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Sensor voltage change

Discussion in 'Electronic Design' started by WAYNEL, Jul 8, 2005.

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  1. WAYNEL

    WAYNEL Guest

    Hi

    I have a humidity sensor that produces a linear responce from 0 to 100%
    RH calibrated at:

    0.958v @ 0% RH
    3.268v @ 75.3% RH

    I have a PID controller that accepts 0 to 1v at the sensor input.

    I am trying to produce 0v for 0% RH and 1v for 100%RH or 0.753 for
    75.3%RH.

    Can anyone propose a very simple circuit that will do something
    similar. e.g using a transistor or opamp.

    Cheers

    WayneL
     
  2. Jon

    Jon Guest

    Use an op-amp.
    Non Inverting input goes to ground
    2 input resistors: R1 = R2 = any practical value.
    1 Feedback resistor R3 = R1/(3.268-0.958) = R1/2.31
    Sensor goes to inverting input through R1.
    Offset voltage of -0.958 goes to Inverting input through R2.
    ..
    The output voltage will be negative. Invert it with an inverting op
    amp stage with a gain of -1. (Input R = Feedback R).
    ..
    You could use any negative value of offset voltage, and then adjust R2
    accordingly. For example, if you want to use -10V as the offset
    voltage, then R2 would = R1 * 10/.958.
     
  3. Fred Bloggs

    Fred Bloggs Guest

    View in a fixed-width font such as Courier.
     
  4. GPG

    GPG Guest

    3.068*R
     
  5. Fred Bloggs

    Fred Bloggs Guest

    It's a DA. He wants Vout=(Vin-0.958)/(3.268-0.958)=(Vin-0.958)*(1/2.31)
     
  6. Hang on Fred..... dVin/dRH is 2.31V per 75.3% change, which
    is 3.068V per 100%.

    For preferred values. 12k/3.9k is a ratio of 3.077, or about
    0.3% error.
     
  7. Fred Bloggs

    Fred Bloggs Guest

    Right- did not realize the OP was asking for both a circuit and a
    transfer function. To the high school dropout:

    View in a fixed-width font such as Courier.

    ..
    .. |
    .. |
    .. |
    .. X + - - - - - - - / E
    .. | / |
    .. | /
    .. | / |
    .. | /
    .. | / |
    .. 3.268 +- - - - - / C
    .. | / | |
    .. | /
    .. Vsensor | / | |
    .. | /
    .. | / | |
    .. | /
    .. | / | |
    .. 0.958 + - - -A - - - - -B - -D
    .. | | |
    .. | |
    .. | | | |
    .. | |
    .. | | | |
    .. 0 ------+----------+-----+-----
    .. 0 75.3 100
    ..
    .. RH->
    ..
    ..
    ..
    ..
    .. ED CB
    .. ABC similar to ADE => -- = --
    .. AD AB
    ..
    .. AD=100 AB=75.3 CB=3.268-0.958=2.31
    ..
    .. 2.31
    .. Then ED= ---- x 100 = 3.068
    .. 75.3
    ..
    .. which makes X= 3.068 + 0.958= 4.026
    ..
    ..
    .. The RH sensor puts out 4.026V at RH=100%
    ..
    ..
    ..
    ..
    ..
    .. DESIRED FUNCTION
    ..
    .. |
    .. |
    .. 1V+- - - - - - /
    .. | / |
    .. | /
    .. | / |
    .. PLC IN | /
    .. | / |
    .. | /
    .. | / |
    .. | /
    .. | / |
    .. | /
    .. 0V+------+------------+---------
    .. 0.958 4.026 Vsensor
    ..
    .. -----+------------+---------
    .. 0% 100% RH
    ..
    ..
     
  8. WAYNEL

    WAYNEL Guest

    Thankyou very much Fred Bloggs.


    WayneL
     
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