Differential Op Amp to convert F to C degrees?

Discussion in 'General Electronics Chat' started by antseezee, Mar 5, 2007.

  1. antseezee

    Thread Starter Active Member

    Sep 16, 2006
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    We have an upcoming lab where we have to design a F & C degrees output thermometer based off of the LM34 chip (10mV / degree Fahrenheit output). Here's the design I came up with to convert to Celsius. Pretty good, eh?

    [​IMG]

    The formula for Fahr --> Cel conversion is:

    (5/9) * (Fdeg - 32)

    Since the chip has 10mV / deg Fahr, I combined a 1/10th * 5/9th gain to be a 5/90th gain. This gain will display the temperature exactly when the VOM is in mV mode. Used a differential amplifier to subtract the Fahrenheit voltage from a constant .320V (the 32 in the equation). And the differential amplifier has the Fahrenheit on the positive input, yielding a positive result (unless a negative temperature, negative voltage).

    I have a question though. Is there any better method of getting a precise .320V on the negative input? The voltage divider, while easy and practical, is not giving me exactly the amount I need. I'm trying to NOT use a separate voltage source as this will require more components. I also noticed that the voltage divider on the - input is affecting the Ri value there (the parallel combination comes out to .979k ohms, + 89k ~= 90k). Thanks.
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    Use a National Semiconductor LM136 2.5 volt reference device. Recalculate your voltage divider. You will want to make sure that the voltage divider gives your reference a source impedance that when added to the resistance R2 equals 90K. That will minimize the error in your output caused by the small input currents associated with the uA741.

    You may want to consider using an opamp with an FET input such as the TL071 or similar.

    hgmjr
     
  3. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
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    Wow, analog computation makes a comeback!.

    Fixed resistor values never quite work out for precise voltages. Substitute a trim pot with a resistance close to the two resistors in the divider, and use the wiper to find the exact voltage out you need.
     
  4. antseezee

    Thread Starter Active Member

    Sep 16, 2006
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    Thanks for the feedback so far. I should have mentioned that we can only do the following:

    Use Op Amps, standard components, and the LM34 chip. No other IC chips are useable, just the LM34. It is an Analog Design course as well, mainly based around the 741 and all of its properties. I was using the +- 15V source since this was the standard voltage source for the 741, and the LM34 is powerable anywhere from 6V to 30V. When I did the calculation, the ratio I would need was 46.8k to 1k to get exactly .320 V from a 15V source.

    I figured the input impedance was to blame. I'm curious though. Should I increase the resistance on all resistors but maintain the same ratio to further rid of the input currents? I know you're suppose to generally use resistors from 1k < x < 1M for the 741.

    I did not know about that potentiometer setup. I will try to give it a whirl - thanks!
     
  5. mrmeval

    Distinguished Member

    Jun 30, 2006
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  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
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    Antseezee, your gain is wrong. From your sensor, it needs to be 5/9, not 5/90. The way you have it set up, when F goes from 0.32 to 2.12, C will go from 0 to 0.100. You want it to go from 0 to 1.00.
     
  7. antseezee

    Thread Starter Active Member

    Sep 16, 2006
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    0
    Thanks for the feedback. Unfortunately, you're right, the design was wrong. In the lab, biggest problem was getting the balanced closed-loop differential amplifier to operate at the correct gain. We tried tuning it but simply went back to a normal differential amplifer. We compensated by putting a voltage divider at the output. Extra work, slightly more components, but it did work.

    Yeah I was pretty foolish to not test the extremes on the design. I simply tested the 32 degrees condition because I knew the answer would be close to 0. Of course, with the gain set wrong, you'd barely be able to tell the difference between 0 and 100u.
     
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