AMC1350: Loading at the input of AMC1350

[hiyo]

Dear Team,
I am using AMC1350 in one of my application. Please see the below circuit.
You can see that the voltage at the input of AMC1350 is 2.294V

The expected voltage at the input of AMC1350 is 2.3766V

I believe the loss of voltage at the input of AMC1350 is due to it's input impedance of 1.25Mega ohm.
I put a resistor of value 1.25Mega ohm in parallel with R2 and obtained a value of 2.294V.

May I know any method to avoid this loss.

 

[Nigel Goodwin]

I've no idea about the AMC1350, but you could either lower the values of the potential divider (which depends on the output impedance of the unknown source) or place a buffer opamp between the divider and AMC1350.

 

[rjenkinsgb]

You need to calculate the divider as if there were a 1.25M resistor across the lower half.

The result should then be correct in circuit, with the AMC1350 acting as the 1.25M

 

[danadak]

Yes input loading will alter your divider results.

You might examine model to see what they used as nominal value, eg. to get more exact
correlation in value simulated. Aslo might examine model to see how they wired its Voffset
and input Z.

Also notice their graph on Zin versus T is not a nominal 1.25 Meg, its a little above 1.24Meg.

Of course if you decide to use a buffer it brings along its own baggage of offset V, drift, PSRR,
and other errors, so be cognizant of that.


Regards, Dana.