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LMV324: Simple conditioning circuit needed

Jerry Bucci
Expert 1540 points

Part Number: LMV324
Other Parts Discussed in Thread: MSP430F437

Hi 

I have a product in design that has a sensor that puts out a voltage range of 1.11V to 1.54V. I need to condition that to 0V to 2.5V for input to a MSP430 2.5V A/D. I simulated the circuit below , but not sure if that is the best way to do it. I have 2 spare sections of LMV324 to use. Here is what I have but I cant seem to get exactly what I want  in simulation by changing the gain and offset. The input impedance of the sensor is the midpoint between two 75 ohm resistors in a bridge.

Any ideas?

Thankst

  • 0
    Guru 140200 points

    Hi Jerry,

    you could do it this way:

    jerry_lmv324_1.TSC

    Kai

  • 0
    Expert 1540 points

    Thanks so much!. Actually I saw that circuit in the TI cookbook.  I'm using 3.3V as the Vcc, but still works okay. I have to tweak it a little to get the ends not to be over 2.5 or under 0. 

    Again thanks for the quick reply!!!!

  • Former Member
    0 Former Member in reply to Jerry Bucci

    Hi Jerry,

    For future reference I would recommend the excel calculator in this post.  It's a great tool for quickly creating a voltage scaler circuit and works in a lot of different cases.

    Hopefully you will find it to be useful at some point in the future.

    Regards,
    Daniel

  • 0 in reply to Former Member
    Expert 1540 points

    Thanks Daniel,

    That was my next question... Is there a calculator for it? Very helpful- Thanks!!

  • 0 in reply to Jerry Bucci
    Guru 140200 points

    You are welcome :-)

    I just see that I have made the calculation with 1.15V and 1.54V instead of 1.11V and 1.54V. I'm sorry for that.

    The calculation is very simple:

    U1, as being a differential amplifier, is responsible for the span of output signal:

    (1.54V - 1.15V) x 56k / 10k = 0.39V x 5.6 = 2.184V

    and U2 for the offset:

    2.5V x 10k / (10k + 12k4) = 1.116V

    This gives

    (1.15V - 1.116V) x 5.6 = 0.19V

    and

    (1.54V - 1.116V) x 5.6 = 2.37V

    And the span is indeed

    2.37V - 0.19V = 2.18V

    An output offset of 0.19V was chosen because the LMV324 cannot go deeper than about 0.2V above the negative supply rail in the worst case. And a span of 2.18V was chosen to allow some headroom of a bit more than 0.1V (concretely spoken 0.13V) at the top of the span to account for manufacturing tolerances of components and some error of ADC.

    As the input signal span is rather small and due to this the gain of differential amplifier rather high, the circuit is a bit sensible against little changes of R5 = 12k4. So you might want to use a trimmer here. You could replace R5 by a series circuit of fixed 10k resistor and 5k multiturn cermet trimmer.

    Kai

  • 0 in reply to kai klaas69
    Expert 1540 points

    Kai,

    Thank you for all this. I am concerned about the sensitivity of the circuit as the sensors can vary a little in both the voltage range and offset. I could reduce the span of the input to the A/D to use less then the full 0-2.5V range to allow for variations. However, if the offset is that sensitive, it's a problem because a pot is not a good option for this low cost design. The MSP430F437 that is being used has spare channels of A/D. Maybe I could run a filtered PWM, to generate a variable DC reference voltage for the input to the bias voltage follower? 

    Daniel pointed to a spread sheet calculator that generates the resistor values required for different input ranges and offsets. I will have to characterize a bunch of sensors to see what the variations are and determine whether one set of resistor values can accommodate the variations first.

    Tx

    Jerry

  • 0 in reply to Jerry Bucci
    Guru 140200 points

    Hi Jerry,

    then just take fixed components and provide enough footroom and headroom to take account for all tolerances. The 12bit ADC of MSP430F437 might still provide sufficient resolution.

    Kai 

  • 0 in reply to kai klaas69
    Expert 1540 points

    yes that's the best solution. Thanks Again.

  • 0 in reply to Ron Michallick
    Expert 1540 points

    yes, Daniel Miller pointed me to it. It is a great tool. 

    Thanks,

    Jerry