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How can you represent negative values using the flash ADC method?

In books and online I have only ever seen positive references for the comparators used in the flash ADC architecture.

For a simple sine wave, how can you represent both its negative values and positive values using the flash ADC method? Or with this method is it not possible

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  • \$\begingroup\$ Where do you see a problem in using reference voltages that are also in the negative range? \$\endgroup\$
    – Curd
    Mar 21 '18 at 20:10
  • \$\begingroup\$ On the potential divider "side" of the circuit I have only seen references with positive voltages. Maybe the question should of been, how to get positive and negative values from a voltage divider? \$\endgroup\$
    – user131618
    Mar 21 '18 at 20:12
  • \$\begingroup\$ Simply by connecting the low side to a negative voltage!?! \$\endgroup\$
    – Curd
    Mar 21 '18 at 20:13
  • \$\begingroup\$ @Curd Yes but than the high side will also be connected to a positive voltage, so which voltage do I put into the equation, the negative supply or positive supply? \$\endgroup\$
    – user131618
    Mar 21 '18 at 20:15
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Negative/positive is just relative to the point you choose to be zero.

In general ADCs can only work with input voltages within their supply voltage. For example an ADC powered from a 5 V source (ground and + 5 V) could accept input voltages between 0 and 5 V.

If we then choose 2.5 V to be the "zero" reference of the input signal anything below 2.5 V would represent a negative voltage, anything above 2.5 V a positive voltage.

But what if I have a sinewave between -1 V and + 1 V ?

Then you can either "level shift" it, I mean add 2.5 V to it. For an audio signal where we're not interested in the DC value, this can be done by AC coupling with a capacitor and some resistors.

Or (more practical for situations where you do want the DC value) you can power the ADC with +2.5 V and -2.5 V.

So basically you "see" a problem where there isn't one. Indeed ADCs by themselves usually cannot work with negative values but there is no need. We can just "shift" the ADC's working point so that all input voltages become positive. This "shifting" of the input voltage is quite trivial.

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  • \$\begingroup\$ Thank you for your explanation, that makes alot of sense \$\endgroup\$
    – user131618
    Mar 21 '18 at 20:19

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