I'm trying to determine the accuracy of the DC circuit below but I am unsure how to account for input bias currents, input offset voltages, open loop output impedance, input resistance and open/closed loop gain errors at the same time for a single problem. All other examples use one or two of these parameters but never all available parameters together. I also have not seen anyone mention anything about when you are operating at a voltage that is not specified in the datasheet. I understand that many of the parameters defined have a minimal effect on the output but I would still like to know the correct way of finding the tolerance.

The following are additional details to the circuit:

The operating temperature range is -40°C to +85°C

The op amp is TLE2064I

The input voltage ranges from +2.4775V to +2.5225V over the temperature range, nominally +2.5V.

I do not have an accuracy requirement, I just need to specify the accuracy of the circuit.

Also to be clear, I may be mixing up the idea of accuracy and tolerance, I am simply interested in the min and max voltages that can appear on the output.


simulate this circuit – Schematic created using CircuitLab

Analysis based on my understanding

Looking up the TLE2064I, I can find the following parameters:

TLE2064I Datasheet table 1

TLE2064I Datasheet table 2 and 3

Datasheet can be found here from TI's website.

The first problem I have is that in this example, I'm operating at +/- 12.5V, not +/- 15V or +/- 5V as defined in the datasheet. Are we supposed to extrapolate the values of each parameter between the 5V and 15V tables the datasheet provides?

If we assume that I can use the 15V table, then I would go through the circuit twice, once to get the minimum output voltage and again to get the maximum out voltage.



First I would get all of the min and max values of each part:

\$R_{(R1, min)} = 100kΩ * (1 - 0.001 + (-40°C - 25°C) * 25ppm/°C) = 99,737.5Ω\$

\$R_{(R1, max)} = 100kΩ * (1 + 0.001 - (-40°C - 25°C) * 25ppm/°C) = 100,262.5Ω\$

\$R_{(R2, min)} = 100kΩ * (1 - 0.001 + (-40°C - 25°C) * 25ppm/°C) = 99,737.5Ω\$

\$R_{(R2, max)} = 100kΩ * (1 + 0.001 - (-40°C - 25°C) * 25ppm/°C) = 100,262.5Ω\$

\$R_{(R3, min)} = 10kΩ * (1 - 0.01 - (-40°C - 25°C) * 100ppm/°C) = 9.84kΩ\$

\$R_{(R3, max)} = 10kΩ * (1 + 0.01 + (-40°C - 25°C) * 100ppm/°C) = 10.16kΩ\$

Bias current & Offset voltage

Second I would go through the circuit trying to account for each of the parameters and their specific effects. I'd start with the input bias currents and make the assumption that the maximum bias current passes through the negative input pin. This current will generate a voltage across one of the resistors although I don't believe we can determine which one so I would end up trying both and taking the worst one as my result. Once we have that voltage I would add the input offset voltage to this value to get an "equivalent" Input offset voltage (EIOS). I'd also only be interested in the maximum offset as the minimum would be the same as the maximum with the opposite polarity.

Assuming the current passes through R1 and into the negative input pin of the op amp:

\$V_{(EIOS, max)} = R_{R1, max} * I_{-bias, max} + V_{os, max} = 100,262.5Ω * 5nA + 7.3mV = 7.8013125mV\$

Input Resistance

Third, I'd struggle to see how to account for the input resistance. The datasheet says it's \$10^{12}Ω\$ but is this going to be viewed as the parallel combination of \$R_{1}\$, \$R_{2}\$ and the input resistance? Is this "doubling dipping" with the input bias currents as that already presents a load to the input?

Gain error

Fourth, I'd struggle to see how to account for the input resistance and gain error. The datasheet states that \$A_{VD} = 20\frac{V}{mV}\$ or \$20,000\frac{V}{V}\$ over the temperature range.

I believe that the following equation is true:

\$A_{Closed} = \frac{A_{Open}}{1 + A_{Open} * \beta}\$

where \$\beta\$ is the feedback factor determined by R1 and R1 for this inverting op amp configuration.

\$\beta_{min} = \frac{R_{R1, min}}{R_{R1, min} + R_{R2, max}} = 0.4986875\$

\$\beta_{max} = \frac{R_{R1, max}}{R_{R1, max} + R_{R2, min}} = 0.5013125\$

\$A_{(CL, min)} = \frac{A_{OL min}}{1 + A_{OL min} \times \beta_{max}} = \frac{20,000\frac{V}{V}}{1 + 20,000\frac{V}{V} \times 0.4986875} = 1.9945648109\$

\$A_{(CL, max)} = \frac{A_{OL max}}{1 + A_{OL max} \times \beta_{min}} = \frac{20,000\frac{V}{V}}{1 + 20,000\frac{V}{V} \times 0.5013125} = 2.0050627835\$

Although this result seems a bit confusing to me, I'm guessing that this gain of 2 is because half of the gain is used to go from ~+2.5V to ~0V and the other half is to go from ~0V to ~-2.5V?

Now that I have the closed loop gain, I would have thought I would have subtracted \$V_{EIOS}\$ from the voltage that would normally be seen across R1 and then multiplied by the closed loop gain but this clearly does not work and I did not even get to try to understand how to handle the open-loop output impedance issue.

I'm struggling to see where exactly things are going wrong. Is this the completely wrong approach? Am I making a mathematical mistake? Is the way I'm approaching the problem incorrect?

  • \$\begingroup\$ In your calculations, you are using more significant figures than reasonably necessary. If you have 0.1% resistors, there is no need to go over 2 to 3 places after the decimal. I suggest you reduce the significant figures so others who will read this question can read it (and answer it) easier. \$\endgroup\$
    – Pxl
    Commented Nov 28, 2023 at 0:59
  • \$\begingroup\$ I am unable to provide a full answer at this time, but a useful resource for you may be This PDF by TI \$\endgroup\$
    – Pxl
    Commented Nov 28, 2023 at 1:01
  • \$\begingroup\$ You can just say the input supply has a precision of ±0.9%. \$\endgroup\$
    – Hearth
    Commented Nov 28, 2023 at 6:04
  • 1
    \$\begingroup\$ "Are we supposed to extrapolate the values of each parameter" It's reasonable to do that between values listed in the datasheet. Outside that range no-one knows. You can measure a sample ic but there's no guarantee that other ics will behave the same way. \$\endgroup\$
    – Graham Nye
    Commented Nov 28, 2023 at 11:29
  • 1
    \$\begingroup\$ If you account for bias currents, you don't need to worry about input resistance. \$\endgroup\$
    – Mattman944
    Commented Nov 29, 2023 at 9:39

2 Answers 2


I always start by including the error terms in the schematic. I believe that it looks like this. Then people can follow your equations easier.

Unless you are using an exotic opamp, or your closed loop gain is very high, the error due to open-loop gain should be insignificant.

Doing this by hand is tedious. I analyzed a similar circuit as part of my Six-Sigma project. People don't normally analyze these by hand anymore, there are modern tools to help.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ I've been tasked with writing a report to determine the "accuracy" of the circuit along with justification for the result. What would the modern tools be that people typically use? I've used LT spice but i need to get the worst case condition. Do you know of any resources that may help? \$\endgroup\$
    – Noahwar97
    Commented Nov 28, 2023 at 19:44
  • 1
    \$\begingroup\$ @Noahwar97 - I worked for a large company that use expensive Mentor Graphics software. I never used it myself, I mostly did digital worst case analysis. I always used a spreadsheet and followed a process similar to what you are doing. You are doing a Extreme Value Analysis, there are other types: diablomountainresearch.com/publications/… Your company (and/or customer) will determine what methods are allowed. \$\endgroup\$
    – Mattman944
    Commented Nov 29, 2023 at 9:35
  • 1
    \$\begingroup\$ Av=-R2/R1, ΔAv/Av=ΔR1/R1+ΔR2/R2=0.2% \$\endgroup\$
    – Franc
    Commented Nov 29, 2023 at 17:52

Sensitivity study for parametric variations. (Idea from the text "Linear and nonlinear circuits" by - Leon O. Chua, Charles Desoer, Ernest T. S. Kuh - McGraw-Hill:

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