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i have an 23 bit ADC with voltage reference of 1.4 V. If my input Analog signal is AC with frequency 50 HZ and its rms ranges from 0.2 Vrms to 1.3 Vrms. Then its peak voltages will vary from 0.28 V to 1.838 V.

Now, considering ADC high value that it can digitize (1.4 V). What does it mean? I mean, if i consider input Analog rms values or peak values to find out that this adc will work for my inputs. Because if i consider rms value then it probably would work, but it ADC is concerned with max peak of incoming Analog signal then my inputs with peaks higher than 1.4 V won't be able to digitized correctly.?

Please guide, what is considered by choosing ADC, is it the rms of Analog or peak.?

And if it is peak, then what would happen if Analog value is higher than adc specified value? Will all values higher than 1.4 V will treated as same digital value??

Thanks in advance..

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  • \$\begingroup\$ What type of ADC do you have and what is the AC input relative to ground? In means that it goes + and -1.8Vpk. \$\endgroup\$ – Jeroen3 Sep 11 '18 at 12:04
  • \$\begingroup\$ Peak of course. \$\endgroup\$ – Marko Buršič Sep 11 '18 at 12:07
  • \$\begingroup\$ Do you want to measure the waveform (i.e. so you can see the 50Hz sine wave) or just the amplitude at say 1Hz? Either way, you're going to need a different ADC and/or some signal conditioning. \$\endgroup\$ – Jack B Sep 11 '18 at 12:10
  • \$\begingroup\$ I want to know that at peak of input signal (when signal value s highest in cycle) is 1.8V. Will it be converted into a valid digital value by this ADC?? \$\endgroup\$ – BetaEngineer Sep 11 '18 at 12:21
  • \$\begingroup\$ Will all values higher than 1.4 V will treated as same digital value?? Maybe, maybe not. The data sheet should be explicit about what happens in those circumstances. These days, most ADCs seem to be well behaved and limit at the maximum code, some even have an overflow bit. \$\endgroup\$ – Neil_UK Sep 11 '18 at 14:00
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An RMS sine wave of 1.3 volts has a peak-to-peak value of 2 x \$\sqrt2\$ x 1.3 volts i.e. 3.677 volts p-p. It is not only that range that has to be accommodated by your ADC but you also need to DC bias the AC signal so that when it is 0 volts AC, the DC level is 3.677 volts / 2 i.e. 1.838 volts.

If you don't do this you will receive a clipped digital version of your input analogue waveform and you may also damage your ADC.

You should also read any prospective ADC data sheet and understand the effects of gain and zero offset errors and take these into account so that you don't get digital clipping.

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  • \$\begingroup\$ Thanks. i guess i still don't know the whole operation of ADC's. I thought the reference voltage of ADC means that e.g 23-bit ADC will have all 23-bits as logic '1' when input analog is at 1.4 V level. i need to study it more. thanks for the insight.. \$\endgroup\$ – BetaEngineer Sep 11 '18 at 12:24
  • \$\begingroup\$ does this vpp matters if we have differential input ADC ? \$\endgroup\$ – kakeh Mar 25 at 4:47
  • \$\begingroup\$ Of course. The peak extremes have to be within the input range. \$\endgroup\$ – Andy aka Mar 25 at 8:18

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