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I have understood how auto-ranging is typically achieved on a digital multimeter, but I'm having trouble finding an example of how (DC) polarity is handled. How does the meter know if a voltage is negative (to show a '-') and not fail from having a negative voltage on the ADC?

I figured it could possibly be done by having ground be in the middle of the ADCs range? (eg by having values: 0 @-2v, 1024 @0v, 2048 @+2v)

So: How is polarity measured/detected in DMMs? (pros and cons of different methods would be useful)

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  • \$\begingroup\$ Considering how negative voltages are common, I don't see how auto polarity would be a good thing. How will the meter know the voltage your measure is actually negative and when its just connect backwards. \$\endgroup\$
    – Passerby
    Commented Mar 31, 2014 at 18:30
  • \$\begingroup\$ I think I didn't make the question clear enough. By "Auto-Polarity" I meant how the meter will tolerate a negative DC voltage, recognize it and show a little '-' symbol. \$\endgroup\$
    – tehwalris
    Commented Mar 31, 2014 at 18:51

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Here is a typical dual-slope integrating meter chip operation (7107):

enter image description here

After the auto-zero phase, the output of the integrator is 0 as far as the comparator is concerned. Then the unknown signal is integrated for 1000 counts. Following the fixed integration period, the state of the comparator will indicate whether the input was positive or negative, and which way the de-integration should proceed (DE+ or DE-). It is de-integrated with the reference towards zero (from either direction) and the time 0-1999 counts is measured and displayed as the reading.

enter image description here

enter image description here

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  • \$\begingroup\$ So what would happen to the waveform if Vin is negative? Does the de-integrate phase slope down for both positive and negative? (And why?) How is the negative input polarity detected? \$\endgroup\$
    – tehwalris
    Commented Mar 31, 2014 at 19:56
  • \$\begingroup\$ If Vin is negative, then the waveform is flipped. It goes down first and then up again to zero. The input polarity is detected by the comparator state at the end of the integration period. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Mar 31, 2014 at 19:57
  • \$\begingroup\$ So it detects the polarity before the integrate phase? (To decide to integrate negative) \$\endgroup\$
    – tehwalris
    Commented Mar 31, 2014 at 19:59
  • \$\begingroup\$ Ah, no, the integration direction is bipolar-- if it bobbles around zero during the integration it could even change direction during the 1000 counts. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Mar 31, 2014 at 20:08
  • \$\begingroup\$ Oh, ok, I think I understand now. At first I was thinking that the first stage (integrate) is with the reference and the second (de-integrate) is with the input. Now I see that it's the opposite (input then reference) and it all makes sense now. \$\endgroup\$
    – tehwalris
    Commented Mar 31, 2014 at 20:38

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