ADS 1178: transfer function question

Niranjan Madhavan

I was testing a Data acquisition system based on 1178.I plotted voltage to count output and found this graph

Test condition: Vrefp: 2.5V, Vrefn=0, Avdd=5V IOVdd=3.3V DVdd= 1.8V
Sampling rate=20khz.

All input voltage is centered at 2.5V. What would be the reason I am getting that saturation and flipping to the negative half? Is it some thing related to my amplifier output?

regards

niranjan

over 11 years ago

0 Greg Hupp over 11 years ago

TI__Mastermind 23235 points

Are you taking into account the output data is two's compliment? (see page 17 of the datasheet)

Are your signals maintaining the necessary common mode and absolute voltages required for the inputs?

0 Niranjan Madhavan over 11 years ago in reply to Greg Hupp

Intellectual 605 points

I am pretty sure about the common mode voltage. It is at 2.5V and the absolute inputs are below Vdd.

Acquiring software I am using is a third party software and have limited knowledge about how they interpret. That is also one of my concern. I can understand getting saturated and clipped, but I am not understanding the change of phase at around 1.335Vpp input.

Common node is 2.5V and Vref p is 5V and VrefN is gnd.

0 Greg Hupp over 11 years ago in reply to Niranjan Madhavan

TI__Mastermind 23235 points

Be careful..Vref cannot be 5V!! See spec table for Vref (max value currently set at 3.1V)

0 Niranjan Madhavan over 11 years ago in reply to Greg Hupp

Intellectual 605 points

Sorry Greg, there was a typo on previous reply.
I checked again physically on BOARD

VrefP is at 2.5V and VrefN is 0V
AVDD 5.0
IOVdd 3.3V1
DVDD 1.8V

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ADS 1178: transfer function question