Sanity check regarding an ADC chip

[Jimbo bob]

Dear enthusiasts,

I am planning to work with a AD7730 ADC. On the page 24 of the datasheet you can read the following (and please by all means have a look if you see necessary):

n some applications, the analog input range may be biased either around system ground or slightly below system ground. In such cases, the AGND of the AD7730 must be biased negative with respect to system ground so the analog input voltage does not go within 1.2 V of AGND. Care should taken to ensure that the differential between either AVDD or DVDD and this biased AGND does not exceed 5.5 V.

More over....

Bipolar input ranges do not imply that the part can handle negative voltages with respect to system ground on its analog inputs unless the AGND of the part is also biased below system ground.

I have also attached a schematic of what I am planning to put together. My problem is that a friend of mine who has some considerable experience in electronics is telling me that I can directly feed my quarter bridge connections to the ADC just like the diagram and the ADC will account for the negative voltage internally. I am failing to see how the ADC can account for that internally. Moreover, the reason in the schematics the bridge is directly connected to the ADC is because it is a load cell for a weight scale and will not show both tension and compression which correspond to both positive and negative voltages relative to the "system's ground". Is it me or do you actually get the same idea as me when you read the datasheet which involves taking care of the negative bridge output before inputting it into the ADC?

Many thanks in advanced!!

 

[Harald Kapp]

and the ADC will account for the negative voltage internally

This may be a misunderstanding on the side of your friend or on your side.

The AD7730 has differential inputs which means that the input voltage is measured from Ain(+) to Ain(-). This voltage cna be positive or negative. However, the absolute voltae on both Ain(+) or Ain(-) has to be at least Vagnd+1.2V.

Example:
Vain(+)=1.50 V (with respect to AGND=0V)
Vain(-)= 1.45 V (with respect to AGND=0V)
Vain(+)-Vai(-) = 1.50 V -1.45 V = 0.05 V = 50 mV

Vain(+)=1.45 V (with respect to AGND=0V)
Vain(-)= 1.50 V (with respect to AGND=0V)
Vain(+)-Vai(-) = 1.45 V -1.5 V0 = -0.05 V = -50 mV

Even though both input voltages are positive in both examples, the diference is positive in the first example and negative in the second example. For a typical bridge circuit this is exactly what you need.

 

[Jimbo bob]

This may be a misunderstanding on the side of your friend or on your side.

The AD7730 has differential inputs which means that the input voltage is measured from Ain(+) to Ain(-). This voltage cna be positive or negative. However, the absolute voltae on both Ain(+) or Ain(-) has to be at least Vagnd+1.2V.

Example:
Vain(+)=1.50 V (with respect to AGND=0V)
Vain(-)= 1.45 V (with respect to AGND=0V)
Vain(+)-Vai(-) = 1.50 V -1.45 V = 0.05 V = 50 mV

Vain(+)=1.45 V (with respect to AGND=0V)
Vain(-)= 1.50 V (with respect to AGND=0V)
Vain(+)-Vai(-) = 1.45 V -1.5 V0 = -0.05 V = -50 mV

Even though both input voltages are positive in both examples, the diference is positive in the first example and negative in the second example. For a typical bridge circuit this is exactly what you need.

Wow!!! Harald you are a saviour.This makes a lot of sense. What I seem to be repeatedly doing is confusing my basic circuitry understanding and misinterpreting information. I overlooked the fact that even though the bridge output can be negative in cases, but the actual leads coming out of the bridge are carrying positive voltages relative to the system ground as they are just voltage dividers between a positive excitation voltage and ground at 0 volts. Therefore those are the voltages which will be fed into AIN+ and AIN- and their difference can be positive and negative, just like the bridge output can be.

Thanks for your time Harald.