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How to switch between low and high gain on inputs?


timmolter

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in the manual (https://reference.digilentinc.com/analog_discovery_2:refmanual) is says "An analog switch (ADG612) allows selecting high-gain versus low-gain (EN_HG_SC1, EN_LG_SC1) signals from the FPGA. The P and N branches of the differential path are switched together." I am measuring low voltages and would like to use the high gain option in order to get more resolution at lower voltages. How can I switch that on using the "Waveforms"?

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Is there a similar software flag for setting the range on the AWG too? I see in the above-linked datasheet:

  • Absolute Resolution(amplitude ≤1V): 166μV
  • Absolute Resolution(amplitude >1V): 665μV
But I don't see a reference to the software params to set this. I'm asking about this in reference to using the SDK. I assume the Waveforms software is handling the switch automatically depending on the selected waveform.
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One more question related to this: How is the oscilloscope gain setting related to the SDK method 'FDwfAnalogInChannelRangeSet'? What does the 'voltsRange' parameter do exactly? Shouldn't it be a boolean selecting high or low gain? And is it symmetric above and below 0V?

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To have a generic parameter for the SDK the (pk2pk) range is specified in voltage units, range = 10 x V/div. 
The Analog Discovery (1&2) has two ranges 5V and 50V. For 5V or less low range is used, for higher range setting high gain (with 1X attenuation option).
The EExplorer has about 64 gains and the future products might have other steps.

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OK, thanks. I have the AD2, so if I set 'voltsRange' below 5, it will use the -2.5 to +2.5V (5V pk2pk) range and if I set 'voltsRange' above 5, it will use the -25 to +25V (50V pk2pk) range? I just want to be absolutely sure I understand.

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