Precision of ADC's

Mahesh,


I'm sorry about the delay. The engineer that normally supports this device is out, so I'll start with a couple of things to check.

First, I'd like to know more about the application. What are they measuring? Is this device in a noisy environment with a lot of potential for EMI? Are they willing to share a schematic? Even a partial schematic may help.

The ADC gives an output value that is represents the input compared to the reference. If the reference is noisy, then the output value will be noisy as well. What reference is being used? What is the noise specification for the reference?

Is there noise on the supply? Often, the supply noise translates to bad readings, not just though any PSR specification, but from how it affects the sensor (and reference) drive.

I mentioned this earlier, but EMI may give extra noise as well. I'd make sure there aren't large wire loops that pick up EMI and cause bad readings. It might help to enclose the board in a metal box that can limit the effect of EMI in the environment.

Lastly, I'd look a the raw data. Check to see if the noise is periodic. There is the chance that there is line noise in the system. See if it is 50 or 60Hz noise based on a large set of continuous data and check to see where line noise might be making its way to the ADC.


Joseph Wu

Umamahesh,


I've looked at the schematic and the ADC used in it is the ADS1252, not the ADS1278.

If this is the correct schematic, I suspect that the problem is the supply that is being used as the reference. As I mentioned in the previous post, if the reference is noisy (in this case the supply), then the measurement will reflect the noise.


Joseph Wu

Mahesh,


The ADC makes a conversion and gives out a digital code proportional to the ratio between the input voltage and the reference. For the ADS1252, the ADC makes a differential measurement (where VIN- can be higher than VIN+, the full range is ±VREF, so with a +5V reference, the ADC can measure from -5V to 5V. The output code becomes:

ADC output = (VIN/VREF)*2^23

If you look at this equation, if VREF has noise, and the input varies a little, then the ADC output will have noise, even if VIN is completely noiseless.

So how much noise will you see? Let's assume that VREF is 5V, with about 2.5mV of noise. Lets also say VIN is 4V.

Ideally, the ADC output would be (4V/5V)*2^23 = 6710886d

Now make the calculation when the reference is 5.0025V and when the reference is 4.9975V

(4V/5.0025V)*2^23 = 6707533d = 66594Dh
(4V/4.9975V)*2^23 = 6714244d = 667384h

Using a 5V reference with a noise of 2.5mV (or 0.05%), this ADC becomes a 11-bit ADC, where originally the noise floor was probably somewhere near 20 bits.

There may be some noise contribution to the circuit with the input filtering (which I think is a bit excessive), and there may be some EMI pick-up because there is a lot of circuitry and a potential for layout problems, but I think the majority of the noise comes from using the supply as the reference. Just to be sure, can you tell me what they are using for the supply? We might be able to guess what the input noise is based on that.

In some cases, using the supply as the reference may not affect the measurement. For example, in a measurement with a bridge, noise on the supply will be transparent, because the drive on the bridge causes a proportional error on the input, which cancels out the error.

Can you describe their application? It might help to know what they are measuring. Is there a reason they are using the ADS1252? It is an older part, and there are many newer devices that might work better in their application.


Joseph Wu