Hall effect sensor - 480-5198-ND - Question

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Discussion Overview

The discussion centers around the operation of a Hall effect sensor, specifically model 480-5198-ND, and how to modify its output behavior in a circuit. Participants explore how to achieve an "OFF" state by default, with the sensor turning "ON" only in the presence of a magnet. The scope includes practical circuit design and component configuration.

Discussion Character

  • Homework-related, Technical explanation

Main Points Raised

  • One participant describes their current setup where the Hall effect sensor is "ON" by default and seeks a solution to invert this behavior.
  • Another participant suggests using a common-emitter NPN transistor stage to invert the output signal from the Hall effect sensor.
  • A request for clarification on how the common-emitter configuration inverts the signal is made, prompting an explanation of the circuit's operation.
  • Further inquiry is made regarding the calculation of resistor values R1 and R2 in the proposed circuit.

Areas of Agreement / Disagreement

Participants are exploring a common approach to modifying the sensor's output behavior, but there is no consensus on the specific resistor values or detailed calculations at this point.

Contextual Notes

The discussion does not provide specific values or methods for calculating resistor values, and assumptions about the circuit configuration and component specifications are not fully detailed.

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Homework Statement


I am using the a hall effect sensor - model 480-5198-ND. I have tested this simple circuit using a LED and it is working. However, it is "ON" by default. Only when the magnet is near the hall effect sensor , it will be "OFF".

How do I make it such that it is OFF by default and ON only when the magnet is near the hall effect sensor?

Homework Equations

The Attempt at a Solution


http://[ATTACH=full]199957[/ATTACH]
 

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The A in my diagram means that it is connected to the LED.
 
Follow it with a common-emitter NPN transistor stage to invert the output signal. Does that make sense?
 
Can I have some explanation why it is able to invert the signal?
 
May I know how do I calculate the value of R1 and R2?
 

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