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Internal resistance of Wavegen



I have been using the Analog Discovery 1 (legacy version) as a portable oscilloscope and function generator. I am using the Wavegen as a function generator to test an analog front end (AFE) which includes an integrated instrumentation amplifier and an ADC (http://www.ti.com/tool/ADS1293EVM). Trying to characterise the performance of the AFE, I used the Analog Discovery's function generator to generate sine waves of different amplitudes (between 1mV and 100 mV pk-pk) and frequencies (between 40 and 200 Hz, in steps of 40 Hz). The gain of the instrumentation amplifier is fixed at 3.5 and the bandwidth of the AFE is programmable (ranging between 170 Hz and 340 Hz).

I find that the gain of the system varies with frequency and is either 1 or slightly higher/lower than 1, depending on input voltage and frequency. I did not include any load resistance in my measurement circuit since the internal impedance of the voltage generator was 0 (as mentioned in https://forum.digilentinc.com/topic/3203-analog-discovery-2-waveform-output-impedance/ and https://forum.digilentinc.com/topic/18292-impedance-for-analog-discover-2-itself/#comment-47859).

I would like to know if it would be necessary to include any load resistance while using the AWG as input to an instrumentation amplifier. If this is not the reason, what could be the reason for this mismatch in gain/amplification factor? I have posted a query in the TI forum (link: https://e2e.ti.com/support/data-converters/f/73/t/898470). I have also attached the WaveForms workspace file I used.

Thanks in advance for helping out with this.

Best regards,



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Hi @ganesh

To properly measure the DUT the ratio between two Scope channels should be used, like it is indicated in the WF Network Analyzer.
The Wavegen output of AD2 has zero impedance but with high loads it could enter in limitation (10-50mA).
In case you are using the BNC adapter notice the 50/0 Ohm and AC/DC jumpers.


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