OPA810: Gain Error

You would need to attach the circuit. The gain error is a resistor tolerance and loop gain question - can't answer LG without noise gain - schematic would be necessary - 

Hi Jan,

the OPA810 is an operational amplifier. Because of this the gain error depends on the set gain, signal frequency and some other parameters. Can you give some more details on your application? Eventually a schematic?

Kai

Hi,

firstly im using the OPA810 as voltage Buffer with a gain of 1 (0dB). 

Secondly im using the OPA810 as active integrator as a noninverting amplifier. the Gain is set to 161. Here you can see the schematic. Thanks for you help.

So continuing this discussion it seems, 

So there is a nominal then +/-spread on the DC gain, to run a response spread over frequency you would need to modify the internal model terms - in any case, at DC, 

1. Gain of 1 nominal error will have (Aol/(Aol+1)) error (with 120dB nominal DC Aol, that is 1e6/(1e6+1) = 0.999999 gain compression, going to the min screen point of 108dB at 25C will reduce this to 0.999996). There will also be a gain compression due to CMRR. 

2. The higher noise gain of 161 will have a LG/(LG+1) gain compression - nominally that would be (with LG= 1e6/161 = 6211) 6211/6212 = 0.999839. This will have a much larger % variation due to resistor tolerances, but the more dominant term here will be the resistor tolerances into the basic gain equation. 

Hi,

thanks for the answer. So you are saying that the main contribution to the gain error is coming from the tolerance of the gain setting resistors? Did i understood that right?

Jan

Well Jan, yes for the high gain stage, but no for the unity gain stage as there are no resistors. 

In the unity gain stage the gain error very small as you calculated it. Am i right?

Jan