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Homework Help: Potential Divider

  1. Apr 29, 2007 #1
    1. The problem statement, all variables and given/known data

    Why is a potential divider required in a sensor circuit?

    2. Relevant equations

    V_out = r1/(r1+r2)*V_in

    3. The attempt at a solution

    Firstly you can get the required voltage across the sensor by changing the resistance of r2 (if r1 is the sensor).

    Apparently, though, using a potential divider also increases the sensitivity of the sensor. ie. a small change in resistance causes a larger change in output, but I don't see how this works. could someone explain?

  2. jcsd
  3. Apr 29, 2007 #2
    any ideas?
  4. Apr 29, 2007 #3


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    Gold Member

    Your answer seems logical - i.e.to get the correct voltage drop across the sensor.

    You can work out the sensitivity by calculating d v_out/dr1 where r1 is the sensor resistance, as a function of r1 and r2.
  5. Apr 30, 2007 #4
    ah thanks! a bit of calc shows that you have maximum sensitivity when r1=r2...
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