ADC Math - reading 30Vdc


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  1. #1
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    Default Re: ADC Math - reading 30Vdc

    There are a couple of ways. One is to use the DIG command to extract the numbers. Another way would be to use the /(divide) and //(modulus) functions.

    So, you print (number /100)
    print decimal point
    print (number // 100)

    I do not have a handy example to show you.

  2. #2
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    Default Re: ADC Math - reading 30Vdc

    I tried the DIG command earlier and I got an odd character.

    Interestingly, with a 8.0V source, I'm getting a reading of 15.4V. I'll have to do some further testing.


    Meanwhile, any good Math for a thermistor?

  3. #3
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    Default Re: ADC Math - reading 30Vdc

    Quote Originally Posted by jmgelba View Post
    I tried the DIG command earlier and I got an odd character.

    Interestingly, with a 8.0V source, I'm getting a reading of 15.4V. I'll have to do some further testing.


    Meanwhile, any good Math for a thermistor?
    Fixed both of these issues. Thanks for the help everyone.
    On to the next part of the project: controlling a digital pot.

  4. #4
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    Default Re: ADC Math - reading 30Vdc

    Also make certain that you have an adequate acquisition time. The ADC is most accurate when the source impedance is < 2K ohms. With a 10 K source, you will need an acquisition time of 11uSec or greater.
    Charles Linquist

  5. #5
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    Default Re: ADC Math - reading 30Vdc

    Thanks Charles. I have had this set to 50uS right from the start. All my ADC channels are working very well.

  6. #6
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    Default Re: ADC Math - reading 30Vdc

    How accurate do you need you 3V? A 100K and 10 K don't make a 10:1 ratio.
    Ohms Law RTotal/ Rx (in this case 10K)
    110K / 10K is a ratio of 11:1

    So dropping 30V across 110K will give 27.27V across the 100K and 2.73V across the 10K


    aajgss

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