AD628 common mode offset on the output

Hi Robin,

How did you measure the 14 mV error that you are seeing? Is it only a component of VCM or is it the total offset voltage of your system?

I agree that the error from the common mode voltage should be around 5 mV as you calculated but there are also other error source that you need to consider like the offset voltage, bias current*Rext, etc that could affect the total accuracy of the system.

Looking at the offset voltage of the output amplifier as shown below, configured at gain of 100 V/V (Rext1= 1MΩ, Rext2=10.2kΩ), it will give you around 15 mV additional offset. 

Effect of bias current to offset voltage is around 3 mV ( 3nA*10kΩ*100V/V)

These error sources could already give you a total offset voltage of around 23 mV. 

Do you have a target accuracy that you want to meet?

Can you tell me more about your applications and specs requirements so that I could recommend suitable parts for your applications.

Best regards,

Emman

Hi,

First of all, thank you for your answer.

In my previous message, when i speak about common mode offset I'm not speaking about global offset wich is bigger but about the offset difference between 0V and 3V of common mode voltage.

Yesterday i made an other test with 10V of common mode voltage -> that leads to a 48mV common mode offset on the output.

About the application and specs:

"

I have to needs :

- measure a voltage on a 300R/1% resistor with a current between 0 and 100uA (accuracy 2%, common mode 5V)

- measure a battery voltage : 0-700mV differential (accuracy 2%, common mode 1V)

"

Best regards,

Robin

Do you have an other part that could suit ?

Best regards,

Robin

Hi Robin,

Thanks for giving more details on your applications. The common mode rejection of the part is very dependent on the matching of the resistors (10kΩ) inside the chip for AD628. Having 300ohms shunt resistor can unbalance the matching of these resistors and therefore degrade the CMRR performance of the part.  One thing you can do to compensate this is to add series resistor equivalent to the shunt resistor to one of the inputs as shown on the figure below.

Do you intend to use one circuit design to measure voltage on shunt resistor and battery voltage? or separate circuit for both measurements?

Another solution is to use higher input impedance difference amplifier like the AD8479 to lessen the effect of the shunt resistor on the CMRR performance. It also has higher overall dc accuracy specs compared to AD628 but it only has a gain of 1V/V and you might need additional gain stage amplifier.

If you can use  supply voltage higher than the expected common mode voltage, the best solution that I can offer is to use an instrumentation amplifier instead like the AD8421. Instrumentation amplifiers are optimized for best CMRR performance but you need to make sure that the VCM is not higher than the supply voltage and also be wary of the input common mode vs output swing of the part.

Best regards,

Emman

Hi,

the 300ohms resistor will be used on the final application.

For now, i'm just using an arbitrary waveform generator (50ohm serie resistance).

I tried to put a 50ohms resistance on the input to compensate : the result is very good.

Thank you for your answers and your time.

Best regards,

Robin