Interfacing DDS to Low-Noise Signal Conditioning - Page 1
Forum section: electronics / projects

author: Evan.Cornell
author_rank: Regular Contributor
timestamp: January 13, 2020, 05:28:07 pm
content: I would like to use AD9102 as a self-test signal generator to a low noise sensor signal conditioning path.The sensor is a capacitive sensor, approximately 1.7nF. The frequency range is 10kHz to 200kHz, thus the sensor impedance varies roughly from 9.36kOhm to 468ohm.I have a low noise amplifier, non-inverting, bipolar rail connected directly to sensor. However, I would like to be able to either disconnect the sensor and inject a voltage signal from AD9102 to the low noise amplifier (signal conditioning + ADC self-test) or keep the sensor connected to LNA and use AD9102 to inject a known current signal into the sensor (basically implementing an impedance analyzer).My question is this: does anyone have any recommendations on how to best accomplish this analog interface? I am looking for extremely low additive noise analog switches (LNA runs from +-3.3V rails), and how to arrange signal conditioning between AD9102 for both voltage injection into LNA and current injection into the sensor.

author: Kleinstein
author_rank: Super Contributor
timestamp: January 13, 2020, 07:17:24 pm
content: The AD9102 is rather high quality for just a test signal. Anyway the interface from a DDS to the amplifier would be very similar with other DDS chips. For switching the signal, there are CMOS switches, that should be suitable. During the normal sensor use I would additionally turn off the DDS to avoid cross talk. CMOS switches are no longer perfect off at the higher frequencies.The DDS chip usually needs a multi stage LC based reconstruction filter. With only relatively low frequencies the filter BW could be a little lower than normal - however one still needs to check the RF suppression. So it may still be an option be stay close to the normal circuit and not try to cut corners very much. A lower frequency low pass may not attenuate as well at the high end.Depending on the amplitude needed one may need some amplification or at least a buffer. With calibrations just a resistor for coupling could be enough. The amplifier would likely need a series resistor anyway to isolated it from to much capacitive load.

author: ogden
author_rank: Super Contributor
timestamp: January 13, 2020, 07:57:20 pm
content: Quote from: Evan.Cornell on January 13, 2020, 05:28:07 pmIf you need excitation signal then AD9102 seems like unnecessary overkill, especially considering freq range. I would like to hear why you think AD9833 which would be my choice, is not good enough for you.

author: Evan.Cornell
author_rank: Regular Contributor
timestamp: January 13, 2020, 08:06:57 pm
content: Quote from: ogden on January 13, 2020, 07:57:20 pmCriteria I neglected to mention previously is that the end-user desires arbitrary waveform generation capability for this block. AD9833 does not appear to support that feature.

author: Evan.Cornell
author_rank: Regular Contributor
timestamp: January 13, 2020, 09:13:29 pm
content: AD9102 provides a differential current output, and the Eval board uses op-amp to to convert to single-ended voltage output. I can take that via analog switch to inject into front-end of my LNA circuit, that I understand well enough.I guess more specifically my question is this: how would I take that single ended voltage signal and convert to a known current level to inject into my capacitance sensor, which doesn't have a constant impedance vs. frequency?

author: ogden
author_rank: Super Contributor
timestamp: January 13, 2020, 09:42:28 pm
content: Textbook approach would be capacitance bridge which is derived from Wheatstone bridge. Terminals of the sensor insulated from ground or power rails? There are other ways to measure impedance. Self-balancing bridge for example. AD even gave chip for that,AD5933. BTW you did not state desired accuracy.