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I have a target board(With Nuc240,Cortex M0) with ADC connected through an OP-Amp(NE5532APSR) as shown in the image.enter image description here

where AINV1 is the input(0-30 V) and AINVI_ADC is the input to Microcontroller(0-3V) and AINV1_GND - Analog input voltage supply ground

I configured the inputs and I'm getting the Digital values in the range 0 to 4096. for the 12 bit ADC. The input to Op-Amp(negative feedback) ranges between 0 to 30 volt for which corresponding microcontroller input will be 0 - 3 volt.In the code,should I consider the input resistance R44 and R48 in calculation or just multiply the ADC reading by 7.324*10^-3 ie. -> (3/4096 * 10) to get the exact input voltage, if not what are the parameters I must consider?

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  • \$\begingroup\$ How is your AINV1_GND referenced to your system GND? And a 10k/1k5 voltage divider doesn't produce 0-3V from 0-30V. \$\endgroup\$
    – brhans
    Jan 11, 2017 at 13:57
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    \$\begingroup\$ A 5532 can't swing rail to rail, if you're running it from 3v7 and gnd it's unlikely to make a 3v output. \$\endgroup\$
    – Colin
    Jan 11, 2017 at 14:06
  • \$\begingroup\$ I agree, but can u answer the question please,what parameter I must consider in calculating the Vin? \$\endgroup\$
    – Arun Joe Cheriyan
    Jan 11, 2017 at 14:09
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    \$\begingroup\$ (1.5/11.5) * ADC COUNT * 7.324*10^-3, the adc count * 7... gives you the 0-3v level going into the adc, and the 1.5/11.5 is from the voltage divider before the buffer. With the problems mentioned though this won't quite work as it should. \$\endgroup\$
    – Colin
    Jan 11, 2017 at 14:17
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    \$\begingroup\$ It's shown as 0 ohms, and the adc should be high impedance input so the drop would be negligible \$\endgroup\$
    – Colin
    Jan 11, 2017 at 14:35

1 Answer 1

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R44 has nothing to do with anything after your voltage follower (which is the term for the morphology you're using). R48 does, as does the output impedance of your amplifier.

Whether you have to worry about it or not depends on what your requirements are. The output impedance of your circuit (which includes R48) forms a voltage divider with the input impedance of your ADC. The output impedance of the op amp will be very low, so lets not worry about it. If R48 is 1kOhm and the input impdedance of your ADC is 9kOhm, then you will read 0.9Vout, where Vout is the output voltage of the amplifier. If R48 is 50 Ohms and the input impdeance remains 9kOhm, you will read 0.994 Vout. For most purposes, that's "close enough" that you needn't worry about it.

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