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While doing an experiment we reached a conclusion which we wrote down as the DMM and the ADC reading being different.

Now after doing some research It does confuse me as a DMM does contain an ADC of it's own. Can someone explain the reason of such?

Our ADC is very simple : An attenuator (voltage divider) connected to an ADC chip.

One of the reasons I can assume is having some inaccuracies in the attenuator but when I asked our prof he told me that there are TWO reasons in fact. Can someone kindly clarify?

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    \$\begingroup\$ Linearity, calibration, quantisation, component accuracy, offset ... \$\endgroup\$ – JIm Dearden Mar 23 '17 at 10:59
  • \$\begingroup\$ What about in terms of pure circuitary. How can you explain it? \$\endgroup\$ – Zaid Al Shattle Mar 23 '17 at 11:05
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The main reason, assuming you hooked up everything correctly and all the equipment was working correctly, is the difference in input impedance between the DMM and your A/D circuit.

The input impedance of the DMM should be clearly listed in its manual. It may even be written write on the unit. A common value is 10 MΩ.

The input impedance of the voltage divider is the sum of the two resistances. The input impedance of a typical CMOS A/D is probably much higher than your divider resistances, so you can ignore its contribution.

Unless you have a very unusual A/D, the sum of your two divider resistors is much less than the input impedance of the DMM. Therefore consider what the voltage source you were measuring does with a 10 MΩ load, and a few kΩ load that your A/D circuit presented.

Of course the dividers add some error too. Look up the accuracy of the resistors and compute the overall accuracy of the divider. Most likely, the accuracy of the DMM is higher than even your bare A/D. Even without circuit loading issues, the DMM is likely significantly more accurate than your A/D circuit. Then the A/D has quantization error too. If it's a 10 bit A/D, for example, then you get about ±0.05% error just from quantization. Likely the resistors swamp that error, though.

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There more than two reasons.

  1. Your measured different points.

  2. Variable input.

  3. Differenct references.

  4. Loading on the circuit.

  5. Operator error.

  6. Faulty instruments.

  7. Faulty recording of measurements.

....

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  • \$\begingroup\$ If ignoring faults in the instruments what part of the circuitary is causing the error so-to-speak. Is it the attenuator or the ADC itself? \$\endgroup\$ – Zaid Al Shattle Mar 23 '17 at 11:05

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