must know if first part of my question is the expected effect/outcome in your speculation first. I am kind of blind at the moment so it'd be nice to at least have a clear goal).
     
  10. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Your circuit needs to
    • read the voltage from the hall sensor
    • if necessary filter the voltage to remove noise (low pass filter) and potentially amplify it for further processing
    • use the filtered and amplified voltage to decide which LED to turn on
    #1 and 2# are easy and suitable for simple analog circuits (resistors, capacitors, operational amplifiers)
    #3 can be done in various ways.:
    • analog using e.g. window comparators to detect in which range the output voltage is and light a corresponding LED
    • digital using e.g. an analog digital converter and a microcontroller to detect the range and light a corresponding LED
     
  11. Hastaga

    Hastaga

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    Jan 10, 2018
    Thank you for your reply. I guess my lecturer has assigned me something that is beyond the usual expectation -_- if op amp is included that means i need two 12 V power source (Throughout my entire semester, we are accustomed to operate on 1 single 5V power source only). I don't think i had ever heard of a window voltage comparator either =_=

    I'll go read up on window comparators and hopefully learn enough of it before the deadline = ="
     
  12. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Look for single supply opamps, preferably rail-to-rail. There are quite a few that can operate from a single 5 V supply.

    A window comparator is described e.g. here.
     
  13. Frankchie

    Frankchie

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    Nov 14, 2017
    At a glance it looks the the sensor idles at 1v with no magnetic field and varys roughly _+ .9v according to magnet polarity. By connecting two LEDs, each one attached between a 1v reference voltage and Vout but with opposite led polarity the LEDs should light independently according to the polarity of the magnet.

    At a glance I did not see the current drive capability for the sensor so I am not sure if it can directly drive LEDs, but given that an op amp seems beyond what the lecturer would expect perhaps my above idea is the solution.

    Perhaps some other folks here can expand on my above idea.
     
  14. Harald Kapp

    Harald Kapp Moderator Moderator

    11,806
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    Nov 17, 2011
    Basically a good and simple idea. At 2.3 mA sink and 200 μA source current not practical. Some amplification will be required.
    Of course one could use bipolar transistors with base-emitter connected as you suggest, then put LEds in the collector circuit of the transistors. This should produce sufficient gain and abviatre the need for opamps.
     
  15. Frankchie

    Frankchie

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    Nov 14, 2017
    Haraid, Thanks for fleshing out my idea.

    The forward voltage requirement of more than 1 volt for most LEDs could also be a problem.

    The original post says that one LED could be driven, probably using the sink capability of the sensor. So maybe only one transistor is necessary to drive the other LED using the sensors source capability.
     
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